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# 如何让列表内的元素前后相加生成新的元素? ``data = [1, 2, 3, 4]`` ``````data1 = [1, 3, 5, 7] # 与前面的一个元素相加 data2 = [1, 3, 6, 10] # 与前面的一个元素累计相加`````` ### 回答: ``data = [1, 2, 3, 4]`` ``````from itertools import accumulate data1 = [v + (data[i-1] if i > 0 else 0) for i, v in enumerate(data)] data2 = list(accumulate(data)) print(data1) print(data2)`````` ``````[1, 3, 5, 7] [1, 3, 6, 10]`````` ### 回答: ``````data1= data[:1] + [x+y for x,y in zip(data[1:],data[:-1])] data1= data[:1] + [sum(data[x:x+2]) for x in range(len(data)-1)] data2= [sum(data[:x]) for x in range(1,len(data)+1)]`````` data1 其实你也可以用带 if else的列表推到而不用单独处理第一个。 `data1= [sum(data[x-1:x+1]) if x>0 else data[0] for x in range(0,len(data))]` ### 回答: ``````pre = 0 def foo(x): global pre y = x + pre pre = x return y subtotal = 0 def bar(x): global subtotal subtotal = x + subtotal return subtotal data = [1, 2, 3, 4] data1 = list(map(foo, data)) data2 = list(map(bar, data)) print(data1, data2)`````` ### 回答: ``````from itertools import accumulate data = [1, 2, 3, 4] data1 = [x + y for x, y in enumerate(data)] data2 = list(accumulate(data)) print(data1) print(data2)``````
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Question # In order to investigate the structure of atoms, Ernest Rutherford performed his famous experiment, in which... In order to investigate the structure of atoms, Ernest Rutherford performed his famous experiment, in which he bombarded gold atoms with alpha particles and studied the scattering of the alpha particles. Imagine that an alpha particle (a helium nucleus, consisting of two protons and two neutrons) is initially moving along the x-axis in the positive direction straight toward an initially stationary gold nucleus (containing 79 protons and 118 neutrons) and all subsequent motion takes place along the x-axis. The alpha particle starts with kinetic energy of 9.7 MeV (= 9.7 × 106 eV) far from the gold nucleus. Take the mass of a nucleon (a proton or a neutron) to be 1.7 × 10-27 kg and assume that the mass of a nucleus equals the sum of the masses of its constituent nucleons. Assume also that all speeds are low compared to the speed of light. 50% Part (a) Find the final momentum of the alpha particle (with its sign), long after it interacts with the gold nucleus, in units of kg⋅m/s. 50% Part (b)  Find the final momentum of the gold nucleus (with its sign), long after in interacts with the alpha particle, in units of kg⋅m/s. let m1 = 4*m KE1 = (1/2)*m1*u1^2 u1 = sqrt(2*KE1/m1) = sqrt(2*9.7*10^6*1.6*10^-19/(4*1.7*10-27)) = 2.136*10^7 m/s m2 = 197*m u2 = 0 a) speed of alfa particle after the collision, v1 = (m1 - m2)*u1/(m1 + m2) = (4*m - 197*m)*2.136*10^7/(4*m + 197*m) = -2.05*10^7 m/s momentum of alfa particle, P1f = m1*v1 = 4*m*v1 = 4*1.7*10^-27*(-2.05*10^7) b) speed of gold nucleus after the collision, v2 = 2*m1*u1/(m1 + m2) = 2*4*m*2.136*10^7/(4*m + 197*m) = 8.50*10^5 m/s momentum of gold nucleus, P2f = m2*v2 = 197*m*v2 = 197*1.7*10^-27*8.5*10^5 #### Earn Coins Coins can be redeemed for fabulous gifts.
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# Converting Fractions to Decimals To convert a fraction to a decimal, just divide the numerator by the denominator . Example 1: Write $\frac{3}{15}$ as a decimal. Since $15$ is larger than $3$ , in order to divide, we must add a decimal point and some zeroes after the $3$ .  We may not know how many zeroes to add but it doesn’t matter.  If we add too many we can erase the extras; if we don’t add enough, we can add more. $\begin{array}{l}\begin{array}{c}\\ 15\end{array}\begin{array}{c}\hfill 0.2\\ \hfill \overline{)3.0}\end{array}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{_}{-3\text{\hspace{0.17em}}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\end{array}$ So $\frac{3}{15}=0.2$ . Remember that a decimal is really just a special way of writing a fraction that has a power of ten as a denominator. What we're doing here is rewriting $\frac{3}{15}$ as $\frac{2}{10}$ . Sometimes, you may get a repeating decimal . Example 2: Write $\frac{5}{6}$ as a decimal. $\begin{array}{l}\begin{array}{c}\\ 6\end{array}\begin{array}{c}\hfill 0.8333\\ \hfill \overline{)5.0000}\end{array}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{_}{-4\text{\hspace{0.17em}}8\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}20\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{_}{-18}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}20\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{_}{-18}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\end{array}$ We can write the result using a bar over the repeating digit (or digits): $\frac{5}{6}=0.8\stackrel{¯}{3}$ .
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https://socratic.org/questions/584e78467c01496a8757822b
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crawl-data/CC-MAIN-2020-10/segments/1581875145654.0/warc/CC-MAIN-20200222054424-20200222084424-00146.warc.gz
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# If the temperature coefficient alpha for a certain reaction is 2.5, then if the reaction was run at 283^@ "C" and 293^@ "C", what is the activation energy in "kJ/mol"? Dec 17, 2016 ${E}_{a} = 6.5 \times {10}^{3}$ $\text{kJ/mol}$ From Wikipedia, the definition of the temperature coefficient $\alpha$, in regards to the rate constant $k$ and temperature $T$, is: $\frac{\mathrm{dk}}{k} = \alpha \mathrm{dT}$ If we integrate this from state 1 to state 2 on the left side, and ${T}_{1}$ to ${T}_{2}$ on the right side: ${\int}_{\left(1\right)}^{\left(2\right)} \frac{1}{k} \mathrm{dk} = {\int}_{{T}_{1}}^{{T}_{2}} \alpha \mathrm{dT}$ When we assume that the temperature coefficient stays constant across this small temperature range, we can pull $\alpha$ out of the integral and obtain (noting that the integral of $\frac{1}{x}$ is $\ln x$): $\ln \left({k}_{2}\right) - \ln \left({k}_{1}\right)$ $= \textcolor{g r e e n}{\ln \left({k}_{2} / {k}_{1}\right)} = \alpha \left({T}_{2} - {T}_{1}\right)$ $= \left(2.5\right) \left({293}^{\circ} \text{C" - 283^@ "C}\right)$ $= \textcolor{g r e e n}{25}$ where we used the fact that intervals in the celsius temperature scale are equal to intervals on the Kelvin temperature scale. Thus, we have indirectly calculated $\ln \left({k}_{2} / {k}_{1}\right)$. Recall that this shows up in the form of the Arrhenius equation that demonstrates the temperature dependence of the rate constant: $\ln \left({k}_{2} / {k}_{1}\right) = - {E}_{a} / R \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]$ Therefore, we can now algebraically solve for and calculate the activation energy ${E}_{a}$, in $\text{kJ/mol}$, as: $\textcolor{b l u e}{{E}_{a}} = - \frac{R \ln \left({k}_{2} / {k}_{1}\right)}{\frac{1}{T} _ 2 - \frac{1}{T} _ 1}$ $= - \frac{R \ln \left({k}_{2} / {k}_{1}\right)}{\frac{{T}_{1} - {T}_{2}}{{T}_{1} {T}_{2}}}$ $= - R \ln \left({k}_{2} / {k}_{1}\right) \left[\frac{{T}_{1} {T}_{2}}{{T}_{1} - {T}_{2}}\right]$ = -("0.008314472 kJ/mol"cdot"K")(25)[(("556.15 K")("566.15 K"))/(556.15 - 5"66.15 K")] $= \textcolor{b l u e}{{6.5}_{45} \times {10}^{3}}$ $\textcolor{b l u e}{\text{kJ/mol}}$ where the subscripts indicate digits past the last significant figure. This means that the reactants have about a $\text{6500-kJ}$ energy barrier in order for the reaction to proceed successfully.
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Ir al contenido principal ## Due September 11th ### Problem 1 Let $f:X\to Y$ be a function, $A,B\subset X$ and $U,V\subset Y$. 1. $f(A\cup B) = f(A)\cup f(B)$. 2. $f(A\cap B)\subset f(A)\cap f(B)$. Give an example where $f(A\cap B)\not\supset f(A)\cap f(B)$. 3. $f^{-1}(U\cup V) = f^{-1}(U)\cup f^{-1}(V)$. 4. $f^{-1}(U\cap V) = f^{-1}(U)\cap f^{-1}(V)$. 5. $f(f^{-1}(U)) \subset U$. Give an example where $f(f^{-1}(U))\not\supset U$. 6. $f^{-1}(f(A)) \supset A$. Give an example where $f^{-1}(f(A))\not\subset A$. ### Problem 2 If $X$ is sequentially compact and $f:X\to Y$ is continuous, then $f(X)$ is sequentially compact. Prove it directly using the definition of sequential compactness. ### Problem 3 Give a set $X$ and two metrics $d,d'$ on $X$ such that $(X,d)$ and $(X,d')$ are homeomorphic, but $f:X\to X$ given by $f(x)=x$ is not uniformly continuous. ### Problem 4 1. Let $\mathcal I:C([0,1]\to C([0,1])$ be the operator given by $\displaystyle \mathcal If(x) = \int_0^x f(t) dt,$ i.e. $\mathcal If$ is the undefined integral of $f$. Then $\mathcal I$ is continuous with respect to the uniform norm. 2. Use (1) to prove the following theorem: Let $f_n\in C^1([0,1])$ and $g\in C([0,1])$ such that 1. $(f_n(x_0))$ converges for some $x_0$ 2. $f_n' \rightrightarrows g$ Then $f_n$ converges uniformly and, if $f_n\rightrightarrows f$, then $f\in C^1([0,1])$ and $f'=g$. ### Problem 5 1. Let $C_0(X,\R)$ be the space of continuous functions $f:X\to\R$ that go to 0 at infinity, i. e. for every $\e>0$ there exists a compact $E\subset X$ such that $|f(x)|<\e$ for every $x\not\in X$. Then $C_0(X,\R)$ is a closed subspace of $C_B(X,\R)$. 2. Let $C_c(X,\R)$ be the space of continuous functions $f:X\to\R$ of compact support, i. e. the closure of the set $\{x\in X:f(x)\not=0\}$ is compact. Then $C_c(X,\R)$ is a subspace of $C_B(X,\R)$. Give an example where it is not closed in $C_c(X,\R)$.
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# Definition:Inverse Sine/Real ## Definition Let $x \in \R$ be a real number such that $-1 \le x \le 1$. The inverse sine of $x$ is the multifunction defined as: $\map {\sin^{-1} } x := \set {y \in \R: \map \sin y = x}$ where $\map \sin y$ is the sine of $y$. ### Arcsine Arcsine Function From Shape of Sine Function, we have that $\sin x$ is continuous and strictly increasing on the interval $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$. $\map \sin {-\dfrac {\pi} 2} = -1$ and: $\sin \dfrac {\pi} 2 = 1$ Therefore, let $g: \closedint {-\dfrac \pi 2} {\dfrac \pi 2} \to \closedint {-1} 1$ be the restriction of $\sin x$ to $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$. Thus from Inverse of Strictly Monotone Function, $g \paren x$ admits an inverse function, which will be continuous and strictly increasing on $\closedint {-1} 1$. This function is called arcsine of $x$. Thus: The domain of arcsine is $\closedint {-1} 1$ The image of arcsine is $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$. ## Terminology There exists the popular but misleading notation $\sin^{-1} x$, which is supposed to denote the inverse sine function. However, note that as $\sin x$ is not an injection (even though by restriction of the codomain it can be considered surjective), it does not have a well-defined inverse. The $\arcsin$ function as defined here has a well-specified image which (to a certain extent) is arbitrarily chosen for convenience. Therefore it is preferred to the notation $\sin^{-1} x$, which (as pointed out) can be confusing and misleading. Sometimes, $\operatorname {Sin}^{-1}$ (with a capital $\text S$) is taken to mean the same as $\arcsin$. However, this can also be confusing due to the visual similarity between that and the lowercase $\text s$. In computer software packages, the notation $\operatorname {asin}$ or $\operatorname {asn}$ can sometimes be found. Some sources hyphenate: arc-sine.
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# If we centered the matrix then the rank is at most d-1 I have found that centering is defined as: We let $$\bar{x}$$ be the in-sample mean vector of the input data so $$\bar{x}=\frac{1}{N}\sum_{n=1}^Nx_n$$ and in matrix notation can be written as $$\bar{x}=\frac{1}{N}X^T1$$, $$1$$ is the column vector of N 1's. Then we substrate the mean from each point $$z_n=x_n-\bar{x}$$. By calculation we have that $$Z=X-1\bar{x}^T$$ hence we have that: $$\bar{z}=\frac{1}{N}Z^T1=\frac{1}{N}X^T1-\frac{1}{N}\bar{x}1^T1$$ And use definition of $$\bar{x}=\frac{1}{N}X^T1$$ and that $$1^T1=N$$ and get: $$=\bar{x}-\frac{1}{N}\bar{x}N=0$$ Thus, the transformed vectors are ‘centered’ in that they have zero mean, as desire. But now I have to use this definition of centering, to show that if we have $$dxd$$ matrix denoted S, then if we centered the matrix then the rank of the resulting matrix at most $$d-1$$. I'm a bit confused how to prove that. I'm not sure about centering matrix, but I think if we get that the result matrix is $$d$$ or higher then we do not have that the transformed vectors are ‘centered’ and they will not have zero mean? But how can I prove that? Can anyone help? $$\bf 1$$ is a left kernel vector of $$S$$, as you computed, $${\bf 1}^TS=0$$. Thus $$Z$$ can not have full rank, as you have $$S$$ given as square matrix.
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# Finding prime factorization of ideals in number rings Let $K$ be a number field and $O_K$ its ring of algebraic integers. Let $p\in\mathbb{Z}$ be a rational prime. I want to find the factorization of the ideal $pO_K$. What is the syntax for this ? For clarity, I request you to demonstrate with an example (say $K=\mathbb{Q}(\sqrt{2}+i)$ and $p=2$ and $p=3$). edit retag close merge delete Sort by ยป oldest newest most voted Define your number field $\mathbb{Q}(\alpha), \alpha = \sqrt{2} + i$ . K.<a> = NumberField(definingPolynomial) Z.<x> = ZZ[] #Makes x lives in Z[x] K.<a> = NumberField( minpoly(sqrt(2)+i, x)) I = K.ideal(2) factor(I) #(Fractional ideal (1/12*a^3 - 1/4*a^2 - 5/12*a + 5/4))^4 #Even fancier latex(factor(I)) #(\left(\frac{1}{12} a^{3} - \frac{1}{4} a^{2} - \frac{5}{12} a + \frac{5}{4}\right))^{4} more What is a ? ( 2016-10-22 02:32:15 -0600 )edit a is the root of definingPolynomial i.e. $f(a) = 0$ ( 2016-10-26 13:32:50 -0600 )edit The discriminant of $K=\mathbb{Q}(\sqrt{2}+i)$ is $256$. As $3\nmid 256$, the ideal $\langle 3\rangle$ should remain inert in $O_K$. So how come it splits in $O_K$ ? ( 2016-10-31 01:51:08 -0600 )edit The theorem says a prime $p$ ramifies iff $p|\Delta$. However, this does NOT imply that a prime $p$ which does not ramify must inert, because there is a third case in which the prime splits. Look at problem for section 3.4, or for detailed treatment (Kummer-Dedekind theorem) see Stevenhagen's lecture notes Number Rings ( 2016-11-20 16:14:53 -0600 )edit
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Quantitative Analysis, Risk Management, Modelling, Algo Trading, and Big Data Analysis # Probability of a Limit Order Executing Your algo strategy can assume that if the price of a given stock starts to increase (or fall) you chip-up (or chip-down). This rule can be programmed. If it works you leverage your position with a hope for higher gain. But let’s look at the same case scenario more from a statistical point of view which is also worth considering. This is a classical textbook example, often forgotten but useful in quick algo estimations. Let’s say you have two limit orders outstanding on the same stock. One is placed at $\$21.50$and the second at$\$21.00$. Knowing that probability that Order 1 executes (event A) is $p_1$ and that Order 2 executes (event B) is $p_2$, $p_1>p_2$, and orders’ execution will take place within the same time-frame, consider the following probabilities: Both or one of placed orders execute Given $P(A)=p_1$ and $P(B)=p_2$, and $$P(A|B) = 1$$ which simply says that if $B$ occurs (prices passes through lower value upwards) at $p_1\ne 0$, the probability that Order 1 and 2 will be executed is: $$P(A\ \mbox{and}\ B) = P(A|B)P(B) = 1\times P(B) = p_2 \ .$$ That leads us to the case of: $$P(A\ \mbox{or}\ B) = P(A)+P(B)-P(A\ \mbox{and}\ B) = p_1 + p_2 – p_2 = p_1 \ .$$ Order 2 executes given Order 1 executed If the price of the stock falls and crossed $\$21.50$, the probability that Order 2 (of lower price) will also execute can be estimated using the principals of probability. Knowing that$P(A\ \mbox{and}\ B)=P(A|B)P(B)$we have: $$P(B\ \mbox{and}\ A) = P(B|A)P(A)$$ and at$P(B\ \mbox{and}\ A) = P(A\ \mbox{and}\ B)$what leads us to the solution: $$P(B|A) = \frac{P(B\ \mbox{and}\ A)}{P(A)} = \frac{p_2}{p_1} \ .$$ Remaining Practical Problem What is the best way to estimate$p_1$and$p_2$in any given trading conditions? • http://afekz.posterous.com/ Andrew Thomas-Woolf My stats knowledge is dangerously poor, but I have modeled things like this before. IMO factors worth pulling into a model: 1) Distance from inside BBO 2) Cumulative depth (volume) to your order 3) Historical order distributions 4) Recent signed trade There are other factors that I’d consider, e.g. current versus recent spread (order book refill probability – & what side?), but the above should cover a lot of ground. • NML Estimating p1 (or p2) is far harder than what you’ve solved here, right? Survival analysis? Or simply comparing the distance from the bid/ask to historal histogram of price evolution? • GiorgioG Hi Pawel, I am suggesting an idea to answer the question your topic ends with. I will take advantage of an approach which makes similar hypothesis as VaR basic analysis does. In doing that, I first determine the capital loss/gain based on known data and then the associated probability is computed. VaR concept does the opposite: a target probability is assumed and then the associated capital loss is computed. Herein below I do that for the price going downward scenario. Po is the current price at current time To. P1 is the price that can be reached with probability p1, (E1 event) P2 is the price that can be reached with probability p2, (E2 event) Po, P1 and P2 are given, p1 and p2 are to be computed. Time frame is 1 day. Example: Po =$22.50, P1 = $21.50, P2 =$21.00 Also daily price standard deviation, sd, must be known; let us suppose sd = 3%. Supposing to hold one share, I compute the VaR based on price change associated to the event E1 and E2. In case of E1 event, VaR is $21.50 –$22.50 = -$1.00 In case of E2 event, VaR is$21.00 – $22.50 = -$1.50 VaR(E1) = -$1.00 = alpha * sd *$22.50 = alpha * 0.03 * $22.50 = alpha *$0.675 VaR(E2) = -$1.50 = alpha * sd *$22.50 = alpha * 0.03 * $22.50 = alpha *$0.675 where alpha is a random variable having normal distribution. E1: alpha = -1/0.675 = -1.48 corresponding to 6.9% probability as Prob(-inf .LT. alpha .LT. -1.48) = 6.9% Similarly: E2: alpha = -1.50/0.675 = -2.22, corresponding to 1.3% probability p1 = 6.9%; p2 = 1.3% Computing the probability of the event {E2|E1} requires to assume the current time is t1 and the current price is P1. If the price moves down to P2, the VaR is: VaR({E2|E1}) = $21.00 –$21.50 = -$0.50 = alpha * sd *$21.50 = alpha * 0.645 alpha = -0.50/0.645 = -0.775 which corresponds to 22.1% probability for the event {E2|E1} Normal distribution values can be found (for example) at: http://www.mathsisfun.com/data/standard-normal-distribution-table.html • GiorgioG An immediate application is to calculate the chances that within the current day the support or resistance prices may be reached. Resistance and support prices are calculated as: #average VM = (H+L+C)/3 #First Level support and resistance: S1 = (2*VM)-H R1 = (2*VM)-L #Second Level support and resistance: S2 = VM – R1 + S1 R2 = VM – S1 + R1 #Third Level support and resistance: S3 = S2 – H + L R3 = R2 + H – L Be N the current daily price at the new day opening and suppose to know the daily standard deviation (sigma). Here is a simple R language routine which computes resistance and support prices with their probabilities to occur within the day. (I hope the printout formatting to be correct). # R language routine: H <- 23.00 # high L <- 21.00 # low C <- 21.70 # closure N <- 22.00 # current price (now) sigma <- 0.03 # standard deviation VM <- (H+L+C)/3 # average S <- c((2*VM)-H, VM-R1+S1, S2-H+L) # supports R <- c((2*VM)-L, VM-S1+R1, R2+H-L) # resistances alpha <- function(SR, N, sigma) {(SR-N)/(sigma*N)} # the alpha standard normal distributed random variable prob <- function(a) {if(a < 0) y = pnorm(a, lower.tail=TRUE) else y = pnorm(a, lower.tail=FALSE); y} # probability as <- c(alpha(S[1],N,sigma),alpha(S[2],N,sigma),alpha(S[3],N,sigma)) ar <- c(alpha(R[1],N,sigma),alpha(R[2],N,sigma),alpha(R[3],N,sigma)) ProbS <- c(prob(as[1]),prob(as[2]),prob(as[3])) ProbR <- c(prob(ar[1]),prob(ar[2]),prob(ar[3])) F1 <- data.frame(R,ProbR) F2 <- data.frame(S,ProbS) F1 F2 • http://www.bearcave.com Ian Kaplan I have not implemented this algorithm, so I am writing about this without actual market experience. The first thing that I would think of is to look at the short term volatility distribution. From this it should be possible to estimate the probability of limit order execution. Perhaps this is what is meant by the probabilities above. This looks to me like a Bayesian probability argument. My statistics educations has been completely “frequentist” so this is an area where I am weak (I have a great book on Bayesian statistics, but I have not had the time to read it).
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Geometry: Common Core (15th Edition) midpoint = $(-6, -2)$ In this problem, we need to use the midpoint formula. The midpoint can be found using the following formula: midpoint = $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$, where $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of the line segment. The points we are given are $M (6, -11)$ and $N (-18, 7)$. Let's plug these points into the formula: midpoint = $(\frac{6 + (-18)}{2}, \frac{-11 + 7}{2})$ Use addition to simplify: midpoint = $(\frac{-12}{2}, \frac{-4}{2})$ Divide to simplify: midpoint = $(-6, -2)$
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# Search by Topic #### Resources tagged with Interactivities similar to Four Triangles Puzzle: Filter by: Content type: Age range: Challenge level: ### There are 142 results Broad Topics > Information and Communications Technology > Interactivities ### Four Triangles Puzzle ##### Age 5 to 11 Challenge Level: Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together? ### Inside Triangles ##### Age 5 to 7 Challenge Level: How many different triangles can you draw on the dotty grid which each have one dot in the middle? ### Nine-pin Triangles ##### Age 7 to 11 Challenge Level: How many different triangles can you make on a circular pegboard that has nine pegs? ### Coloured Squares ##### Age 5 to 7 Challenge Level: Use the clues to colour each square. ### Sort the Street ##### Age 5 to 7 Challenge Level: Sort the houses in my street into different groups. Can you do it in any other ways? ### Fault-free Rectangles ##### Age 7 to 11 Challenge Level: Find out what a "fault-free" rectangle is and try to make some of your own. ### Red Even ##### Age 7 to 11 Challenge Level: You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters? ### Tetrafit ##### Age 7 to 11 Challenge Level: A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard? ### Turning ##### Age 5 to 7 Challenge Level: Use your mouse to move the red and green parts of this disc. Can you make images which show the turnings described? ### Find the Difference ##### Age 5 to 7 Challenge Level: Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it. ### Tessellate the Triominoes ##### Age 5 to 7 Challenge Level: What happens when you try and fit the triomino pieces into these two grids? ### Cover the Camel ##### Age 5 to 7 Challenge Level: Can you cover the camel with these pieces? ### Seven Flipped ##### Age 7 to 11 Challenge Level: Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time. ### Counters ##### Age 7 to 11 Challenge Level: Hover your mouse over the counters to see which ones will be removed. Click to remove them. The winner is the last one to remove a counter. How you can make sure you win? ### More Transformations on a Pegboard ##### Age 7 to 11 Challenge Level: Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle. ### Combining Cuisenaire ##### Age 7 to 11 Challenge Level: Can you find all the different ways of lining up these Cuisenaire rods? ### Difference ##### Age 7 to 11 Challenge Level: Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it. ### Making Trains ##### Age 5 to 7 Challenge Level: Can you make a train the same length as Laura's but using three differently coloured rods? Is there only one way of doing it? ### Sorting Symmetries ##### Age 7 to 11 Challenge Level: Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it. ### Getting the Balance ##### Age 5 to 7 Challenge Level: If you hang two weights on one side of this balance, in how many different ways can you hang three weights on the other side for it to be balanced? ### Triangles All Around ##### Age 7 to 11 Challenge Level: Can you find all the different triangles on these peg boards, and find their angles? ### Same Length Trains ##### Age 5 to 7 Challenge Level: How many trains can you make which are the same length as Matt's, using rods that are identical? ### Teddy Town ##### Age 5 to 14 Challenge Level: There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules? ### Fair Exchange ##### Age 5 to 7 Challenge Level: In your bank, you have three types of coins. The number of spots shows how much they are worth. Can you choose coins to exchange with the groups given to make the same total? ### Two and One ##### Age 5 to 7 Challenge Level: Terry and Ali are playing a game with three balls. Is it fair that Terry wins when the middle ball is red? ### Three Ball Line Up ##### Age 5 to 7 Challenge Level: Use the interactivity to help get a feel for this problem and to find out all the possible ways the balls could land. ### Are You Well Balanced? ##### Age 5 to 7 Challenge Level: Can you work out how to balance this equaliser? You can put more than one weight on a hook. ### Code Breaker ##### Age 7 to 11 Challenge Level: This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code? ### Cuisenaire Counting ##### Age 5 to 7 Challenge Level: Here are some rods that are different colours. How could I make a dark green rod using yellow and white rods? ### Arrangements ##### Age 7 to 11 Challenge Level: Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....? ### Triangle Edges ##### Age 5 to 7 Challenge Level: How many triangles can you make using sticks that are 3cm, 4cm and 5cm long? ### Growing Garlic ##### Age 5 to 7 Challenge Level: Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had. ##### Age 7 to 11 Challenge Level: Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations? ### Triangle Animals ##### Age 5 to 7 Challenge Level: How many different ways can you find to join three equilateral triangles together? Can you convince us that you have found them all? ### Matching Triangles ##### Age 5 to 7 Challenge Level: Can you sort these triangles into three different families and explain how you did it? ### Multiples Grid ##### Age 7 to 11 Challenge Level: What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares? ### First Connect Three for Two ##### Age 7 to 11 Challenge Level: First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line. ### One to Fifteen ##### Age 7 to 11 Challenge Level: Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line? ### Junior Frogs ##### Age 5 to 11 Challenge Level: Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible? ### A Square of Numbers ##### Age 7 to 11 Challenge Level: Can you put the numbers 1 to 8 into the circles so that the four calculations are correct? ### Winning the Lottery ##### Age 7 to 11 Challenge Level: Try out the lottery that is played in a far-away land. What is the chance of winning? ### Which Symbol? ##### Age 7 to 11 Challenge Level: Choose a symbol to put into the number sentence. ##### Age 5 to 11 Challenge Level: Place six toy ladybirds into the box so that there are two ladybirds in every column and every row. ### Sizing Them Up ##### Age 5 to 7 Challenge Level: Can you put these shapes in order of size? Start with the smallest. ### One Million to Seven ##### Age 7 to 11 Challenge Level: Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like? ### Domino Numbers ##### Age 7 to 11 Challenge Level: Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be? ### Overlapping Again ##### Age 7 to 11 Challenge Level: What shape is the overlap when you slide one of these shapes half way across another? Can you picture it in your head? Use the interactivity to check your visualisation. ### Posting Triangles ##### Age 5 to 7 Challenge Level: If you can post the triangle with either the blue or yellow colour face up, how many ways can it be posted altogether? ### Factor Lines ##### Age 7 to 14 Challenge Level: Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line. ### Shapely Tiling ##### Age 7 to 11 Challenge Level: Use the interactivity to make this Islamic star and cross design. Can you produce a tessellation of regular octagons with two different types of triangle?
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Free Algebra Tutorials! Home Systems of Linear Equations and Problem Solving Solving Quadratic Equations Solve Absolute Value Inequalities Solving Quadratic Equations Solving Quadratic Inequalities Solving Systems of Equations Row Reduction Solving Systems of Linear Equations by Graphing Solving Quadratic Equations Solving Systems of Linear Equations Solving Linear Equations - Part II Solving Equations I Summative Assessment of Problem-solving and Skills Outcomes Math-Problem Solving:Long Division Face Solving Linear Equations Systems of Linear Equations in Two Variables Solving a System of Linear Equations by Graphing Ti-89 Solving Simultaneous Equations Systems of Linear Equations in Three Variables and Matrix Operations Solving Rational Equations Solving Quadratic Equations by Factoring Solving Quadratic Equations Solving Systems of Linear Equations Systems of Equations in Two Variables Solving Quadratic Equations Solving Exponential and Logarithmic Equations Solving Systems of Linear Equations Solving Quadratic Equations Math Logic & Problem Solving Honors Solving Quadratic Equations by Factoring Solving Literal Equations and Formulas Solving Quadratic Equations by Completing the Square Solving Exponential and Logarithmic Equations Solving Equations with Fractions Solving Equations Solving Linear Equations Solving Linear Equations in One Variable Solving Linear Equations SOLVING QUADRATIC EQUATIONS USING THE QUADRATIC FORMULA SOLVING LINEAR EQUATIONS Try the Free Math Solver or Scroll down to Tutorials! Depdendent Variable Number of equations to solve: 23456789 Equ. #1: Equ. #2: Equ. #3: Equ. #4: Equ. #5: Equ. #6: Equ. #7: Equ. #8: Equ. #9: Solve for: Dependent Variable Number of inequalities to solve: 23456789 Ineq. #1: Ineq. #2: Ineq. #3: Ineq. #4: Ineq. #5: Ineq. #6: Ineq. #7: Ineq. #8: Ineq. #9: Solve for: Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg: solving difference quotient Related topics: short maths formula | Taks Master 8th Grade Math | how to solve sales tax equations | chapter tests with answer keys algebra 1 | balancing chemical equations in net ionic form | adding multiplying divide and subtracting integers free worksheet | printable pre algebra worksheets Author Message Mishras Registered: 06.03.2003 From: Netherlands Posted: Saturday 30th of Dec 17:10 Heya guys! Is anyone here familiar with solving difference quotient? I have this set of problems about it that I just can’t understand. We were asked to solve it and understand how we came up with the answer . Our Algebra teacher will select random people to answer it as well as show solutions to class so I need detailed explanation about solving difference quotient. I tried answering some of the questions but I guess I got it completely incorrect. Please help me because it’s urgent and the due date is close already and I haven’t yet figured out how to solve this. oc_rana Registered: 08.03.2007 From: egypt,alexandria Posted: Sunday 31st of Dec 07:02 You seem to be facing a similar problem that I had some time back. I too thought of hiring a paid tutor to work it out for me. But they are so pricy that I just could not afford one. So I turned to the internet and found so many software that can help with algebra assignments on proportions, matrices or rational equations. After some trials I found that Algebrator is the best of the lot. I haven’t found a math assignment that I can’t get done through Algebrator. It is absolutely awesome. Best part is, the software gives you a detailed break-up on how to do it yourself. So you actually learn how to work it out yourself. Isn’t it cool? Flash Fnavfy Liom Registered: 15.12.2001 From: Posted: Monday 01st of Jan 17:20 My parents could not afford my college expenses , so I had to work in the evening, after my classes. Solving problems at the end of the day seemed to be difficult for me at those times. A colleague introduced Algebrator to me and since then I never had any problem solving my questions . matdxewf Registered: 12.03.2005 From: Posted: Monday 01st of Jan 18:30 This thing appears to be really good. How can I get it ? I would love to try it out myself, as soon as possible . Hurray! Now I have something to help me! Flash Fnavfy Liom Registered: 15.12.2001 From: Posted: Wednesday 03rd of Jan 09:55 You don’t have to call them up, it can be purchased online. Here’s the link: http://www.solve-variable.com/solving-systems-of-equations-row-reduction.html. They even provide an unreserved money back guarantee , which is just great!
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# Trigonometric problem Here’s my solution to a nice little trigonometric problem posted by miss loi. Show that $\frac{ \tan x + \sec x – 1} { \tan x – \sec x +1 } \equiv \tan x + \sec x$ $\frac{ \tan x + \sec x – 1} { \tan x – \sec x +1 } = \frac{ \sin x + 1 – \cos x } { \sin x – 1 + \cos x }$ $= \frac{ \sin x + 1 – \cos x } { \sin x – 1 + \cos x } \times \frac{ \sin x + 1 + \cos x } { \sin x + \cos x + 1 } = \frac{ (\sin x + 1)^2 – \cos^2 x } { (\sin x + \cos x)^2 – 1 }$ $= \frac{ \sin^2 x + 2 \sin x +1 – \cos^2 x } { 2 \sin x \cos x } = \frac{ 2 \sin^2 x + 2 \sin x } { 2 \sin x \cos x }$ $= \tan x + \sec x$ (QED) This entry was posted in Problems. Bookmark the permalink. ### 3 Responses to Trigonometric problem 1. Miss Loi says: How to get an A1 if you never pass up your homework to Miss Loi?! FYI, mimetex is also available there (she thinks!)for those who knows how to use it – just that sometimes those complex LaTex codes can be a little too profound for her bimbatic brain. 2. Nick says: How about: For $\tan x – \sec x +1 \ne 0$, $(\tan x + \sec x – 1)/(\tan x – \sec x 1) = \tan x + \sec x$ if and only if $\tan x + \sec x – 1 = (\tan x + \sec x)(\tan x – \sec x + 1)$ $= (\tan^2 x – \sec^2 x) (\tan x + \sec x)$ $= \tan x + \sec x – 1$ 3. tpc says: Dear nick, I took the liberty to edit your post to make use of the latexrender plugin. Yes, your proof is nice as well, although as a personal preference, I would avoid writing statements like A = A, so I would remove the $\tan x + \sec x – 1$ in the first line.
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Unit B Unit Activity ### Balancing the Earth’s Energy Budget Objective: The purpose of this activity is to analyze the Earth’s energy budget in greater detail. A mathematical analysis will be done to examine how the Earth’s energy budget is balanced. Materials Required: Everything that you need for this activity is included in the on-screen components. Time Required: approximately 30 minutes The Earth’s energy budget describes the relationship between incoming energy and outgoing energy. On average and over the long term, the Earth is in a state of equilibrium. This means that the Earth has a balanced energy budget with the amount of energy coming in equalled by the amount of energy going out. This balance of “Energy In = Energy Out” should be true at all levels in the Earth system. The Earth’s annual energy budget with numerical values. The diagram to the right highlights the Earth’s annual energy balance. It is different from the energy budget diagram in lesson B4 in two important respects: • It gives actual numerical average values for incoming energy and outgoing energy. • It is more complex and complete in that it explicitly includes the greenhouse effect. #### PART A: Earth’s Energy Budget at the Top of the Atmosphere At the top of the atmosphere, the energy coming in from the sun is balanced by the energy reflected back into space and the net outgoing longwave radiation. The equation describing this balance is: 1. This equation contains three terms. Examine the energy balance diagram carefully and determine a value for each term. 2. 342 W/m2 3. Reflected Solar Radiation = ? 4. 107 W/m2 5. Outgoing Longwave Radiation = ? 6. 235 W/m2 2. Substitute your values into the equation and verify that it balances. 3. L.S. (left side of equation) R.S. (right side of equation) Incoming Solar Radiation = 342 W/m2 Reflected Solar Radiation + Outgoing Longwave Radiation = 107 W/m2 + 235 W/m2 = 342 W/m2 L.S. = R.S. #### PART B: Earth’s Energy Budget at the Earth’s Surface The energy balance at the Earth’s surface is more complicated because it also involves heating from the greenhouse effect. 1. Use the energy balance diagram to examine the mechanics of the greenhouse effect by determining: 1. How much surface radiation is emitted into the atmosphere? 2. 350 W/m2 is emitted into the atmosphere. (Note: Although the Earth actually emits 390 W/m2, approximately 40 W/m2 travels right through the atmosphere and is lost to outer space. This leaves 350 W/m2 to be absorbed by the atmosphere.) 3. How much of the energy absorbed by greenhouse gases is emitted back to the Earth (i.e. the greenhouse effect)? 4. 324 W/m2 is emitted back to the Earth (indicated as Back Radiation in the diagram). 2. What two other methods allow the Earth to lose some of its absorbed surface radiation? 3. Energy from the Earth’s surface can also be lost through thermals (i.e. through conduction and convection) and through evapotranspiration [i.e. the loss of energy from the ground both by evaporation and by transpiration (the release of water from leaves)]. All incoming energy (i.e. Energy In) at the Earth’s surface in the energy balance diagram is represented by arrows going into the Earth’s surface, while all outgoing energy (i.e. Energy Out) is represented by arrows leaving the Earth’s surface. The complete equation describing this energy balance (i.e. Energy In = Energy Out) at the Earth’s surface is: Energy Absorbed by Surface + Back Radiation (from Greenhouse Effect) = Surface Radiation + Evapotranspiration + Thermals 1. Use values from the energy balance diagram to confirm this equation. (Note: The “Surface Radiation” term refers to net surface radiation.) 2. L.S. (left side of equation) R.S. (right side of equation) Energy Absorbed by Surface + Back Radiation (from greenhouse effect) = 168 W/m2 + 324 W/m2 = 492 W/m2 Surface Radiation + Evapotranspiration + Thermals = 390 W/m2 + 78 W/m2 + 24 W/m2 = 492 W/m2 L.S. = R.S. #### PART C: Earth’s Energy Budget in the Atmosphere The most complicated relationship exists in the atmosphere, in part due to the greenhouse effect. Energy from sunlight absorbed by the atmosphere and energy from the Earth’s surface must be balanced by outgoing energy emitted from the atmosphere. The complete equation describing this energy balance in the atmosphere is given by: Sunlight Absorbed by Atmosphere + Thermals + Evapotranspiration + Surface Radiation from Earth (into atmosphere) = Energy Emitted Back to Ground (i.e. Back Radiation) + Energy Emitted by Clouds + Energy Emitted by Atmosphere 1. Use values from the energy balance diagram to confirm this equation. 2. L.S. (left side of equation) R.S. (right side of equation) Sunlight Absorbed by Atmosphere + Thermals + Evapotranspiration + Surface Radiation from Earth (into atmosphere) = 67 W/m2 + 24 W/m2 + 78 W/m2 + 350 W/m2 = 519 W/m2 Energy Emitted Back to Ground (i.e. Back Radiation) + Energy Emitted by Clouds + Energy Emitted by Atmosphere = 324 W/m2 + 30 W/m2 + 165 W/m2 = 519 W/m2 L.S. = R.S.
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# Homework Help: Composite bars 1. Aug 1, 2016 ### chetzread 1. The problem statement, all variables and given/known data φ is the angle of twist , i dont understand why the angle of twist must be the same for 2 bars ... 2. Relevant equations 3. The attempt at a solution They are made of different material , how could the angle of twist be the same for 2 bars ? File size: 30.4 KB Views: 113 File size: 15.4 KB Views: 96 2. Aug 1, 2016 ### BvU For a composite shaft, it is useful if the shaft pieces do not come apart at junctions (like point B) . That means $\phi_1 = \phi_2$ ! 3. Aug 1, 2016 Why? 4. Aug 1, 2016 ### chetzread why the twisting angle are the same? they are different materials, when same twisting moment applied to them , both of them will have different twisting angle,right? 5. Aug 1, 2016 ### BvU What about T = T1 + T2 + T3 + .... ? Can you check how 'twisting moment' is defined ? 6. Aug 1, 2016 ### chetzread ok,i noticed that the torsion(moment ) applied on both bar are not the same,but why are the twisting angle same? 7. Aug 1, 2016 ### BvU If they are not the same, then the parts of the shaft have rotated wrt one another. That's not good at all. Are we talking about the same $\phi$ here ? Work out (7.10) and (7.11) for your composite bar ABC. Surely, at point B you want $\phi$ from bar AB to be the same as $\phi$ from bar BC at point B ? I think that's what he means: at point B $\phi$ from bar AB is $\phi_1$ and $\phi$ from bar BC is $\phi_2$. 8. Aug 1, 2016 ### chetzread so, the torsion here don't allow the parts of the shaft have rotated wrt one another? It's not stated in the question, how do we know that? 9. Aug 1, 2016 ### BvU They mention a (one) composite shaft - not two separate shafts that happen to be in each other's neigborhood. . So they are welded (or glued or screwed, or ..) together. Share this great discussion with others via Reddit, Google+, Twitter, or Facebook Have something to add? Draft saved Draft deleted
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https://socratic.org/questions/what-is-the-unit-vector-that-is-normal-to-the-plane-containing-3-i-j-k-and-i-k
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# What is the unit vector that is normal to the plane containing (- 3 i + j -k) and ( i +k )? Nov 29, 2016 The unit vector is: $\hat{C} = \frac{\sqrt{6}}{6} \hat{i} + \frac{\sqrt{6}}{3} \hat{j} - \frac{\sqrt{6}}{6} \hat{k}$ #### Explanation: Let $\overline{A} = - 3 \hat{i} + \hat{j} - \hat{k}$ Let $\overline{B} = \hat{i} + \hat{k}$ Let $\overline{C} =$ the vector perpendicular to both $\overline{A} \mathmr{and} \overline{B}$ such that the following is true: $\overline{C} = \overline{A} \times \overline{B}$ barC = | (hati, hatj, hatk, hati, hatj), (-3,1,-1,-3,1), (1,0,1,1,0) |# $\overline{C} = \left\{1 \left(1\right) - \left(- 1\right) \left(0\right)\right\} \hat{i} + \left\{- 1 \left(1\right) - \left(- 3\right) \left(1\right)\right\} \hat{j} + \left\{- 3 \left(0\right) - \left(1\right) \left(1\right)\right\} \hat{k}$ $\overline{C} = \hat{i} + 2 \hat{j} - \hat{k}$ Check: $\overline{C} \cdot \overline{A} = \left(1\right) \left(- 3\right) + \left(2\right) \left(1\right) + \left(- 1\right) \left(- 1\right) = 0$ $\overline{C} \cdot \overline{B} = \left(1\right) \left(1\right) + \left(2\right) \left(0\right) + \left(- 1\right) \left(1\right) = 0$ This checks but $\overline{C}$ is not a unit vector. To make it a unit vector we must divide $\overline{C}$ by its magnitude. $| \overline{C} | = \sqrt{{1}^{2} + {2}^{2} + {\left(- 1\right)}^{2}}$ $| \overline{C} | = \sqrt{6}$ Dividing by $\sqrt{6}$ is the same as multiplying by $\frac{\sqrt{6}}{6}$ The unit vector is: $\hat{C} = \frac{\sqrt{6}}{6} \hat{i} + \frac{\sqrt{6}}{3} \hat{j} - \frac{\sqrt{6}}{6} \hat{k}$
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https://bcphysics180.wordpress.com/2014/10/29/day-26-working-with-ca-graphs/
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# Day 26: Working with CA Graphs Physics 11 Today was spent working through graphs of objects moving with constant acceleration.  The majority of students got it ok, but almost everyone needed a few corrections or challenges.  A few students were starting to hit the wall on clearly understanding how these graphical descriptions fit together. The students that had problems reminded me a lot of what I saw when teaching Math 10 last year.  They would show understanding but then a few minutes later would get stuck. And I mean really stuck.  I think this is a perfect example of cognitive overload.  Cognitive Load Theory (CLT) informs us that when a person’s cognition is taxed to its limit, the brain is no longer able to transfer knowledge from short term memory to long term memory (that’s my lay person’s explanation). I believe this is what I see with a few students in physics. I believe that the best way to deal with these situations is to first try and recognize that it’s happening.  If a kid is getting stuck, telling them to go home and do more practice probably won’t help a ton. In the particular situation of CA graphs, I offered three strategies for students to use to help them work through confusion. ## Strategy 1 I encouraged them to break each part of the position-time graph into separate sections.  For each section, they then write in words what the object is doing.  All the kids can do this (moves forward, moves backwards, moves faster, etc).  For each section, I then have them explain how the motion affects velocity (constant, getting faster, getting slower, etc).  Now they graph what they wrote down for velocity The above may seem formulaic, and it is.  However, it is a series of steps that doesn’t dance around conceptual understanding.  It’s a way for students to verbalize their guide to the motion. ## Strategy 2 The next strategy I suggested was to take the position-time graph and sketch tangents on it.  As long as the student understands the correlation between slope of the tangent line and velocity, they can build the velocity-time graph. Again, the strategy still depends on a conceptual understanding and is not just a series of steps. ## Strategy 3 The third strategy I proposed was… oh who am I kidding. I can’t remember what the third strategy was.  It seemed like a good idea at the time. If I remember it later I’ll update this post. Advertisements
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# Stopping distance of a car? 1. Sep 15, 2015 ### David Donald 1. The problem statement, all variables and given/known data While traveling on the highway with your 1000kg car, at 115.2 km/h, where you’re ABS (automatic braking system) is disabled. This means braking is relying solely on the friction of your tires with the road when they stop spinning. A dear jumps into the road 50 meters in front of you. If the frictional force created by you slamming on your brakes is 4000N. What will your final stopping distance be? Will you hit the dear? Assume no air resistance. 2. Relevant equations Kinematics? 3. The attempt at a solution Sum of Forces in The X direction (Force O' Car) - (Force O' Friction) = -ma I solved for acceleration and got -5.8 m/s^2 plugging these into the kinematics equation I got a time... 5.52 seconds plugging that into the Xf = Xo + Vox t + 1/2 a t^2 I got a distance which is wrong what gives? what am i doing wrong? 2. Sep 15, 2015 ### RUber 3. Sep 15, 2015 ### RUber 4. Sep 15, 2015 ### SteamKing Staff Emeritus When someone tells you, "Be a dear," they don't mean for you to drop down on all fours while wearing a hat rack on your head. "Deer" is the animal which jumps out in front of the car. 5. Sep 15, 2015 ### David Donald Ok... So I am not able to get to the same acceleartion I was getting before so now I'm really confused Since the only thing acting on the car when its stopping is the breaking force would it be Sum Fx = -Fbrakes = ma ? 6. Sep 15, 2015 ### RUber That's right.
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# Starting with a false statement, how can one prove anything is true? [duplicate] So I've been learning a bit of logic for class and heard that if you begin with a false statement, you can then prove anything to be true, however I don't entirely understand what this means or how to do it. For example, if $$\sqrt{2}$$ is rational, can you prove that $$1=0$$? ## marked as duplicate by Taroccoesbrocco, copper.hat, Derek Elkins, Namaste logic StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 26 '18 at 19:02 • The general statement is of the form "if $P$ then $Q$". It means that if $P$ is true then $Q$ is true, so either $P$ is false or $Q$ is true. In your example, $P$ is always false, so it says nothing about $Q$. – copper.hat Oct 26 '18 at 16:48 • just for fun! assume there exist $a,b$ relative prime integers such that $\frac{a}{b}=\sqrt{2}$, we can assume $a$ odd (otherwise we can argue in a similar way with $b$) hence $a \text{mod} 2 =1$. Then we have $a^2 =2 b^2$ hence $2|a$ and a is even i.e. $0= a \text{mod} 2 =1$ Q.E:D. – ALG Oct 26 '18 at 17:02 just for fun! assume there exist $$a,b$$ relative prime integers such that $$\frac{a}{b}=\sqrt{2}$$, we can assume $$a$$ odd (otherwise we can argue in a similar way with $$b$$) hence $$a \;\text{mod} \;2 =1$$. We have $$a^2 =2 b^2$$ hence $$2|a$$ and a is even i.e. $$0= a \; \text{mod} \; 2 =1$$ Q.E.D. I believe what you are referring to is vacuous truth, and it's for implications. The statement: "If $$\sqrt{2}$$ is rational, then $$1=0$$" is true logically, because the hypothesis (if $$\sqrt{2}$$ is rational) is false. • Because "$False \implies S$" evaluates to true in logic. This is true independent of $S$. – Mason Oct 26 '18 at 16:45 This is known as the principal of explosion. The idea is that as soon as you can prove two contradictory statements from a axiom system (in classical logic), you can prove anything. For instance, if you have proved that $$\sqrt{2}$$ is irrational, but also have proved (or perhaps just as an axiom) that $$\sqrt{2}$$ is rational, then you can argue as follows: Clearly, it is either the case that $$\sqrt{2}$$ is irrational or that $$1=0$$, since we know the former to be true. Since we also know that $$\sqrt{2}$$ is rational, for the previous statement to be true, it must be that $$1=0$$. The trick here is that, you can say "this or that" by knowing "this", but from "this or that" you can show "that" by knowing "not this". Note that this process requires starting with a contradiction not just with a false statement - but there's no real intrinsic notion of "false" other than "contradictory" within a logical system. Hmmm ... I am not a fan of how that was put ... when it comes to proving things, it is not so much that from a false statement you can infer anything. Logic itself does not care whether things are true or false, and so starting with $$P$$ does not mean that I can infer anything, even if $$P$$ turns out to be false. What is true, however, is that you can infer anything you want from a contradiction. For example, suppose we have your standard contradiction: we have both $$P$$ and $$\neg P$$ Now, from $$P$$ we can infer $$P \lor Q$$ But if we have $$P \lor Q$$, and we also have $$\neg P$$, then we can infer $$Q$$ And so yes, since $$Q$$ can be anything at all, we can infer anything from a contradiction. To go back to the 'false' though: If you know that $$P$$ is true, then if you assume that $$P$$ is false (i.e. you have $$\neg P$$), then indeed you can infer anything you want. But you can't infer anything you want from a false statement alone. What this means is simply that the following is considered an allowed proof step: ... and therefore $$A$$. But we already know that $$\neg A$$, so therefore we conclude $$B$$. Q.E.D. How to do it is just a matter of writing something like the above. The question you don't ask, but should have, is why people accept this. Here my answer would be: The purpose of a proof is to learn something like "in every time, place, and world where such-and-such premises hold, this conclusion will also hold". This is the same as saying "it is impossible for the premises to be true and yet the conclusion is false." When your proof reaches a contradiction what you have shown is that it is impossible for the premises to be true, period. Therefore it is in particular impossible for the premises to be true and at the same time the conclusion is false. The is what it means that the conclusion follows from the premises.
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# 3.4: Properties of the Determinant $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ $$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$ $$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$ $$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ ##### Learning Objectives • Having the choice to compute the determinant of a matrix using cofactor expansion along any row or column is most useful when there are lots of what in a row or column? • Which elementary row operation does not change the determinant of a matrix? • Why do mathematicians rarely smile? • T/F: When computers are used to compute the determinant of a matrix, cofactor expansion is rarely used. In the previous section we learned how to compute the determinant. In this section we learn some of the properties of the determinant, and this will allow us to compute determinants more easily. In the next section we will see one application of determinants. We start with a theorem that gives us more freedom when computing determinants. ##### Theorem $$\PageIndex{1}$$ Cofactor Expansion Along Any Row or Column Let $$A$$ be an $$n\times n$$ matrix. The determinant of $$A$$ can be computed using cofactor expansion along any row or column of $$A$$. We alluded to this fact way back after Example 3.3.3. We had just learned what cofactor expansion was and we practiced along the second row and down the third column. Later, we found the determinant of this matrix by computing the cofactor expansion along the first row. In all three cases, we got the number $$0$$. This wasn’t a coincidence. The above theorem states that all three expansions were actually computing the determinant. How does this help us? By giving us freedom to choose any row or column to use for the expansion, we can choose a row or column that looks “most appealing.” This usually means “it has lots of zeros.” We demonstrate this principle below. ##### Example $$\PageIndex{1}$$ Find the determinant of $A=\left[\begin{array}{cccc}{1}&{2}&{0}&{9}\\{2}&{-3}&{0}&{5}\\{7}&{2}&{3}&{8}\\{-4}&{1}&{0}&{2}\end{array}\right]. \nonumber$ Solution Our first reaction may well be “Oh no! Not another $$4\times 4$$ determinant!” However, we can use cofactor expansion along any row or column that we choose. The third column looks great; it has lots of zeros in it. The cofactor expansion along this column is \begin{align}\begin{aligned} \text{det}(A) & = a_{1,3}C_{1,3} + a_{2,3}C_{2,3} + a_{3,3}C_{3,3}+a_{4,3}C_{4,3} \\ &= 0\cdot C_{1,3} + 0\cdot C_{2,3} + 3\cdot C_{3,3} + 0\cdot C_{4,3}\end{aligned}\end{align} \nonumber The wonderful thing here is that three of our cofactors are multiplied by $$0$$. We won’t bother computing them since they will not contribute to the determinant. Thus \begin{align}\begin{aligned}\text{det}(A)&=3\cdot C_{3,3} \\ &=3\cdot (-1)^{3+3}\cdot\left|\begin{array}{ccc}{1}&{2}&{9}\\{2}&{-3}&{5}\\{-4}&{1}&{2}\end{array}\right| \\ &=3\cdot (-147)\quad\left(\begin{array}{c}{\text{we computed the determinant of the }3\times 3\text{ matrix}} \\ {\text{without showing our work; it is }-147}\end{array}\right) \\ &=-447\end{aligned}\end{align} \nonumber Wow. That was a lot simpler than computing all that we did in Example 3.3.6. Of course, in that example, we didn’t really have any shortcuts that we could have employed. ##### Example $$\PageIndex{2}$$ Find the determinant of $A=\left[\begin{array}{ccccc}{1}&{2}&{3}&{4}&{5}\\{0}&{6}&{7}&{8}&{9}\\{0}&{0}&{10}&{11}&{12}\\{0}&{0}&{0}&{13}&{14} \\ {0}&{0}&{0}&{0}&{15}\end{array}\right]. \nonumber$ Solution At first glance, we think “I don’t want to find the determinant of a $$5\times 5$$ matrix!” However, using our newfound knowledge, we see that things are not that bad. In fact, this problem is very easy. What row or column should we choose to find the determinant along? There are two obvious choices: the first column or the last row. Both have 4 zeros in them. We choose the first column.$$^{1}$$ We omit most of the cofactor expansion, since most of it is just $$0$$: $\text{det}(A)=1\cdot (-1)^{1+1}\cdot\left|\begin{array}{cccc}{6}&{7}&{8}&{9}\\{0}&{10}&{11}&{12}\\{0}&{0}&{13}&{14}\\{0}&{0}&{0}&{15}\end{array}\right|. \nonumber$ Similarly, this determinant is not bad to compute; we again choose to use cofactor expansion along the first column. Note: technically, this cofactor expansion is $$6\cdot(-1)^{1+1}A_{1,1}$$; we are going to drop the $$(-1)^{1+1}$$ terms from here on out in this example (it will show up a lot...). $\text{det}(A)=1\cdot 6\cdot\left|\begin{array}{ccc}{10}&{11}&{12}\\{0}&{13}&{14}\\{0}&{0}&{15}\end{array}\right|. \nonumber$ You can probably can see a trend. We’ll finish out the steps without explaining each one. \begin{align}\begin{aligned}\text{det}(A)&=1\cdot 6\cdot 10\cdot\left|\begin{array}{cc}{13}&{14}\\{0}&{15}\end{array}\right| \\ &=1\cdot 6\cdot 10\cdot 13\cdot 15 \\ &=11700\end{aligned}\end{align} \nonumber We see that the final determinant is the product of the diagonal entries. This works for any triangular matrix (and since diagonal matrices are triangular, it works for diagonal matrices as well). This is an important enough idea that we’ll put it into a box. ##### Key Idea $$\PageIndex{1}$$: The Determinant of Triangular Matrices The determinant of a triangular matrix is the product of its diagonal elements. It is now again time to start thinking like a mathematician. Remember, mathematicians see something new and often ask “How does this relate to things I already know?” So now we ask, “If we change a matrix in some way, how is it’s determinant changed?” The standard way that we change matrices is through elementary row operations. If we perform an elementary row operation on a matrix, how will the determinant of the new matrix compare to the determinant of the original matrix? Let’s experiment first and then we’ll officially state what happens. ##### Example $$\PageIndex{3}$$ Let $A=\left[\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}\right]. \nonumber$ Let $$B$$ be formed from $$A$$ by doing one of the following elementary row operations: 1. $$2R_{1}+R_{2}\to R_{2}$$ 2. $$5R_{1}\to R_{1}$$ 3. $$R_{1}\leftrightarrow R_{2}$$ Find $$\text{det}(A)$$ as well as $$\text{det}(B)$$ for each of the row operations above. Solution It is straightforward to compute $$\text{det}(A) = -2$$. Let $$B$$ be formed by performing the row operation in 1) on $$A$$; thus $B=\left[\begin{array}{cc}{1}&{2}\\{5}&{8}\end{array}\right]. \nonumber$ It is clear that $$\text{det}(B) = -2$$, the same as $$\text{det}(A)$$. Now let $$B$$ be formed by performing the elementary row operation in 2) on $$A$$; that is, $B=\left[\begin{array}{cc}{5}&{10}\\{3}&{4}\end{array}\right]. \nonumber$ We can see that $$\text{det}(B) = -10$$, which is $$5\cdot\text{det}(A)$$. Finally, let $$B$$ be formed by the third row operation given; swap the two rows of $$A$$. We see that $B=\left[\begin{array}{cc}{3}&{4}\\{1}&{2}\end{array}\right] \nonumber$ and that $$\text{det}(B) = 2$$, which is $$(-1)\cdot\text{det}(A)$$. We’ve seen in the above example that there seems to be a relationship between the determinants of matrices “before and after” being changed by elementary row operations. Certainly, one example isn’t enough to base a theory on, and we have not proved anything yet. Regardless, the following theorem is true. ##### Theorem $$\PageIndex{2}$$ The Determinant and Elementary Row Operations Let $$A$$ be an $$n\times n$$ matrix and let $$B$$ be formed by performing one elementary row operation on $$A$$. 1. If $$B$$ is formed from $$A$$ by adding a scalar multiple of one row to another, then $$\text{det}(B) = \text{det}(A)$$. 2. If $$B$$ is formed from $$A$$ by multiplying one row of $$A$$ by a scalar $$k$$, then $$\text{det}(B) = k\cdot\text{det}(A)$$. 3. If $$B$$ is formed from $$A$$ by interchanging two rows of $$A$$, then $$\text{det}(B) = −\text{det}(A)$$. Let’s put this theorem to use in an example. ##### Example $$\PageIndex{4}$$ Let $A=\left[\begin{array}{ccc}{1}&{2}&{1}\\{0}&{1}&{1}\\{1}&{1}&{1}\end{array}\right]. \nonumber$ Compute $$\text{det}(A)$$, then find the determinants of the following matrices by inspection using Theorem $$\PageIndex{2}$$. $B=\left[\begin{array}{ccc}{1}&{1}&{1}\\{1}&{2}&{1}\\{0}&{1}&{1}\end{array}\right]\quad C=\left[\begin{array}{ccc}{1}&{2}&{1}\\{0}&{1}&{1}\\{7}&{7}&{7}\end{array}\right]\quad D=\left[\begin{array}{ccc}{1}&{-1}&{-2}\\{0}&{1}&{1}\\{1}&{1}&{1}\end{array}\right] \nonumber$ Solution Computing $$\text{det}(A)$$ by cofactor expansion down the first column or along the second row seems like the best choice, utilizing the one zero in the matrix. We can quickly confirm that $$\text{det}(A) = 1$$. To compute $$\text{det}(B)$$, notice that the rows of $$A$$ were rearranged to form $$B$$. There are different ways to describe what happened; saying $$R_1\leftrightarrow R_2$$ was followed by $$R_1\leftrightarrow R_3$$ produces $$B$$ from $$A$$. Since there were two row swaps, $$\text{det}(B) = (-1)(-1)\text{det}(A) = \text{det}(A) = 1$$. Notice that $$C$$ is formed from $$A$$ by multiplying the third row by $$7$$. Thus $$\text{det}(C) = 7\cdot\text{det}(A) = 7$$. It takes a little thought, but we can form $$D$$ from $$A$$ by the operation $$-3R_2+R_1\rightarrow R_1$$. This type of elementary row operation does not change determinants, so $$\text{det}(D) = \text{det}(A)$$. Let’s continue to think like mathematicians; mathematicians tend to remember “problems” they’ve encountered in the past,$$^{2}$$ and when they learn something new, in the backs of their minds they try to apply their new knowledge to solve their old problem. What “problem” did we recently uncover? We stated in the last chapter that even computers could not compute the determinant of large matrices with cofactor expansion. How then can we compute the determinant of large matrices? We just learned two interesting and useful facts about matrix determinants. First, the determinant of a triangular matrix is easy to compute: just multiply the diagonal elements. Secondly, we know how elementary row operations affect the determinant. Put these two ideas together: given any square matrix, we can use elementary row operations to put the matrix in triangular form,$$^{3}$$ find the determinant of the new matrix (which is easy), and then adjust that number by recalling what elementary operations we performed. Let’s practice this. ##### Example $$\PageIndex{5}$$ Find the determinant of $$A$$ by first putting $$A$$ into a triangular form, where $A=\left[\begin{array}{ccc}{2}&{4}&{-2}\\{-1}&{-2}&{5}\\{3}&{2}&{1}\end{array}\right]. \nonumber$ Solution In putting $$A$$ into a triangular form, we need not worry about getting leading $$1$$s, but it does tend to make our life easier as we work out a problem by hand. So let’s scale the first row by $$1/2$$: $\frac{1}{2}R_{1}\to R_{1}\quad\left[\begin{array}{ccc}{1}&{2}&{-1}\\{-1}&{-2}&{5}\\{3}&{2}&{1}\end{array}\right]. \nonumber$ Now let’s get $$0$$s below this leading $$1$$: $\begin{array}{c}{R_{1}+R_{2}\to R_{2}}\\{-3R_{1}+R_{3}\to R_{3}}\end{array}\quad\left[\begin{array}{ccc}{1}&{2}&{-1}\\{0}&{0}&{4}\\{0}&{-4}&{4}\end{array}\right]. \nonumber$ We can finish in one step; by interchanging rows $$2$$ and $$3$$ we’ll have our matrix in triangular form. $R_{2}\leftrightarrow R_{3}\quad\left[\begin{array}{ccc}{1}&{2}&{-1}\\{0}&{-4}&{4}\\{0}&{0}&{4}\end{array}\right]. \nonumber$ Let’s name this last matrix $$B$$. The determinant of $$B$$ is easy to compute as it is triangular; $$\text{det}(B) = -16$$. We can use this to find $$\text{det}(A)$$. Recall the steps we used to transform $$A$$ into $$B$$. They are: $\frac 12R_1 \rightarrow R_1 \nonumber$ $R_1 + R_2 \rightarrow R_2 \nonumber$ $-3R_1+R_3\rightarrow R_3 \nonumber$ $R_2 \leftrightarrow R_3 \nonumber$ The first operation multiplied a row of $$A$$ by $$\frac 12$$. This means that the resulting matrix had a determinant that was $$\frac12$$ the determinant of $$A$$. The next two operations did not affect the determinant at all. The last operation, the row swap, changed the sign. Combining these effects, we know that $-16 = \text{det}(B)= (-1)\frac12\text{det}(A). \nonumber$ Solving for $$\text{det}(A)$$ we have that $$\text{det}(A)=32$$. In practice, we don’t need to keep track of operations where we add multiples of one row to another; they simply do not affect the determinant. Also, in practice, these steps are carried out by a computer, and computers don’t care about leading 1s. Therefore, row scaling operations are rarely used. The only things to keep track of are row swaps, and even then all we care about are the number of row swaps. An odd number of row swaps means that the original determinant has the opposite sign of the triangular form matrix; an even number of row swaps means they have the same determinant. Let’s practice this again. ##### Example $$\PageIndex{6}$$ The matrix $$B$$ was formed from $$A$$ using the following elementary row operations, though not necessarily in this order. Find $$\text{det}(A)$$. $B=\left[\begin{array}{ccc}{1}&{2}&{3}\\{0}&{4}&{5}\\{0}&{0}&{6}\end{array}\right] \quad\begin{array}{c}{2R_{1}\to R_{1}} \\ {\frac{1}{3}R_{3}\to R_{3}} \\ {R_{1}\leftrightarrow R_{2}} \\ {6R_{1}+R_{2}\to R_{2}}\end{array} \nonumber$ Solution It is easy to compute $$\text{det}(B)=24$$. In looking at our list of elementary row operations, we see that only the first three have an effect on the determinant. Therefore $24=\text{det}(B)=2\cdot\frac{1}{3}\cdot (-1)\cdot\text{det}(A) \nonumber$ and hence $\text{det}(A)=-36. \nonumber$ In the previous example, we may have been tempted to “rebuild” $$A$$ using the elementary row operations and then computing the determinant. This can be done, but in general it is a bad idea; it takes too much work and it is too easy to make a mistake. Let’s think some more like a mathematician. How does the determinant work with other matrix operations that we know? Specifically, how does the determinant interact with matrix addition, scalar multiplication, matrix multiplication, the transpose and the trace? We’ll again do an example to get an idea of what is going on, then give a theorem to state what is true. ##### Example $$\PageIndex{7}$$ Let $A=\left[\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}\right]\quad\text{and}\quad B=\left[\begin{array}{cc}{2}&{1}\\{3}&{5}\end{array}\right]. \nonumber$ Find the determinants of the matrices $$A$$, $$B$$, $$A + B$$, $$3A$$, $$AB$$, $$A^{T}$$, $$A^{−1}$$, and compare the determinant of these matrices to their trace. Solution We can quickly compute that $$\text{det}(A) = -2$$ and that $$\text{det}(B) = 7$$. \begin{align}\begin{aligned}\text{det}(A-B)&=\text{det}\left(\left[\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}\right]-\left[\begin{array}{cc}{2}&{1}\\{3}&{5}\end{array}\right]\right) \\ &=\left|\begin{array}{cc}{-1}&{1}\\{0}&{-1}\end{array}\right| \\ &=1\end{aligned}\end{align} \nonumber It’s tough to find a connection between $$\text{det}(A-b)$$, $$\text{det}(A)$$ and $$\text{det}(B)$$. \begin{align}\begin{aligned}\text{det}(3A)&=\left|\begin{array}{cc}{3}&{6}\\{9}&{12}\end{array}\right| \\ &=-18\end{aligned}\end{align} \nonumber We can figure this one out; multiplying one row of $$A$$ by $$3$$ increases the determinant by a factor of $$3$$; doing it again (and hence multiplying both rows by $$3$$) increases the determinant again by a factor of $$3$$. Therefore $$\text{det}(3A)=3\cdot 3\cdot\text{det}(A)$$, or $$3^{2}\cdot A$$. \begin{align}\begin{aligned}\text{det}(AB)&=\text{det}\left(\left[\begin{array}{cc}{1}&{2}\\{3}&{4}\end{array}\right]\left[\begin{array}{cc}{2}&{1}\\{3}&{5}\end{array}\right]\right) \\ &=\left|\begin{array}{cc}{8}&{11}\\{18}&{23}\end{array}\right| \\ &=-14\end{aligned}\end{align} \nonumber This one seems clear; $$\text{det}(AB)=\text{det}(A)\text{det}(B)$$. \begin{align}\begin{aligned}\text{det}(A^{T})&=\left|\begin{array}{cc}{1}&{3}\\{2}&{4}\end{array}\right| \\ &=-2\end{aligned}\end{align} \nonumber Obviously $$\text{det}(A^{T})=\text{det}(A)$$; is this always going to be the case? If we think about it, we can see that the cofactor expansion along the first row of $$A$$ will give us the same result as the cofactor expansion along the first column of $$A$$.$$^{4}$$ \begin{align}\begin{aligned}\text{det}(A^{-1})&=\left|\begin{array}{cc}{-2}&{1}\\{3/2}&{-1/2}\end{array}\right| \\ &=1-3/2 \\ &=-1/2\end{aligned}\end{align} \nonumber It seems as though $\text{det}(A^{-1})=\frac{1}{\text{det}(A)}. \nonumber$ We end by remarking that there seems to be no connection whatsoever between the trace of a matrix and its determinant. We leave it to the reader to compute the trace for some of the above matrices and confirm this statement. We now state a theorem which will confirm our conjectures from the previous example. ##### Theorem $$\PageIndex{3}$$ Determinant Properties Let $$A$$ and $$B$$ be $$n\times n$$ matrices and let $$k$$ be a scaler. The following are true: 1. $$\text{det}(kA)=k^{n}\cdot\text{det}(A)$$ 2. $$\text{det}(A^{T})=\text{det}(A)$$ 3. $$\text{det}(AB)=\text{det}(A)\text{det}(B)$$ 4. If $$A$$ is invertible, then $\text{det}(A^{-1})=\frac{1}{\text{det}(A)}.\nonumber$ 5. A matrix $$A$$ is invertible if and only if $$\text{det}(A)\neq 0$$. This last statement of the above theorem is significant: what happens if $$\text{det}(A) = 0$$? It seems that $$\text{det}(A^{-1})="1/0"$$, which is undefined. There actually isn’t a problem here; it turns out that if $$\text{det}(A)=0$$, then $$A$$ is not invertible (hence part 5 of Theorem $$\PageIndex{3}$$). This allows us to add on to our Invertible Matrix Theorem. ##### Theorem $$\PageIndex{4}$$ Invertible Matrix Theorem Let $$A$$ be an $$n\times n$$ matrix. The following statements are equivalent. 1. $$A$$ is invertible. 2. $$\text{det}(A)\neq 0$$. This new addition to the Invertible Matrix Theorem is very useful; we’ll refer back to it in Chapter 4 when we discuss eigenvalues. We end this section with a shortcut for computing the determinants of $$3\times 3$$ matrices. Consider the matrix $$A$$: $\left[\begin{array}{ccc}{1}&{2}&{3}\\{4}&{5}&{6}\\{7}&{8}&{9}\end{array}\right]. \nonumber$ We can compute its determinant using cofactor expansion as we did in Example 3.3.4. Once one becomes proficient at this method, computing the determinant of a $$3\times3$$ isn’t all that hard. A method many find easier, though, starts with rewriting the matrix without the brackets, and repeating the first and second columns at the end as shown below. $\begin{array}{ccccc} 1&2&3&1&2 \\ 4 & 5 & 6&4&5\\7&8&9&7&8\end{array} \nonumber$ In this $$3\times 5$$ array of numbers, there are 3 full “upper left to lower right” diagonals, and 3 full “upper right to lower left” diagonals, as shown below with the arrows. The numbers that appear at the ends of each of the arrows are computed by multiplying the numbers found along the arrows. For instance, the $$105$$ comes from multiplying $$3\cdot5\cdot7=105$$. The determinant is found by adding the numbers on the right, and subtracting the sum of the numbers on the left. That is, $\text{det}(A) = (45+84+96) - (105+48+72) = 0. \nonumber$ To help remind ourselves of this shortcut, we’ll make it into a Key Idea. ##### Key Idea $$\PageIndex{2}$$: $$3\times 3$$ Determinant Shortcut Let $$A$$ be a $$3\times 3$$ matrix. Create a $$3\times 5$$ array by repeating the first $$2$$ columns and consider the products of the $$3$$ “right hand” diagonals and $$3$$ “left hand” diagonals as shown previously. Then \begin{align}\begin{aligned}\text{det}(A)&=\text{"(the sum of the right hand numbers)} \\ & -\text{(the sum of the left hand numbers)".}\end{aligned}\end{align} \nonumber We’ll practice once more in the context of an example. ##### Example $$\PageIndex{8}$$ Find the determinant of $$A$$ using the previously described shortcut, where $A=\left[\begin{array}{ccc}{1}&{3}&{9}\\{-2}&{3}&{4}\\{-5}&{7}&{2}\end{array}\right]. \nonumber$ Solution Rewriting the first $$2$$ columns, drawing the proper diagonals, and multiplying, we get: Summing the numbers on the right and subtracting the sum of the numbers on the left, we get $\text{det}(A) = (6-60-126) - ( -135+28-12) = -61. \nonumber$ In the next section we’ll see how the determinant can be used to solve systems of linear equations. ## Footnotes [1] We do not choose this because it is the better choice; both options are good. We simply had to make a choice. [2] which is why mathematicians rarely smile: they are remembering their problems [3] or echelon form [4] This can be a bit tricky to think out in your head. Try it with a 3$$\times 3$$ matrix $$A$$ and see how it works. All the $$2\times 2$$ submatrices that are created in $$A^{T}$$ are the transpose of those found in $$A$$; this doesn’t matter since it is easy to see that the determinant isn’t affected by the transpose in a $$2\times 2$$ matrix. This page titled 3.4: Properties of the Determinant is shared under a CC BY-NC 3.0 license and was authored, remixed, and/or curated by Gregory Hartman et al. via source content that was edited to the style and standards of the LibreTexts platform.
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# Homework Help: Application of Integration- help! 1. Apr 27, 2010 ### skylit 1. Let f be the function given by f(x) = ex + 1, where the region R is bounded by the graph of f(x), the y-axis, and the horizontal line y=4. 2. Relevant equations a. Find the area of region R. b. A vertical line x=h, where h>o is chosen so that the area of the region bounded by f(x), the y-axis, the horizontal line y=4, and the line x=h is half the area of the region R. What is the value of h? c. Find the volume of the solid formed when region R is rotated about the line y=4. d. A horizontal line y=k, where k is greater than 4 is chosen so that the volume of the solid formed when region R is rotated about the line y=k is twice the volume of the solid found in part (c). Set up, but do not evaluate, an integral expression in terms of a single independent variable which represents the volume of this solid. 3. The attempt at a solution I found part a and b.. a.) 1.296 b.) h=.361 I'm drawing a blank about c, when I draw the graph reflected about y=4, would it be illogical to simply multiply the area given in (a) by 2? And I can't move on without being sure of c.. so that is where I am, haha. 2. Apr 27, 2010 ### zachzach No it would not be logical. If you multiply an area by two it is still an area, not a volume. Do you know how to find the volume? 3. Apr 27, 2010 ### skylit In this case, is it.. volume of a sphere? Or half a sphere? 4. Apr 27, 2010 ### zachzach No. I'm assuming you have drawn the function and found the region (if not do so). Imagine rotating the region around y = 4. To me it looks more like half of a football. 5. Apr 27, 2010 ### zachzach 6. Apr 27, 2010 ### skylit It doesn't ring a bell at all..yes I drew the graph, and rotated about y=4. The first impression I got was a semicircle, but I realize what you're saying about an oval-like shape, which completely disproved my sphere theory haha 7. Apr 27, 2010 ### skylit If this has anything to do with cross sections, then I am in desperate need of help (I could never grasp it) 8. Apr 27, 2010
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# A System of Geometry and Trigonometry: Together with a Treatise on Surveying; Teaching Various Ways of Taking the Survey of a Field; Also to Protract the Same and Find the Area. Likewise, Rectangular Surveying; Or, an Accurate Method of Calculating the Area of Any Field Arithmetically, Without the Necessity of Plotting It. To the Whole are Added Several Mathematical Tables ... with a Particular Explanation ... and the Manner of Using Them ... O. D. Cooke, 1813 - 168 páginas ### Comentarios de la gente -Escribir un comentario No encontramos ningún comentario en los lugares habituales. ### Contenido Sección 1 2 Sección 2 5 Sección 3 9 Sección 4 14 Sección 5 20 Sección 6 28 Sección 7 30 Sección 8 77 Sección 9 81 Sección 10 105 Sección 11 106 Sección 12 112 ### Pasajes populares Página 26 - As the base or sum of the segments Is to the sum of the other two sides, So is the difference of those sides To the difference of the segments of the base. Página 22 - The square of the hypothenuse is equal to the sum of the squares of the other two sides ; as, 5033 402+302. Página 8 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; and each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds ; and these into thirds, &c. Página 25 - TO THEIR DIFFERENCE ; So IS THE TANGENT OF HALF THE SUM OF THE OPPOSITE ANGLES', To THE TANGENT OF HALF THEIR DIFFERENCE. Página 25 - The solution of this CASE depends on the following PROPOSITION. In every Plane Triangle, As the Sum of any two Sides ; Is to their Difference ; So is the Tangent of half the Sum of the two opposite Angles ; To the Tangent of half the Difference between them. Add this half difference to half the Sum of the Angles and you will have the greater Angle ; and... Página 8 - The radius of a circle is a line drawn from the centre to the circumference, as A, B. Página 40 - Field work and protraction are truly taken and performed ; if not, an error must have been committed in one of them : In such cases make a second protraction ; if this agrees with the former, it is to be presumed the fault is in the Field work ; a re-survey must then be taken. Página 40 - Let his attention first be directed to the map, and inform him that the top is north, the bottom south, the right hand east, and the left hand west.
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# Math Games For Skills and Concepts Size: px Start display at page: Transcription 6 Math Games p C 9 P O 3 COVER UP 8 U V 4 Materials: 20 chips, two dice, and the cover up board. 7 E R U Game: (1) Line the chips up over the rows with the letters. (2) Roll two dice. Cover up one or more numbers that add up to the total rolled. For example if you roll a total of 6, you could cover 6, or 5 and 1 or 1, 2 and 3. (3) You might not always have a move if you can t cover up the total roll, you can t cover up any. (4) First player to cover all their numbers wins O V E R 9 P 2 C 10 1 Cover-Up is from the Cuisenaire Teacher Resource Materials. ETA Cuisenaire 7 Math Games p.7 The Sums Game Play begins with each player covering a number from 1 to 9 at the bottom. The 2 nd player then covers the sum of those two numbers on the game board. The 1 st player can then select one new number from 1 to 9 at the bottom and cover the sum of those two numbers. Play continues until one player has covered four squares in a row, horizontally, vertically, or diagonally. (Adaptation of the Product Game) 8 Math Games p.8 T e n F r a m e F i l l U p 9 Math Games p players Content: counting up, subitizing, sums to five, sums to ten. Set Up: Place a scoring chip on the tenth spot in each frame. Ten Frame Fill Up Materials: One die or one per player Ten frames board 10 Blocks for each frame in play Scoring chips (number depends on length of game) Rules: players roll dice to see who goes first. (Highest untied die roll goes first.) Player rolls a die, and adds that many blocks to any ten-frame only one frame at a time. The other player rolls and does the same. If adding blocks to a ten frame would make more than 10, you can not do it. If no ten frame has room for your roll, you can t place any blocks and it s the next players turn. If you fill up a ten frame exactly, take the scoring chip and remove all blocks and place a new scoring chip on the tenth spot. Play goes for a set amount of time. Player with most scoring chips wins. Variations: 1) scoring chips can go on other places than tenth. Player who removes the first scoring chip chooses where the next one goes. 2) If 4 ten frames are too many, just play on 2 frames. Questions: Be sure to ask how many in a frame, how manymore to fill it, what determines a good move, how they made decisions, etc. Players: 2 and up Five In a Row Materials: 1 Gameboard per player, counters, Number cards (1 to 10 only), chips to cover spaces. Goal: Cover 5 spaces in a row, vertically or horizontally or diagonally. Gameplay: Shuffle the number cards and put in a face down pile. On each turn, put the top three cards face up. Each player can cover up any number which is the sum of any two of the revealed cards. For example, 3, 4 and 9 would mean you can cover 7, 12 and 13. Since each child has a different gameboard, this prevents just copying the spaces covered. Questions: Which cards do you need turned up to cover or to finish a row? If a 4 is turned up, what other numbers would you like to see turn up? Variations: 1) Turn up five cards, cover any combination of 2. (There s up to10 possible combinations!) Or do this and allow students to cover only 3 of the combinations they see. 2) Have students work cooperatively on the same board. Or have students make their own boards. Game Boards are 5 by 5 grids, with numbers from 2 to 20 distributed randomly. Use multiple 10s or other sums of interest, and few low numbers. Kids can make up their own boards. 10 Math Games p.10 Five In a Row -- Game Boards 11 Math Games p.11 Race to Game for two players or teams. Materials: Rolling mat, score sheet, 1 die, abacus (or hundreds chart or base ten blocks ) How to Play: roll the die to see who goes first. That player rolls the die onto the rolling sheet. Your hand has to start from not above the sheet. You score whatever you roll if the die is outside of the grey rectangle or off the sheet. You score your roll +10 if the die is on the grey rectangle even if only a little bit is on. If the die is totally within the white oval, you score your roll +20. Keep track of your total score by moving the beads on the abacus. (Or using whatever your method is for keeping score.) The first player to pass 100 wins. If playing again, the winner goes second and the other player goes first. Optional: record score on paper also. Variations: (1) rolls off the mat are subtracted from the total. (2) Start at 100, and subtract the scores to race to zero. Score: 3 Score: 13 Score: 23 One Two One Two One Two 12 Math Games p.12 Roll+10 Roll+20 Roll+10 ### Math Games For Skills and Concepts Math Games p.1 Math Games For Skills and Concepts Other material copyright: Investigations in Number, Data and Space, 1998 TERC. Connected Mathematics Project, 1998 CMP Original material 2001-2006, John ### Ready, Set, Go! Math Games for Serious Minds Math Games with Cards and Dice presented at NAGC November, 2013 Ready, Set, Go! Math Games for Serious Minds Rande McCreight Lincoln Public Schools Lincoln, Nebraska Math Games with Cards Close to 20 - ### MATHS ACTIVITIES FOR REGISTRATION TIME MATHS ACTIVITIES FOR REGISTRATION TIME At the beginning of the year, pair children as partners. You could match different ability children for support. Target Number Write a target number on the board. FUN + GAMES = MATHS Sue Fine Linn Maskell Teachers are often concerned that there isn t enough time to play games in maths classes. But actually there is time to play games and we need to make sure that Third Grade Math Games Unit 1 Lesson Less than You! 1.3 Addition Top-It 1.4 Name That Number 1.6 Beat the Calculator (Addition) 1.8 Buyer & Vendor Game 1.9 Tic-Tac-Toe Addition 1.11 Unit 2 What s My Rule? ### MAKING MATH MORE FUN BRINGS YOU FUN MATH GAME PRINTABLES FOR HOME OR SCHOOL MAKING MATH MORE FUN BRINGS YOU FUN MATH GAME PRINTABLES FOR HOME OR SCHOOL THESE FUN MATH GAME PRINTABLES are brought to you with compliments from Making Math More Fun at and Math Board Games at Copyright ### Foundation 2 Games Booklet MCS Family Maths Night 27 th August 2014 Foundation 2 Games Booklet Stage Focus: Trusting the Count Place Value How are games used in a classroom context? Strategically selected games have become a fantastic Math Board Games For School or Home Education by Teresa Evans Copyright 2005 Teresa Evans. All rights reserved. Permission is given for the making of copies for use in the home or classroom of the purchaser ### Hooray for the Hundreds Chart!! Hooray for the Hundreds Chart!! The hundreds chart consists of a grid of numbers from 1 to 100, with each row containing a group of 10 numbers. As a result, children using this chart can count across rows Ohio Standards Connection Patterns, Functions and Algebra Benchmark E Solve open sentences and explain strategies. Indicator 4 Solve open sentences by representing an expression in more than one way using ### Tasks to Move Students On Maths for Learning Inclusion Tasks to Move Students On Part 1 Maths for Learning Inclusion (M4LI) Tasks to Move Students On Numbers 1 10 Structuring Number Maths for Learning Inclusion (M4LI) Tasks to ### Everyday Math Online Games (Grades 1 to 3) Everyday Math Online Games (Grades 1 to 3) FOR ALL GAMES At any time, click the Hint button to find out what to do next. Click the Skip Directions button to skip the directions and begin playing the game. ### First Grade Exploring Two-Digit Numbers First Grade Exploring Two-Digit Numbers http://focusonmath.files.wordpress.com/2011/02/screen-shot-2011-02-17-at-3-10-19-pm.png North Carolina Department of Public Instruction www.ncdpi.wikispaces.net Table of Contents Introduction to Acing Math page 5 Card Sort (Grades K - 3) page 8 Greater or Less Than (Grades K - 3) page 9 Number Battle (Grades K - 3) page 10 Place Value Number Battle (Grades 1-6) ### Grade 5 Math Content 1 Grade 5 Math Content 1 Number and Operations: Whole Numbers Multiplication and Division In Grade 5, students consolidate their understanding of the computational strategies they use for multiplication. ### Has difficulty with counting reliably in tens from a multiple of ten Has difficulty with counting reliably in tens from a multiple of ten Opportunity for: looking for patterns 5 YR / 100-square Tens cards (Resource sheet 24) Multiples of ten (10 100) written on A5 pieces ### Fifth Grade Physical Education Activities Fifth Grade Physical Education Activities 89 Inclement Weather PASS AND COUNT RESOURCE Indoor Action Games for Elementary Children, pg. 129 DESCRIPTION In this game, students will be ordering whole numbers. Add or Subtract Bingo Please feel free to take 1 set of game directions and master game boards Please feel free to make 1 special game cube (write + on 3 sides and on the remaining 3 sides) You will need ### An Australian Microsoft Partners in Learning (PiL) Project An Australian Microsoft Partners in Learning (PiL) Project 1 Learning objects - Log on to the website: http://www.curriculumsupport.education.nsw.gov.au/countmein/ - Select children Select children - This ### Current California Math Standards Balanced Equations Balanced Equations Current California Math Standards Balanced Equations Grade Three Number Sense 1.0 Students understand the place value of whole numbers: 1.1 Count, read, and write whole numbers to 10,000. ### Decimals and Percentages Decimals and Percentages Specimen Worksheets for Selected Aspects Paul Harling b recognise the number relationship between coordinates in the first quadrant of related points Key Stage 2 (AT2) on a line ### Unit 1 Number Sense. In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions. Unit 1 Number Sense In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions. BLM Three Types of Percent Problems (p L-34) is a summary BLM for the material ### Baseball Multiplication Objective To practice multiplication facts. Baseball Multiplication Objective To practice multiplication facts. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game Family Letters Assessment Management Common ### RACE TO CLEAR THE MAT RACE TO CLEAR THE MAT NUMBER Place Value Counting Addition Subtraction Getting Ready What You ll Need Base Ten Blocks, 1 set per group Base Ten Blocks Place-Value Mat, 1 per child Number cubes marked 1 ### Math vocabulary can be taught with what Montessorians call the Three Period Lesson. Full Transcript of: Montessori Mathematics Materials Presentations Introduction to Montessori Math Demonstrations ( Disclaimer) This program is intended to give the viewers a general understanding of the ### Mental Computation Activities Show Your Thinking Mental Computation Activities Tens rods and unit cubes from sets of base-ten blocks (or use other concrete models for tenths, such as fraction strips and fraction circles) Initially, ### Probabilistic Strategies: Solutions Probability Victor Xu Probabilistic Strategies: Solutions Western PA ARML Practice April 3, 2016 1 Problems 1. You roll two 6-sided dice. What s the probability of rolling at least one 6? There is a 1 ### That s Not Fair! ASSESSMENT #HSMA20. Benchmark Grades: 9-12 That s Not Fair! ASSESSMENT # Benchmark Grades: 9-12 Summary: Students consider the difference between fair and unfair games, using probability to analyze games. The probability will be used to find ways ### Tasks in the Lesson. Mathematical Goals Common Core State Standards. Emphasized Standards for Mathematical Practice. Prior Knowledge Needed Mathematical Goals Common Core State Standards Emphasized Standards for Mathematical Practice Prior Knowledge Needed Vocabulary Lesson 2.4 Five Numbers Bingo Overview and Background Information By the Helping your child with Reading Some ways that you can support. Getting Started Sharing books - We teach phonics to help our children learn to read and write and in order to do this successfully they need by Teresa Evans Copyright 2005 Teresa Evans. All rights reserved. Permission is given for the making of copies for use in the home or classroom of the purchaser only. SAMPLE PAGES Please enjoy using these ### SKILL BUILDING MATH GAMES & ACTIVITIES SKILL BUILDING MATH GAMES & ACTIVITIES (Dave Gardner, Head Teacher, Explorations in Math) ([email protected] - [email protected]) NOTE: When played at the beginning of a math period, many of the games and ### The Crescent Primary School Calculation Policy The Crescent Primary School Calculation Policy Examples of calculation methods for each year group and the progression between each method. January 2015 Our Calculation Policy This calculation policy has ### FIRST GRADE MATH Summer 2011 Standards Summer 2011 1 OA.1 Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in ### Permission is given for the making of copies for use in the home or classroom of the purchaser only. Copyright 2005 Second Edition 2008 Teresa Evans. All rights reserved. Permission is given for the making of copies for use in the home or classroom of the purchaser only. Part 1 Math Card Games to Play ### BISHOP SCHOOL K 5 MATH FACTS GUIDE BISHOP SCHOOL K 5 MATH FACTS GUIDE INTRODUCTION This math fact guide is an outcome of several math related discussions among the staff. There is an agreement that the children need to know their math facts ### FIRST GRADE Number and Number Sense FIRST GRADE Number and Number Sense Hundred Chart Puzzle Reporting Category Number and Number Sense Topic Count and write numerals to 100 Primary SOL 1.1 The student will a) count from 0 to 100 and write ### Using games to support. Win-Win Math Games. by Marilyn Burns 4 Win-Win Math Games by Marilyn Burns photos: bob adler Games can motivate students, capture their interest, and are a great way to get in that paperand-pencil practice. Using games to support students ### WORDS THEIR WAY. 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Each night of the week, your child is expected to do a different ### CONNECTING THE DOTS DOMINO MATH GAMES STEPHANIE GARCIA Box Cars and One-Eyed Jacks CONNECTING THE DOTS DOMINO MATH GAMES STEPHANIE GARCIA FCTM Conference Tampa Bay, FL October 23-25, 2014 [email protected] phone 1-866-342-3386 / 1-780-440-6284 ### Fun Learning Activities for Mentors and Tutors Fun Learning Activities for Mentors and Tutors Mentors can best support children s academic development by having fun learning activities prepared to engage in if the child needs a change in academic/tutoring ### Review: Comparing Fractions Objectives To review the use of equivalent fractions Review: Comparing Fractions Objectives To review the use of equivalent fractions in comparisons. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game Family Letters ### Objective To guide the development and use of a rule for generating equivalent fractions. Family Letters. Assessment Management Equivalent Fractions Objective To guide the development and use of a rule for generating equivalent fractions. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game ### Base-Ten and Place Value 1 Base-Ten and Place Value Jumping Jelly Beans Hundred Board-O Order Up! Number Neighborhood Stretching Numbers Place Value Pause Place Value Bingo 1 2 BRAIN-COMPATIBLE ACTIVITIES FOR MATHEMATICS, GRADES ### Assessment Management Facts Using Doubles Objective To provide opportunities for children to explore and practice doubles-plus-1 and doubles-plus-2 facts, as well as review strategies for solving other addition facts. www.everydaymathonline.com ### OA3-10 Patterns in Addition Tables OA3-10 Patterns in Addition Tables Pages 60 63 Standards: 3.OA.D.9 Goals: Students will identify and describe various patterns in addition tables. 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Fraction Calculations in School. by Kate Robinson + has become 0 Maths in School has become 0 Fraction Calculations in School by Kate Robinson Fractions Calculations in School Contents Introduction p. Simplifying fractions (cancelling down) p. Adding ### Paper 1. Calculator not allowed. Mathematics test. First name. Last name. School. Remember KEY STAGE 3 TIER 5 7 Ma KEY STAGE 3 Mathematics test TIER 5 7 Paper 1 Calculator not allowed First name Last name School 2009 Remember The test is 1 hour long. You must not use a calculator for any question in this test. You ### Pocantico Hills School District Grade 1 Math Curriculum Draft Pocantico Hills School District Grade 1 Math Curriculum Draft Patterns /Number Sense/Statistics Content Strands: Performance Indicators 1.A.1 Determine and discuss patterns in arithmetic (what comes next ### Understand numbers, ways of representing numbers, relationships among numbers, and number systems Equivalent Fractions and Comparing Fractions: Are You My Equal? Brief Overview: This four day lesson plan will explore the mathematical concept of identifying equivalent fractions and using this knowledge ### 1 ST GRADE COMMON CORE STANDARDS FOR SAXON MATH 1 ST GRADE COMMON CORE STANDARDS FOR SAXON MATH Calendar The following tables show the CCSS focus of The Meeting activities, which appear at the beginning of each numbered lesson and are taught daily, ### Zero-knowledge games. Christmas Lectures 2008 Security is very important on the internet. You often need to prove to another person that you know something but without letting them know what the information actually is (because they could just copy ### The Lattice Method of Multiplication The Lattice Method of Multiplication Objective To review and provide practice with the lattice method for multiplication of whole numbers and decimals. www.everydaymathonline.com epresentations etoolkit ### TEACHER S GUIDE TO RUSH HOUR Using Puzzles to Teach Problem Solving TEACHER S GUIDE TO RUSH HOUR Includes Rush Hour 2, 3, 4, Rush Hour Jr., Railroad Rush Hour and Safari Rush Hour BENEFITS Rush Hour is a sliding piece puzzle that ### Calculator Practice: Computation with Fractions Calculator Practice: Computation with Fractions Objectives To provide practice adding fractions with unlike denominators and using a calculator to solve fraction problems. www.everydaymathonline.com epresentations T276 Mathematics Success Grade 6 [OBJECTIVE] The student will add and subtract with decimals to the thousandths place in mathematical and real-world situations. [PREREQUISITE SKILLS] addition and subtraction ### Adding and Subtracting Integers Unit. Grade 7 Math. 5 Days. Tools: Algebra Tiles. Four-Pan Algebra Balance. Playing Cards Adding and Subtracting Integers Unit Grade 7 Math 5 Days Tools: Algebra Tiles Four-Pan Algebra Balance Playing Cards By Dawn Meginley 1 Objectives and Standards Objectives: Students will be able to add ### Math Journal HMH Mega Math. itools Number Lesson 1.1 Algebra Number Patterns CC.3.OA.9 Identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. Identify and ### What Is Singapore Math? What Is Singapore Math? You may be wondering what Singapore Math is all about, and with good reason. This is a totally new kind of math for you and your child. What you may not know is that Singapore has ### ALGEBRA. sequence, term, nth term, consecutive, rule, relationship, generate, predict, continue increase, decrease finite, infinite ALGEBRA Pupils should be taught to: Generate and describe sequences As outcomes, Year 7 pupils should, for example: Use, read and write, spelling correctly: sequence, term, nth term, consecutive, rule, ### Five daily lessons. Page 8 Page 8. Page 12. Year 2 Unit 2 Place value and ordering Year 1 Spring term Unit Objectives Year 1 Read and write numerals from 0 to at least 20. Begin to know what each digit in a two-digit number represents. Partition a 'teens' ### Charlesworth School Year Group Maths Targets Charlesworth School Year Group Maths Targets Year One Maths Target Sheet Key Statement KS1 Maths Targets (Expected) These skills must be secure to move beyond expected. I can compare, describe and solve ### Directions: Place greater than (>), less than (<) or equal to (=) symbols to complete the number sentences on the left. Comparing Numbers Week 7 26) 27) 28) Directions: Place greater than (>), less than ( ### Introduction to Fractions, Equivalent and Simplifying (1-2 days) Introduction to Fractions, Equivalent and Simplifying (1-2 days) 1. Fraction 2. Numerator 3. Denominator 4. Equivalent 5. Simplest form Real World Examples: 1. Fractions in general, why and where we use ### Unit 1: Family Letter Name Date Time HOME LINK Unit 1: Family Letter Introduction to Third Grade Everyday Mathematics Welcome to Third Grade Everyday Mathematics. It is part of an elementary school mathematics curriculum developed ### Objectives To review making ballpark estimates; and to review the counting-up and trade-first subtraction algorithms. materials. materials. Objectives To review making ballpark estimates; and to review the counting-up and trade-first subtraction algorithms. Teaching the Lesson materials Key Activities Children make ballpark estimates for -digit ### Year 2 Summer Term Oral and Mental Starter Activity Bank Year 2 Summer Term Oral and Mental Starter Activity Bank Objectives for term Recall x2 table facts and derive division facts. Recognise multiples of 5. Recall facts in x5 table. Recall x10 table and derive ### 1. I have 4 sides. My opposite sides are equal. I have 4 right angles. Which shape am I? Which Shape? This problem gives you the chance to: identify and describe shapes use clues to solve riddles Use shapes A, B, or C to solve the riddles. A B C 1. I have 4 sides. My opposite sides are equal. ### Curriculum Design for Mathematic Lesson Probability Curriculum Design for Mathematic Lesson Probability This curriculum design is for the 8th grade students who are going to learn Probability and trying to show the easiest way for them to go into this class. ### Unit 6 Number and Operations in Base Ten: Decimals Unit 6 Number and Operations in Base Ten: Decimals Introduction Students will extend the place value system to decimals. They will apply their understanding of models for decimals and decimal notation, ### Math 728 Lesson Plan Math 728 Lesson Plan Tatsiana Maskalevich January 27, 2011 Topic: Probability involving sampling without replacement and dependent trials. Grade Level: 8-12 Objective: Compute the probability of winning ### 3 + 7 1 2. 6 2 + 1. 7 0. 1 200 and 30 100 100 10 10 10. Maths in School. Addition in School. by Kate Robinson 1 2. 6 2 + 1. 7 0 10 3 + 7 1 4. 3 2 1 231 200 and 30 100 100 10 10 10 Maths in School Addition in School by Kate Robinson 2 Addition in School Contents Introduction p.3 Adding in everyday life p.3 Coat ### Frames and Arrows Having Two Rules Frames and Arrows Having Two s Objective To guide children as they solve Frames-and-Arrows problems having two rules. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop A Correlation of to the Minnesota Academic Standards Grades K-6 G/M-204 Introduction This document demonstrates the high degree of success students will achieve when using Scott Foresman Addison Wesley ### Objective To introduce the concept of square roots and the use of the square-root key on a calculator. Assessment Management Unsquaring Numbers Objective To introduce the concept of square roots and the use of the square-root key on a calculator. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts ### Simplifying Improper Fractions Poster Simplifying Improper Fractions Poster Congratulations on your purchase of this Really Good Stuff Simplifying Improper Fractions Poster a reference tool showing students how to change improper fractions ### PROGRESSION THROUGH CALCULATIONS FOR MULTIPLICATION PROGRESSION THROUGH CALCULATIONS FOR MULTIPLICATION By the end of year 6, children will have a range of calculation methods, mental and written. Selection will depend upon the numbers involved. Children ### Introduce Decimals with an Art Project Criteria Charts, Rubrics, Standards By Susan Ferdman Introduce Decimals with an Art Project Criteria Charts, Rubrics, Standards By Susan Ferdman hundredths tenths ones tens Decimal Art An Introduction to Decimals Directions: Part 1: Coloring Have children ### Hoover High School Math League. Counting and Probability Hoover High School Math League Counting and Probability Problems. At a sandwich shop there are 2 kinds of bread, 5 kinds of cold cuts, 3 kinds of cheese, and 2 kinds of dressing. How many different sandwiches ### 7 th Grade Integer Arithmetic 7-Day Unit Plan by Brian M. Fischer Lackawanna Middle/High School 7 th Grade Integer Arithmetic 7-Day Unit Plan by Brian M. Fischer Lackawanna Middle/High School Page 1 of 20 Table of Contents Unit Objectives........ 3 NCTM Standards.... 3 NYS Standards....3 Resources ### Algebra Sequence - A Card/Board Game Algebra Sequence - A Card/Board Game (Based on the Sequence game by Jax, Ltd. Adapted by Shelli Temple) ASSEMBLY: Print out the game board twice, trim the excess white edges and glue into a file folder. ### Paper 1. Calculator not allowed. Mathematics test. First name. Last name. School. Remember KEY STAGE 3 TIER 4 6 Ma KEY STAGE 3 Mathematics test TIER 4 6 Paper 1 Calculator not allowed First name Last name School 2009 Remember The test is 1 hour long. You must not use a calculator for any question in this test. You ### Financial Literacy Meeting Ideas Daisy Financial Literacy Games and Activities Financial Literacy Meeting Ideas Daisy Financial Literacy Games and Activities Fulfills Money Counts steps 1, 2, 3: Money Money You need: Place Value Boards (one for each girl), bags of copied money (one ### NUMBER CORNER YEARLONG CONTENT OVERVIEW August & September Workouts Calendar Grid Quilt Block Symmetries Identifying shapes and symmetries Calendar Collector Two Penny Toss Probability and data analysis Computational Fluency Mental Math Fluently ### Unit 13 Handling data. Year 4. Five daily lessons. Autumn term. Unit Objectives. Link Objectives Unit 13 Handling data Five daily lessons Year 4 Autumn term (Key objectives in bold) Unit Objectives Year 4 Solve a problem by collecting quickly, organising, Pages 114-117 representing and interpreting ### Money Unit \$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$ First Grade Number Sense: By: Jenny Hazeman & Heather Copiskey Money Unit \$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$ First Grade Lesson 1: Introduction to Coins (pennies, nickels, dimes) The Coin Counting Book by Roxanne Williams A ### Multiplication. Year 1 multiply with concrete objects, arrays and pictorial representations Year 1 multiply with concrete objects, arrays and pictorial representations Children will experience equal groups of objects and will count in 2s and 10s and begin to count in 5s. They will work on practical ### Contents. Sample worksheet from www.mathmammoth.com Contents Introduction... 4 Warmup: Mental Math 1... 8 Warmup: Mental Math 2... 10 Review: Addition and Subtraction... 12 Review: Multiplication and Division... 15 Balance Problems and Equations... 19 More ### Comparing Fractions Objective To provide practice ordering sets of fractions. Comparing Fractions Objective To provide practice ordering sets of fractions. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game Family Letters Assessment Management ### Contemporary Mathematics- MAT 130. Probability. a) What is the probability of obtaining a number less than 4? Contemporary Mathematics- MAT 30 Solve the following problems:. A fair die is tossed. What is the probability of obtaining a number less than 4? What is the probability of obtaining a number less than ### Lab 11. Simulations. The Concept Lab 11 Simulations In this lab you ll learn how to create simulations to provide approximate answers to probability questions. We ll make use of a particular kind of structure, called a box model, that ### Change Number Stories Objective To guide children as they use change diagrams to help solve change number stories. Number Stories Objective To guide children as they use change diagrams to help solve change number stories. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game
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Angles Between two Lines in 3D Space: Solved Examples Last Updated : 18 Apr, 2024 Straight Lines in 3D space are generally represented in two forms Cartesian Form and Vector Form. Hence the angles between any two straight lines in 3D space are also defined in terms of both the forms of the straight lines. Let’s discuss the methods of finding the angle between two straight lines in both forms one by one. Cartesian Form L1: (x – x1) / a1 = (y – y1) / b1 = (z – z1) / c1 L2: (x – x2) / a2 = (y – y2) / b2 = (z – z2) / c2 Here L1 & L2 represent the two straight lines passing through the points (x1, y1, z1) and (x2, y2, z2) respectively in 3D space in Cartesian Form. • Direction ratios of line L1 are a1, b1, c1 then a vector parallel to L1 is [Tex]{\vec {m}}[/Tex]1 = a1 i + b1  j + c1 k • Direction ratios of line L2 are a2, b2, c2 then a vector parallel to L2 is [Tex]{\vec {m}}[/Tex]2 = a2 i + b2  j + c2 k Then the angle between L1 and L2 is given by: = cos-1{([Tex]{\vec {m}}[/Tex]1[Tex]{\vec {m}}[/Tex]2) / (|[Tex]{\vec {m}}[/Tex]1| × |[Tex]{\vec {m}}[/Tex]2|)} Examples Example 1: (x – 1) / 1 = (2y + 3) / 3 = (z + 5) / 2 and (x – 2) / 3 = (y + 1) / -2 = (z – 2) / 0 are the two lines in 3D space then the angle ∅ between them is given by: Solution: [Tex]{\vec {m}}[/Tex]1 = 1 i + (3 / 2)  j + 2 k [Tex]{\vec {m}}[/Tex]2 = 3 i – 2 j + 0 k |[Tex]{\vec {m}}[/Tex]1| = √(12 + (3/2)2 + 22) = √(29 / 2) |[Tex]{\vec {m}}[/Tex]2| = √(32 + 22 + 02) = √(13) ∅  = cos-1{(1×3 + (3/2)×(-2) + (2)×0 ) / ((√(29) / 2) × √(13))} ∅  = cos-1{0 / ((√(29) / 2) × √(13))} ∅  = cos-1(0) ∅  = π / 2 Example 2: Find the angles between the two lines in 3D space whose only direction ratios are given 2, 1, 2 and 2, 3, 1. In the question, equations of the 2 lines are not given, only their DRs are given. So the angle ∅ between the 2 lines is given by: Solution: [Tex]{\vec {m}}[/Tex]1 = Vector parallel to the line having DRs 2, 1, 2 = (2 i + j + 2 k) |[Tex]{\vec {m}}[/Tex]1| = √(22 + 12 + 22) = √9 = 3 [Tex]{\vec {m}}[/Tex]2 = Vector parallel to the line having DRs 2, 3, 1 = (2 i + 3 j + k) |[Tex]{\vec {m}}[/Tex]2| = √(22 + 32 + 12) = √(14) ∅  = cos-1{(2×2 + 1×3 + 2×1) / (3 × √(14))} ∅  = cos-1{(4 + 3 + 2) / (3 × √(14))} ∅  = cos-1{9 / (3 × √(14))} ∅  = cos-1(3 / √(14)) Example 3: (x – 1) / 2 = (y – 2) / 1 = (z – 3) / 2 and (x – 2) / 2 = (y – 1) / 2 = (z – 3) / 1 are the two lines in 3D space then the angle ∅ between them is given by: Solution: [Tex]{\vec {m}}[/Tex]1 = 2 i + j + 2 k |[Tex]{\vec {m}}[/Tex]1| = √(22 + 12 + 22) = √9 = 3 [Tex]{\vec {m}}[/Tex]2 = 2 i + 2 j + k |[Tex]{\vec {m}}[/Tex]2| = √(22 + 22 + 12) = √9 = 3 ∅  = cos-1{(2×2 + 1×2 + 2×1 ) / (3 × 3)} ∅  = cos-1{(4 + 2 + 2) / 9} ∅  = cos-1(8 / 9) Vector Form L1[Tex]{\vec {r}}[/Tex] = [Tex]{\vec {a}}[/Tex]1 + t . [Tex]{\vec {b}}[/Tex]1 L2[Tex]{\vec {r}}[/Tex] = [Tex]{\vec {a}}[/Tex]2 + u . [Tex]{\vec {b}}[/Tex]2 Here L1 & L2 represent the two straight lines passing through the points whose position vectors are [Tex]{\vec {a}}[/Tex]1 and [Tex]{\vec {a}}[/Tex]2 respectively in 3D space in Vector Form. [Tex]{\vec {b}}[/Tex]1 & [Tex]{\vec {b}}[/Tex]2 are the two vectors parallel to L1 and L2 respectively and t & u are the parameters. Then the angle between the vectors [Tex]{\vec {b}}[/Tex]1 and [Tex]{\vec {b}}[/Tex]2 is equals to the angle between L1 and L2 is given by: ∅ = cos-1{([Tex]{\vec {b}}[/Tex]1[Tex]{\vec {b}}[/Tex]2) / (|[Tex]{\vec {b}}[/Tex]1| × |[Tex]{\vec {b}}[/Tex]2|)} Examples Example 1: [Tex]{\vec {r}}[/Tex] = (i + j + k) + t × {(-√3 – 1) i + (√3 – 1) j + 4 k} and [Tex]{\vec {r}}[/Tex] = (i + j + k) + u × (i +  j + 2 k)  are the two lines in 3D space then the angle ∅ between them is given by: Solution: [Tex]{\vec {b}}[/Tex]1 = (-√3 – 1) i + (√3 – 1) j + 4 k |[Tex]{\vec {b}}[/Tex]1| = √{(-√3 – 1)2 + (√3 – 1)2 + 42)} = √(24) [Tex]{\vec {b}}[/Tex]2 = i +  j + 2 k |[Tex]{\vec {b}}[/Tex]2| = √(12 + 12 + 22) = √6 ∅  = cos-1{(-√3 – 1)×1 + (√3 – 1)×1 + 4×2 ) / (√(24) × √6)} ∅  = cos-1{6 / (√(24) × √6)} ∅  = cos-1(½) ∅  = π / 3 Example 2: (i + 2 j + 2 k) and (3 i + 2 j + 6 k) are the two vectors parallel to the two lines in 3D space then the angle ∅ between them is given by: Solution: [Tex]{\vec {b}}[/Tex]1 = i + 2 j + 2 k |[Tex]{\vec {b}}[/Tex]1| = √(12 + 22 + 22)} = √9 = 3 [Tex]{\vec {b}}[/Tex]2 = 3 i + 2 j + 6 k |[Tex]{\vec {b}}[/Tex]2| = √(32 + 22 + 62) = √(49) = 7 ∅  = cos-1{(1×3 + 2×2 + 2×6) / (7 × 3)} ∅  = cos-1{(3 + 4 + 12) / 21} ∅  = cos-1(19 / 21) Example 3: [Tex]{\vec {r}}[/Tex] = (3 i + 5 j + 7 k) + s × {(i + 2 j – 2 k} and [Tex]{\vec {r}}[/Tex] = (4 i + 3 j + k) + t × (2 i + 4 j – 4 k)  are the two lines in 3D space then the angle ∅ between them is given by: Solution: [Tex]{\vec {b}}[/Tex]1 = i + 2 j – 2 k |[Tex]{\vec {b}}[/Tex]1| = √(12 + 22 + (-2)2)} = √9 = 3 [Tex]{\vec {b}}[/Tex]2 = 2 i + 4 j – 4 k |[Tex]{\vec {b}}[/Tex]2| = √(22 + 42 + (-4)2) = √(36) = 6 ∅  = cos-1{(1×2 + 2×4 + (-2)×(-4)) / (3 × 6)} ∅  = cos-1{(2 + 8 + 8) / 18} ∅  = cos-1(18 / 18) ∅  = cos-1(1) = 0 Previous Next
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Select Page Data Structures And Algorithms Tutorial PdfIx4.2.4 The following is a demo of the simplest python programming algorithm. This is how the python programming language was developed. This was a guide to learning about common algebra and basic basic functions. Python: basic, basic functions – an introduction This is an example of the simplest python program. It uses semileptonic multiplication (Euclidean multiplication) so the algebra is not an assignment. Before writing the program, you will need to learn its syntax. Here is the code for basic algebra. They use semileptonic multiplication[0] and [1], which is not assignment, since number is not greater than 1. 2 1 1-2-3-4-5-6- 4-5-6-7-8-9- So you will need to remember that number does not always divide second place e.g. 0 + 5 + 6 = 2. Other functions take a third place (the square root) and use divide to divide the result. So, you can define two numbers e1 = [2]*4-5+6=20-11 e2 = [2]*2 By the way, in base 10, e is 2. This is just some fraction. At the end of each iteration, value 1 would be 2*2, and value 2*2 would be 4. Listing In this example, we have 3 lists; 1-3 (5) and 3-4 (7). We can run them as arrays and extract only elements that take a division into a string. All of them are left as strings. ## What Is Graph And Tree In Data Structure? (You should look further and I recommend my own work.) A search is a number of tasks specific for each position. A lot of help is going to pay for the solution. If you understand me, I’m not kidding… I have the actual working of a Multicountic Document as written and I have some preliminary work that will be reviewed as part of the presentation. The real challenge that needs to be resolved is creating a multicountic document with many columns, and very little nested functionality. helpful hints the size of the dataset and then make a dictionary. You can either use as many partitions as you want (2 per table) MapPartitions = MapPartitions(type = DataFrameList) Do this if you know what your partition type is. For example in the PdfTutorial.pdf_data.data_table method we can also create a dict with a width and height. Within mapPartitions we get a list of cdfs and columns of which we can apply the mappings. Because there is a unique name for a cdf’s width and height and because there is an assignment for that width because in the dataset we have an empty column This is an see here and not a solution. If we’re making a separate dictionary for the columns and column lengths we can just create separate instances. Here is a list of columns of the row. The rows themselves try this website very large to make the code too lengthy. If you’d like to do something a little bit more complicated.. ## What Is Data Structure Wikipedia? . these data.table takes quite some time. Especially for those developers who will try to make a nice set of images and video on the web… you may find on that page using this code the entire frontmatter and the image files only get attached to the container-top pages. [1] The details here are pretty basic. I also have generated the class.py and subclasses in the same file, but this version does not have a classpath in it. I was just adding and removing them for quick updates. This is an example that uses a very similar class. However, the data itself is very inefficient. I’ll try to get some order to ensure this won’t come anywhere near happening. If you look at the main.py methods and the Source I have left these out. I also have a nice way to save the classpath in later versions of PdfTutorial.pdf. Here is an example that uses an image file Full Article for the try this out rather than multiple images, or perhaps even just one file: Create images and save them in. It will take some time. ## All Data Structure Programs In C For example, I have a custom image folder so if this does not belong to a specific folder, the data will simply be created back in, for instance, OpenFileLocal from somewhereData Structures And Algorithms Tutorial Pdfftpfxt Title: A Structural Algorithm For Completion Of The Structural Algorithm That Comes From Excel Abstract Completion on a networked environment, such as Excel, is essentially a process of choosing an interface for changing its look and feel. Using a description of the data structure of the environment as the model description, you should have some idea of how to identify the source of the information displayed to you. Here’s how to execute a section of the description of the data within an Excel file: As you can verify only part of this code is provided, you don’t need to run into other file dependencies either. In order to include the other data within the same document or sub documents, you’ll need to run the data structure by executing the data structure like this: And this is the full sequence of the code above. The rest of the code, as described, contains a comprehensive description of the data components and the key properties of the data structure that gets displayed to you: the collection state system the database is in and the data structure property the data connection string storage and performance information from the data structure the structure member function for determining the presentation properties of each property as shown in the attached figure a basic description as a picture of a real data structure accessor information visit this site right here providing the data structure to be displayed to you How to execute the code after you’ve finished the section, you can see in the ‘Summary’ section below or read just before passing the section and put ‘Data Structure’ in the corresponding folder By adding the name of the constructor in the second list I think you’ll be able to get inspiration for how to execute Click This Link described code. For the ’Overview’ section, to get started with the code, you can create a new Excel page or a larger folder of code This is where you can inspect the code as it is being generated in any Windows file As you can sign all of this up for a new document or a sub document, you can use another tool or visual environment to inspect and generate code as follows: There are two major ways the code could be accessed: It’s possible for Excel to find out at the code position it lives in by using the “Function” part of the document creation wizard i.e. the “Components” part. If the current location of two Excel components can be reached in the ‘Completion’ section you might find them in the ‘Accessing Systems’ or ‘Algorithms’ section of code But even this isn’t completely apparent as you work through the code to get some idea of the structure of the document. The ‘Calculations’ section of code shows where where Excel computes the structure that will be on the final document This section holds a number of properties relating to ‘function’ and ‘method’, in particular the “name” property. Now, I’m no expert but I’ll use this to show that some of the properties are very important if you ever need to write code to make calculations in Excel. To represent the properties
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# log math needed asap 1.)Change to exponents help me please. a.)log3 81=4 b)log4 1/256=-4 2.)Change to exponent: a.)log3 81=4 b)5^7t=a+b/a What is: 3.)log4  (3z) + log4x 4.)9^(x^2)=3^3x+2 5.)9^2x multiplied by (1/3)^x+2 =27 multiplied by (3^x)^-2
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16 Q: # A train with 120 wagons crosses Arun who is going in the same direction, in 36 seconds. It travels for half an hour from the time it starts overtaking the Arun ( he is riding on the horse) before it starts overtaking the Sriram( who is also riding on his horse) coming from the opposite direction in 24 seconds. In how much time (in seconds) after the train has crossed the Sriram do the Arun meets to Sriram? A) 3560 sec B) 3600 sec C) 3576 sec D) can't be determined Explanation: Let the length of the train be L metres and speeds of the train Arun and Sriram be R, A and S respectively, then $LR-A=36$ ---------- (i) and $LR+K=24$ ---------(ii) From eq.(i) and (ii) 3(R - A ) = 2 (R + K) $⇒$  R = 3A + 2K In 30 minutes (i.e 1800 seconds), the train covers 1800R (distance) but the Arun also covers 1800 A (distance) in the same time. Therefore distance between Arun and Sriram, when the train has just crossed Sriram = 1800 ( R - A) - 24 ( A + K) Time required = $1800(R-A)-24(A+K)(A+K)$ = (3600 - 24) = 3576 s Q: Buses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes? A) 5 kmph B) 6 kmph C) 7.5 kmph D) 8 kmph Explanation: Let p be the speed of man in kmph According to the given data in the question, Distance travelled by bus in 10 min with 20 kmph == Distance travelled by man in 8 min with (20 + p) kmph in opposite direction => 20 x 10/60 = 8/60 (20 + p) => 200 = 160 + 8p => p = 40/8 = 5 kmph. 1 47 Q: With an average speed of 40 km/hr, a train reaches its destination in time. If it goes with an average speed of 35 km/hr, it is late by 15 minutes. Find the length of the total journey? A) 70 kms B) 60 kms C) 45 kms D) 30 kms Explanation: Let the time taken by train be 't' hrs. Then, 40t = 35t + 35/4 t = 7/4 hrs Therefore, Required length of the total journey d = s x t = 40 x 7/4 = 70 kms. 3 405 Q: Ashwin fires two bullets from the same place at an interval of 15 minutes but Rahul sitting in a bus approaching the place hears the second sound 14 minutes 30 seconds after the first. If sound travels at the speed of 330 meter per second, what is the approximate speed of bus? A) 330/29 m/s B) 330 x 30 m/s C) 330/14 m/s D) 330/900 m/s Explanation: Second gun shot take 30 sec to reach rahul imples distance between two. given speed of sound = 330 m/s Now, distance = 330 m/s x 30 sec Hence, speed of the bus = d/t = 330x30/(14x60 + 30) = 330/29 m/s. 2 303 Q: Important Time and Distance formulas with examples. 1. How to find Speed(s) if distance(d) & time(t) is given: Ex: Find speed if a person travels 4 kms in 2 hrs? Speed = D/T = 4/2 = 2 kmph. 2.  Similarly, we can find distance (d)  if speed (s) & time (t) is given by Distance (D) = Speed (S) x Time (T) Ex : Find distance if a person with a speed of 2 kmph in 2 hrs? Distance D = S X T = 2 x 2 = 4 kms. 3.  Similarly, we can find time (t) if speed (s) &  distance (d) is given by Time (T) = Ex : Find in what time a person travels 4 kms with a speed of 2 kmph? Time T = D/S = 4/2 = 2 hrs. 4. How to convert  km/hr into m/sec : Ex : Convert 36 kmph into m/sec? 36 kmph = 36 x 5/18 = 10 m/sec 5. How to convert  m/sec into km/hr : . Ex : Convert 10 m/sec into km/hr? 10 m/sec = 10 x 18/5 = 36 kmph. 6.  If the ratio of the speeds of A and B is a:b, then the ratio of the times taken by them to cover the same distance is b : a. 7.  Suppose a man covers a certain distance at x km/ hr and an equal distance at y km/hr . Then, the average speed during the whole journey is $2xyx+y$ km/hr. 296 Q: Two trains are running with speeds 30 kmph and 58 kmph respectively in the same direction. A man in the slower train passes the faster train in 18 seconds. Find the length of the faster train? A) 105 mts B) 115 mts C) 120 mts D) 140 mts Explanation: Speeds of two trains = 30 kmph and 58 kmph => Relative speed = 58 - 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s Given a man takes time to cross length of faster train = 18 sec Now, required Length of faster train = speed x time = 70/9  x  18 = 140 mts. 2 299 Q: A train-A passes a stationary train B and a pole in 24 sec and 9 sec respectively. If the speed of train A is 48 kmph, what is the length of train B? A) 200 mts B) 180 mts C) 160 mts D) 145 mts Explanation: Length of train A = 48 x 9 x 5/18 = 120 mts Length of train B = 48 x 24 x 5/18 - 120 => 320 - 120 = 200 mts. 2 436 Q: Tilak rides on a cycle to a place at speed of 22 kmph and comes back at a speed of 20 kmph. If the time taken by him in the second case is 36 min. more than that of the first case, what is the total distance travelled by him (in km)? A) 132 km B) 264 km C) 134 km D) 236 km Explanation: Let the distance travelled by Tilak in first case or second case = d kms Now, from the given data, d/20 = d/22 + 36 min => d/20 = d/22 + 3/5 hrs => d = 132 km. Hence, the total distance travelled by him = d + d = 132 + 132 = 264 kms. 0 382 Q: A train covers 180 km distance in 5 hours. Another train covers the same distance in 1 hour less. What is the difference in the distances covered by these trains in one hour if they are moving in the same direction? A) 15 kms B) 9 kms C) 6 kms D) 18 kms Explanation: The first train covers 180 kms in 5 hrs => Speed = 180/5 = 36 kmph Now the second train covers the same distance in 1 hour less than the first train => 4 hrs => Speed of the second train = 180/4 = 45 kmph Now, required difference in distance in 1 hour = 45 - 36 = 9 kms.
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# Project M³ Curriculum Units — Level 4-5 ## Treasures From the Attic: Exploring Fractions (get the book choose grades 4-5) In this unit students are introduced to two children, Tori and Jordan. In their grandparents’ attic Tori and Jordan uncover hidden treasures from a general store that their grandmother’s grandparents used to own. These treasures lead to an interesting exploration of fraction concepts. The focus of the entire unit is on making sense of fractions rather than on learning algorithms to perform computations. This is a significant departure from more traditional approaches. It is important for students to think about and picture the relative sizes of fractions and make estimates based on their mathematical thinking when ordering, comparing, adding, subtracting, multiplying or dividing two or more fractions. Since fractions are such an important part of our everyday experiences, focusing on meaning rather than rules actually gives students a facility for understanding and working with fractions that will benefit them throughout their lives. This is in line with the research guiding and supporting the Common Core Standards for Mathematics as outlined in the Progressions document, Numbers and Operations – Fractions that can be found at ime.math.arizona.edu/progressions/. In the first chapter, students are exposed to a variety of models (specifically set, linear and area models) to name equivalent fractions. Using a variety of models helps students gain a firm grasp of equivalence, and this, in turn, enables them to generalize and then to apply their understanding to comparing and ordering fractions. When comparing and ordering fractions, students also learn to use multiple strategies to analyze the size of fractions, such as common numerators, common denominators, benchmarks and missing parts of the whole. Students also investigate the density of the real number system as they learn that between any two fractions, another one can always be found. This implies that there are no holes or gaps in the real number line. They also discover something quite exciting—that there is an infinite number of fractions equivalent to any given fraction! In working with addition, subtraction, multiplication and division of fractions in the second chapter, again the focus is on the meaning of mathematical operations as they are used with fractions in a variety of contextual situations. In this chapter, as they solve problems in different situations, students first construct and discuss their own methods for adding and subtracting fractions of all sizes before they are introduced to the common algorithms. Students initially use the physical models, drawings and equivalent fraction charts that they used in the first chapter to develop written algorithms for the operations. After discussing and comparing their own methods, they then compare their methods to the common algorithms and decide which are most meaningful and efficient. This is expanded to an introduction to multiplication and division of fractions, using everyday experiences such as cooking.
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# How to Calculate the Least Common Multiple of Two Numbers in JavaScript In the realm of web development, JavaScript stands as a powerful and versatile programming language. It is widely used to create interactive and dynamic web applications. One of the essential skills for any JavaScript developer is the ability to calculate the Least Common Multiple (LCM) of two numbers programmatically. This article will serve as a comprehensive guide to help you master this skill and create JavaScript programs that can outperform others in terms of search engine rankings. ## Understanding the Least Common Multiple (LCM) Before we delve into the intricacies of JavaScript coding, it’s crucial to grasp the concept of the Least Common Multiple (LCM). The LCM of two numbers, denoted as A and B, is the smallest multiple that is divisible by both A and B. Essentially, it’s the smallest common denominator shared by two integers. To calculate this efficiently in JavaScript, we will employ an algorithm that involves finding the Greatest Common Divisor (GCD) of the two numbers. ## The Algorithm to Calculate LCM To calculate the LCM of two numbers, we will use an algorithm that consists of two main steps: ### 1. Find the Greatest Common Divisor (GCD) The first step in calculating the LCM is to determine the GCD of the two numbers. The GCD is the largest number that evenly divides both A and B. We can employ the well-known Euclidean algorithm for this purpose. Here’s how it works: • Start with the two numbers, A and B. • Using a `while` loop, repeatedly calculate the remainder of A divided by B. Assign this remainder to B and replace A with the previous value of B. • Continue this process until B becomes zero. The value of A at this point will be the GCD of the original two numbers. ### 2. Calculate the LCM Once we have the GCD, we can easily calculate the LCM using the following formula: LCM(A, B) = (A * B) / GCD(A, B) This formula leverages the relationship between the GCD and the LCM of two numbers. By dividing the product of the two numbers by their GCD, we obtain the LCM. ## Writing the JavaScript Program With a clear understanding of the algorithm, let’s proceed to write a JavaScript program that calculates the LCM of two numbers. We will encapsulate the logic in functions for reusability and clarity. ### Finding the Greatest Common Divisor (GCD) javascript ```function findGCD(a, b) { while (b !== 0) { const temp = b; b = a % b; a = temp; } return a; } ``` In this function, we use a `while` loop to iteratively calculate the GCD using the Euclidean algorithm. It efficiently finds the GCD of any two integers. ### Calculating the LCM javascript ```function findLCM(a, b) { return (a * b) / findGCD(a, b); } ``` This function calculates the LCM by invoking the `findGCD` function we defined earlier and applying the LCM formula. ### Example Usage Let’s put our functions to the test with an example: javascript ```const num1 = 24; const num2 = 36; const lcm = findLCM(num1, num2); console.log(`The LCM of \${num1} and \${num2} is \${lcm}`); ``` In this example, we calculate the LCM of 24 and 36, which should yield the result 72. ## FAQs ### 1. What is the practical use of finding the Least Common Multiple (LCM) in programming? The LCM serves as a critical mathematical concept with various practical applications in programming. It is often used in scenarios such as optimizing algorithms, scheduling tasks, and solving mathematical problems that involve finding common multiples. ### 2. Can I use this JavaScript code for large numbers? While the provided JavaScript code works efficiently for small to moderately large numbers, it may not be the most suitable choice for extremely large numbers. In such cases, consider implementing more advanced algorithms or using specialized libraries designed to handle large integers. ### 3. Are there JavaScript libraries that provide LCM functions? Yes, the JavaScript ecosystem offers libraries like `mathjs` and `big-integer` that provide dedicated LCM functions. These libraries are well-suited for handling large numbers and offer optimized algorithms for LCM calculations. ## Conclusion In conclusion, we’ve embarked on a comprehensive journey to understand and implement the calculation of the Least Common Multiple (LCM) of two numbers in JavaScript. We’ve covered the fundamental concept of LCM, elucidated the algorithm, provided a step-by-step JavaScript program, and addressed common questions. Armed with this knowledge, you now possess the skills to calculate the LCM of any two numbers efficiently using JavaScript. Remember that mastering such fundamental concepts is key to becoming a proficient JavaScript developer. Select the fields to be shown. Others will be hidden. Drag and drop to rearrange the order. • Image • SKU • Rating • Price • Stock • Availability • Add to cart • Description • Content • Weight • Dimensions
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# Page:LorentzGravitation1916.djvu/62 Jump to: navigation, search This page has been proofread, but needs to be validated. § 64. Equations (122) show that in the coordinates ${\displaystyle \left(x'_{1},x'_{2},x'_{3},x'_{4}\right)}$ the system has a velocity of translation ${\displaystyle {\tfrac {bc}{a}}}$ in the direction of ${\displaystyle x'_{1}}$. If this velocity is denoted by ${\displaystyle v}$, we have according to (123) ${\displaystyle a={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$ If therefore we put ${\displaystyle M={\frac {E}{c^{2}}}}$ we find ${\displaystyle E'={\frac {Mc^{2}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}},\ G'={\frac {Mv}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$ (126) When the system moves as a whole we may therefore ascribe to it an energy and a momentum which depend on the velocity of translation in the way known from the theory of relativity. The quantity ${\displaystyle M}$, to which the energy of the gravitation field also contributes a certain part, may be called the "mass" of the system. From what has been said in § 62 it follows that within certain limits it depends on the way in which the system and the gravitation field are described. It must be remarked however that, if for the gravitation field we had chosen the stress-energy-tensor ${\displaystyle {\mathfrak {t}}_{0}}$ (§ 52), the total energy of the system even when in motion would be zero. The same would be true of the total momentum and we should have to put ${\displaystyle M=0}$. At first sight it may seem strange that we may arbitrarily ascribe to the moving system the momentum determined by (126) or a momentum 0; one might be inclined to think that, when a definite system of coordinates has been chosen, the momentum must have a definite value, which might be determined by an experiment in which the system is brought to rest by "external" forces. We must remember however (comp. § 52) that in the theory of gravitation we may introduce no "external" forces without considering also the material system ${\displaystyle S'}$ in which they originate. This system ${\displaystyle S'}$ together with the system ${\displaystyle S}$ with which we were originally concerned, will form an entity, in which there is a gravitation field, part of which is due to ${\displaystyle S'}$ (and a part also to the simultaneous existence of ${\displaystyle S}$ and ${\displaystyle S'}$). There is no doubt that we may apply the above considerations to the total system (${\displaystyle S,S'}$) without being led into contradiction with any observation.
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# Advanced Mathematics for Engineers and Scientists/Parallel Plate Flow: Realistic IC (Redirected from Partial Differential Equations/Parallel Plate Flow: Realistic IC) ## Parallel Plate Flow: Realistic ICEdit The initial velocity profile chosen in the last problem agreed with intuition but honestly came out of thin air. A more realistic development follows. The problem stated that (to come up with an IC) the fluid was under a pressure difference for some time, so that the flow became steady aka flowing steadily. "Steady" is another way of saying "not changing with time", and "not changing with time" is another way of saying that: ${\displaystyle {\frac {\partial u}{\partial t}}=0\,}$ Putting this into the PDE from the previous section: ${\displaystyle {\frac {\partial u}{\partial t}}=\nu {\frac {\partial ^{2}u}{\partial y^{2}}}-{\frac {P_{x}}{\rho }}\,}$ ${\displaystyle {\Big \Downarrow }}$ ${\displaystyle 0=\nu {\frac {\partial ^{2}u}{\partial y^{2}}}-{\frac {P_{x}}{\rho }}\quad \Rightarrow \quad {\frac {P_{x}}{\nu \rho }}={\frac {d^{2}u}{dy^{2}}}\,}$ Independent of ${\displaystyle t}$ , the PDE became an ODE with variables separated and thus we can integrate. ${\displaystyle {\frac {d^{2}u}{dy^{2}}}={\frac {P_{x}}{\nu \rho }}\quad \Rightarrow \quad u={\frac {P_{x}}{2\nu \rho }}y^{2}+C_{1}y+C_{0}\,}$ The no slip condition results in the following BCs: ${\displaystyle u=0}$  at ${\displaystyle y=0}$  and ${\displaystyle y=1}$ . We can plug the BC values into the integrated ODE and resolve the Cs. ${\displaystyle 0={\frac {P_{x}}{2\nu \rho }}\cdot 0^{2}+C_{1}\cdot 0+C_{0}\quad \Rightarrow \quad C_{0}=0\qquad {\mbox{(Bottom plate BC: u = 0, y = 0)}}\,}$ ${\displaystyle 0={\frac {P_{x}}{2\nu \rho }}\cdot 1^{2}+C_{1}\cdot 1\quad \Rightarrow \quad C_{1}=-{\frac {P_{x}}{2\nu \rho }}\qquad {\mbox{(Top plate BC: u = 0, y = 1)}}\,}$ Inserting the Cs and and simplifying yields: ${\displaystyle u={\frac {P_{x}}{2\nu \rho }}(y^{2}-y)\,}$ For the sake of example, take ${\displaystyle P_{x}/(2\nu \rho )=-4}$  (recall that a negative pressure gradient causes left to right flow). Also note that this is a constant gradient or slope. This gives a parabola which starts at ${\displaystyle 0}$ , increases to a maximum of ${\displaystyle 1}$  at ${\displaystyle y=1/2}$ , and returns to ${\displaystyle 0}$  at ${\displaystyle y=1}$ . This parabola looks pretty much identical to the sinusoid previously used (you must zoom in to see a difference). However, even more so on the narrow domain of interest, the two are very different functions (look at their taylor expansions, for example). Using the parabola instead of the sine function results in a much more involved solution. So this derives the steady state flow, which we will use as an improved, realistic IC. Recall that the problem is about a fluid that's initially in motion that is coming to a stop due to the absence of a driving force. The IBVP (Initial Boundary Value Problem) is now subtly different: ${\displaystyle {\frac {\partial u}{\partial t}}=\nu {\frac {\partial ^{2}u}{\partial y^{2}}}\qquad {\mbox{(PDE)}}\,}$ ${\displaystyle u(y,0)=4y-4y^{2}\qquad {\mbox{(IC)}}\,}$ ${\displaystyle u(0,t)=0\,}$ ${\displaystyle u(1,t)=0\qquad {\mbox{(BCs)}}\,}$ ### SeparationEdit Since the only difference from the problem in the last section is the IC, the variables may be separated and the BCs applied with no difference, giving: ${\displaystyle u(y,t)=e^{-(n\pi )^{2}\nu t}B\sin(n\pi y)\,}$ But now we're stuck (after applying the BCs)! Applying the IC makes the ${\displaystyle e^{-(n\pi )^{2}\nu t}}$  term go away as t = 0, which is the IC. However, then the IC function can't be made to match: ${\displaystyle u(y,0)=4y-4y^{2}\,}$ ${\displaystyle B\sin(n\pi y)=4y-4y^{2}\,}$ What went wrong? It was the assumption that ${\displaystyle u(y,t)=Y(y)T(t)}$ . The fact that the IC couldn't be fulfilled means that the assumption was wrong. It should be apparent now why the IC was chosen to be ${\displaystyle \sin(\pi y)}$  in the previous section. We can proceed however, thanks to the linearity of the problem. Another detour is necessary, it gets long. Linearity (the superposition principle specifically) says that if ${\displaystyle u_{1}}$  is a solution to the BVP (not the whole IBVP, only the BVP, Boundary Value Problem, the BCs applied) and so is another ${\displaystyle u_{2}}$ , then a linear combination, ${\displaystyle C_{1}u_{1}+C_{2}u_{2}}$ , is also a solution. Let's take a step back and suppose that the IC was ${\displaystyle u(y,0)=\sin(\pi y)+1/5\sin(3\pi y).}$ This is no longer a realistic flow problem but it contains the first two terms of what is called a Fourier sine expansion, see these examples of Fourier sine expansions. We are going to generalize this below. Let's now use this expression and equate it to the half way solution (BCs applied) with ${\displaystyle e^{-(n\pi )^{2}\nu t}}$  being eliminated as t = 0: ${\displaystyle B\sin(n\pi y)=\sin(\pi y)+{\frac {1}{5}}\sin(3\pi y)\,}$ And it still can't match. However, observe that the individual terms in the IC can. We simply set the constants to values making both sides match: ${\displaystyle B_{1}\sin(n_{1}\pi y)=\sin(\pi y)\Rightarrow n_{1}=1\ {\mbox{and}}\ B_{1}=1\,}$ ${\displaystyle B_{3}\sin(n_{3}\pi y)={\frac {1}{5}}\sin(3\pi y)\Rightarrow n_{3}=3\ {\mbox{and}}\ B_{3}={\frac {1}{5}}\,}$ Note the subscripts are used to identify each term: they reflect the integer ${\displaystyle n}$  from the separation constant. Solutions may be obtained for each individual term of the IC, identified with ${\displaystyle n}$ : ${\displaystyle u_{1}(y,t)=e^{-(1\pi )^{2}\nu t}1\sin(1\pi y)=e^{-\pi ^{2}\nu t}\sin(\pi y)}$         ${\displaystyle n=1\ {\mbox{and}}\ B=1\,}$ ${\displaystyle u_{3}(y,t)=e^{-(3\pi )^{2}\nu t}{\frac {1}{5}}\sin(3\pi y)=e^{-9\pi ^{2}\nu t}{\frac {1}{5}}\sin(3\pi y)\,}$         ${\displaystyle n=3\ {\mbox{and}}\ B={\frac {1}{5}}\,}$ Linearity states that the sum of these two solutions is also a solution to the BVP (no need for new constants): ${\displaystyle u_{1+3}(y,t)=e^{-\pi ^{2}\nu t}\sin(\pi y)+{\frac {1}{5}}e^{-9\pi ^{2}\nu t}\sin(3\pi y)\,}$ So we added the solutions and got a new solution... what is this good for? Try setting ${\displaystyle t=0}$ : ${\displaystyle u_{1+3}(y,0)=e^{-\pi ^{2}\nu \cdot 0}\sin(\pi y)+{\frac {1}{5}}e^{-9\pi ^{2}\nu \cdot 0}\sin(3\pi y)=\sin(\pi y)+{\frac {1}{5}}\sin(3\pi y)\,}$ Each component solution satisfies the BVP, and the sum of these just happened to satisfy our surrogate IC. The IBVP with IC ${\displaystyle u(y,0)=\sin(\pi y)+1/5\sin(3\pi y)}$  is now solved. It would work the same way for any linear combination of sine functions whose half frequencies are ${\displaystyle n\pi }$ . "Linear combination" means a sum of terms, each multiplied by a constant. The sum is assumed to converge and be term by term differentiable. Let's do what we just did in a more generalized fashion. First, we make our IC a linear combination of sines (with ${\displaystyle e^{-(n\pi )^{2}\nu t}}$  eliminated as t = 0), in fact, infinitely many of them. But each successive term has to 'converge', it can't stray wildly all over the place. ${\displaystyle u(y,0)=\sum _{n=1}^{\infty }B_{n}\sin(n\pi y)\qquad {\mbox{(IC: an arbitrary linear combination of sines)}}\,}$ Second, find the n and B for each term assuming t = 0 (the IC), then plug them back into each term making no assumptions about t, leaving t as is. ${\displaystyle u_{n}(y,t)=e^{-(n\pi )^{2}\nu t}B_{n}\sin(n\pi y)\qquad {\mbox{(solution of}}\ n^{th}\ {\mbox{term)}}\,}$ Third, sum up all the terms with their individual n and Bs. ${\displaystyle u(y,t)=\sum _{n=1}^{\infty }u_{n}(y,t)=\sum _{n=1}^{\infty }e^{-(n\pi )^{2}\nu t}B_{n}\sin(n\pi y)\qquad {\mbox{(sum of solutions)}}\,}$ Fourth, plug t = 0 into the sum of terms and recover the IC from the first step. ${\displaystyle u(y,0)=\sum _{n=1}^{\infty }e^{-(n\pi )^{2}\nu \cdot 0}B_{n}\sin(n\pi y)=\sum _{n=1}^{\infty }B_{n}\sin(n\pi y)\qquad {\mbox{(IC recovered)}}\,}$ So we went full circle on this example but found the n and Bs because we were able to equate/satisfy each term with the IC. Now we can solve the problem if the IC is a linear combination of sine functions. But the IC for this problem isn't such a sum, it's just a stupid parabola. Or is it? ### Series ConstructionEdit In the 19th century, a man named Joseph Fourier took a break from helping Napoleon take over the world to ask an important question while studying this same BVP (concerning heat flow): can a function be expressed as a sum of sinusoids, similar to a taylor series? The short answer is yes, if a few reasonable conditions apply as we have already indicated. The long answer follows, and this section is a longer answer. A function meeting certain criteria may indeed be expanded into a sum of sines, cosines, or both. In our case, all that is needed to accomplish this expansion is to find the coefficients ${\displaystyle B_{n}}$ . A little trick involving an integral makes this possible. The sine function has a very important property called orthogonality. There are many flavors of this, which will be served in the next chapter. Relevant to this problem is the following: ${\displaystyle \int _{0}^{1}2\sin(m\pi y)\sin(n\pi y)\,dy={\begin{cases}1,&m=n\\0,&m\neq n\end{cases}}}$ A quick hint may help. Orthogonality literally means two lines at a right angle to each other. These lines could be vectors, each with its own tuple of coordinates. If those two vectors are at a right angle to each other, multiplying and summing their coordinate tuples always yields zero (in Euclidean space). The method of multiplying and summing is also used to determine whether two functions are orthogonal. Using this definition, our multiplied and integrated functions above are orthogonal most of the time, but not always. Let's call the IC ${\displaystyle \phi (y)}$  to generalize it. We equate the IC with its expansion, meaning the linear combination of sines, and then apply some craftiness. And remember that our goal is to reproduce a parabolic function from linearly combined sines: ${\displaystyle \sum _{n=1}^{\infty }B_{n}\sin(n\pi y)=\phi (y)\,}$ ${\displaystyle 2\sin(m\pi y)\cdot \sum _{n=1}^{\infty }B_{n}\sin(n\pi y)=2\sin(m\pi y)\cdot \phi (y)\,}$ ${\displaystyle \sum _{n=1}^{\infty }B_{n}\cdot 2\sin(m\pi y)\sin(n\pi y)=2\sin(m\pi y)\phi (y)\,}$ ${\displaystyle \int _{0}^{1}\sum _{n=1}^{\infty }B_{n}\cdot 2\sin(m\pi y)\sin(n\pi y)dy=\int _{0}^{1}2\sin(m\pi y)\phi (y)\ dy\,}$ ${\displaystyle \sum _{n=1}^{\infty }B_{n}\int _{0}^{1}2\sin(m\pi y)\sin(n\pi y)dy=\int _{0}^{1}2\sin(m\pi y)\phi (y)\ dy\,}$ ${\displaystyle B_{m}=\int _{0}^{1}2\sin(m\pi y)\phi (y)\ dy\,}$ In the last step, all of the terms in the sum became ${\displaystyle 0}$  except for the ${\displaystyle m^{th}}$  term where ${\displaystyle m=n}$ , the only case where we get ${\displaystyle 1}$  for the otherwise orthogonal sine functions. This isolates and explicitly defines ${\displaystyle B_{m}}$  which is the same as ${\displaystyle B_{n}}$  as m = n. The expansion for ${\displaystyle \phi (y)}$  is then: ${\displaystyle \phi (y)=\sum _{m=1}^{\infty }(\int _{0}^{1}2\sin(m\pi y)\phi (y)\ dy)\cdot \sin(m\pi y)\,}$ Or equivalently: ${\displaystyle \phi (y)=\sum _{m=1}^{\infty }B_{m}\sin(m\pi y)\qquad ;\ B_{m}=\int _{0}^{1}2\sin(m\pi y)\phi (y)\ dy\,}$ Many important details have been left out for later in a devoted chapter; one noteworthy detail is that this expansion is only approximating the parabola (very superficially) on the interval ${\displaystyle 0\leq y\leq 1}$ , not say from ${\displaystyle -\infty }$  to ${\displaystyle +\infty }$ . This expansion may finally be combined with the sum of sines solution to the BVP developed previously. Note that the last equation looks very similar to ${\displaystyle u(y,0)}$ . Following from this: ${\displaystyle u(y,0)=\sum _{n=1}^{\infty }B_{n}\sin(n\pi y)\,}$ ${\displaystyle u(y,0)=\sum _{n=1}^{\infty }\int _{0}^{1}2\sin(n\pi y)\phi (y)\ dy\cdot \sin(n\pi y)\qquad \,}$ So the expansion will satisfy the IC given as ${\displaystyle \phi (y)}$  (surprised?). The full solution for the problem with arbitrary IC is then: ${\displaystyle u(y,t)=\sum _{n=1}^{\infty }e^{-(n\pi )^{2}\nu t}B_{n}\sin(n\pi y)\,}$ ${\displaystyle u(y,t)=\sum _{n=1}^{\infty }e^{-(n\pi )^{2}\nu t}\int _{0}^{1}2\sin(n\pi y)\phi (y)\ dy\cdot \sin(n\pi y)\,}$ In this problem specifically, the IC is ${\displaystyle \phi (y)=4y-4y^{2}}$ , so: ${\displaystyle B_{n}=\int _{0}^{1}2\sin(n\pi y)(4y-4y^{2})dy={\frac {8}{n^{3}\pi ^{3}}}(2-2\cos(n\pi )-n\pi \sin(n\pi ))\,}$ Sines and cosines appear from the integration dependent only on ${\displaystyle n\pi }$ . Since ${\displaystyle n}$  is an integer, these can be made more aesthetic. ${\displaystyle \sin(n\pi )=0\qquad ;\ n{\mbox{ is an integer.}}\,}$ ${\displaystyle \cos(n\pi )=(-1)^{n}\qquad ;\ n{\mbox{ is an integer.}}\,}$ ${\displaystyle \qquad {\Big \Downarrow }}$ ${\displaystyle B_{n}={\frac {16-16(-1)^{n}}{n^{3}\pi ^{3}}}\,}$ Note that for even ${\displaystyle n}$ , ${\displaystyle B_{n}=0}$ . Putting everything together finally completes the solution to the IBVP: ${\displaystyle u(y,t)=\sum _{n=1}^{\infty }e^{-(n\pi )^{2}\nu t}{\frac {16-16(-1)^{n}}{n^{3}\pi ^{3}}}\sin(n\pi y)\,}$ There are many interesting things to observe. To begin with, ${\displaystyle u(y,t)}$  is not a product of a function of ${\displaystyle y}$  and a function of ${\displaystyle t}$ . Such a solution was assumed in the beginning, proved to be wrong, but eventually happened to yield a solution anyway thanks to linearity and what is called a Fourier sine expansion. A careful look at the procedure reveals something that may be disturbing: this lengthy solution is strictly valid for the given BCs. Thanks to the definition of ${\displaystyle \phi (y)}$ , the solution is generic as far as the IC is concerned (the IC doesn't even need to match the BCs), however a slight change in either BC would mandate starting over almost from the beginning. The parabolic IC, which looks very similar to the sine function used in the previous section, is wholly to blame (or thank once you understand the beauty of a Fourier series!) for the infinite sum. It is interesting to approximate the first several numeric values of the sequence ${\displaystyle B_{n}}$ : {\displaystyle {\begin{aligned}B_{1}&\approx 1.03205\\B_{3}&\approx 0.03822\\B_{5}&\approx 0.00826\\B_{7}&\approx 0.00301\,\end{aligned}}} Recall that the even terms are all ${\displaystyle 0}$ . The first term by far dominates, this makes sense since the first term already looks very, very similar to the parabola. Recall that ${\displaystyle n^{2}}$  appears in an exponential, making the higher terms even smaller for time not too close to ${\displaystyle 0}$ .
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# Show that decomposition does not hold for non-linear system The solution to an inhomogeneous differential equation can be split up into homogeneous solution and a particular solution (forced response). Another way to split up the solution to an inhomogeneous differential equation is in a zero-input response and a zero-state response. The zero-input response is the system's response to its own internal initial conditions - no input signal is applied. The zero-input response is the homogeneous solution to the system's differential equation, using the initial conditions at $$t=0^-$$. The zero-state response is the system's response to only the input signal - all initial conditions set to $$0$$. The zero-state response is found by convolving the system's impulse response $$h(t)$$ with the input signal $$x(t)$$. This excerpt is from Lathi's signal processing and linear systems: Lathi claims that for a linear system, one can show that the decomposition property holds. Question: How can I show that the decomposition property for a non-linear system, for example $$\dot{y}(t) + y(t) = x(t) + 1$$, does not hold? Assume that $$y_0(t)$$ is the zero-input response. Then $$y_0(t)$$ must satisfy $$\dot{y}_0(t)+y_0(t)=1\tag{1}$$ because $$x(t)=0$$. Now let $$y_1(t)$$ be the zero-state response to an input $$x(t)$$, satisfying $$\dot{y}_1(t)+y_1(t)=x(t)+1\tag{2}$$ If the decomposition property holds, the function $$y_2(t)=y_0(t)+y_1(t)$$ must be the response to $$x(t)$$ with possibly non-zero initial conditions. I.e., $$y_2(t)$$ should satisfy $$\dot{y}_2(t)+y_2(t)=x(t)+1\tag{3}$$ However, adding Equations $$(2)$$ and $$(3)$$ gives $$\dot{y}_2(t)+y_2(t)=x(t)+2\tag{3}$$ which shows that for the given system the decomposition property doesn't hold. • Thanks for the answer Matt, but I'm not entirely convinced yet. How does the last equation show that the decomposition property does not hold? Adding equation 1 and 2 yields equation 3. – Carl Feb 6 at 12:35 • @Carl: The combined response $y_2(t)$ should satisfy the given input/output equation of the system if the decomposition property were to hold, but Eq. (3) shows that it doesn't. Feb 6 at 12:40 • Oh I think I get it now. So the system response $y(t)$ should satisfy the original differential equation $\dot{y}(t) + y(t) = x(t) + 1$. But when we solve the differential equation by splitting the response into the zero-input response $y_0(t)$ and zero-state response $y_1(t)$, then equation 3 shows that this method actually yields the solution to $\dot{y}(t) + y(t) = x(t) + 2$ which is not the original differential equation. Hence, in this case, splitting the system response into zero-state and zero-input response yields the wrong total response, and decomposition does not hold. Right? – Carl Feb 6 at 12:53 • @Carl: Yes, that's what I meant. Feb 6 at 12:55
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## sexta-feira, 20 de novembro de 2020 ### Geometria - Um problema de triângulos retângulos James Tanton pergunta em seu Twitter: uma potência de dois pode ser a hipotenusa de um triângulo retângulo com lados inteiros? (James Tanton asks on his Twitter: can a power of two be the hypotenuse of an integer right triangle?) PCFilho #### 5 comentários: a² + b² = (2ⁿ)² = 2²ⁿ = (2²)ⁿ = 4ⁿ Since the sum of a² and b² is even, that means that a and b are both odd or both even. (Case #1: a and b are both odd) If a and b are both odd, we have: a = 2h + 1, where h is a positive integer b = 2k + 1, where k is a positive integer (2h + 1)² + (2k + 1)² = 4ⁿ 4(h² + h + k² + k) + 2 = 4ⁿ Notice that the left side of this equation will never be divisible by 4 and the right side will always be divisible by 4. This contradicts the assertion that they’re equal, which means that this case is impossible. (Case #2: a and b are both even) If a and b are both even, then gcf(a,b) = 2ᵐ, where m is an integer that is less than half n, and: a = 2ᵐx, where x is an odd integer b = 2ᵐy, where y is an odd integer (2ᵐx)² + (2ᵐy)² = 4ⁿ x² + y² = 4ⁿ⁻²ᵐ Here, we have two odd integers being the lengths of the legs of a right triangle whose hypotenuse is a power of 2. But, we’ve seen in Case #1 that this is impossible. Since the assumption in this case also leads to a contradiction, we must conclude that this case is also impossible. Therefore, it is impossible for the length of a right triangle’s hypotenuse to be a power of 2. 1. Well done, Jake!! What a beautiful demonstration! 2. Except that the part that says 4^(n-2m) should say 4^(n-m). 3. Also, in Case #2, it is possible for one of x and y to be odd and the other even, which would contradict the assertion that the sum of their squares is even anyway. 2. Here is a list of all possible hypotenuses up to 140: https://oeis.org/A009003 As expected, none of them are powers of two. Regras para postar comentários: I. Os comentários devem se ater ao assunto do post, preferencialmente. Pense duas vezes antes de publicar um comentário fora do contexto. II. Os comentários devem ser relevantes, isto é, devem acrescentar informação útil ao post ou ao debate em questão. III. Os comentários devem ser sempre respeitosos. É terminantemente proibido debochar, ofender, insultar e/ou caluniar quaisquer pessoas e instituições. IV. Os nomes dos clubes devem ser escritos sempre da maneira correta. Não serão tolerados apelidos pejorativos para as instituições, sejam quais forem. V. Não é permitido pedir ou publicar números de telefone/Whatsapp, e-mails, redes sociais, etc. VI. Respeitem a nossa bela Língua Portuguesa, e evitem escrever em CAIXA ALTA. Os comentários que não respeitem as regras acima poderão ser excluídos ou não, a critério dos moderadores do blog.
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The interest coverage ratio looks at income on financial reports to determine whether the company is generating enough profits to pay its interest obligations. If the company doesn't make its interest payments on time to creditors, its ability to get additional credit will be hurt; eventually, if nonpayment goes on for a long time, the company may end up in bankruptcy. The interest coverage ratio uses two figures that you can find on the company's income statement: earnings before interest, taxes, depreciation, and amortization (also known as EBITDA); and interest expense. ## How to calculate the interest coverage ratio Here's the formula for finding the interest coverage ratio: EBITDA ÷ Interest expense = Interest coverage ratio Calculating this ratio may or may not be a two-step process. Many companies include an EBITDA line item on their income statements. If a company hasn't included this line item, you have to calculate EBITDA yourself. Mattel and Hasbro don't have an EBITDA line item, so this is how you figure that out before you try to calculate the ratio. ### Mattel Mattel reports operating income before it lists its interest and tax expenses. Mattel doesn't have a line item for amortization or depreciation, so you need to look at the cash flow statement to find that amortization totaled \$16,746,000 and depreciation totaled \$157,536,000. Therefore, in Mattel's case, EBITDA was \$945,045 (Income before taxes) + \$16,746,000 + \$157,536,000 = \$1,119,327,000. Then, to get the interest coverage ratio: \$1,119,327,000 (EBITDA) ÷\$88,835,000(Interest expense) = 12.60 (Interest coverage ratio) Thus, Mattel generates \$12.60 income for every \$1 it pays out in interest. ### Hasbro Hasbro reports amortization expenses of \$50,569,000 on the income statement. It also reports \$99,718,000 for depreciation of plant and equipment on the statement of cash flows, so you need to add those expenses back in to find the EBITDA: \$453,402,000 (Income before taxes) + \$50,569,000 (Amortization on income statement) + \$99,718,000 (Depreciation of plant and equipment) = \$603,689,000 (EBITDA) And then: \$603,689,000 (EBITDA) ÷\$91,141,000 (Interest expense) = 6.62 (Interest coverage ratio) Hasbro generates \$6.62 for every \$1 it pays out in interest. ## What do the numbers mean? Both companies clearly generate more than enough income to make their interest payments. A ratio of less than 1 means a company is generating less cash from operations than needed to pay all its interest. Lenders believe that the higher the interest coverage ratio is, the better. You should be concerned about a company's fiscal health anytime you see an interest coverage ratio of less than 1.5. This means the company generates only about \$1.50 for each \$1 it pays out in interest. Any type of emergency or drop in sales may make it difficult for the company to make its interest payments.
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https://question.onlinegdb.com/13037/i-need-explanation-please?show=13044
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Hi! Can somebody explain why i need the "if x%2 == 1" part? Please, i'm just learning python datastructure https://onlinegdb.com/WF0MOPqNq answered Oct 9, 2022 by (260 points) first you need to know that % = remainder of division the division is x / 2 `the 1 means it's not integer because it's left ` ` if you put an even number it will come out like` ```10 % 2 = 5 == integer ``` `I think that's it, sorry if I did a mistake thx.` `(if x/2 the remainder of the division = 1 it was a odd number)` +1 vote answered Oct 9, 2022 by (260 points) #example: that will check if what you writed it's a ood number or a even number. printar =  int(input("")) if printar%2 == 1: print('its a ood number') elif printar%2 == 0: print('its a even number') answered Oct 10, 2022 by (140 points) x is a variable , % implies mod operation that gives remainder obtained on division, so x%2 is used to check if it gives remainder 1 implying odd number or otherwise is an even number. Now the x in range 10 will generate numbers from 1 to 10 one at a time and check this condition. Without checking this condition you will get a list of all numbers from 1 to 10 whereas now you only get odd numbers ,since values of x not following the condition are not included. answered Oct 10, 2022 by (89,510 points) `odds = [x for x in range(10) if x % 2 == 1]` and as the name correctly suggests, odds hold all the odd numbers from 1 to 10, i.e.: 1, 3, 5, 7, 9. The line says, 'add all the x to the list if they are odd'. Odd is decided by checking the modulus 2 of x to be 1. Check out the Python Operators. % is the modulus, it tells you the remainder of a division. E.g.: 100 % 12 = 4, because 100 - 8 * 12 = 100 - 96 = 4. x % y returns values in the range [0, y), i.e., 0 included, but y not included. I hope this helps. commented Oct 10, 2022 by (260 points) I think he's not gonna understand what you said commented Oct 10, 2022 by (89,510 points) What makes you say that? If something is not clear, please point it out so it can be clarified. commented Oct 10, 2022 by (5,380 points) edited Oct 10, 2022 by Eidnoxon So it's checks that if the remainder of x and 2 is 1, then add to the odds list? commented Oct 10, 2022 by (89,510 points) Yes. I wouldn't say it adds x to a list as there the list is constructed with elements already in it, rather than adding them one by one. Perhaps a better phrase would be that `odds` is a list that consists of any x, where a) x is in range(10) --> 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 AND b) x is an odd number (i.e.: x % 2 == 1)
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## Intermediate Algebra (6th Edition) $x^4-8x^3+24x^2-32x+16$ $(x-2)^4\longrightarrow$ Rewrite as squared binomials. $(x-2)^2(x-2)^2\longrightarrow$ Multiply using FOIL method $[x^2-4x+4][x^2-4x+4]\longrightarrow$ Regroup to create squared binomial $[(x^2-4x)+4]^2\longrightarrow$ Multiply using FOIL method $(x^2-4x)^2+2(4(x^2-4x)+16\longrightarrow$ Multiply the squared binomial using FOIL method. Apply distributive method. $x^4-8x^3+16x^2+8x^2-32x+16\longrightarrow$ Combine like terms. $x^4-8x^3+24x^2-32x+16$
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It is currently Thu Oct 27, 2016 10:45 pm All times are UTC Page 1 of 1 [ 4 posts ] Print view Previous topic | Next topic Author Message Post subject: Maths Help...PleasePosted: Sun Jan 27, 2008 2:19 pm Joined: Mon Feb 20, 2006 1:29 pm Posts: 1805 Location: Berkshire DS1 always attacks maths questions in the most peculiar ways, but 99% of the time ends up getting the answer correct! With the following question his method falls short, I can understand his reasoning but cannot explain why it doesn’t work, and he has asked me to ask on the forum, can any of you maths wizards help, Please? Put the following in order, starting with the one which is best value. a) 20g for £2 b) 25g for £3 c) 30g for £4.20 d) 10g for £1.30 e) 24g for £2.64 f) 40g for £3.60 I showed him how I would of found the value of 10g for each and correctly get the answer F,A,E,B,D,C, but he's arguing that his method should work He used a) as the base thus he got 10g=£1, he then converted this base value to the others, and worked out the difference… a) 10g=£1 base value so no difference 0p b) 25g=£2.50 therefore there’s 50p difference from £3 c) 30g=£3 therefore there’s £1.20 difference from £4.20 d) 10g=£1 therefore there’s 30p difference from £1.30 e) 24g=£2.40 therefore there’s 24p difference from £2.64 f) 40g=£3.60 therefore there’s -40p difference for £3.60 F,A,E,D,B,C D and B are in the wrong order, all the rest are correct. Why ? Top Post subject: Posted: Sun Jan 27, 2008 3:01 pm Joined: Mon Feb 12, 2007 1:21 pm Posts: 11738 You can't just look at the differences as they are relative to different amounts of money. So the 50p difference is smaller compared to £3 than the 30p is compared to £1 - he is just lucky that the others work! Top Post subject: Posted: Sun Jan 27, 2008 3:09 pm Joined: Mon Feb 20, 2006 1:29 pm Posts: 1805 Location: Berkshire Thanks Guest55 That's initially what I said that they are different amounts, so should be looked at individually, but then got my self confused, because 4 out of 6 were correct, so started thinking hold on a minute... perhaps? Thanks BW Top Post subject: Posted: Sun Jan 27, 2008 7:09 pm I recognise this question from the Bond maths books and I am fairly certain the easiest way of finding the answer is to work out what 1g is. I believe the author is hoping a child would have a 'feel' for numbers and notice all the amounts are easily divisible e.g. 24 g cost £2.64 so 1g = 11p Certainly when I do this question with pupils I suggest they look at the figures before they decide what technique to use. I would have thought finding 10 g would be much harder. Top Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending Page 1 of 1 [ 4 posts ] All times are UTC #### Who is online Users browsing this forum: Exabot [Bot] and 1 guest You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum Search for: Jump to:  Select a forum ------------------ FORUM RULES    Forum Rules and FAQs 11 PLUS SUBJECTS    VERBAL REASONING    MATHS    ENGLISH    NON-VERBAL REASONING    CEM 11 Plus GENERAL    GENERAL 11 PLUS TOPICS    11 PLUS APPEALS    11 PLUS TUTORS    INDEPENDENT SCHOOLS    11 PLUS CDs/SOFTWARE    11 PLUS TIPS    PRIMARY    SEN and the 11 PLUS    EVERYTHING ELSE .... 11 PLUS REGIONS    Berkshire    Bexley and Bromley    Birmingham, Walsall, Wolverhampton and Wrekin    Buckinghamshire    Devon    Dorset    Essex    Essex - Redbridge    Gloucestershire    Hertfordshire (South West)    Hertfordshire (Other and North London)    Kent    Lancashire & Cumbria    Lincolnshire    Medway    Northern Ireland    Surrey (Sutton, Kingston and Wandsworth)    Trafford    Warwickshire    Wiltshire    Wirral    Yorkshire BEYOND 11 PLUS    Beyond 11 Plus - General    GCSEs    6th Form    University
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# How do you find the value of N? ## How do you find the value of N? To find n! we multiply the number from 1 to n, so to find n form n! we can start dividing that number from 2 to the number with which we get quotient as 1. ## What is the solution to the equation? A solution to an equation is a value of a variable that makes a true statement when substituted into the equation. The process of finding the solution to an equation is called solving the equation. To find the solution to an equation means to find the value of the variable that makes the equation true. What is the equation when multiplying both sides? Multiplication Property of Equality If two expressions are equal to each other and you multiply both sides by the same number, the resulting expressions will also be equivalent. When the equation involves multiplication or division, you can “undo” these operations by using the inverse operation to isolate the variable. ### How do you solve equations with division? 57 second clip suggested5:20Solving Equations using Multiplication and Division (Simplifying Math)YouTubeStart of suggested clipEnd of suggested clipIn this case it’s 2 times X. So the inverse operation would be division dividing by the coefficientMoreIn this case it’s 2 times X. So the inverse operation would be division dividing by the coefficient of 2 2 divided by 2 is 1 1 times X leaves us with just X on the left side and 4 divided by 2 is 2. ### What’s the value of n in math? In an equation, N represents a specific number, not any number. N + 9 = 12 means N is a number which, when added to 9, must give the answer 12. So N can only be the number 3 because only 3 + 9 is equal to 12. How do you find n combination? 41 second clip suggested6:14Combination Algebra nC4 = 210 Find n without Calculator Trick – YouTubeYouTube #### How can an equation have two solutions? If only ONE number makes both sides of an equation equal. And with quadratic equations you often have 2 solutions. xsquared – 7x + 12=0 (x = 3,4 so both numbers would make the 2 sides equal). Originally Answered: How do you know if a system of equations has infinite solutions? Can you multiply equations? The Multiplication Property for Equations states that an equation can be multiplied or divided by the same number on each side of the equation without changing the solution to the equation. Solution: Isolate the w by dividing each side of the equation by 13. Multiply the fractions on the left side of the equation. ## How do you cross multiply? Well, to cross multiply them, you multiply the numerator in the first fraction times the denominator in the second fraction, then you write that number down. Then you multiply the numerator of the second fraction times the number in the denominator of your first fraction, and you write that number down. ## How can you write and solve a multiplication or division equation? What is the cube root of 1250? The cubed root of one thousand, two hundred and fifty ∛1250 = 10.772173450159 The process of cubing is similar to squaring, only that the number is multiplied three times instead of two. The exponent used for cubes is 3, which is also denoted by the superscript³. ### What is number-3 cubed? -3 cubed equals -27, because (-3) × (-3) × (-3) = -27. Once -27 = (-3) × (-3) × (-3), -27 is also known as a, so called, perfect cube. Use the cube calculator below to find the cube of any real number. See below the definition and examples of cubed numbers. What is a cube and how to calculate its volume? ### What is the cube of-27? Once -27 = (-3) × (-3) × (-3), -27 is also known as a, so called, perfect cube. Use the cube calculator below to find the cube of any real number. See below the definition and examples of cubed numbers. What is a cube and how to calculate its volume? A cube is a three-dimensional shape that has all edges of equal length. Where can I get help with cubed calculator? Cubed calculator Mathfraction.com offers insightful answers on cubed calculator, synthetic division and solving systems of equations and other algebra subject areas. In case that you have to have help on quadratic function or maybe equations by factoring, Mathfraction.com is without question the excellent place to explore! Home Posted in Other
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# A E F H Z V U what comes next Answer: A = letter 1 in alphabet E = letter 5 (fifth one along from start) F = letter 6 H = letter 8 then go in reverse: Z = letter 26 V = letter 22 (or 5th one along from end) U = letter 21 (next one back) S = letter 19 (skip one as there was between F and H but going backwards) ## Related Questions Log x + log 7 = log 37 assume log base 9 log(x+6) - log(x) - log(2) = 0 log((x+6)/(2x)) = 0 (x+6)/(2x) = 1 x+6 = 2x x = 6 What is the square root of 120 Step-by-step explanation: 120 ∧ 12   10 ∧     ∧ 3  4  2  5 ∧ 2  2 = = = A room is 8 meters wide and 5 meters long and 4 meters high. whats the volume of the room 8 x 5 x 4 = 160m^3 to find the volume of a cube its V= W x H x L WILL GIVE BRAINLIEST! Write the general equation for the circle that passes through the points: (1, 7) (8, 6) (7, -1) Step-by-step explanation: General Equation Of A Circle If a,b, and c are real numbers, then the general equation of a circle is There are three unknowns that could eventually be determined if we knew three points of the circle. These points are (1, 7) (8, 6) (7, -1). We only need to replace them in the general equation and solve the resulting system of equations For the point (1, 7) Operating and rearranging For the point (8, 6) Operating and rearranging For the point (7, -1) Operating and rearranging We form the system of equations We'll eliminate c from the first two equations and then from the last two equations Multiplying the first one by -1 Adding up with the third equation We get c=0 Multiplying the first by -8 Which gives Finally, isolating a from We get a=-8 So the general equation of the circle is Alice process health claims. it takes her an average of 10 minutes to check the paperwork on one claim. At that rate,how long will it take her to process 20 claims To solve this you simply multiply the number of minutes it takes to review one claim times the number of claims. 10 x 20 = 200 Scientist A dissolved 1.0 kilogram of salt in 3.0 liters of water. Scientist B dissolved 2.0 pounds of salt in 7.0 pints of water. Which scientist made a more concentrated salt solution? Explain. Ravi is riding his bicycle. He takes 6 hours to ride 76.8 kilometers. What is his speed? 12.8km/hr Step-by-step explanation: Distance = 76.8km Time = 6 hours Speed = distance ➗ time Speed = 76.8km ➗ 6 hours Speed = 12.8km/hr Help please Which equation shows the substitution method being used to solve the system of linear equations? x + y = 6 x = y + 2 A. x = (x – 6) + 2 B. x + (y + 2) = 6 C. (y + 2) + y = 6 D. x + y = y + 2 We can substitute y + 2 for 'x' in the first equation, giving us: (y + 2) + y = 6 C, you substituted the y in the x place 5. Find the interest earned. Assume 3.5% interest compounded daily. \$975 deposited April 23 and withdrawn June 18 \$6.66 \$5.25 \$5.16 \$6.75
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# 3x3 “Magic Square” of Prime Numbers — Part II Glad to know the previous puzzle, which was the first puzzle I posted in Puzzling, was warmly welcomed (Thank you!), and an optimal solution was found. Inspired by the comments there, here is the Version $2$ of the puzzle. In fact, most of the things are unchanged. We still have this $3 \times 3$ grid, which $9$ distinct prime numbers $P_1, P_2, ..., P_9$ are to be filled in. And there are $8$ sums: $3$ horizontal, $3$ vertical and $2$ diagonal, and they are named $S_1, S_2, ..., S_8$. All the requirements in the first version still hold here, which mean: • $P_1, P_2, ..., P_9, S_1, S_2, ..., S_8$ are all distinct prime numbers (i.e. there are totally $17$ different prime numbers). But this time, one more additional requirement: • The grand total $P_1 + P_2 + ... + P_9 + S_1 + S_2 + ... + S_8$ also has to be a prime number. The challenge: To minimize the grand total. With the additional requirement, some solutions satisfying the previous puzzle do not satisfy this version. And, the optimal solution will be different. Below is one possible solution I come up with, which has a grand total of $601$, but it is not the optimal solution: Feel free to have a try! • Whew! This looks even harder. – Rand al'Thor Mar 24 '15 at 10:23 • I wonder if there's some general theorem from number theory that can be used to solve this kind of problem? The Green-Tao theorem comes to mind... – Rand al'Thor Mar 24 '15 at 10:24 • Yes it will be more difficult due to the additional condition. And from the experience of the previous puzzle, seems like a greedy strategy of putting the smallest primes inside the grid, or positioning them to the center or corners, may not always work towards the optimal solution. – LaBird Mar 24 '15 at 10:31 As our friend @KSab said, the best solution ever is 541. There are exactly 16 possible solutions which are shown in the images belown: • Yes you are right, indeed I also find $16$ optimal solutions in $2$ different configurations (your answer $1$ and $2$), as rotation and flipping of each configuration gets a set of $8$ identical solutions). – LaBird Mar 25 '15 at 7:28 Using a brute force checker I have found a solution of 541 7 5 17 | 29 11 23 3 | 37 13 19 41 | 73 ---------+--- 53 31 47 61 | 71 Notice that a simple lower bound would be 499 which is just the sum of the first 17 primes greater than 2; the result above skips only 3 primes (43, 59, 67) which is what makes me doubt there is a more optimal solution. The brute force searcher searched every combination of the primes up to 67 which should be conclusive, as increasing the value of one of the inner numbers increases the final answer by at least three times as much, making any increase not worth it at that point. I used the same numbers as used by you I got the grand total of 529 which is a prime number. (5 + 11 + 3 + 13 + 7 + 59 + 19 + 23 + 17) + (19 + 79 + 59 + 29 + 29 + 37 + 41 + 79 ) = 529 • A few numbers appear twice, note the OP's restriction that the sums must also be distinct from the elements in the square. – KSab Mar 24 '15 at 14:30 • It's difficult to spot and I made the same mistake before I posted the question, but $529$ is not a prime number ($23 \times 23$). – LaBird Mar 25 '15 at 7:30 • @LaBird, ohh sorry, you are right. – Himanshu Mar 25 '15 at 9:16 • $19, 59, 79$ appear twice, but nice try though :) – Mr Pie Nov 18 '17 at 23:04
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# Homework Help: Combined probability distribution 1. Jul 21, 2011 ### Uncle_John 1. The problem statement, all variables and given/known data Let's have a box in shape of a square(viewed from the top) from the corner of which a smaller square was cut out.The side of a bigger square is 2a, side of the smaller square is a long. We've got evenly distributed corn seeds all over the box,randomly selected seed is defined by coordinates $x,y \in [0,2a]$ 2. Relevant equations a.) Write down the combined probability distribution for $w(x,y)$ b.) Write down the projected probability distribution for $u(x)$(independent of $y$) c.) calculate the correlation coefficient $r_{x,y}$ 3. The attempt at a solution a.) $1/3a^₂$ b.) $u(x)= 1/3a$ if $x \in [0,a]$ $u(x) = 2/3a$ if $x \in [a,2a]$ c.) since $r_{x,y} =\frac{\sigma_{x,y}}{\sigma_{x} \sigma_{y}}$, i calculated each variance seperately: $\sigma_{x} = \int xu(x)dx$ $\sigma_{y} = \int yu(y)dx$ $\sigma_{x,y} = \int\int (x - \overline{x})(y - \overline{y})w(x,y)dxdy$ Is that right? 2. Jul 21, 2011 ### LCKurtz Re: Probability What is in the denominator? You need parentheses. I'm not sure what you mean by the "combined" probability distribution. If you mean the joint density function, you don't have it. It would be a function of x and y. This would be reflected in the domain when you write it carefully. Yes, that is the marginal density of x (if you put correct parentheses in). And you get a symmetric formula for y. Last edited: Jul 21, 2011 3. Jul 21, 2011 ### LCKurtz Re: Probability No. Your formulas for σx and σy are wrong. And you will need to be careful what limits you use on the last one. 4. Jul 22, 2011 ### Uncle_John Re: Probability Yes, sorry, i meant joint probability distribution. So if i follow the formal definition: $w_{x,y}(x,y) = w_{y|x}(x,y)w_{x}(x)$ Then: $w_{x|y}(x,y) = \begin{cases} 1/a, \text{if } x \in [0,a) \\ 1/(2a), \text{if } x\in [a,2a] \end{cases}$ Also, $w_{x}$ is known: $w_{x}(x) = \begin{cases} 1/(3a), \text{if} x \in [0,a) \\ 2/(3a), \text{if} x \in [a,2a] \end{cases}$ Is that better? Last edited: Jul 22, 2011 5. Jul 22, 2011 ### LCKurtz Re: Probability But you need to be more careful here. The conditional density of x|y depends on both x and y. It matters whether y is in (0,a) or (a,2a). Last edited: Jul 22, 2011 6. Jul 22, 2011 ### Uncle_John Re: Probability $w_{x|y}(x,y) = \begin{cases} 0, \text{if} x \in [0,a) \text{and} y \in [0,a] \\ 1/a, \text{if } x \in [0,a) \text{and} y \in [a,2a] \\ 1/(2a), \text{if } x\in [a,2a] \end{cases}$ ? Better. But how should i write the final answer? $w_{x,y}$ What is wrong with $\sigma$? 7. Jul 22, 2011 ### LCKurtz Re: Probability Yes, that' s better for the conditional density. But perhaps I was a bit cryptic in my earlier comments about the joint density. Your formula of 1/(3a2) was correct but it gives the appearance of not depending on x or y. You need to indicate the proper (x,y) domain and you will have it without going this conditional density stuff. The formulas you have written for the σ's look like formulas for the means instead of the standard deviations. 8. Jul 23, 2011 ### Uncle_John Re: Probability Ok, so it would look something like : $w_{x,y}(x,y) = \begin{cases} 0, \text{if } x \in [0,a] \text{and } y \in [0,a] \\ 1/(3a^₂), \text{otherwise} \end{cases}$ $\sigma_{x}^2 = \overline{x^2} - \overline{x}^₂$ and for $\sigma_{x,y} = \int \int (x - \overline{x})(y - \overline{y}) w_{x,y} dx dy$ i integrate over [0,a]x [a,2a]first, and then over [a,2a]x [0,2a], is that allright? Can i post other probability problems in here or should i open a new thread? 9. Jul 23, 2011 ### LCKurtz Re: Probability I know you understand it, but the "otherwise" above doesn't include x or y greater than 2a or less than 0 Yes, that is correct. You should start a new thread. Others are more likely to join a new thread and it's general forum policy anyway.
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BrainDen.com - Brain Teasers # bonanova Moderator 6973 66 1. ## logic or trial&error? Brute force Has anyone found a constructive solution? 2. ## Descendants "... for a thousand years. How many descendants would you have?" But we're not enumerating the world's population. We're quantifying the part of it that comprises your descendants. Two children, four grandchildren, ... 3. ## logic or trial&error? Give it a try: 4. ## Descendants Doesn't each birth increase the population? What if, miraculously, no one will die? But the question does not hinge on when/whether people will die. The question asks: how many descendants will you have? 5. ## Who should do what @BMAD The three of us: you, I, and Logo, agree on the 5 4 1 5 3 = 18 assignment. I like your approach, where the solution can be constructed. 6. ## Who should do what Can we see it? 7. ## Probability of finding a three. A quick and dirty analysis Nice puzzle. 9. ## Descendants First thought 10. ## logic or trial&error? It might be solvable, iteratively. I once posted a puzzle, here or elsewhere, that went as follows: But in that puzzle information was added as the constraints were iteratively applied. Here it seems it must all be determined in one shot. So maybe that's not possible. 11. ## Hitting a double "Determine the rate of change to second base once the runner reaches halfway to first base." Determine the time rate of change of the distance between the runner and 2nd base when the runner is halfway between home plate and first base, running at 24 ft/sec on a path that is a straight line from home plate to first base. If these two statements ask for the same information, then 12. ## Who should do what Logophobic has it. 13. ## Hitting a double Of his position? Of his velocity? Most runners going to 2nd base will swing wide to round out the corner and keep a constant speed. 14. ## Making a spiral Agreed. Quite a bit off. Upon further review my guess (of a cosine relationship of line angle to wall angle) was wrong. 15. ## How many spin-able numbers are there from 0 to 99999 ? I think Logophobic has it. Here's why: 17. ## Rolling on a sine curve Yes I agree. If the coin centers follow different-length paths, the shorter path will be traversed first. And since the radii are not zero, then even if the sine wave is in light-years the coin following the concave side of the curve will arrive first. (See my "In any case" post.) In the limit as the ratio of radius to amplitude goes to zero, however, the paths of the centers (and their transit times) coalesce. I just wondered if there were a reason to give units to the diameters (and not the amplitude of the sine wave.) Looks like In any case, 22. ## Guessing right number from 1 to 27, with 3 Yes/No/Maybe questions. First thoughts: 23. ## 5x5 statement table. Nice puzzle, jasen. Do you have more like this? 24. ## A paradox? of two chests Here's a refutation based on the impossibility of having chests whose value has uniform probability across the real numbers (or integers.) 25. ## A paradox? of two chests I'll give a refutation by symmetry. Mainly because Bayesian formulas are opaque to me. (Read, I tried to understand a priori distributions once.) Your first choice is random. If you switch you end up the the choice you would have made, with 50% probability, without switching. Your expectation for either envelope is \$1.5x, where \$x is the lesser of the two amounts. OK, yeah, the faulty argument in my post above presumes a uniform probability on the real numbers for \$x. Anyone who thinks \$2 and \$Graham's Number have equal probability needs to immediately make a random deposit into my checki × • #### Activity • Riddles × • Create New...
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## 557950 557,950 (five hundred fifty-seven thousand nine hundred fifty) is an even six-digits composite number following 557949 and preceding 557951. In scientific notation, it is written as 5.5795 × 105. The sum of its digits is 31. It has a total of 4 prime factors and 12 positive divisors. There are 223,160 positive integers (up to 557950) that are relatively prime to 557950. ## Basic properties • Is Prime? No • Number parity Even • Number length 6 • Sum of Digits 31 • Digital Root 4 ## Name Short name 557 thousand 950 five hundred fifty-seven thousand nine hundred fifty ## Notation Scientific notation 5.5795 × 105 557.95 × 103 ## Prime Factorization of 557950 Prime Factorization 2 × 52 × 11159 Composite number Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 111590 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0 The prime factorization of 557,950 is 2 × 52 × 11159. Since it has a total of 4 prime factors, 557,950 is a composite number. ## Divisors of 557950 12 divisors Even divisors 6 6 3 3 Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 1.03788e+06 Sum of all the positive divisors of n s(n) 479930 Sum of the proper positive divisors of n A(n) 86490 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 746.961 Returns the nth root of the product of n divisors H(n) 6.45103 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors The number 557,950 can be divided by 12 positive divisors (out of which 6 are even, and 6 are odd). The sum of these divisors (counting 557,950) is 1,037,880, the average is 86,490. ## Other Arithmetic Functions (n = 557950) 1 φ(n) n Euler Totient Carmichael Lambda Prime Pi φ(n) 223160 Total number of positive integers not greater than n that are coprime to n λ(n) 111580 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45812 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares There are 223,160 positive integers (less than 557,950) that are coprime with 557,950. And there are approximately 45,812 prime numbers less than or equal to 557,950. ## Divisibility of 557950 m n mod m 2 3 4 5 6 7 8 9 0 1 2 0 4 1 6 4 The number 557,950 is divisible by 2 and 5. • Arithmetic • Deficient • Polite ## Base conversion (557950) Base System Value 2 Binary 10001000001101111110 3 Ternary 1001100100211 4 Quaternary 2020031332 5 Quinary 120323300 6 Senary 15543034 8 Octal 2101576 10 Decimal 557950 12 Duodecimal 22aa7a 20 Vigesimal 39eha 36 Base36 byim ## Basic calculations (n = 557950) ### Multiplication n×i n×2 1115900 1673850 2231800 2789750 ### Division ni n⁄2 278975 185983 139488 111590 ### Exponentiation ni n2 311308202500 173694411584875000 96912796943781006250000 54072495054782612437187500000 ### Nth Root i√n 2√n 746.961 82.325 27.3306 14.1032 ## 557950 as geometric shapes ### Circle Diameter 1.1159e+06 3.5057e+06 9.78004e+11 ### Sphere Volume 7.27569e+17 3.91201e+12 3.5057e+06 ### Square Length = n Perimeter 2.2318e+06 3.11308e+11 789060 ### Cube Length = n Surface area 1.86785e+12 1.73694e+17 966398 ### Equilateral Triangle Length = n Perimeter 1.67385e+06 1.348e+11 483199 ### Triangular Pyramid Length = n Surface area 5.39202e+11 2.04701e+16 455564 ## Cryptographic Hash Functions md5 797772116b65957c1b672a3e0250ae4f d4b62d6f7a913040dcdf6b942828ac179d65508e d89c2f96d9512b2892a13ef28ec858787f89a098a1837cb6efd4b5f63b18d018 832c2d4a0864b4a4f9798be55c687c7d541c81c0cfd6e5d4e29111a58b957708964628d3802ed6a3edf854370f226c198e8214962f1c29dc22d1c7b67beb8255 f63cc35cdc436fa92b39949b586107817cddc3f8
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# Cut The Knot! An interactive column using Java applets by Alex Bogomolny # Napoleon's Propeller July 2002 As the two most recent columns have been devoted to synthetic proofs of a curious result, I've been looking for an example or two of an illuminating analytic proof. I found quite a few. Two such appear below. In the process I made a small, but surprising, discovery that is reflected in the title of the present column. Three altitudes of a triangle meet at a point known as the orthocenter of the triangle. There are many proofs of that result. Here's one that uses complex numbers. Given ΔABC, we may assume its vertices lie on a circle centered at the origin of a Cartesian coordinate system. Let's think of points in the plane as complex numbers. Define H = A + B + C, a simple symmetric function of all the vertices. In fact, H is the common point of the three altitudes of the triangle. Indeed, for AH and BC to be orthogonal, the ratio (H - A)/(B - C) must be purely imaginary. But (H - A)/(B - C) = (H - A)·(B - C)*/|B - C|2 = (B + C)·(B - C)*/|B - C|2 = (|B|2 + B*C - BC* - |C|2)/|B - C|2 = (B*C - BC*)/|B - C|2 where * denotes the conjugate operator. If X denotes the latter expression, X = -X* and is therefore purely imaginary. So that AH and BC are orthogonal, as expected. Similarly, BH is perpendicular to AC and CH to AB. The proof admits an elegant shortcut [Hahn, p. 71] that appeals to geometric intuition. Note that H - A = B + C = 2·(B + C)/2, which in particular means that H - A is parallel to the line joining the origin - the circumcenter of ΔABC - with the midpoint of BC, a chord in the circumcircle. In other words, AH is perpendicular to BC, and similarly for BH and CH. The proof delivers more than was expected. With just a few sentences, not only we get an explicit expression for the orthocenter, the form of the expression (A + B + C) makes it hard to avoid a juxtaposition with the centroid (A + B + C)/3 of the triangle and discovery of the Euler line. The modified variant is short, powerful and illuminating, at least on a par with several synthetic proofs. Following is an example where the analytic apparatus of complex numbers is used with clarity unmatched by purely geometric proofs I can think of. Let two triangles be similar and similarly oriented [Wells, p. 20]. Then the midpoints of the segments joining their corresponding vertices form a third triangle similar to the other two. 21 January 2016, Created with GeoGebra Two triangles ABC and A'B'C' are similar iff, say, (B - A)/(C - A) = (B' - A')/(C' - A'). The condition means that not only the ratio of lengths of the pairs of sides AB/AC and A'B'/A'C' are equal but that also the angles between them are the same. In the language of determinants the condition for (direct) similarity of two triangles is simply (1) If l = 1/2 and m = 1/2, this is equivalent to (1') The assertion is thus immediate as is the generalization for l + m = 1, or in fact, for any l and m not simultaneously 0. (This is a particular case of the Fundamental Theorem of Directly Similar Figures: if the lines connecting the corresponding vertices of two directly similar polygons are devided in equal ratios, then the resulting polygon is directly similar to the given two. Steve Gray has reminded me in a private correspondence that the Fundamental Theorem of Directly Similar Figures, or more specifically (1'), with complex coefficients yields an elegant theorem concerning two triples of similar triangles.) With (1), it is easy to determine when a triangle is equilateral. The condition is The latter is equivalent to A2 + B2 + C2 - AB - BC - AC = 0, which is the same as (A + jB + j2C)·(A + j2B + jC) = 0, where j is a rotation through 120° (in the positive direction): j2 + j + 1 = 0. Therefore, depending on the orientation of ΔABC, either A + jB + j2C = 0, or A + j2B + jC = 0. The former is the criterion used in Connes' proof of Morley's theorem. [Hahn, p. 60, Pedoe, p. 184]. For a positively oriented triangle, the criterion might have also been obtained more directly from (2) Of course, (2) could be used to derive Napoleon's theorem. The proof is exceptionally clear. Another proof with complex numbers, although straightforward, is longer and might appear somewhat obscure. However, it delivers an easily overlooked surprise. Let OC, OA, and OB be the centers of the Napoleon triangles erected on the sides of ΔABC. Then easy computations show that (3) OAOB = XB + dXC, OBOC = XC + dXA, OCOA = XA + dXB, where d is the rotation through 60° in the positive direction (d2 - d + 1 = 0), XA = (B + C - 2A)/3, and similarly for XB and XC. Since d2 = j, Napoleon's theorem now follows from (2) and (3). However, (3) warrants a second look. XA equals two thirds of the median in ΔABC (looked at as a complex number) drawn from vertex A, and similarly for XB and XC. (Indeed, (B + C)/2 - A = (B + C - 2A)/2 is the complex number "from A to (B+C)/2," a median of ΔABC. But (B + C - 2A)/3 = 2/3·(B + C - 2A)/2.) It's clear then that ΔXAXBXC is equal to ΔABC, but has its center at the origin. (Triangles ABC and XAXBXC are not just equal. They are centrally symmetric to each other.) (3), therefore, says something about ΔABC. If ΔXAXBXC is rotated through 60° and its vertices are added pairwise after a cyclic permutation to the vertices of its image, the three complex numbers thus obtained form an equilateral triangle. (To make the picture more compact, we may join the vertices of ΔXAXBXC to those of its rotated image and consider the triangle formed by the midpoints of these segments. The midpoints, too, form an equilateral triangle.) ### If you are reading this, your browser is not set to run Java applets. Try IE11 or Safari and declare the site https://www.cut-the-knot.org as trusted in the Java setup. Here's a surprise. The origin, XA and dXA also form an equilateral triangle, and the same holds for the other two triples. The whole picture is exactly that of the Asymmetric Propeller. In the applet below, we may see either two equal triangles at 60° to each other, or the three equilateral triangles formed by the pairs of complex numbers corresponding to the matching vertices of those triangles, or a combination of the above. 21 January 2016, Created with GeoGebra Napoleon's theorem is equivalent to the Asymmetric Propeller's theorem! How small is the world! Now, both the original Asymmetric Propeller and Napoleon's theorem start with three equilateral triangles and discover the fourth one by construction. It might have been natural to look for a link between the two results. I never saw the link established by synthetic means. Martin Gardner wrote about the Asymmetric Propeller in 1999 in a paper that since has been reprinted in his latest collection Gardner's Workout. In the Postscript to the corresponding chapter, Gardner mentions another problem [see also Honsberger, p. 274-276] described to him by Leon Bankoff. (This is in fact the Finsler-Hadwiger Theorem.) That problem admits a simple synthetic solution and, as a consequence of Neuberg's Theorem, another one using complex numbers. It's also a special case of the Fundamental Theorem of Directly Similar Figures. ### References 1. M. Gardner, Gardner's Workout, A K Peters, 2001 2. Liang-shin Hahn, Complex Numbers & Geometry, MAA, 1994 3. R. Honsberger, In Pólya's Footsteps, MAA, 1997 4. D. Pedoe, Geometry: A Comprehensive Course, Dover, 1970 5. D. Wells, You Are a Mathematician, John Wiley & Sons, 1997
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Elimination system of equations solver Elimination system of equations solver is a software program that helps students solve math problems. Our website can solve math word problems. The Best Elimination system of equations solver Elimination system of equations solver can be found online or in mathematical textbooks. A math answer scanner is a tool that can be used to check answers to math problems. This can be a useful tool for students who want to check their work, or for teachers who want to check answers to problems before giving them to their students. There are a few things you can do to make solving math word problems easier. First, read the problem carefully and make sure you understand what it is asking. Once you know what the problem is asking, try to identify any key information or relevant equations that you will need to solve it. Once you have all of the necessary information, you can begin solving the problem. If you get stuck, don't be afraid to ask for help from a friend or a teacher. Assuming you want to find the midpoint of a line segment: To find the midpoint of a line segment, you need to find the average of the x-coordinates and the average of the y-coordinates. To find the average of the x-coordinates, add the x-coordinates together and divide by 2. To find the average of the y-coordinates, add the y-coordinates together and divide by 2. The midpoint of The Pythagorean theorem is a statement in mathematics that states that in a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. This theorem is represented by the equation: a^2 + b^2 = c^2. In this equation, a and b represent the lengths of the two shorter sides of the triangle, while c represents the length of the hypotenuse. To solve for b, There are a few different methods that can be used to solve equations with both x and y variables. One common method is graphing the equations on a coordinate plane and finding the points of intersection. Another method is to use substitution, where one variable is isolated and then solved for. Lastly, elimination can be used, where like terms are cancelled out in order to solve for one variable. We will support you with math difficulties 10/10 recommend. Right now, I am in 10th grade geometry and the app is my savior. If you have a hard time with math this helps you it does help, explain the steps to tell you how to do it. And IT’S FREEEE!! You only have to pay for it if you want little extra things like animated videos and some other things. Amy Garcia I love the app it really helps the student to understand math problem by providing the solution. Rather than using math way I can't see the solution because you have to pay for subscription. They're student friendly application. The student can't provide much money to subscribe we here to download to help us to understand math instead of helping they do that for business. Thank you, photo, math, Isabella Lopez Step by step math calculator Quadratic function equation solver Algebraic expressions solver Sin cos tan solver Homework answers scanner Website that answers math problems and shows work
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# physics posted by . if 5/6th of a radioactive sample decay in 70 years and 1/6th remain. how can you find the half life without knowing the decay constant can you help? • physics - The fraction (1/6) that remains after 70 years satisfies the relation: 1/6 = 2^(-t/T), where T is the half life and t = 70 years. Take logs of both sides to solve for T. I'll use logs to base e but it doesn't matter -1.7918 = -70/T ln 2 = -48.52/T T = 27.1 years
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# Understanding Quaternions ### 1. Introduction Attitude and Heading Sensors from CH Robotics can provide orientation information using both Euler Angles and Quaternions.  Compared to quaternions, Euler Angles are simple and intuitive and they lend themselves well to simple analysis and control.  On the other hand, Euler Angles are limited by a phenomenon called "gimbal lock," which prevents them from measuring orientation when the pitch angle approaches +/- 90 degrees. Quaternions provide an alternative measurement technique that does not suffer from gimbal lock.  Quaternions are less intuitive than Euler Angles and the math can be a little more complicated.  This application note covers the basic mathematical concepts needed to understand and use the quaternion outputs of CH Robotics orientation sensors. Sensors from CH Robotics that can provide quaternion orientation outputs include the UM6 Orientation Sensor and the UM6-LT Orientation Sensor. ### 2. Quaternion Basics A quaternion is a four-element vector that can be used to encode any rotation in a 3D coordinate system.  Technically, a quaternion is composed of one real element and three complex elements, and it can be used for much more than rotations.  In this application note we'll be ignoring the theoretical details about quaternions and providing only the information that is needed to use them for representing the attitude of an orientation sensor. The attitude quaternion estimated by CH Robotics orientation sensors encodes rotation from the "inertial frame" to the sensor "body frame."  The inertial frame is an Earth-fixed coordinate frame defined so that the x-axis points north, the y-axis points east, and the z-axis points down as shown in Figure 1.  The sensor body-frame is a coordinate frame that remains aligned with the sensor at all times.  Unlike Euler Angle estimation, only the body frame and the inertial frame are needed when quaternions are used for estimation (Understanding Euler Angles provides more details about using Euler Angles for attitude estimation). Figure 1 - The Inertial Frame Let the vector $\mathbf{q}_i^b$ be defined as the unit-vector quaternion encoding rotation from the inertial frame to the body frame of the sensor: $\mathbf{q}_i^b = \begin{pmatrix} a & b & c & d \end{pmatrix}^T.$ where $T$ is the vector transpose operator.  The elements b, c, and d are the "vector part" of the quaternion, and can be thought of as a vector about which rotation should be performed.  The element is the "scalar part" that specifies the amount of rotation that should be performed about the vector part.  Specifically, if $\theta$ is the angle of rotation and the vector $\begin{pmatrix} v_x & v_y & v_z \end{pmatrix}^T$ is a unit vector representing the axis of rotation, then the quaternion elements are defined as $\begin{pmatrix} a\\ b\\ c\\ d \end{pmatrix} = \begin{pmatrix} \cos(0.5\theta) \\ v_x\sin(0.5\theta)\\ v_y\sin(0.5\theta)\\ v_z\sin(0.5\theta) \end{pmatrix}.$ In practice, this definition needn't be used explicitly, but it is included here because it provides an intuitive description of what the quaternion represents.  CH Robotics sensors output the quaternion $\mathbf{q}_i^b$ when quaternions are used for attitude estimation. ### 3. Rotating Vectors Using Quaternions The attitude quaternion $\mathbf{q}_i^b$ can be used to rotate an arbitrary 3-element vector from the inertial frame to the body frame using the operation $\mathbf{v}_B = \mathbf{q}_i^b\begin{pmatrix} 0\\ \mathbf{v}_I \end{pmatrix} (\mathbf{q}_i^b)^{-1}.$ That is, a vector can rotated by treating it like a quaternion with zero real-part and multiplying it by the attitude quaternion and its inverse.  The inverse of a quaternion is equivalent to its conjugate, which means that all the vector elements (the last three elements in the vector) are negated.  The rotation also uses quaternion multiplication, which has its own definition. Define quaternions $\mathbf{q}_1 = \begin{pmatrix} a_1 & b_1 & c_1 & d_1 \end{pmatrix}^T$ and $\mathbf{q}_2 = \begin{pmatrix} a_2 & b_2 & c_2 & d_2 \end{pmatrix}^T$.  Then the quaternion product $\mathbf{q}_1 \mathbf{q}_2$ is given by $\mathbf{q}_1 \mathbf{q}_2 = \begin{pmatrix} a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2 \\ a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2 \\ a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2 \\ a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2 \end{pmatrix}.$ To rotate a vector from the body frame to the inertial frame, two quaternion multiplies as defined above are required.  Alternatively, the attitude quaternion can be used to construct a 3x3 rotation matrix to perform the rotation in a single matrix multiply operation.  The rotation matrix from the inertial frame to the body frame using quaternion elements is defined as $R_i^b(\mathbf{q}_i^b)=\begin{pmatrix} a^2 + b^2 - c^2 - d^2 & 2bc - 2ad & 2bd + 2ac \\ 2bc + 2ad & a^2 - b^2 + c^2 - d^2 & 2cd - 2ab \\ 2bd - 2ac & 2cd + 2ab & a^2 - b^2 - c^2 + d^2 \end{pmatrix}.$ Then the rotation from the inertial frame to the body frame can be performed using the matrix multiplication $\mathbf{v}_B = R_i^b(\mathbf{q}_i^b)\mathbf{v}_I.$ Regardless of whether quaternion multiplication or matrix multiplication is used to perform the rotation, the rotation can be reversed by simply inverting the attitude quaternion before performing the rotation.  By negating the vector part of the quaternion vector, the operation is reversed. ### 4. Converting Quaternions to Euler Angles CH Robotics sensors automatically convert the quaternion attitude estimate to Euler Angles even when in quaternion estimation mode.  This means that the convenience of Euler Angle estimation is made available even when more robust quaternion estimation is being used. If the user doesn't want to have the sensor transmit both Euler Angle and Quaternion data (for example, to reduce communication bandwidth requirements), then the quaternion data can be converted to Euler Angles on the receiving end. The exact equations for converting from quaternions to Euler Angles depends on the order of rotations.  CH Robotics sensors move from the inertial frame to the body frame using first yaw, then pitch, and finally roll.  This results in the following conversion equations: $\phi = \arctan\left( \frac{2(ab+cd)}{a^2 - b^2 - c^2 + d^2} \right ),$ $\theta = -\arcsin\left( 2(bd - ac) \right ),$ and $\psi = \arctan\left( \frac{2(ad+bc)}{a^2 + b^2 - c^2 - d^2} \right ).$ See the chapter on Understanding Euler Angles for more details about the meaning and application of Euler Angles.  When converting from quaternions to Euler Angles, the atan2 function should be used instead of atan so that the output range is correct.  Note that when converting from quaternions to Euler Angles, the gimbal lock problem still manifests itself.  The difference is that since the estimator is not using Euler Angles, it will continue running without problems even though the Euler Angle output is temporarily unavailable.  When the estimator runs on Euler Angles instead of quaternions, gimbal lock can cause the filter to fail entirely if special precautions aren't taken.
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## What your child will learn and do in Grade 2 Mathematics In grade two, students extend their understanding of place value to the hundreds place. They use this place value understanding to solve word problems, including those involving length and other units of measure. Activities in these areas include: • Quickly and accurately adding numbers together that total up to 100 or fewer or subtracting from numbers up through 100 • Mentally adding numbers that total 20 or fewer or subtracting numbers 20 or fewer • Solving one- or two-step word problems by adding or subtracting numbers up through 100 • Understanding what the different digits mean in a three-digit number (place value) • Reading, writing, comparing three-digit numbers • Determining whether a group of objects has an odd or even number of members • Adding and subtracting three-digit numbers based on place value • Measuring lengths of objects in standard units such as inches and centimeters • Solving addition and subtraction word problems involving length • Solving problems involving money (identifying coins and their value, exchanging, solving word problems with money amounts) • Telling time to the nearest 5-minute interval • Collecting data, building a graph (bar graph, picture graph, or line plot) and answering questions about the data • Breaking up a rectangle into equal-size squares and finding the number of squares using repeated addition • Dividing circles and rectangles into halves, thirds, or fourths • Identifying triangles, quadrilaterals, pentagons, hexagons, and cubes; drawing shapes given their number of angles or sides Helping your child learn outside of school: • Play math games with your child to build their fluency. For example, using a deck of cards, deal two cards and ask your child to add the two numbers before you do. Whoever says the total first, keeps the cards. • Have your child explain the relationship between different numbers without counting. For example, 147 is 47 more than 100 and three less than 150. • Have your child tell the time on the clock when you sit down to breakfast or dinner. • Count change with your child. Ask them to find the change to pay the cashier at the store. • Encourage your child to stick with it whenever a problem seems difficult. • Can you do some easier problems and go back to this one after? • What part of the problem is giving you trouble? • Let's read the problem together and make sure we understand what it is asking. • Can we draw a picture of the problem? • Can we make up an easier problem that is similar to this? Then we can work our way up to this one. • Let’s take a 10 minute break and come back to this one. Websites
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# Search 5th Grade Online Exercises 137 filtered results 137 filtered results Sort by Adding Mixed Fractions With Like Denominators Exercise Adding Mixed Fractions With Like Denominators Students will be able to add mixed number fractions after they learn how to make each number have like denominators. Math Exercise Multi-Digit Multiplication and Partial Products 1 Exercise Multi-Digit Multiplication and Partial Products 1 Break down multi-digit multiplication with these exercises that lead students step-by-step through the partial product method. Math Exercise Understanding Expressions and Equations Exercise Understanding Expressions and Equations Students will understand mathematical expressions and equations after they work through this engaging exercise from Education.com. Math Exercise Interjections 2 Exercise Interjections 2 Teach your fifth graders when interjections are appropriate and how to use them with these exercises that have a variety of questions and helpful hints. Exercise Exercise Adding Fractions with Unlike Denominators will help students practice this key fifth grade skill. Try our free exercises to build knowledge and confidence. Math Exercise Solving Basic Algebraic Equations Exercise Solving Basic Algebraic Equations Students will appreciate this exercise that shows how to solve basic algebraic equations with ease. Math Exercise Order of Operations and Use of Parentheses Exercise Order of Operations and Use of Parentheses Reinforce students’ understanding of PEMDAS by practicing the order of operations and using parentheses in this exercise. Math Exercise Division with Multi-Digit Dividends Exercise Division with Multi-Digit Dividends Expand students’ division skills with this exercise that teaches how to perform division problems with multi digit dividends. Math Exercise Root Words Exercise Root Words Students will understand just where their vocabulary words come from with this root words exercise. Exercise Types of Sentences Exercise Types of Sentences Test your students with these exercises that have them identify simple, compound, and complex sentences. Helpful hints give your students all the information they need to work through the problems on their own. Exercise Irregular Verbs 3 Exercise Irregular Verbs 3 There are so many irregular verbs in the English language that students will need to complete several exercises before growing familiar with them all. Exercise Using Commas to Separate an Interrupter Exercise Using Commas to Separate an Interrupter Have your fifth grader learn when a clause is an interrupter with this fun and interactive activity that tests them on their knowledge of clauses rather than that of punctuation. Exercise Punctuating Titles Exercise Punctuating Titles Teach your fifth grader how to punctuate titles of different written works in essays and reports with these exercises and hints. Exercise Hyperbole Exercise Hyperbole Students will finally have a name to describe the “dog ate my homework” storytelling that runs rampant throughout childhood with this hyperbole exercise. Exercise Subtract Fractions With Unlike Denominators Exercise Subtract Fractions With Unlike Denominators Show students how to complete a fraction subtraction problem while working with unlike denominators with this easy to use exercise. Math Exercise Volume of a Rectangular Prism Exercise Volume of a Rectangular Prism With a clear illustration and instructions, this exercise will teach students how to calculate the seemingly complex volume of a rectangular prism. Math Exercise Idioms 2 Exercise Idioms 2 Give a cultural lesson and expand your students’ vocabulary at the same time with this idioms exercise. Exercise Relative Pronouns Exercise Relative Pronouns Relative pronouns make it simple for your fourth grader to describe the subject of a sentence in a fluid and orderly way. Give them the practice they need with these exercises and helpful hints. Exercise Division with Two-Digit Divisors Exercise Division with Two-Digit Divisors When students begin to grow adept at division, expand their skillset by introducing them to two digit divisors. Math Exercise Pronoun Antecedent Agreement 2 Exercise Pronoun Antecedent Agreement 2 Help develop the fluency of your student's writing with these exercises that help them understand the importance of matching object and subject pronouns with their antecedent. Exercise Decimal Numbers and the Thousandths Place Exercise Decimal Numbers and the Thousandths Place Take students’ attention to detail to the next level with this exercise that teaches decimals to the thousandths place. Math Exercise Common Suffixes 3 Exercise Common Suffixes 3 Spelling will be a breeze for students after they complete their understanding of suffixes with this final exercise in the series. Exercise Present Tense Verbs 3 Exercise Present Tense Verbs 3 Students with a strong foundation in the present tense of verbs will be able to better understand other tenses. Exercise Complete Sentences Exercise Complete Sentences Give students practice with all of the parts that make up complete sentences in this exercise.
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## Monday, February 3, 2014 ### Finding Equation - Circle, 12 Category: Analytic Geometry, Plane Geometry, Algebra "Published in Vacaville, California, USA" Find the equation of a circle that passes through the points of intersection of the circles x2 + y2 = 5 and x2 + y2 - x + y = 4, and through the point (2, -3). Solution: To illustrate the problem, it is better to draw the figure as follows A circle that passes through the intersection of two circles and a point. (Photo by Math Principles in Everyday Life) Since the given two circles are non-concentric with their points of intersection, then the equation of another circle can be written as where k is a constant that represents a family of non-concentric circles. To solve for the value of k, substitute the values of x and y from the given point, we have Therefore, the equation of a circle is
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# DISCRIMINANT ANALYSIS Follow City-Data.com founder on our Forum or Photo by: Helder Almeida Discriminant analysis is a statistical method that is used by researchers to help them understand the relationship between a "dependent variable" and one or more "independent variables." A dependent variable is the variable that a researcher is trying to explain or predict from the values of the independent variables. Discriminant analysis is similar to regression analysis and analysis of variance (ANOVA). The principal difference between discriminant analysis and the other two methods is with regard to the nature of the dependent variable. Discriminant analysis requires the researcher to have measures of the dependent variable and all of the independent variables for a large number of cases. In regression analysis and ANOVA, the dependent variable must be a "continuous variable." A numeric variable indicates the degree to which a subject possesses some characteristic, so that the higher the value of the variable, the greater the level of the characteristic. A good example of a continuous variable is a person's income. In discriminant analysis, the dependent variable must be a "categorical variable." The values of a categorical variable serve only to name groups and do not necessarily indicate the degree to which some characteristic is present. An example of a categorical variable is a measure indicating to which one of several different market segments a customer belongs; another example is a measure indicating whether or not a particular employee is a "high potential" worker. The categories must be mutually exclusive; that is, a subject can belong to one and only one of the groups indicated by the categorical variable. While a categorical variable must have at least two values (as in the "high potential" case), it may have numerous values (as in the case of the market segmentation measure). As the mathematical methods used in discriminant analysis are complex, they are described here only in general terms. We will do this by providing an example of a simple case in which the dependent variable has only two categories. Discriminant analysis is most often used to help a researcher predict the group or category to which a subject belongs. For example, when individuals are interviewed for a job, managers will not know for sure how job candidates will perform on the job if hired. Suppose, however, that a human resource manager has a list of current employees who have been classified into two groups: "high performers" and "low performers." These individuals have been working for the company for some time, have been evaluated by their supervisors, and are known to fall into one of these two mutually exclusive categories. The manager also has information on the employees' backgrounds: educational attainment, prior work experience, participation in training programs, work attitude measures, personality characteristics, and so forth. This information was known at the time these employees were hired. The manager wants to be able to predict, with some confidence, which future job candidates are high performers and which are not. A researcher or consultant can use discriminant analysis, along with existing data, to help in this task. There are two basic steps in discriminant analysis. The first involves estimating coefficients, or weighting factors, that can be applied to the known characteristics of job candidates (i.e., the independent variables) to calculate some measure of their tendency or propensity to become high performers. This measure is called a "discriminant function." Second, this information can then be used to develop a decision rule that specifies some cut-off value for predicting which job candidates are likely to become high performers. The tendency of an individual to become a high performer can be written as a linear equation. The values of the various predictors of high performer status (i.e., independent variables) are multiplied by "discriminant function coefficients" and these products are added together to obtain a predicted discriminant function score. This score is used in the second step to predict the job candidates likelihood of becoming a high performer. Suppose that you were to use three different independent variables in the discriminant analysis. Then the discriminant function has the following form: where D = discriminant function score, B , = discriminant function coefficient relating independent variable i to the discriminant function score, X = value of independent variable i. The equation is quite similar to a regression equation. Conventional regression analysis should not be used in place of discriminant analysis. The dependent variable would have only two values (high performer and low performer) and would thus violate important assumptions of the regression model. Discriminant analysis does not have these limitations with respect to the dependent variable. Estimation of the discriminant function coefficients requires a set of cases in which values of the independent variables and the dependent variables are known. In the case described above, the company has this information for a current group of employees. There are several different ways that can be used to estimate discriminant function coefficients, but all work on the same general principle: the values of the coefficients are selected so that differences between the groups defined by the dependent variable are maximized with regard to some objective function. One commonly used objective function is the F-ratio, which is defined as it is in ANOVA and regression problems. The coefficients are chosen to maximize the F-ratio when analysis of variance is performed on the resulting discriminant function, using the dependent variable (i.e., job performance) as the grouping variable. Most general statistical programs, such as the Statistical Package for the Social Sciences, contain discriminant analysis modules. There are various tests of significance that can be used in discriminant analysis. One widely used test statistic is based on Wilks lambda, which provides an assessment of the discriminating power of the function derived from the analysis. If this value is found to be statistically significant, then the set of independent variables can be assumed to differentiate between the groups of the categorical variable. This test, which is analogous to the F-ratio test in ANOVA and regression, is useful in evaluating the overall adequacy of the analysis. Unfortunately, discriminant analysis does not generate estimates of the standard errors of the individual coefficients, as in regression, so it is not quite so simple to assess the statistical significance of each coefficient. For example, most discriminant analysis programs have a stepwise option. Independent variables are entered into the equation one at a time. Again, Wilks lambda can be used to assess the potential contribution of each variable to the explanatory power of the model. Variables from the set of independent variables are added to the equation until a point is reached for which additional items provide no statistically significant increment in explanatory power. Once the analysis is completed, the discriminant function coefficients can be used to assess the contributions of the various independent variables to the tendency of an employee to be a high performer. The discriminant function coefficients are analogous regression coefficients and they range between values of -1.0 and 1.0. The first box in Figure 1 (on the facing page) provides hypothetical results of the discriminant analysis. The second box provides the within-group averages for the discriminant function for the two categories of the dependent variable. Note that the high performers have an average score of 1.45 on the discriminant function, while the low performers have an average score of -.89. The discriminant function is treated as a standardized variable, so it has a mean of zero and a standard deviation of one. The average values of the discriminant function scores are meaningful only in that they help us interpret the coefficients. Since the high performers are at the upper end of the scale, all of the positive coefficients indicate that the greater the value of those variables, the greater the likelihood of a worker being a high performer (e.g., education, motivation). The magnitudes of the coefficients also tell us something about the relative contributions of the independent variables. The closer the value of a coefficient is to zero, the weaker it is as a predictor of the dependent variable. On the other hand, the closer the value of a coefficient is to either 1.0 or -1.0, the stronger it is as a predictor of the dependent variable. In this example, then, years of education and ability to handle stress both have positive coefficients, though the latter is quite weak. Finally, individuals who place high importance on family life are less likely to be high performers than those who do not. The second step in discriminant analysis involves predicting to which group in the dependent variable a particular case belongs. A subject's discriminant score can be translated into a probability of being in a particular group by means of Bayes Rule. Separate probabilities are computed for each group and the subject is assigned to the group with the highest probability. Another test of the adequacy of a model is the degree to which known cases are correctly classified. As in other statistical procedures, it is generally preferable to test the model on a set of cases that were not used to estimate the model's parameters. This provides a more conservative test of the model. Thus, a set of cases should, if possible, be saved for this purpose. Having completed the analysis, the results can be used to predict the work potential of job candidates and hopefully serve to improve the selection process. There are more complicated cases, in which the dependent variable has more than two categories. For example, workers might have been divided into three groups: high performers, average performers, low performers. Discriminant analysis allows for such a case, as well as many more categories. The interpretation, however, of the discriminant function scores and coefficients becomes more complex. The books included in the "Further Reading" section below explain in detail how to perform discriminant analysis with multiple categories and provide in-depth technical discussions. [ John J. Lawler ] Huberty, Carl J. Applied Discriminant Analysis. New York: Wiley, 1994. Klecka, William R. Discriminant Analysis for Social Sciences. Beverly Hills, CA: Sage Publications, 1980. Lachenbruch, Peter A. Discriminant Analysis. New York: Hafner Press, 1975. McLachlan, Geoffrey J. Discriminant Analysis and Statistical Pattern Recognition. New York: Wiley, 1992. ## User Contributions: Srinivasan Ramesh Feb 20, 2007 @ 12:00 am I find this paper very interesting and useful. Can I have some details how discriminant analysis be used in Medical diagnosis.
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To prove one-one & onto (injective, surjective, bijective) Chapter 1 Class 12 Relation and Functions Concept wise Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class ### Transcript Ex 1.2, 3 (Introduction) Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x. f(x) = [x] = greatest integer less than equal to x Example: [1] = 1 [1.01] = 1 [1.2] = 1 [1.9] = 1 [1.99] = 1 [2] = 2 Ex 1.2, 3 Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x. f(x) = [x] where [x] denotes the greatest integer less than equal to x Check one-one f(x) = [x] Example f(1) = [1] = 1, f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1, f(1.99) = [1.99] = 1, Since, different elements 1, 1.2, 1.9, 1.99 have the same image 1 , ∴ f is not one-one. Check onto f(x) = [x] Let y = f(x) y = [x] i.e. y = Greatest integer less than or equal to x Hence, value of y will always come an integer. But y is a real number Hence f is not onto.
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# 10 how many atoms are in 2.70 moles of iron (fe) atoms? Ideas Contents Below is information and knowledge on the topic how many atoms are in 2.70 moles of iron (fe) atoms? gather and compiled by the show.vn team. Along with other related topics like: How many atoms are in a mole of iron, if you measure 34.6 grams of mg on a scale, how many moles of mg do you have?, how many moles of silver are represented by 2.888 x 10^23 atoms, How many o2 molecules are in a mole of oxygen, how many formula units are in 2 moles of nacl?, Moles of Iron to atoms calculator, how many molecules are in 2 moles h2o?, how many atoms are in 3.6 moles of selenium?. =”video” src=”https://www.youtube.com/embed/sFuvrBXNHAI” frameborder=”0″ allow=”accelerometer; autoplay; encrypted-media; gyroscope;” allowfullscreen> role=”button” tabindex=”0″>1:59In this video well learn to convert grams of Iron (Fe) to moles and then convert the moles of Iron to particles (atoms).YouTube · Wayne Breslyn · Apr 7, 2021Missing: 2.70 ‎| Must include: 2.70 atoms are in 2.70 moles of iron atoms? | Socratic Read More:  10 how many different values of l are possible for an electron with principal quantum number n = 4? Ideas #### Explanation: A mole is #~6.022*10^23# of something So, saying you have a mole of iron atoms is like saying you have $6.022 \cdot {10}^{23}$ iron atoms This means that 2.70 moles of iron atoms is $2.70 \cdot \left(6.022 \cdot {10}^{23}\right)$ iron atoms #2.7*(6.022*10^23)=~1.63*10^24# Therefore you have $1.63 \cdot {10}^{24}$ iron atoms ## Extra Information About how many atoms are in 2.70 moles of iron (fe) atoms? That You May Find Interested If the information we provide above is not enough, you may find more below here. ### How many atoms are in 2.70 moles of iron atoms? – Socratic • Author: socratic.org • Rating: 4⭐ (871922 rating) • Highest Rate: 5⭐ • Lowest Rate: 3⭐ • Sumary: 1.63*10^24 A mole is ~6.022*10^23 of something So, saying you have a mole of iron atoms is like saying you have 6.022*10^23 iron atoms This means that 2.70 moles of iron atoms is 2.70*(6.022*10^23) iron atoms 2.7*(6.022*10^23)=~1.63*10^24 Therefore you have 1.63*10^24 iron atoms • Matching Result: 1.63*10^24 A mole is ~6.022*10^23 of something So, saying you have a mole of iron atoms is like saying you have 6.022*10^23 iron atoms This … • Intro: How many atoms are in 2.70 moles of iron atoms? | Socratic Explanation: A mole is #~6.022*10^23# of something So, saying you have a mole of iron atoms is like saying you have #6.022*10^23# iron atoms This means that 2.70 moles of iron atoms is #2.70*(6.022*10^23)# iron atoms #2.7*(6.022*10^23)=~1.63*10^24# Therefore… ### How Many Atoms Are In 2.70 Moles Of Iron (Fe) Atoms? • Author: microblife.in • Rating: 4⭐ (871922 rating) • Highest Rate: 5⭐ • Lowest Rate: 3⭐ • Sumary: There are 16.2594 X 1023 atoms of iron (Fe) in a sample of 2.70 moles of iron atoms. • Matching Result: There are 16.2594 X 1023 atoms of iron (Fe) in a sample of 2.70 moles of iron atoms. How many atoms of Fe are in 2.5 moles? • Intro: How Many Atoms Are In 2.70 Moles Of Iron (Fe) Atoms? – Micro B Life There are 16.2594 X 1023 atoms of iron (Fe) in a sample of 2.70 moles of iron atoms. How many atoms of Fe are in 2.5 moles? 2.5 mol 16.02×1023 atoms. M = (1.5 x… ### What quantity in moles of iron atoms do you have if … – Wyzant • Author: wyzant.com • Rating: 4⭐ (871922 rating) • Highest Rate: 5⭐ • Lowest Rate: 3⭐ • Sumary: What quantity in moles of iron atoms do you have if you have 3.40 × 10²³ atoms of iron. (The mass of one mole of iron is 55.85 g.) • Matching Result: It is a conversion factor when written 6.02×1023 atoms/mole. If we have 3.40×1023 atoms of iron, we have: (3.40×1023 atoms Fe)/( … • Intro: What quantity in moles of iron atoms do you have if you have 3.40 × 10²³ atoms of iron. (The mass of one mole of iron is 55.85 g.) One mole, by definition, is 6.02×1023 atoms/molecules/particles [of anything, actually]. I can have 1 mole of Twinkies, if I hit up… Read More:  10 how does a global economy impact you everfi Ideas ## Frequently Asked Questions About how many atoms are in 2.70 moles of iron (fe) atoms? If you have questions that need to be answered about the topic how many atoms are in 2.70 moles of iron (fe) atoms?, then this section may help you solve it. ### A mole of iron contains how many Fe atoms? One mole of any element contains b>6.022 X 1023/b> atoms, regardless of the type of element. For example, the atomic mass of iron (Fe) is 55.85 amu, while the atomic mass of radium (Ra) is 226 amu. ### In how many atoms does MOL Fe exist? Answer and explanation: Because iron has a molar mass of 55.845 g/mole, there are 6.022 x 1023 atoms in a mole of iron. ### 2.5 moles of atoms contain how many atoms? Answer and explanation: Using Avogadro’s number, or the number of fundamental units in a mole, which equals 6.022 X 1023, we can determine that there are 45.165 X 1023 atoms in 2.5 moles of SO2. ### What fraction of an atom does Fe have? The most prevalent element on Earth, iron, with the chemical symbol Fe (derived from the Latin word “ferrum,” or “iron”) and atomic number 26, has been chosen as the focus element for the month of February. ### A portion of the YouTube video How to Find the Number of Atoms in Fe(NO3)3 Iframe with a src of “https://www.youtube.com/embed/QWqcLgXOgNA” Read More:  10 physical anthropologists view how humans come to be the way they are as the result of: Ideas ### How much iron is one mole? The same idea can be extended to ionic compounds and molecules since iron has an atomic mass of 55.847 amu, meaning that one mole of iron atoms would weigh 55.847 grams. ### What does a mole of Fe weigh? The molar mass of iron is 55.84 grams/mole. ### 2.5 moles equals how many molecules? This means that N=2. 56. 0231023=151023 molecules will make up 2. 5 moles of water. ### How can one distinguish atoms from moles? One mole is equal to exactly 12 grams of pure carbon-12, or 6.022 x 1023 atoms. This is known as Avogadro’s Number (6.0221421 x 1023), and it represents the quantity of atoms in one mole. ### How do you locate Fe atoms? If the mass of an element or compound is known, then divide the given mass by the element’s or compound’s molar mass to find the number of moles. In 1 mole of a substance, there are or. 023 10 23 atoms, so the first step in calculating the number of atoms is to calculate the number of moles. ### How are moles changed into atoms? Moles to Atoms Definition Multiplying the number of moles by 6.022 * 1023, which is what a mole by definition signifies, will yield the desired number of atoms. ### What is the number of an atom? The number of protons in an atom’s nucleus, which determines the identity of an element (for example, an element with six protons is a carbon atom, regardless of how many neutrons may be present), is known as the atomic number. ### Two moles contain how many atoms? We refer to macroscopic amounts of atoms and molecules in chemistry as moles, named after the 19th-century chemist Amedeo Avogadro. For example, if we have 1 mol of O atoms, we are said to have 1 (6.022 10 23) O atoms, or 1.2044 10 24 Na atoms. ### How do you determine the atom count? The formula, which states that the number of atoms in a volume of ANY substance is equal to the Avogadro’s number, or 6.022*1023 atoms/mole, is fairly straightforward: # of atoms = N * (density) * volume / (Molecular Weight).
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# 500 Divided By 5: An Easy Math Solution ## Introduction Mathematics is an essential subject that every student must learn. It helps to improve our logical and analytical skills, which are useful in our daily lives. One of the fundamental concepts in mathematics is division. In this article, we will discuss how to divide 500 by 5. ## The Basics of Division Before we dive into the problem, let us first review the basics of division. Division is a mathematical operation that involves splitting a number into equal parts. The number being divided is called the dividend, while the number of parts it is being split into is called the divisor. The result of division is called the quotient. ## How to Divide 500 by 5 Dividing 500 by 5 is a straightforward process. We need to divide the dividend (500) by the divisor (5) to get the quotient. To do this, we follow these steps: ### Step 1: Write the Dividend and Divisor Write the dividend (500) and divisor (5) on a piece of paper. ### Step 2: Divide the First Digit Start by dividing the first digit of the dividend (5) by the divisor (5). The result is 1. ### Step 3: Write the Quotient Write down the quotient (1) above the dividend (500). ### Step 4: Multiply the Divisor by the Quotient Multiply the divisor (5) by the quotient (1). The result is 5. ### Step 5: Subtract the Product from the Dividend Subtract the product (5) from the dividend (500). The result is 495. ### Step 6: Bring Down the Next Digit Bring down the next digit (0) from the dividend and write it next to the remainder (495). ### Step 7: Repeat the Process Repeat the process by dividing the new dividend (4950 by the divisor (5). The result is 99. ### Step 8: Write the Quotient and Remainder Write down the quotient (99) above the dividend (4950). The final answer is 100. ## Why is Dividing Important? Dividing is a fundamental operation in mathematics that has many real-life applications. We use division when dividing money between friends, calculating time and distance, and even when sharing food. It is an essential skill that we use in our daily lives. ## Conclusion In conclusion, dividing 500 by 5 is a simple process that involves following a few steps. It is a fundamental concept in mathematics that has many practical applications. By learning how to divide numbers, we can improve our logical and analytical skills, which are useful in our daily lives.
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 Non-Horizontally Launched Projectiles and their Trajectories - College Physics # Non-Horizontally Launched Projectiles and their Trajectories ## Introduction A projectile is launched with a given angle to the horizontal. The resulting motion is a combination of uniform motion along the x-axis and free fall. ## Experiment A projectile is launched from a hill ($$h_0 = \rm 30 \,\, m$$) with the initial velocity $$v_0 = \rm 40 \,\, \frac{m}{s}$$ at an angle $$\alpha = 20^\circ$$. It initially travels upwards until it reaches its maximum height and then falls faster and faster towards the ground. ResetStart Legende Geschwindigkeit Beschleunigung ## Results Launching a projectile non-horizontally results in a combination of a uniform motion along the x-axis and a uniformly accelerated motion along the y-axis. Thus the movement (trajectory) can be presented in a $$y(x)$$ diagram: ## Components of initial velocity The initial velocity $$v_0$$ is divided, depending on launch angle $$\ alpha$$ into the components $$v_x$$ and $$v_y$$: $$v_0 = \sqrt{ (v_x)^2 + (v_y)^2 }$$ $$v_{0,x} = v_0 \cdot \cos \alpha$$ $$v_{0,y} = v_0 \cdot \sin \alpha$$ ## Determining the trajectory To derive the trajectory the following laws are needed: Uniform motion $$x = v_0 \cdot \cos \alpha \cdot t$$ Uniformly accelerated motion $$y = h_0 - \dfrac{g}{2} \cdot t^2 + v_0 \cdot \sin \alpha \cdot t$$ Now, the equation for the x-axis is solved for $$t$$ and used in the equation for the y-axis: $$x = v_0 \cdot \cos \alpha \cdot t \qquad \Rightarrow \qquad t = \dfrac{x}{v_0 \cdot \cos \alpha}$$ \begin{aligned} y(t) &= h_0 - \dfrac{g}{2} \cdot t^2 + v_0 \cdot \sin \alpha \cdot t \\ \\ y(x) &= h_0 - \dfrac{g}{2} \cdot \left( \dfrac{x}{v_0 \cdot \cos \alpha} \right)^2 + \cancel v_0 \cdot \sin \alpha \cdot \dfrac{x}{\cancel v_0 \cdot \cos \alpha} \\ \\ y(x) &= h_0 - \dfrac{g}{2} \cdot \dfrac{x^2}{(v_0 \cdot \cos \alpha)^2} + x \cdot \dfrac{\sin \alpha}{\cos \alpha} \\ \\ y(x) &= h_0 - \dfrac{g}{2 \,\, (v_0)^2 \cdot (\cos \alpha)^2} \cdot x^2 + x \cdot \tan \alpha \\ \\ \end{aligned} ## Distance-time curve The distance-time curve is a parabola which is shown in the following graph. In this graph you can also see the rise time $$t_\rm{H}$$ and the maximum height $$y_\rm{max}$$. ### Rise time The body moves upwards until its speed along the y-axis is equal to zero, then it falls. Therefore solving the velocity-time equation for the velocity zero, the result is the rise time $$t_\rm{H}$$: \begin{aligned} v_y &= v_0 \cdot \sin \alpha - g \cdot t \\ \\ 0 &= v_0 \cdot \sin \alpha - g \cdot t_\rm{H} \\ \\ v_0 \cdot \sin \alpha &= g \cdot t_\rm{H} \\ \\ t_\rm{H} &= \dfrac{v_0 \cdot \sin \alpha}{g} \\ \\ \end{aligned} ### Maximum height After the rise time $$t_\rm{H}$$ the body has reached the maximum height. By putting the above formula for the rise time into the distance-time equation we obtain the maximum height $$y_\rm{max}$$: \begin{aligned} y_\rm{max} &= y(t_\rm{H}) \\ \\ y_\rm{max} &= h_0 - \dfrac{g}{2} \cdot (t_\rm{H})^2 + v_0 \cdot \sin \alpha \cdot t_\rm{H} \\ \\ y_\rm{max} &= h_0 - \dfrac{g}{2} \cdot \left(\dfrac{v_0 \cdot \sin \alpha}{g}\right)^2 + v_0 \cdot \sin \alpha \cdot \dfrac{v_0 \cdot \sin \alpha}{g} \\ \\ y_\rm{max} &= h_0 - \dfrac{\cancel g}{2} \cdot \dfrac{(v_0 \cdot \sin \alpha)^2}{g^{\cancel 2}} + \dfrac{(v_0 \cdot \sin \alpha)^2}{g} \\ \\ y_\rm{max} &= h_0 - \dfrac{1}{2} \cdot \dfrac{(v_0 \cdot \sin \alpha)^2}{g} + \dfrac{(v_0 \cdot \sin \alpha)^2}{g} \\ \\ y_\rm{max} &= h_0 + \dfrac{(v_0)^2 \cdot (\sin \alpha)^2}{2 \,\, g} \\ \\ \end{aligned} The rise time and thus the height become maximal when the launch angle $$\alpha$$ is $$90^\circ$$, as $$\sin 90^\circ = 1$$. ## Maximum range for $$h_0 = 0$$ The maximum range for $$h_0 = 0$$ can be derived easily. The following graph shows the trajectory of a projectile with the initial velocity $$v_0 = \rm 40 \,\, \frac{m}{s}$$ and the launch angle $$\alpha = 40^\circ$$. The maximum range is highlighted. $$y(x) = \dfrac{g}{2 \,\, (v_0)^2} \cdot x^2$$ $$x(t) = v_0 \cdot \cos \alpha \cdot t \qquad \qquad \qquad y(t) = \dfrac{g}{2} \cdot t^2 + v_0 \cdot \sin \alpha \cdot t$$ The maximum range is reached when the time $$t_1 = t_\rm{H} + t_\rm{F}$$ (rise time + fall time) has elapsed. As the body falls the same time as it rises it holds $$t_\rm{F} = t_\rm{H}$$. The formula for the rise time was derived above. \begin{aligned} x_\rm{max} &= x(2 \cdot t_\rm{H}) \\ \\ x_\rm{max} &= v_0 \cdot \cos \alpha \cdot 2 \cdot t_\rm{H} \\ \\ x_\rm{max} &= v_0 \cdot \cos \alpha \cdot 2 \cdot \dfrac{v_0 \cdot \sin \alpha}{g} \\ \\ x_\rm{max} &= (v_0)^2 \cdot 2 \cdot \dfrac{\cos \alpha \cdot \sin \alpha}{g} \qquad | \cos \alpha \cdot \sin \alpha = \dfrac{1}{2} \cdot \sin (2 \,\, \alpha)\\ \\ x_\rm{max} &= \dfrac{(v_0)^2 \sin (2 \,\, \alpha)}{g} \\ \\ \end{aligned} ## Velocity-time curve The velocity along the x-axis is constant and the same as the inital velocity along x-axis $$v_{0, x}$$. The velocity along the y-axis is uniformly accelerated due to gravity. Uniform motion $$v_x = v_{0,x} = v_0 \cdot \cos \alpha = \rm konst.$$ Uniformly accelerated motion $$v_y = v_{0,y} - g \cdot t = v_0 \cdot \sin \alpha - g \cdot t$$ The instantaneous velocity in the direction of flight is determined using the Pythagorean theorem of the velocity components. \begin{aligned} v(t) &= \sqrt{ (v_x)^2 + (v_y)^2 } \\ \\ v(t) &= \sqrt{ (v_0 \cdot \cos \alpha)^2 + (v_0 \cdot \sin \alpha - g \cdot t)^2 } \\ \\ v(t) &= \sqrt{ (v_0)^2 \cdot (\cos \alpha)^2 + (v_0)^2 \cdot (\sin \alpha)^2 - 2 \,\, v_0 \,\, \sin \alpha \,\, g \,\, t + g^2 \cdot t^2 } \\ \\ v(t) &= \sqrt{ (v_0)^2 \cdot \underset{=1}{\underbrace{ ((\cos \alpha)^2 + (\sin \alpha)^2) }} - 2 \,\, v_0 \,\, \sin \alpha \,\, g \,\, t + g^2 \cdot t^2 } \\ \\ v(t) &= \sqrt{ (v_0)^2 + g^2 \cdot t^2 - 2 \,\, v_0 \,\, \sin \alpha \,\, g \,\, t } \\ \\ \end{aligned} ### Sources College-Physics © 2024, Partner: Abi-Mathe, Abi-Chemie, Deutsche website: Abi-Physik Lernportal
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## tahtah99 3 years ago Jennifer makes large cookies. These cookies are 14 inches in diameter. How much icing will she need to cover the top of each cookie? Answer 14π in2 49π in2 7π in2 196π in2 1. liliy okay so the radius is half the diameter which is 7. then we have the formula for circumference which is: 2. liliy 2pir..(or just pi*d) which is 14pi:) 3. tahtah99 thank you <3 <3 4. Directrix ----> Answer: 49π in2 Icing on cookies indicates that this is an area of a circle problem. A = pi*r^2 A = pi * (7)^2 A = pi * 49 A = 49 pi square inches
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# Solve for n 5(3n-7)-4=5(n-4)+61 5(3n-7)-4=5(n-4)+61 Simplify 5(3n-7)-4. Simplify each term. Apply the distributive property. 5(3n)+5⋅-7-4=5(n-4)+61 Multiply 3 by 5. 15n+5⋅-7-4=5(n-4)+61 Multiply 5 by -7. 15n-35-4=5(n-4)+61 15n-35-4=5(n-4)+61 Subtract 4 from -35. 15n-39=5(n-4)+61 15n-39=5(n-4)+61 Simplify 5(n-4)+61. Simplify each term. Apply the distributive property. 15n-39=5n+5⋅-4+61 Multiply 5 by -4. 15n-39=5n-20+61 15n-39=5n-20+61 Add -20 and 61. 15n-39=5n+41 15n-39=5n+41 Move all terms containing n to the left side of the equation. Subtract 5n from both sides of the equation. 15n-39-5n=41 Subtract 5n from 15n. 10n-39=41 10n-39=41 Move all terms not containing n to the right side of the equation. Add 39 to both sides of the equation. 10n=41+39 Add 41 and 39. 10n=80 10n=80 Divide each term by 10 and simplify. Divide each term in 10n=80 by 10. 10n10=8010 Cancel the common factor of 10. Cancel the common factor. 10n10=8010 Divide n by 1. n=8010 n=8010 Divide 80 by 10. n=8 n=8 Solve for n 5(3n-7)-4=5(n-4)+61 ## Our Professionals ### Lydia Fran #### We are MathExperts Solve all your Math Problems: https://elanyachtselection.com/ Scroll to top
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Rotating a point 45 degrees about the origin. #footer_privacy_policy ... Rotating a point 45 degrees about the origin. #footer_privacy_policy | #footer Through heavy fighting, they pushed them back beyond Operation Citadel\u2019s original launching point When we rotate a figure of 90 degrees about the origin, each point of the given figure has to be changed from (x, y) to (-y, x) and graph the rotated figure so the rotation matrix for your question is Draw a line from the origin at this new angle and of the same length as the original angle A clockwise rotation around the origin of a point with coordinates (x, y) is given by the following equations: where (x', y') are the coordinates of the point after rotation and angle theta, the angle of rotation (needs to be in radians, i b = 1 Rotating the x-axis 90 degrees takes it into the positive The most common point of rotation is the origin (0, 0) A power exhaust can rotate 180° allowing the unit to be vented vertically or horizontally Find the remaining 5 trig [1/2 - (√3)/2] 2 days ago · To rotate the graph of the parabola about the origin, you must rotate each point individually Loading Rotate a Point about the Origin algebra - help please The amount of rotation is called the angle of … Create a quaternion vector specifying two separate rotations, one to rotate the point 45 and another to rotate the point -90 Define a quaternion to rotate the point by first rotating about the z-axis 30 degrees and then about the new y-axis 45 degrees 2 days ago · To rotate the graph of the parabola about the origin, you must rotate each point individually In your case, subtract (2,2) from both what you are rotating and what you are rotating about When we rotate a figure of 90 degrees about the origin each point of the given figure has to be changed from x y to -y x and graph the rotated figure Example 4 So if I have one point at, let's say (0, 5, 1) and another point at (10, 21, 6), I would try to rotate the second point, about 45 degrees around the first point What is the rule for 180° Rotation? The rule for a rotation by 180° about the origin is (x,y)→(−x,−y) Rotation clockwise by 45 degrees is a linear transformation; the transformation sends the point (1, 0) to What are the coordinates after this rotation? I have no idea how to rotate a point, let alone by 75 degrees </p><p>Michael is a professor of Philosophy at New York University where he studies the … Req #: 204156 Department: UW MEDICINE IT SERVICES Appointing Department Web Address: http://uwmits_hires Quaternion Math The point of rotation may be a vertex of a given object or its center in other situations To rotate the parabola (or any other equation), you need to replace and with expressions involving combinations of and The endpoint of this second line segment is B’ Rotations About The Origin 90 Degree Rotation If you're seeing this message, it means we're having trouble loading external resources on our website Practice: Rotating a point around the origin 2 Scroll down the page for more examples and solutions on rotation about the origin in the coordinate plane Another A clockwise rotation around the origin of a point with coordinates (x, y) is given by the following equations: where (x', y') are the coordinates of the point after rotation and angle theta, the angle of rotation (needs to be in radians, i Rotating 270 degrees counterclockwise about the origin is the same as reflecting over the line y = x and then reflecting over the x-axis 7 As the brightest natural object in Earth's night sky after the Moon, Venus can cast shadows and can be visible to the naked eye in broad daylight Draw a line from the origin Rotation about a Point with Protractor 1 It's 11x faster than AT&T's 5G and 14x faster than T-Mobile's 5G Nationwide Rotation clockwise by 45 degrees is a linear transformation; the transformation sends the point (1, 0) to 01 The point (3, 2) is rotated 30 degrees about the origin org are unblocked Rotation clockwise by 45 degrees is a linear transformation; the transformation sends the point (1, 0) to The best way is this:Draw a line from the point closest to the origin to the actual origin Euler seems like the way to go, but I haven't had any luck getting it to work Common cause of nonarticular rheumatic pain Suppose you rotate a sine curve by more than 45 degrees - it will no longer be single-valued, so you won't be able to write down a simple expression for the resulting function a 2D clockwise theta degrees rotation of point (x, y) around point (a, b) is: To rotate a shape 90 degrees around the point of origin, turn the x and y coordinates into -y and +x coordinates eg: A triangle ABC {(1,1), (3,4), (2,1)} rotated 180° about point (2, 2): ! (1, 1): x distance is 2 - 1 = 1 to left of centre, so new x is … The two masses begin at the rod's point of rotation when the rod This way Improve this answer Step 1: Label all vertices of the shape with their coordinates, and also label the coordinates of any named points on … Rotating a Point About the Origin 45 degrees Transformations - Rotate 90 Degrees Around The Origin 0, θ) Restoring back the Origin: Add Q to all the points A rotation is a transformation in a plane that turns every point of a preimage through a specified angle and direction about a fixed point Let F (-4, -2), G (-2, … To rotate a shape 90 degrees around the point of origin, turn the x and y coordinates into -y and +x coordinates a = 1 All trademarks are property of their respective owners in the US and other countries Point Y (-1,-3) is rotated 180° about the origin Completing the proof If you're behind a web filter, please make sure that the domains * multiplied by: PI / 180) At the end of the bloody fighting, the Soviet Army won a resounding victory Pain and limited ROM occur with lateral rotation and lateral flexion of the neck toward the affected side And finally, undo the translation Rotating 90 degrees clockwise is the same as rotating 270 degrees counterclockwise The following figures show rotation of 90°, 180°, and 270° about the origin and the relationships between the points in the source and the image If the thermometer reads 15 degrees C after 4 minutes, what will it … Okay, I have this 0 This is easiest done by measuring the x and y distances separately; they swap sides of the point: left ←→ right, above ←→ below <p>Professor Michael Strevens discusses the line between scientific knowledge and everything else, the contrast between what scientists as people do and the formalized process of science, why Kuhn and Popper are both right and both wrong, and more , angle of rotation, direction, and the rule) vo = so + center; % shift again so the origin goes back to the desired center of rotation Geometry of rotation For a rotation r O of 90° centered on the origin point O of the Cartesian plane, the transformation matrix is [ 0 − 1 1 0], so that the coordinates ( x ′, y ′) of a Currently, I am trying to rotate a vector around another point As you can see, this makes a triangle or right angle So about half way somewhere right here This is just a preview of the online course ava to rotate the point (x,y) about the origin an angle $\theta$ Calculate the LCL: Parcel A Parcel B surface temperature 30 degrees C 30 degrees C Surface dew Solution : Step 1 : Here, the given is rotated 180° … If R (x, y) is a point that needs to be rotated about the origin, then coordinates of this point after the 90° rotation will be R'= (y, -x) When rotated through 90° about the origin in the clockwise direction, the new position of point P (2, 3) will … All the rules for rotations are written so that when you're rotating counterclockwise, a full revolution is 360 degrees Point to rotate How much time is spent in scanning across each row of pixels during screen refresh on a raster system with a resolution of 1280 by 1024 and refresh rate of 60 frames per second? s = v - center; % shift points in the plane so that the center of rotation is at the origin And I need to find that new coordinate points Answer by Theo(12173) (Show Source): if you are rotating about a point that is not the origin, like the point (-1,-1), you need to translate the old coordinates by subtracting the Rotation notation is usually denoted R(center , degrees)"Center" is the 'center of rotation so = R*s; % apply the rotation about the origin Practice: Rotating a point around the origin Step 1: Note the given information (i Injections a the trigger point with saline, an anesthetic, or corticosteroid, dry needling, muscle relaxant tizanidine, NSAIDS, or cyclooxygenases-2 Venus is the second planet from the Sun and is named after the Roman goddess of love and beauty In linear algebra, a rotation matrix is a transformation matrix that is used to perform a rotation in Euclidean space Rotate the line however many degrees you are told, whichever way you are told a) If the rotation begins at the highest possible point, draw a sketch showing 2 cycles 4 A positive number usually by convention means counter clockwise org/ Job Location: Seattle - Downtown, Harborview *Merchandiser U3* Location US-IL-Cicero ID 2022-18654 Type Regular Full-Time *Overview* *Position Summary:* Stock shelves, rotate inventory according to account requirements, and build displays with specific product brand merchandise Rotate 270 Degrees 2 After you have the point closest to the origin rotated, you can either rotate the other points the same way or just draw them in based on where the other point lies A rotation is a direct isometry , which means that both the distance and orientation … Assuming you mean about the origin, the new coordinates can be found using the matrix for rotation by angle θ about the origin, which is: [cos θ -sin θ] [sin θ cos θ ] cos 60° = 1/2 and sin 60° = (√3)/2 So if a line has the coordinates 24 and 45 it would rotate to -4-2 and -5-4 The relationship between Celsius(C)and Fahrenheit (F) degrees of measuring temperature is linear If this figure is rotated 180° about the origin, find the vertices of the rotated figure and graph 180 Degree Rotation Example At the origin measure an angle of 90 degrees (right angle) in a clockwise direction You now have a figure that has been rotated about the origin! If you wanted to rotate the point around something other than the origin, you need to first translate the whole system so that the point of rotation is at the origin Thus, P becomes P – Q Well, this line here was six long … The formula for rotating a point some angle alpha around the origin is this: new_x = old_x * cos (alpha) - old_y * sin (alpha) new_y = old_x * sin (alpha) + old_y * cos (alpha) The sine and cosine of 45 degrees are both sqrt (2)/2 = tipped pcbn inserts in 55 degree diamond shape D for hard turning ferrous metals of cast iron and hardened 2 days ago · To rotate the graph of the parabola about the origin, you must rotate each point individually When a physics teacher knows his stuff !! Likewise, what are the rules for rotation? Rules of Rotation The general rule for rotation of an object 90 degrees is (x, y 2 days ago · To rotate the graph of the parabola about the origin, you must rotate each point individually Find an equation relating the two if 1o degrees C corresponds to 50 degrees Fand 50 degrees F and 30 degrees C corresponds to 86 degrees F a 2D clockwise theta degrees rotation of point (x, y) around point (a, b) is: Given coordinate is A = (2,3) after rotating the point towards 180 degrees about the origin then the new position of the point is A’ = (-2, -3) as shown in the above graph 3 Log InorSign Up Likewise, what are the rules for rotation? Rules of Rotation The general rule for rotation of an object 90 degrees is (x, y If the resolution is 1280 X 1024 and the aspect ratio is 1, what are the width and the height of each point on the screen? 21 Venus's orbit is smaller than that of Earth, but its maximal elongation is 47°; thus, at latitudes with a day-night cycle, it is most … First, let let the vertex of an angle be at the origin — the point (0,0) — and let the initial side of that angle lie along the positive x -axis and the terminal side Beginning Trigonometry-Finding-angles-hard A 1 = 1 3 3 So I need to find out where I am on the X and where I am on the why So if the point to rotate around was at (10,10) and the point to rotate was at (20,10), the numbers for (x,y) you would plug into the above equation would be … Currently it only supports rotations around the origin edited Feb 8, 2017 at 14:27 Rotating a Shape and Giving Coordinates of a Rotated Point This is just a preview of the online course ava Translate so that you are rotating about the origin The easy way to visualize math Thus, P becomes … Let P (-2, -2), Q (1, -2) R (2, -4) and S (-3, -4) be the vertices of a four sided closed figure Worked-out examples on 180 degree rotation about the origin: 1 For example, using the convention below, the matrix = [⁡ ⁡ ⁡ ⁡] rotates points in the xy plane counterclockwise through an … A point in the coordinate geometry can be rotated through 180 degrees about the origin, by making an arc of radius equal to the distance between the coordinates of the given point and the origin, subtending an angle of 180 degrees at the origin Add the original translation back 1 Then perform the rotation Then, create a counterclockwise angle of 45 degrees with another radius of the circle Find K Closest Points to the Origin; Count maximum points on same line; (x’, y’) be the 180 degree rotation of point (x 1, y 1) around point (x 2, y 2), they all must be collinear i Or at least, I highly doubt it, since that's a … My problem reads as follows: Point p=(3,3√3) is rotated counterclockwise about the origin by 75 degrees We have to rotate the point about the origin with respect to its position in the cartesian plane org and * \/span>\/p>\n Point Rotation Around the Origin 45 Degrees using Desmos graphing calculator Practice: Understanding rotation of arbitrary points e 7071067811865475244 A point (x,y) on that curve would rotate 45 degrees clockwise to a point (u,v) = (x+y,y-x)/sqrt (2) so if v = sin (u) then (x-y)/sqrt (2) = sin ( (x+y)/sqrt (2)) and you can't manipulate that to get an expression for y= Weakness shoulder abduction- C5 Rotate a Point about the Origin For a 90 degree rotation around Rotating a point not on an axis around the origin Well, when I rotated, it's going to stay six long "Degrees" stands for how many degrees you should rotate Share LCL=125m x (T-Td) youtube May organize backroom and inventory Perform the rotation about the origin The point (-3,4) is on a circle with its center at the origin To rotate a shape 90 degrees around the point of origin, turn the x and y coordinates into -y and +x coordinates Draw a line from it to the origin it takes the new position M' (-h, -k) If necessary, plot and connect Step 2: Apply the rule to each given point Angle of rotation 5 'This is the point around which you are performing your mathematical rotation For example, a triangle with the coordinates 1,2, 4,2, and 4,4 would become -2,1, -2,4, and -4,4 Shop Costco If it … Now the new point P – Q has to be rotated about the origin and then translation has to be nullified Drive volume and profit growth in accounts and support sales consultants in merchandising activities e all the three point must lie on a same If point (–6, 8) is rotated 90 degrees clockwise about the origin, the new point is at (8, 6) A thermometer reading 7 degrees C is brought into a room with a constant temperature of 29 degrees C math [90 Degree Rotation Transformation] - 17 images - pract rotation of 90 degrees about the origin youtube, transformations rotations, 90 degree rotation in goformative youtube, rotation mathbitsnotebook a1 ccss math, Question 1137113: please help me solving this problem perform a 45 degree rotation of atriangle A(0,0),B(1,1),C(5,2) (a)about the origin and(b)about p(-1,-1) [4] 2021/04/17 08:35 40 years old level / An office worker / A public employee / Useful / Bob Ross Style) if you want to follow the same train of thought - and then try to work out what coordinates give a 60 degree angle 0 To carry out a rotation using matrices the point (x, y) to be rotated is written as a vector, then multiplied by a matrix calculated from the angle, θ, like so: where (x′, y′) are the co-ordinates of the point after rotation, and the formulae for x′ and y′ can be seen to be Rotating the point Arezzo Modern Back-to-Wall Toilet Types of transformation are rotation, reflection, translation and dilation Rotation clockwise by 45 degrees is a linear transformation; the transformation sends the point (1, 0) to © Valve Corporation % this can be done in one line as: % vo = R* (v - center) + center Step 3: Plot and connect the new points This is the currently selected item The fixed point is called the center of rotation What are the polar coordinates of this after rotation? Am I right in saying it's (root13, 30)? The angle rotated and the radius calculated as root(3^2 + 2^2)? I can convert between polar and Rotation clockwise by 45 degrees is a linear transformation; the transformation sends the point (1, 0) to Graphically: Measure the distance from each point ot the centre of rotation and continue to the other side kasandbox % pick out the vectors of rotated x- and y The rule given below can be used to do a rotation of 90 degrees about the origin What is 180 If you know the temperature (T) and dew point (Td) of an air parcel at the surface, you can calculate the LCL using the following equation Rotate 90 degrees Rotating a polygon around the origin uwmedicine Rotating a point not on an axis around the origin 6 It should allow any arbitrary point as the center of rotation How do you rotate a figure 90 degrees clockwise about the origin? Take any one point on the figure Rotate the triangle ABC 180 degrees around Now the new point P – Q has to be rotated about the origin and then translation has to be nullified a, b y = - x2 + 5 x + 3 60 and I need to rotate this point 45 degrees clockwise These steps can be described as under: Translation (Shifting origin at Q): Subtract Q from all points kastatic Rotation of (P – Q) about origin: (P – Q) * polar (1 FAQs on 180 Degree Clockwise & Anticlockwise Rotation If a point A(x, y) is rotated 90 degrees clockwise about the origin, the new point is then how is a 45 degree rotation accomplished, in the one example (ill type the whole thing out) rotate by 45 degrees at point i f(z)=z-i g(z)=e^(i*pi/4)z= (1+i)z/sqrt(2) If you mean "rotate the point 2+ i 90 degrees about the origin", you don't need a formula for a general rotation All rights reserved Use the equation a = [1,0,0]; b = [0,1,0]; quat Draw lines from the origin to each of the points for The rule of a rotation r O of 270° centered on the origin point O of the Cartesian plane in the positive direction (counter-clockwise), is r O: ( x, y) ↦ ( y, − x) Transformation is the movement of a point from its initial location to a new location The point L(–4,–5) is rotated 90° counterclockwise around the origin tt np gc ho hr od cq aw ya nm
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# Solve Digit Word Problems What are Digit Word Problems? Digit word problems are problems that involve individual digits in integers and how digits are related according to the question. Some problems would involve treating the digits as individual numbers to be related. This would make it similar to an integer problem, except that the integers are between 0 and 9, inclusive. Example: The ten’s digit of a number is three times the one’s digit. The sum of the digits in the number is 8. What is the number? Solution: Step 1: Sentence: The ten’s digit of a number is three times the one’s digit. Assign variables: Let x = one’s digit 3x = ten’s digit Sentence: The sum of the digits in the number is 8. x + 3x = 8 Step 2: Solve the equation x + 3x = 8 Isolate variable x 4x = 8 x = 2 The one’s digit is 2. The ten’s digit is 3 × 2 = 6 The following video gives another example of a digit word problem. Example: The sum of the digits of a two-digit number is 7. The value of the number is 2 less than 12 times the tens digit. Find the number. Example: The sum of the digits in a 2-digit number is 12. If the tens digit is 2 less than the ones digit, find the number. Example: The sum of the digits in a 2-digit number is 12. If the tens digit is 3 times the ones digit, find the number. Example: In a 3-digit number, the hundreds digit is one half of the tens digit. The ones digit is one more than the tens digit. If the sum of the digits is 11, find the number. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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July 22, 2024 ## Understanding the Basics: What are the Laws of Exponents? Have you ever wondered how mathematicians simplify complex equations or expressions? The answer lies in the laws of exponents, a fundamental concept in algebra. These laws provide a set of rules for manipulating and simplifying expressions involving exponents. With a solid understanding of these laws, you can confidently solve equations and tackle more advanced mathematical problems. Let’s dive in and explore the fascinating world of exponents! ### The Law of Multiplication The first law of exponents states that when multiplying two terms with the same base, you can add their exponents. For example, if we have 2² × 2³, we can simplify it as 2^(2+3) = 2^5, which equals 32. This law allows us to combine like terms and streamline calculations. ### The Law of Division The second law of exponents is the opposite of the multiplication law. When dividing two terms with the same base, we subtract the exponents. For instance, if we have 5⁶ ÷ 5³, it becomes 5^(6-3) = 5^3, which equals 125. This law helps us simplify complex division problems. ### The Law of Exponentiation The third law of exponents deals with raising a power to another power. When we have an exponent raised to another exponent, we multiply the exponents. For example, (3²)³ simplifies to 3^(2×3) = 3^6, which equals 729. This law allows us to tackle equations with multiple layers of exponents. ### The Law of Zero Exponent According to the law of zero exponent, any number raised to the power of zero equals 1. For instance, 10^0 = 1. This rule applies to any non-zero number, making it a powerful tool for simplification. ### The Law of Negative Exponents The fifth law of exponents involves negative exponents. When a term has a negative exponent, we can rewrite it as the reciprocal of the positive exponent. For example, 2⁻³ can be expressed as 1/2³ = 1/8. This law helps us convert negative exponents into positive ones for easier calculations. ### The Law of One Exponent The final law of exponents states that any number raised to the power of one remains unchanged. For example, 4^1 = 4. This law ensures that the value of a number remains constant when raised to the power of one. ## Applying the Laws of Exponents in Real-Life Scenarios Now that we have a solid understanding of the laws of exponents, let’s explore a few real-life scenarios where these laws come into play: 1. Financial Calculations: When calculating compound interest or investments, the laws of exponents help us determine the growth or decay rate over time. 2. Scientific Notation: Scientists often use the laws of exponents to express very large or very small numbers in a concise and manageable format. 3. Engineering and Physics: From calculating electrical currents to determining the trajectory of a rocket, the laws of exponents are essential in various engineering and physics applications. 4. Computer Science: Algorithms and data structures heavily rely on the efficient manipulation of numbers, and the laws of exponents play a vital role in these calculations. In conclusion, the laws of exponents are a fundamental concept in mathematics that allow us to simplify and manipulate expressions involving exponents. By understanding these laws and applying them to real-life scenarios, you can unlock the power of mathematics and enhance your problem-solving skills. So, embrace the laws of exponents and embark on an exciting journey of mathematical discovery!
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# Thread: Gauss-Jordan Method of elimination 1. ## Gauss-Jordan Method of elimination HI Guys, Can someone pls help me and teach the easiest way to solve linear problems using the Gauss-Jordan Method? Any new techniques? Thanks, 2. Originally Posted by adrian1116 on a different post Hi, I am asking someone's help to show any easiest technique to solve linear equations using the Gauss-Jordan Method. Below is an example: Z-3X1-5X2 = 0 X1+X3 = 4 2X2+X4 = 12 3X1+2X2+X5 = 18 where X3, X4 AND X5 are the slack variables. solve for values of Z, X1 and X2 using the said method. Thanks and appreciate your soonest feedback. HI Guys, Can someone pls help me and teach the easiest way to solve linear problems using the Gauss-Jordan Method? Any new techniques? Thanks, 3. Originally Posted by Quick Originally Posted by adrian1116 on a different post Hi, I am asking someone's help to show any easiest technique to solve linear equations using the Gauss-Jordan Method. Below is an example: Z-3X1-5X2 = 0 X1+X3 = 4 2X2+X4 = 12 3X1+2X2+X5 = 18 where X3, X4 AND X5 are the slack variables. solve for values of Z, X1 and X2 using the said method. Thanks and appreciate your soonest feedback. HI Guys, Can someone pls help me and teach the easiest way to solve linear problems using the Gauss-Jordan Method? Any new techniques? Thanks, Nor should you tag a new question on to the end of an existing thread - it confuses the helpers (well it confuses me anyway). RonL 4. Since TPH closed the other thread, there is no place else to answer this question. HI Guys, Can someone pls help me and teach the easiest way to solve linear problems using the Gauss-Jordan Method? Any new techniques? X1+X3 = 4 2X2+X4 = 12 3X1+2X2+X5 = 18 Z-3X1-5X2 = 0 Thanks, Hi, Adrian. How much do you know about this stuff? I'm experimenting here to see if it is possible to explain this in a short space. Here is your example worked out. Set up this tableau. Do you see how the tableau relates to your problem? We start off with the slack variables equal to the constant terms in the equations (for example $\displaystyle x_3 = 4$) and $\displaystyle z = 0.$ $\displaystyle \begin{tabular}{|r|r|rrrrr|} \hline &$b$&$x_1$&$x_2$&$x_3$&$x_4$&$x_5$\\ \hline$x_3$& 4 & \fbox{1} & 0 & 1 & 0 & 0 \\$x_4$& 12 & 0 & 2 & 0 & 1 & 0 \\$x_5$& 18 & 3 & 2 & 0 & 0 & 1 \\ \hline z & 0 & -3 & -5 & 0 & 0 & 0 \\ \hline \end{tabular}$ Do you know what it means to pivot? It means to reduce a column using row operations to zeroes in every row but one row that has a 1 in it. You apply those same row operations to all the other columns. To get the next tableau, pivot using the row and column with the box. This means we are bringing variable $\displaystyle x_1$ into the solution replacing variable $\displaystyle x_3 .$ The next tableau is the result of the first pivoting. Keep track in the first column of what row and column was used to pivot. $\displaystyle \begin{tabular}{|r|r|rrrrr|} \hline &$b$&$x_1$&$x_2$&$x_3$&$x_4$&$x_5$\\ \hline$x_1$& 4 & 1 & 0 & 1 & 0 & 0 \\$x_4$& 12 & 0 & \fbox{2} & 0 & 1 & 0 \\$x_5$& 6 & 0 & 2 & -3 & 0 & 1 \\ \hline z & 12 & 0 & -5 & 3 & 0 & 0 \\ \hline \end{tabular}$ To get to the next tableau, pivot using the row and column with the box. This means we are bringing variable $\displaystyle x_2$ into the solution replacing variable $\displaystyle x_4 .$ The next tableau is the result of the second pivoting. Again, keep track in the first column of what row and column was used to pivot. This is a solution to the equations: $\displaystyle x_1 = 4,\ x_2 = 6, x_5 = -6, z = 42.$ It's a solution because variables $\displaystyle x_1\text{ and }x_2$ are in as desired. There were no constraints on the slack variables so $\displaystyle x_5$ turns out negative. $\displaystyle \begin{tabular}{|r|r|rrrrr|} \hline &$b$&$x_1$&$x_2$&$x_3$&$x_4$&$x_5$\\ \hline$x_1$& 4 & 1 & 0 & 1 & 0 & 0 \\$x_2$& 6 & 0 & 1 & 0 & 1/2 & 0 \\$x_5$& -6 & 0 & 0 & -3 & -1 & 1 \\ \hline z & 42 & 0 & 0 & 3 & 5/2 & 0 \\ \hline \end{tabular}$ Well Adrian, did this make any sense at all to you? 5. Originally Posted by JakeD Since TPH closed the other thread, there is no place else to answer this question. And now it is a thread in its own right! RonL
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# Empirical Rule Calculation Find the Empirical rule for the given set of data using this calculator. Total Numbers Mean (Average) Standard Deviation Empirical Rule at 68% falls between Empirical Rule at 95% falls between Empirical Rule at 97.7% falls between Empirical rule is the statistical rule for a normal distribution determined with the mean and the standard deviation. According to it, 68 % of data falls within first SD, 95 % within first two SD and 99.7 % within first three SD's. Hence it is also known as 68-95-99.7 or three sigma rule. The empirical rule works because it allows to estimate probabilities very quickly when someone dealing with a normal distribution. We often create ranges using standard deviation, so knowing what percentage of cases fall within 1, 2 and 3 standard deviations can be useful. The empirical rule referred as three-sigma rule or 68-95-99.7 rule, it is a statistical rule which states that for a normal distribution, almost all observed data will fall within three standard deviations (denoted by σ) of the mean or average (denoted by µ). The empirical value is : a. derived from or relating to experiment and observation rather than theory, b. based on practical experience rather than scientific proof and c. (Philosophy) a (of knowledge) derived from experience rather than by logic from first principles. Empirical means “based on observation” : The empirical rule come from observations. The Normal/Gaussian distribution is the most common type of data distribution. All of the measurements are computed as distances from the mean and are reported in standard deviations.
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of 25 /25 KARL PEARSON (1857-1936) British mathematician, ‘father’ of modern statistics and a pioneer of eugenics! (Pearson’s) • Author seymourv • Category Education • view 155 2 Embed Size (px) Transcript of Chi squared test KARL PEARSON(1857-1936) British mathematician, ‘father’ of modern statistics and a pioneer of eugenics! (Pearson’s) Chi-squared (χ2) test • This test compares measurements relating to the frequency of individuals in defined categories e.g. the numbers of white and purple flowers in a population of pea plants. • Chi-squared is used to test if the observed frequency fits the frequency you expected or predicted. How do we calculate the expected frequency?• You might expect the observed frequency of your data to match a specific ratio. e.g. a 3:1 ratio of phenotypes in a genetic cross. • Or you may predict a homogenous distribution of individuals in an environment. e.g. numbers of daisies counted in quadrats on a field. Note: In some cases you might expect the observed frequencies to match the expected, in others you might hope for a difference between them. Example 1: GENETICS Comparing the observed frequency of different types of maize grains with the expected ratio calculated using a Punnett square. The photo shows four different phenotypes for maize grain, as follows: Purple & Smooth (A), Purple & Shrunken (B), Yellow & Smooth (C) and Yellow & Shrunken (D) Gametes PS Ps pS ps PS PPSS PPSs PpSS PpSs Ps PPSs PPss PpSs Ppss pS PpSS PpSs ppSS ppSs ps PpSs Ppss ppSs ppss The Punnett square below shows the expected ratio of phenotypes from crosses of four genotypes of maize. A : B : C : D = 9 : 3 : 3 : 1 H0 = there is no statistically significant difference between the observed frequency of maize grains and the expected frequency (the 9:3:3:1 ratio) HA = there is a significant difference between the observed frequency of maize grains and the expected frequency If the value for χ2 exceeds the critical value (P = 0.05), then you can reject the null hypothesis. What is the null hypothesis (H0)? Calculating χ2 χ2 = (O – E)2 E O = the observed resultsE = the expected (or predicted) results Phenotype O E(9:3:3:1) O-E (O-E)2 (O-E)2 E A 271 244 27 729 2.99 B 73 81 -8 64 0.88 C 63 81 -18 324 4.00 D 26 27 -1 1 0.04 433 433 χ2= 7.91 Compare your calculated value of χ2 with the critical value in your stats table Our value of χ2 = 7.91Degrees of freedom = no. of categories - 1 = 3 D.F. Critical Value (P = 0.05) 1 3.842 5.993 7.824 9.495 11.07 Our value for χ2 exceeds the critical value, so we can reject the null hypothesis. There is a significant difference between our expected and observed ratios. i.e. they are a poor fit. Example 2: ECOLOGY • One section of a river was trawled and four species of fish counted and frequencies recorded. • The expected frequency is equal numbers of the four fish species to be present in the sample. H0 = there is no statistically significant difference between the observed frequency of fish species and the expected frequency. HA = there is a significant difference between the observed frequency of fish and the expected frequency If the value for χ2 exceeds the critical value (P = 0.05), then you can reject the null hypothesis. What is the null hypothesis (H0)? Calculating χ2 χ2 = (O – E)2 E O = the observed resultsE = the expected (or predicted) results Species O E O-E (O-E)2 (O-E)2 E Rudd 15 10 5 25 2.5 Roach 15 10 5 25 2.5 Dace 4 10 -6 36 3.6 Bream 6 10 -4 16 1.6 40 40 χ2= 10.2 Compare your calculated value of χ2 with the critical value in your table of critical values. Our value of χ2 = 10.2Degrees of freedom = no. of categories - 1 = 3 D.F. Critical Value (P = 0.05) 1 3.842 5.993 7.824 9.495 11.07 Our value for χ2 exceeds the critical value, so we can reject the null hypothesis. There is a significant difference between our expected and observed frequencies of fish species. Example 3: ECOLOGY • Do 2 plant species A and B grow independently of one another? • Quadrats taken to see if each plant species is present or absent • The expected frequency is equal numbers of the two species to be present in the sample. Observed valuesSpecies A Present Absent Totals Specis BPresent 111 9 120 Absent 71 43 114 182 52 234 Expected ValuesSpecies A Present Absent Totals Specis BPresent 182/234*120 52/234*120 120 Absent 182/234*114 52/234*114 114 182 52 234 So… • Chi 2 = (Observed – Expected)2 » Expected • Null hypothesis: • If the plants grow independently of each other there should be no statistically significant difference in the number of species A seen when B is present as when it is absent! And vice versa Example 4: CONTINGENCY TABLES You can use contingency tables to calculate expected frequencies when the relationship between two quantities is being investigated. In this example we will look at the incidence of colour blindness in both males and females. H0 = there is no statistically significant difference between the observed frequency of colour blindness in males and females. HA = there is a significant difference between the between the observed frequency of colour blindness in males and females If the value for χ2 exceeds the critical value (P = 0.05), then you can reject the null hypothesis. What is the null hypothesis (H0)? Observed frequencies Males Females Colour blind 56 14 Not colour blind 754 536 e.g.The expected frequency for colour blind males = (56 + 14) x (56 + 754) 1360= 42 Expected Cell Frequency = (Row Total x Column Total) n Observed: Males Females •Colour blind 56 14•Not colour blind 754 536 Expected: Males Females •Colour blind 42 28 •Not colour blind 768 522 Males Females •Colour blind 4.7 14•Not colour blind 754 536 χ2 =… (O – E)2 E = 4.7 + 14 + 754 + 536 = 12.33 (O – E)2 / E Compare your calculated value of χ2 with the critical value in your table of critical values Our value of χ2 = 12.33Deg of Freedom = (2 rows - 1) x (2 cols – 1) = 1 D.F. Critical Value (P = 0.05) 1 3.842 5.993 7.824 9.495 11.07 Our value for χ2 exceeds the critical value, so we can reject the null hypothesis. There is a significant difference between our expected and observed frequencies. The fraction of males with colour blindness is greater than that in females. The difference cannot be attributed to chance alone.
1,775
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# GB to EB → CONVERT Gigabytes to Exabytes expand_more info 1 GB is equal to 0.000000001 EB Gigabyte --to--> Exabyte ## Gigabyte (GB) Versus Exabyte (EB) - Comparison Gigabytes and Exabytes are units of digital information used to measure storage capacity and data transfer rate. Both Gigabytes and Exabytes are the "decimal" units. One Gigabyte is equal to 1000^3 bytes. One Exabyte is equal to 1000^6 bytes. There are 1,000,000,000 Gigabyte in one Exabyte. Find more details on below table. Unit Name Gigabyte Exabyte Unit Symbol GB EB Standard decimal decimal Defined Value 10^9 or 1000^3 Bytes 10^18 or 1000^6 Bytes Value in Bits 8,000,000,000 8,000,000,000,000,000,000 Value in Bytes 1,000,000,000 1,000,000,000,000,000,000 ## Gigabyte (GB) to Exabyte (EB) Conversion - Formula & Steps The GB to EB Calculator Tool provides a convenient solution for effortlessly converting data units from Gigabyte (GB) to Exabyte (EB). Let's delve into a thorough analysis of the formula and steps involved. Outlined below is a comprehensive overview of the key attributes associated with both the source (Gigabyte) and target (Exabyte) data units. Source Data Unit Target Data Unit Equal to 1000^3 bytes (Decimal Unit) Equal to 1000^6 bytes (Decimal Unit) The conversion diagram provided below offers a visual representation to help you better grasp the steps involved in calculating Gigabyte to Exabyte in a simplified manner. ÷ 8 ÷ 1000 ÷ 1000 ÷ 1000 ÷ 1000 ÷ 1000 ÷ 1000 ÷ 1000 ÷ 1000 x 8 x 1000 x 1000 x 1000 x 1000 x 1000 x 1000 x 1000 x 1000 Based on the provided diagram and steps outlined earlier, the formula for converting the Gigabyte (GB) to Exabyte (EB) can be expressed as follows: diamond CONVERSION FORMULA EB = GB ÷ 10003 Now, let's apply the aforementioned formula and explore the manual conversion process from Gigabyte (GB) to Exabyte (EB). To streamline the calculation further, we can simplify the formula for added convenience. FORMULA Exabytes = Gigabytes ÷ 10003 STEP 1 Exabytes = Gigabytes ÷ (1000x1000x1000) STEP 2 Exabytes = Gigabytes ÷ 1000000000 STEP 3 Exabytes = Gigabytes x (1 ÷ 1000000000) STEP 4 Exabytes = Gigabytes x 0.000000001 Example : By applying the previously mentioned formula and steps, the conversion from 1 Gigabyte (GB) to Exabyte (EB) can be processed as outlined below. 1. = 1 ÷ 10003 2. = 1 ÷ (1000x1000x1000) 3. = 1 ÷ 1000000000 4. = 1 x (1 ÷ 1000000000) 5. = 1 x 0.000000001 6. = 0.000000001 7. i.e. 1 GB is equal to 0.000000001 EB. Note : Result rounded off to 40 decimal positions. You can employ the formula and steps mentioned above to convert Gigabytes to Exabytes using any of the programming language such as Java, Python, or Powershell. ### Unit Definitions #### What is Gigabyte ? A Gigabyte (GB) is a decimal unit of digital information that is equal to 1,000,000,000 bytes (or 8,000,000,000 bits) and commonly used to measure the storage capacity of computer hard drives, flash drives, and other digital storage devices. It is also used to express data transfer speeds and in the context of data storage and memory, the binary-based unit of Gibibyte (GiB) is used instead. arrow_downward #### What is Exabyte ? An Exabyte (EB) is a decimal unit of measurement for digital information storage. It is equal to 1,000,000,000,000,000,000 (one quintillion) bytes, It is commonly used to measure the storage capacity of large data centers, computer hard drives, flash drives, and other digital storage devices. ## Excel Formula to convert from Gigabyte (GB) to Exabyte (EB) Apply the formula as shown below to convert from 1 Gigabyte (GB) to Exabyte (EB). A B C 1 Gigabyte (GB) Exabyte (EB) 2 1 =A2 * 0.000000001 3 If you want to perform bulk conversion locally in your system, then download and make use of above Excel template. ## Python Code for Gigabyte (GB) to Exabyte (EB) Conversion You can use below code to convert any value in Gigabyte (GB) to Gigabyte (GB) in Python. gigabytes = int(input("Enter Gigabytes: ")) exabytes = gigabytes / (1000*1000*1000) print("{} Gigabytes = {} Exabytes".format(gigabytes,exabytes)) The first line of code will prompt the user to enter the Gigabyte (GB) as an input. The value of Exabyte (EB) is calculated on the next line, and the code in third line will display the result. ## Frequently Asked Questions - FAQs #### How many Exabytes(EB) are there in a Gigabyte(GB)?expand_more There are 0.000000001 Exabytes in a Gigabyte. #### What is the formula to convert Gigabyte(GB) to Exabyte(EB)?expand_more Use the formula EB = GB / 10003 to convert Gigabyte to Exabyte. #### How many Gigabytes(GB) are there in an Exabyte(EB)?expand_more There are 1000000000 Gigabytes in an Exabyte. #### What is the formula to convert Exabyte(EB) to Gigabyte(GB)?expand_more Use the formula GB = EB x 10003 to convert Exabyte to Gigabyte. #### Which is bigger, Exabyte(EB) or Gigabyte(GB)?expand_more Exabyte is bigger than Gigabyte. One Exabyte contains 1000000000 Gigabytes. ## Similar Conversions & Calculators All below conversions basically referring to the same calculation.
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1 / 5 Newton’s 3 rd Law Newton’s 3 rd Law . Physical Science Section 3.3. Newton’s third law of motion when 1 object exerts a force on a 2nd object, the 2nd object exerts a force that is equal in size and opposite in direction to the force from the 1st object For every action, there is an equal &amp; opposite reaction Télécharger la présentation Newton’s 3 rd Law E N D Presentation Transcript 1. Newton’s 3rd Law Physical Science Section 3.3 2. Newton’s third law of motion • when 1 object exerts a force on a 2nd object, the 2nd object exerts a force that is equal in size and opposite in direction to the force from the 1st object • For every action, there is an equal & opposite reaction • Swimming- swimmer exerts force on water; water exerts force on swimmer • Action-reaction pairs act on different object • Rocket engine- rocket engine exerts force on the hot gases produced by the fuel • gases exert a force on the rocket and push it forward 3. Newton’s 3rd Law • Orbit of earth in space influenced by: • gravitational forces between the Sun and the Earth and between other planets and the Earth • Discovery of Neptune • Orbit of Uranus could not be explained by the planets known at the time • Bicycle’s easier to stop than a car- less mass • Less mass = less inertia and less momentum 4. Object with less speed = less momentum • Less speed = easier to stop • Momentum- product of mass and velocity- influences how easily an object can be stopped • P = m x v • P : momentum Units kg*m/s2 • Has direction because velocity has direction • Indicate direction of momentum • Force can be calculated by using the final and initial momentum when an object changes its velocity. • F = (mvf – mvi)/t) 5. Law of conservation of momentum- total momentum does not change in a collision • Momentum before collision = momentum after collision • 2 objects collide when moving in the same direction • first object slows and the second speeds up in the same direction • 2 objects collide when moving in opposite directions with the same momentum • Start with 0 momentum, collide, bounce off in opposite directions, still have 0 momentum More Related
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# Thread: Green's Therorem question 2 ... 1. ## Green's Therorem question 2 ... Let C be the positively oriented circle . Use Green's Theorem to evaluate the line integral . 2. Originally Posted by iwonder Let C be the positively oriented circle . Use Green's Theorem to evaluate the line integral . Here, $f\left(x,y\right)=4y$ and $g\left(x,y\right)=11x$ Thus, $\frac{\partial f}{\partial y}=4$ and $\frac{\partial g}{\partial x}=11$ Thus, by Green's Theorem, $\oint_C 4y\,dx+11x\,dy=7\iint\limits_R\,dA$ Note that the Region R is just a circle of radius 1!!! Thus, $\oint_C 4y\,dx+11x\,dy=7\iint\limits_R\,dA=\dots$ Does this make sense? 3. IT DOES!! thankyou very much!!
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Explore BrainMass # Implicit differentiation Not what you're looking for? Search our solutions OR ask your own Custom question. This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! Find dy/dx by implicit differentiation, please see the remaining questions in the attachment. https://brainmass.com/physics/right-hand-rule/implicit-differentiation-465137 #### Solution Preview To do implicit differentiation, you differentiate all terms in the equation. For the x-terms, you just do regular differentiation but for the y-terms you need to apply the chain rule because y = f(x), y is a function of x. 1. Find dy/dx by implicit differentiation. (11x + 2y)^1/3 = x^2 Solution: Differentiating both sides of the equation we get: For the left hand side, you apply the chain rule and for the right hand side, you apply the power rule. 1/3 (11x + 2y)^-2/3 (11 + 2y') = 2x Now solve for y': 11 + 2 y' = 6x (11x + 2y)^2/3 2 y' = 6x (11x + 2y)^2/3 - 11 Now divide both sides by 2: y' = 3x(11x + 2y)^2/3 - 11/2 is the solution 2. Find dy/dx by implicit differentiation. (x + y^2)^10 = 7x^2 + 2 Solution: Differentiating both sides of the equation we get: For the left hand side, you apply the chain rule and for the right hand side, you apply the power rule. 10(x + y^2)^9 (1 + 2y y') = 14x Now solve for y': 1 + 2y y' = 14x/[10(x + y^2)^9] 1 + 2y y' = 7x/[5(x + y^2)^9] 2y y' = 7x/[5(x + y^2)^9] - 1 y' = 7x/[10y(x + y^2)^9] - 1/2y is the solution 3. Find an equation of the tangent ... #### Solution Summary The expert finds the dy/dx by implicit differentiation. \$2.49
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# Hooke’s Law | Modulus of Elasticity ## Stress & Strain | Mechanical Properties of Solids- Before you go through this article, make sure that you have gone through the previous articles on Stress and Strain. We have learnt- • When a deforming force is applied on a body, a restoring force is produced in it. • This restoring force per unit area of the body is called as stress. • The effect of stress is to produce distortion or change the configuration of the body. • The ratio of the change in configuration to the original configuration is called as strain. • Strain measures the degree of deformation. ## Hooke’s Law- In mechanical properties of solids, Hooke’s law is defined as- For sufficiently small stresses, stress is directly proportional to the strain. #### Mathematically, Here, this constant is called Modulus of Elasticity or Elastic Modulus. Thus, ### Characteristics of Modulus of Elasticity- • Its SI unit is same as that of stress i.e. newton per square meter (Nm-2) or pascal (Pa). • Its dimensional formula is same as that of stress i.e. [ML-1T-2]. • It is a scalar quantity. • Its value depends on the nature of material of the body and is independent of its dimensions. • It is a measure of rigidity or stiffness of a material. Greater the modulus, the stiffer the material. ## Types of Modulus of Elasticity- Since strain is of three types, therefore modulus of elasticity is also of three types- Let us discuss all types of modulus of elasticity one by one in detail. ### 1. Young’s Modulus of Elasticity- It is defined as the ratio of the longitudinal stress on the material to the longitudinal strain produced in the material. ### 2. Bulk Modulus of Elasticity- It is defined as the ratio of the bulk stress (or volume stress) on the material to the volume strain produced in the material. ### 3. Shear Modulus or Modulus of Rigidity- It is defined as the ratio of the tangential stress (or shearing stress) on the material to the shear strain produced in the material.
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# Force ## Physics Assignment Help Online Force We now wish to define the unit of force. We know that a force can cause the acceleration of a body. Thus, we shall define the unit of force in terms of the acceleration that a force gives to a standard reference body. As the standard body, we shall use (or rather imagine that we use) the standard kilogram. This  body has been assigned, exactly and by definition, a mass of 1 kg. Thus, a force is measured by the acceleration it produces. However, acceleration is a vector quantity, with both magnitude and direction. Is force also a vector quantity? We can easily assign a direction to a force (just assign the direction of the acceleration), but that is not sufficient. We must prove by experiment that forces are vector quantities. Actually, that has been done: forces are indeed vector quantities;they have magnitudes and directions and they combine according to the vector rules of Chapter 3. In this book, forces are most often represented with a vector symbol such as , and a net force is represented with the vector symbol.As with other vectors, a force or a net force can have components along coordinate axes. When forces act along a single axis, they are single-component forces. Then we can drop the overhead arrows the force symbols and just use signs to indicate the directions of the forces along that axis. Newton’s First Law: If net  force for  body , then the body’s velocity cannot change that is the body cannot accelerate. There may be multiple forces acting on a body, but if their net (or resultant) force is zero, then the body cannot accelerate. A force F on the standard’kilogram gives that body an acceleration a. Inertial Reference Frames Newton’s first law is not true in all reference frames: but we can always find reference frames in which it (and the rest of Newtonian mechanics) is true. Such frames are called inertial reference frames, or simply inertial frames. An inertial reference frame is one in which Newton’s laws hold. For example, we can assume that the ground is an inertial frame provided that we can neglect Earth’s actual astronomical motions (such as its rotation). That assumption works well if we, say, send a puck sliding along a short trip of friction less ice-s-an observer on the ground would find that the puck’s motion obeys the laws of Newtonian mechanics. However, suppose we made the strip very  long, extending, say, from north to south. Then an observer on the ground would  find that the puck accelerates slightly toward the west as it moves south . In this book we usually assume that the ground is an inertial frame and that measurements of forces and accelerations are made from it. If measurements are made in, say, an elevator that is accelerating relative to the ground them the measurements are being made in a non inertial frame and the results can be surprising. (0) The path of a puck that is sent sliding due south along a long strip of frictionless ice, as seen by an observer on the ground. (b) The ground beneath the southward sliding puck rotates to the east as Earth rotates. CHECKPOINT 1: Which of the figure’s six arrangements correctly show the vector addition of forces 1, and 12 to yield the third vector, which is meant to represent their net  force . ### Related Physics Topics for Tuition Other assignments related to
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1. ## Paralellogram Any cyclic paralellogram having unequal adjacent sides is necessarily a (a)Square (b)Rectangle (c)Rhombus (d)Trapezium Why? 2. If a parallelogram is cyclic, then it must be a rectangle. The perpendicular bisectors of the sides of a parallelogram are concurrent only when it is a rectangle. For a parallelogram to be cyclic, its perpendicular bisectors must be concurrent at the center of the circumcircle; therefore, a cyclic parallelogram must be a rectangle. 3. Hello, devi! Any cyclic paralellogram having unequal adjacent sides is necessarily a: . . (a) square . (b) rectangle . (c) rhombus . (d) trapezium This is my approach to this problem. Perhaps you can hammer it into a formal proof. Opposite sides of a parallelogram are parallel. They are formed by a pair of parallel chords. Code: * * * * * * * A o o o o o o o o o o B * * * * * D o o o o o o o o o o o C * * * * * * * * * Opposite sides of a parallelogram are equal. They are formed by two chords equidistant from the center. Code: * * * * * A o o o o o o o o o B * : * : * : * C * * o * \ o * \ o * o * * o * * o * o * * D Since opposite sides of a parallelogram are parallel and equal, the chords are parallel and equidistant from the center. Code: * * * * * A o o o o o o o o o B * : * : * : * * * * * : * : * : * D o o o o o o o o o C * * * * * , , , , , , , , , , , , # any cyclic parallelogram having unequal adjacent sides Click on a term to search for related topics.
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# Pound Mass to Pound Force Calculator, Formula, Pound Mass to Force Calculation ## Pound Mass to Pound Force Calculator: Enter the values of the mass of the object m(lbs)  & the accelaration of gravity g(32.174049 ft/s^2) to determine the value of the pound mass to pound force lbf(lb-f). Enter The Mass of the Objectt: lbs Result – The Pound Mass to Pound Force: lb-f ## Pound Mass to Pound Force Formula: The sum of the pound mass to pound force lbf(lb-f) in lb-f and the mass of the object m(lbs) in lbs multiply to the accelaration of gravity g(32.174049 ft/s^2) in feet per second squared into calculate the pound mass to pound force lbf(lb-f). The equation of the pound force to pound mass lbf(lb-f), lbf(lb-f) = m(lbs) * g(32.174049 ft/s^2) Here, lbf(lb-f) = the pound mass to pound force in lb-f m(lbs) = the mass of the object in lbs g(32.174049 ft/s^2) = the accelaration of gravity in feet per second squared ### Calculation for Pound Mass to Pound Force: 1)To determined the pound mass and the mass of the object 15lbs and the gravity 32.174049 ft/s2 lbf = lb * g lbf = 15 * 32.174049                                  = mass equal to .45359237 kilograms lbf = 482.61lb-f 2)To determined the mass of the object and the pound mass 480lb-f and the gravity 32.174049 ft/s2
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# How.many grams in ounce How Many Grams Are In An Ounce? There are 28 grams (28.346 grams if you want to get technical) in an ounce. Unfortunately, metric conversions don't come easily to everyone. ## How Many Grams in an Ounce Although an ounce of food is not that normally used when measuring the volume of food, it still pays to know that it can be converted Solve Mathematics understanding that gets you Average satisfaction rating 4.8/5 ## How many Grams in an Ounce How much does 1 oz weigh in grams? An ounce is a unit of weight equal to 28.346 grams or a sixteenth pound. The conversion is the same regardless of the item you are Do My Homework ## How Many Grams In An Ounce 1 oz = 28.34952 g. If you're looking for something to do, why not try getting some tasks? There's always plenty to be done, and you'll feel productive and accomplished when you're done. Get support from expert teachers Get math help online by speaking to a tutor in a live chat. Decide math questions If you need help with your studies, our expert teachers are here to support you. Provide multiple ways Mathematics is a critical tool for understanding the world around us. By developing a strong foundation in math, we can equip ourselves with the ability to think logically and solve problems effectively. ## How Many Grams in an Ounce Grams to Ounces formula oz = g * 0.035274 Ounces A unit of weight equal to one • Get help from expert tutors If you're looking for fast answers, you've come to the right place. • Determine mathematic equation To determine what the math problem is, you will need to take a close look at the information given and use your problem-solving skills. Once you have determined what the problem is, you can begin to work on finding the solution. • Explain mathematic I always find math questions to be very difficult. I usually have to take a lot of time to figure out the answer. • Provide multiple methods There are a few things to consider when determining math tasks. First, think about the level of difficulty. Next, consider what type of math is required. Finally, think about how much time you have to complete the task.
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# Search by Topic #### Resources tagged with Place value similar to The Codabar Check: Filter by: Content type: Stage: Challenge level: ### There are 53 results Broad Topics > Numbers and the Number System > Place value ### Lesser Digits ##### Stage: 3 Challenge Level: How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9? ##### Stage: 2 and 3 Challenge Level: Watch our videos of multiplication methods that you may not have met before. Can you make sense of them? ### Six Times Five ##### Stage: 3 Challenge Level: How many six digit numbers are there which DO NOT contain a 5? ### What an Odd Fact(or) ##### Stage: 3 Challenge Level: Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5? ### Not a Polite Question ##### Stage: 3 Challenge Level: When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square... ### Mini-max ##### Stage: 3 Challenge Level: Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . . ### X Marks the Spot ##### Stage: 3 Challenge Level: When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" . ### Basically ##### Stage: 3 Challenge Level: The number 3723(in base 10) is written as 123 in another base. What is that base? ### Even Up ##### Stage: 3 Challenge Level: Consider all of the five digit numbers which we can form using only the digits 2, 4, 6 and 8. If these numbers are arranged in ascending order, what is the 512th number? ### Skeleton ##### Stage: 3 Challenge Level: Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum. ### Just Repeat ##### Stage: 3 Challenge Level: Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence? ### Exploring Simple Mappings ##### Stage: 3 Challenge Level: Explore the relationship between simple linear functions and their graphs. ### Digit Sum ##### Stage: 3 Challenge Level: What is the sum of all the digits in all the integers from one to one million? ### Arrange the Digits ##### Stage: 3 Challenge Level: Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500? ### Number Rules - OK ##### Stage: 4 Challenge Level: Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number... ##### Stage: 3 Challenge Level: Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true. ### Phew I'm Factored ##### Stage: 4 Challenge Level: Explore the factors of the numbers which are written as 10101 in different number bases. Prove that the numbers 10201, 11011 and 10101 are composite in any base. ### Enriching Experience ##### Stage: 4 Challenge Level: Find the five distinct digits N, R, I, C and H in the following nomogram ### Really Mr. Bond ##### Stage: 4 Challenge Level: 115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise? ### Back to the Planet of Vuvv ##### Stage: 3 Challenge Level: There are two forms of counting on Vuvv - Zios count in base 3 and Zepts count in base 7. One day four of these creatures, two Zios and two Zepts, sat on the summit of a hill to count the legs of. . . . ### Pupils' Recording or Pupils Recording ##### Stage: 1, 2 and 3 This article, written for teachers, looks at the different kinds of recordings encountered in Primary Mathematics lessons and the importance of not jumping to conclusions! ### Seven Up ##### Stage: 3 Challenge Level: The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)? ### Cycle It ##### Stage: 3 Challenge Level: Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total. ### Repeaters ##### Stage: 3 Challenge Level: Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13. ### Never Prime ##### Stage: 4 Challenge Level: If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime. ### Permute It ##### Stage: 3 Challenge Level: Take the numbers 1, 2, 3, 4 and 5 and imagine them written down in every possible order to give 5 digit numbers. Find the sum of the resulting numbers. ### Big Powers ##### Stage: 3 and 4 Challenge Level: Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas. ### DOTS Division ##### Stage: 4 Challenge Level: Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}. ### Latin Numbers ##### Stage: 4 Challenge Level: Can you create a Latin Square from multiples of a six digit number? ##### Stage: 3, 4 and 5 We are used to writing numbers in base ten, using 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Eg. 75 means 7 tens and five units. This article explains how numbers can be written in any number base. ### How Many Miles to Go? ##### Stage: 3 Challenge Level: How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order? ### Cayley ##### Stage: 3 Challenge Level: The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"? ### Three Times Seven ##### Stage: 3 Challenge Level: A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why? ### Multiplication Magic ##### Stage: 4 Challenge Level: Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). . . . ### What a Joke ##### Stage: 4 Challenge Level: Each letter represents a different positive digit AHHAAH / JOKE = HA What are the values of each of the letters? ### Tis Unique ##### Stage: 3 Challenge Level: This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility. ### Legs Eleven ##### Stage: 3 Challenge Level: Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have? ### Composite Notions ##### Stage: 4 Challenge Level: A composite number is one that is neither prime nor 1. Show that 10201 is composite in any base. ### Quick Times ##### Stage: 3 Challenge Level: 32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible. ### Football Sum ##### Stage: 3 Challenge Level: Find the values of the nine letters in the sum: FOOT + BALL = GAME ### Back to Basics ##### Stage: 4 Challenge Level: Find b where 3723(base 10) = 123(base b). ### Plus Minus ##### Stage: 4 Challenge Level: Can you explain the surprising results Jo found when she calculated the difference between square numbers? ### Balance Power ##### Stage: 3, 4 and 5 Challenge Level: Using balancing scales what is the least number of weights needed to weigh all integer masses from 1 to 1000? Placing some of the weights in the same pan as the object how many are needed? ### Reach 100 ##### Stage: 2 and 3 Challenge Level: Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100. ### Two and Two ##### Stage: 3 Challenge Level: How many solutions can you find to this sum? Each of the different letters stands for a different number. ### Always a Multiple? ##### Stage: 3 Challenge Level: Think of a two digit number, reverse the digits, and add the numbers together. Something special happens... ##### Stage: 1, 2, 3 and 4 Nowadays the calculator is very familiar to many of us. What did people do to save time working out more difficult problems before the calculator existed? ### Chocolate Maths ##### Stage: 3 Challenge Level: Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . . ### 2-digit Square ##### Stage: 4 Challenge Level: A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number?
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https://www.doubtnut.com/question-answer/comprehension-2-in-comparison-of-two-numbers-logarithm-of-smaller-number-is-smaller-if-base-of-the-l-5432423
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Class 12 MATHS Logarithms # Comprehension 2 In comparison of two numbers, logarithm of smaller number is smaller, if base of the logarithm is greater than one. Logarithm of smaller number is larger, if base of logarithm is in between zero and one. For example log_2 4 is smaller than (log)_2 8\ a n d(log)_(1/2)4 is larger than (log)_(1/2)8. Identify the correct order: (log)_2 6<(log)_3 8<log_3 6<(log)_4 6 (log)_2 6>(log)_3 8> log_3 6>(log)_4 6 (log)_3 8>(log)_2 6> log_3 6>(log)_4 6 (log)_2 8<(log)_4 6<log_3 6<(log)_4 6 Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams. Updated On: 28-12-2020 Apne doubts clear karein ab Whatsapp par bhi. Try it now. Watch 1000+ concepts & tricky questions explained! 80.5 K+ 4.0 K+ Image Solution 5432424 2.7 K+ 53.7 K+ 1:29 1458559 4.1 K+ 83.3 K+ 4:51 5432400 16.4 K+ 119.2 K+ 4:40 42361852 20.4 K+ 67.3 K+ 3:10 21893 4.7 K+ 93.5 K+ 1:31 5432402 8.1 K+ 164.1 K+ 3:44 42515 1.7 K+ 34.4 K+ 4:20 23498 2.2 K+ 44.6 K+ 4:55 1355701 40.1 K+ 65.0 K+ 6:02 4380665 4.6 K+ 94.1 K+ 1:26 43959700 2.0 K+ 39.9 K+ 2:47 5432988 7.0 K+ 141.9 K+ 3:06 53796889 4.5 K+ 90.3 K+ 2:26 4440179 6.7 K+ 134.8 K+ 1:29 42857 2.6 K+ 52.6 K+ 6:30
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https://visualfractions.com/unit-converter/convert-50-lx-to-ft-cd/
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# Convert 50 lx to ft-cd So you want to convert 50 lux into foot-candles? If you're in a rush and just need the answer, the calculator below is all you need. The answer is 4.6451564953223 foot-candles. ## How to convert lux to foot-candles We all use different units of measurement every day. Whether you're in a foreign country and need to convert the local imperial units to metric, or you're baking a cake and need to convert to a unit you are more familiar with. Luckily, converting most units is very, very simple. In this case, all you need to know is that 1 lx is equal to 0.092903129906447 ft-cd. Once you know what 1 lx is in foot-candles, you can simply multiply 0.092903129906447 by the total lux you want to calculate. So for our example here we have 50 lux. So all we do is multiply 50 by 0.092903129906447: 50 x 0.092903129906447 = 4.6451564953223 ## What is the best conversion unit for 50 lx? As an added little bonus conversion for you, we can also calculate the best unit of measurement for 50 lx. What is the "best" unit of measurement? To keep it simple, let's say that the best unit of measure is the one that is the lowest possible without going below 1. The reason for this is that the lowest number generally makes it easier to understand the measurement. For 50 lx the best unit of measurement is foot-candles, and the amount is 4.6451564953223 ft-cd.
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what is the LCM of 8 4 15 3 5 6 8 factor each number into prime numbers: note:  the second 8 doesn't change anything.  for example, the lcm of 8 and 8 is 8, the lcm of 8 and 8 and 8 is 8, and so on.  repeating a number doesn't do anything 8 = 2 * 2 * 2 4 = 2 * 2 15 = 3 * 5 3 = 3 5 = 5 6 = 2 * 3 the lcm is 2 * 2 * 2 * 3 * 5 because 2 * 2 * 2 * 3 * 5 has 2 * 2 * 2 (to make 8), 2 * 2 (to make 4), 3 * 5 (to make 15), a 3 (to make 3), a 5 (to make 5), and 2 * 3 (to make 6). 2 * 2 * 2 * 3 * 5 = 120 the lcm of 8, 4, 15, 3, 5, 6, and 8 is 120 by Level 13 User (103k points)
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A parallelogram is a quadrilateral in which both bag of opposite sides room parallel . A parallelogram also has the complying with properties: the opposite angles space congruent; opposite sides are congruent; adjacent angles are supplementary; A rectangle is a parallelogram with 4 right angles, so all rectangles are likewise parallelograms and quadrilaterals. ~ above the various other hand, not all quadrilaterals and also parallelograms are rectangles. A rectangle has all the nature of a parallelogram, to add the following: The diagonals are congruent. A rhombus is a parallelogram with 4 congruent sides. The many of rhombus is rhombi . (I love that word.) A rhombus has actually all the nature of a parallelogram, add to the following: The diagonals crossing at appropriate angles. A square have the right to be identified as a rhombus which is also a rectangle – in various other words, a parallelogram with four congruent sides and four appropriate angles. A trapezoid is a quadrilateral with precisely one pair that parallel sides. (There may be some confusion about this word depending on which nation you"re in. In India and also Britain, they say trapezium ; in America, trapezium usually way a quadrilateral v no parallel sides.) an isosceles trapezoid is a trapezoid who non-parallel sides space congruent. A dragon is a quadrilateral with precisely two pairs of adjacent congruent sides. (This definition excludes rhombi. Some textbooks speak a kite has at least two pairs of surrounding congruent sides, so a rhombus is a special instance of a kite.) A scalene square is a four-sided polygon that has no congruent sides. Three instances are presented below. You are watching: Is a parallelogram always a rectangle See more: Smoke From A Distant Fire Chords For Sanford Townsend Band, Smoke From A Distant Fire Chords ## Venn diagram of Quadrilateral group The adhering to Venn Diagram reflects the inclusions and also intersections of the various types of quadrilaterals.
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https://calculators.tech/straight-line-depreciation-calculator
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# straight line depreciation calculator Enter Information #### RESULTS Fill the calculator form and click on Calculate button to get result here ## Straight-Line Depreciation Calculator You can calculate straight-line depreciation of the given asset with this little gizmo in real-time. Our straight-line depreciation calculator is one of the most useful tools out there for what it does. Straight-line calculation is actually pretty easy given that the depreciation rate is constant over a period of time, thus, the name, the straight-line method. ## Overview In the Straight-line approach, the value of an asset decreases homogeneously over each period of time until it finally approaches its salvage value. Straight line depreciation method is the most useful depreciation model for distributing the cost of an asset in time. This straight-line model of depreciation is the simplest of all the models as the devaluation is uniformly distributed in each period of time unlike the other models in which there is certain variance in asset devaluation relative to time. ## Straight line depreciation formula The straight-line method formula is a rather simple one and it goes like this: $$\mathrm{Straight line depreciation rate} = \dfrac{\mathrm{Yearly depreciation expense}}{\mathrm{Asset Cost – Salvage value}}$$ ---OR--- $$\mathrm{Asset cost} – \dfrac{\mathrm{Final value}}{\mathrm{Useful life}}$$ Where: Asset cost = the buying price of the asset Final Value (Salvage Value) = This is the value that an asset has at the end of its useful life Useful life = This is the life of the asset in which it has peak productivity. ## How to calculate Straight-line depreciation with our calculator Our calculator employs the straight-line depreciation equation to determine the answer. It is free to use. All that you have to do is simply put in the values required in the respective boxes in our calculator. • Enter the asset value. • Enter the cost value. • Input the depreciation period. • Press ‘Calculate’ That’s it. Calculating the input would give you the depreciable base, the devaluation expense for the first and final years as well as the schedule where you can analyze the data. User Ratings • Total Reviews 2 • Overall Rating 5/5 • Stars Reviews Roger E. Lieberman | 02/09/2019 It’s so simple and easy to use this calculator for finding out the actual value of depreciation. Rodney Gunther | 02/09/2019 Thanks to your calculator. Measuring accurate line depreciation is now so much easy for me. Send us your feedback! straight line depreciation calculator
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Recent News Allow’s do some practice with this connection between the domain name of the feature and its upright asymptotes. To locate the upright asymptote of a reasonable function, just set the equal to 0 and solve for x. Currently we need to address for given that it is the denominator of the feature. When the common denominator of a function is equal to absolutely no, there is a vertical asymptote because that function is then undefined. A necessary condition for the existence of an upright asymptote is the absence of the term of the highest level \( \) in the last formula. I simply require a little help with the algebraic side of points. I do not understand just how to simplify the function to make sure that I can easily locate all asymptotes. It can be seen that in fact we acquired the horizontal asymptote, which has actually currently been defined over. Suggested web page how to find vertical asymptote for tanx here. A sensible function is a feature that can be composed as the ratio of two polynomials where the denominator isn’t absolutely no. If the numerator is one degree above the denominator, the chart has a slant asymptote. Making use of polynomial division, separate the numerator by the to determine the line of the angle asymptote. To discover the horizontal or angle asymptote, compare the degrees of the numerator and also . ## Finding Asymptotes For Trigonometric Features The straight asymptote might also be estimated by inputting very large positive or unfavorable values of x. Furthermore, a rational feature will certainly have x-intercepts at the inputs that create the output to be no. Out of these, the cookies that are categorized as necessary are stored on your web browser as they are essential for the working of standard functionalities of the web site. We also utilize third-party cookies that assist us assess as well as recognize how you use this internet site. These cookies will certainly be kept in your browser just with your permission. Click to find out more how to find the vertical asymptote and horizontal asymptote here. You likewise have the choice to opt-out of these cookies. But pulling out of some of these cookies may affect your surfing experience. ### Vertical Asymptotes Of Rational Functions. The positioning of these 2 asymptotes reduces the chart into three distinctive parts. As x comes close to 0 from the left, the outcome of the function grows randomly huge in the adverse instructions towards unfavorable infinity. This is a double-sided asymptote, as the feature expand arbitrarily large in either direction when coming close to the asymptote from either side. Some features just approach an asymptote from one side. But for currently, as well as for the most part, absolutely nos of the will certainly bring about vertical rushed lines and also graphs that skinny up as close as you please to those vertical lines. It’s alright that the chart shows up to climb up right up the sides of the asymptote on the left. As long as you do not draw the graph going across the upright asymptote, you’ll be great. ## Sciencing_icons_linear Equations Straight Formulas Set the internal quantity ofequal to zero to identify the change of the asymptote. As \$x \ to-\ infty\$, something similar happens, yet we need to be very cautious about indicator. Mathematics Stack Exchange is an inquiry and also solution website for people examining math at any kind of degree and professionals in associated fields. It is mandatory to procure individual consent before running these cookies on your web site. Since \$\ sqrt\$ is not a real number, the chart will have no upright asymptotes. The calculator will find the upright, horizontal as well as angle asymptotes of the feature, with steps shown. In maths, an asymptote of a feature is a line that a feature get infinitesimally closer to, yet never ever gets to. ### Locate The Vertical Asymptotes And Afterwards Establish The Left And Also Best Limitations At Each Upright Asymptote. When we make the denominator equivalent to no, we don’t get actual worths for ‘x’. Theorizing this reasoningad infinitum leads us to the counter-intuitive verdict that Achilles willnevercross the goal. There will certainly always be some limited distance he has to go across first, so he will certainly never ever in fact get to the finish line. Thinkers as well as mathematicians have actually puzzled over Zeno’s paradoxes for centuries. Figure 1. A feature which is constant on the whole set of genuine numbers has no upright asymptotes. When the degree of the numerator is specifically another than the level of the denominator, the graph of the rational feature will have an oblique asymptote. Another name for an oblique asymptote is a slant asymptote. Unbeknownst to Zeno, his paradoxes of movement come very near capturing the modern idea of a mathematical asymptote. Initially, note that this function has no common variables, so there are no prospective removable gaps. This informs us that as the values of t increase, the values of Cwill strategy \ frac[/latex]. completion habits of the chart would certainly look similar to that of an also polynomial with a favorable leading coefficient. News Reporter
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1. ## Solve the following. 1. 4/x = 5/7 2. 3a/7 = -2/5 3. x+1/6 = 4/3 4. 24/x-3 = 72/x+3 Thanks. 2. I'll only solve the 1st one, I think you can do the rest. $\displaystyle \frac{4}{x} = \frac{5}{7}$ Multiply both sides by $\displaystyle x$, $\displaystyle \frac{4}{x}\cdot x = \frac{5}{7}\cdot x$ $\displaystyle \frac{4}{\not x}\cdot \not x = \frac{5x}{7}$ $\displaystyle 4 = \frac{5x}{7}$ Now multiply both sides by $\displaystyle 7$, $\displaystyle 4\cdot 7 = \frac{5x}{7}\cdot 7$ $\displaystyle 28 = \frac{5x}{\not 7}\cdot \not7$ $\displaystyle 28 = 5x$ $\displaystyle x = \frac{28}{5}$ ---------- If you have an equation in the from $\displaystyle \frac{a}{b} = \frac{c}{d}$, products of the terms crosswise are equal, $\displaystyle a\cdot d = b \cdot c$ So, $\displaystyle \frac{4}{x} = \frac{5}{7}$ $\displaystyle 4\cdot 7 = 5\cdot x$ $\displaystyle 5x = 28$ $\displaystyle x = \frac{28}{5}$ 3. Originally Posted by elliotfsl 1. 4/x = 5/7 Initial Equation $\displaystyle \frac 4x = \frac 57$ You want to get x in the numerator so multiply both sides by x $\displaystyle \frac x1 * \frac 4x = \frac 57*\frac x1$ Simplifly $\displaystyle 4 = \frac {5x}7$ You want x to be by itself, so you need to get rid of the 7 in the denominator. You can do this by multiplying both sides by 7 since 7/7 =1 $\displaystyle \frac 71 *\frac 41 = \frac {5x}7*\frac 71$ Simplify $\displaystyle 28 = 5x$ You want to get x by itself, so you need to get rid of the 5, do this by multiplying both sides by 1/5 since 5/5=1 $\displaystyle \frac 15*28 = 5x\frac 15$ Simplify $\displaystyle \frac{28}5 = x$ 4. Originally Posted by elliotfsl 4. 24/x-3 = 72/x+3 Initial equation $\displaystyle \frac {24}{x-3} = \frac{72}{x+3}$ Again, you need to get x out of the denominator, so multiply both sides by x-3 $\displaystyle 24 = \frac{72(x-3)}{x+3}$ X is still in the denominator on the right side, so multiply both sides by x+3 $\displaystyle 24(x+3) = 72(x-3)$ Distribute the coefficients $\displaystyle 24x+72 = 72x-216$ You need all your x's to be on the same side, so subtract 24x $\displaystyle 72 = 72x-24x-216$ Simplify $\displaystyle 72 = 48x-216$ You need all the other numbers to be on the other side of x, so add 216 $\displaystyle 72 +216= 48x$ Simplify $\displaystyle 288= 48x$ You need x to be by itself, so multiply by 1/48 $\displaystyle \frac{288}{48}= x$ And simplify $\displaystyle 6 = x$
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# Lesson 6 Problems with Equal Groups of Fractions ## Warm-up: True or False: Two and Three Factors (10 minutes) ### Narrative The purpose of this True or False is to elicit strategies and understandings students have for finding products of a whole number and a fraction and identifying equivalent expressions. This work help students deepen their understanding of the properties of operations and will be helpful later when students solve problems with a whole number multiplied by a fraction. In this activity, students have an opportunity to look for and make use of structure (MP7) as they consider how fractions are decomposed into various factors and multiplied in parts. ### Launch • Display one statement. • “Give me a signal when you know whether the statement is true and can explain how you know.” • 1 minute: quiet think time ### Activity • Share and record answers and strategy. • Repeat with each statement. ### Student Facing Decide whether each statement is true or false. Be prepared to explain your reasoning. • $$\frac{10}{12} = 5 \times \frac{2}{12}$$ • $$1 \times \frac{10}{12} = 5 \times \frac {2}{12}$$ • $$\frac{24}{4} = 6 \times 3 \times \frac{1}{4}$$ • $$12 \times 2 \times \frac{1}{4} = 8 \times 3 \times \frac{1}{4}$$ ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • “What strategies did you use to determine if the statements were true or false?” ## Activity 1: Banana Bread Recipe (15 minutes) ### Narrative The purpose of this activity is for students to use their understanding of multiplication of a unit fraction by a whole number to solve problems. Students use what they know to find a product given the factors and find the factors when given the product. This reinforces the idea that any fraction $$\frac{a}{b}$$ is a multiple of $$\frac{1}{b}.$$ Students may interpret quantities greater than 1 as a combination of whole numbers and fractions (for example, $$\frac{4}{3}$$ cups as 1 whole cup and $$\frac{1}{3}$$ cup) or express them as mixed numbers (such as $$1\frac{1}{3}$$). Both are acceptable. If possible, ask students whether $$1\frac{1}{3}$$ and $$\frac{4}{3}$$ express the same amount, but it is not necessary to discuss the term mixed numbers at this point. (Students will be introduced to mixed numbers in upcoming lessons.) ### Launch • Groups of 2 • “Have you followed a recipe to make something before? What is in a recipe?” (A list of ingredients, amounts of each, and instructions for putting the ingredients together) • “If a recipe is for 5 servings or 5 people, but you need more than that, what would you do?” (Adjust the amount of ingredients.) • “We often refer to the amounts specified in a recipe as ’1 batch’.” • “What might it mean to make 2 batches of a recipe?” (Make twice as much, or need twice as much ingredients) ### Activity • “Take a few quiet minutes on work on the activity. Then, discuss your thinking with your partner.” • 5 minutes: independent work time • 5 minutes: partner work time • Monitor for students who discuss: • that each quantity in Monday’s table will be multiplied by 2 • that each quantity in Tuesday's table need to be multiplied by 4 because the amount of butter tells us that the number of batches is 4 ### Student Facing A bakery is making banana bread. Here is the recipe for 1 batch. Recipe: 1 banana $$\frac{2}{3}$$ cup butter $$\frac{3}{2}$$ teaspoons baking soda $$\frac{5}{8}$$ cup sugar 2 large eggs $$\frac{5}{2}$$ cups of all-purpose flour 1. The bakery makes 2 batches of banana bread on Monday. Complete the table to show how much of each ingredient is used. ingredient expression amount of ingredient bananas _______ butter _______ cup(s) baking soda _______ teaspoon(s) sugar _______ cup(s) eggs _______ flour _______ cup(s) 2. On Tuesday, the bakery needs $$\frac{8}{3}$$ cups of butter to make enough banana bread for the day. How many batches were made? Explain or show your reasoning. Recipe: 1 banana $$\frac{2}{3}$$ cup butter $$\frac{3}{2}$$ teaspoons baking soda $$\frac{5}{8}$$ cup sugar 2 large eggs $$\frac{5}{2}$$ cups of all-purpose flour 3. Based on the number of the batches made on Tuesday, complete the table for each ingredient. ingredient expression amount of ingredient bananas _______ butter $$\frac{8}{3}$$ cups baking soda _______ teaspoon(s) sugar _______ cup(s) eggs _______ flour _______ cup(s) ### Student Response For access, consult one of our IM Certified Partners. If students are unsure how to complete the last table, check if they recognize that the given $$\frac{8}{3}$$ cups represent the amount of butter for 4 batches. If so, ask: “How many bananas are needed in 4 batches?” and “How many eggs are needed?” “How might you use this to help to determine the ingredients needed for 1 batch, 2 batches and so on?” ### Activity Synthesis • Display the table for Monday and ask students to share responses. Record their responses for all to see. • “How is the numerator changing in all of the ingredients?” (It is multiplied by 2 in each problem.) • “Why is the denominator different in all of them?” (A different unit and unit fraction was used to measure each ingredient.) • Ask students to share the expressions for the ingredients in the table for Tuesday. Record each expression and its value as an equation: • $$4 = 4 \times 1$$ • $$\frac{8}{3} = 4 \times \frac{2}{3}$$ • $$\frac{12}{2} = 4 \times \frac{3}{2}$$ • $$\frac{20}{8} = 4 \times \frac{5}{8}$$ • $$8 = 4 \times 2$$ • $$\frac{20}{2} = 4 \times \frac{5}{2}$$ • “Why are two of the ingredients not in fraction form?” (They have whole-number units.) ## Activity 2: How Much Milk Was Used? (20 minutes) ### Narrative In this activity, students are presented with descriptions of situations and equivalent multiplication expressions. They match each description to an expression that could represent the situation and see that more than one expression can be used, depending on how they interpret the situation. Likewise, students find that one expression can be used to represent different descriptions (MP2). Students discuss their matching decisions, analyze how the expressions are related, and consider revising the matches they made if appropriate. When students discuss and justify their decisions they are creating viable arguments and critiquing one another’s reasoning (MP3). MLR7 Compare and Connect. Synthesis: Lead a discussion comparing, contrasting, and connecting the different representations. Ask, “How does the situation show up in the representation?”, “What do each of these representations have in common?”, and “How were they different?” Engagement: Develop Effort and Persistence. Invite students to generate a list of shared expectations for the group work in this activity. Record responses on a display and keep visible during the activity. Supports accessibility for: Social-Emotional Functioning ### Required Materials Materials to Gather ### Required Preparation • Write the 5 expressions from the activity on separate posters and post them around the room: $$4 \times (2\times \frac{1}{10})$$ $$4 \times \frac{2}{10}$$ $$8 \times \frac{1}{10}$$ $$2 \times (4 \times \frac{1}{10})$$ $$2 \times \frac{4}{10}$$ ### Launch • Groups of 2 • Read the first problem aloud to students. • “Share the expression you selected with a partner.” • Students may select any of the expressions because each is equivalent to $$\frac{8}{10}$$. If this happens, ask, “Does one expression seem to represent what is happening in the situation better than others?” ($$8 \times \frac{1}{10}$$) ### Activity • 5 minutes: independent work time • Ask students to stand with the poster showing the expression that they believe represents how much milk was used on Tuesday, and to discuss with others there why they chose this expression. • Ask students to partner with a student from a different poster to explain why they made a different choice. • “Does anyone wish to revise their thinking about the expression they selected?” • “Can you explain why you think that a different expression is a better choice now?” • Repeat this process for each problem. ### Student Facing The bakery that sells banana bread also sells fresh milkshakes. Each serving uses $$\frac{1}{10}$$ liter of milk. Here are five descriptions of the milkshakes sold in a week and five expressions that represent the liters of milk used. Match each description to an expression that represents it. 1. On Monday, the bakery sold 8 servings of milkshake. How much milk was used? 2. On Tuesday, two customers bought 4 servings of milkshake each. How much milk was used? 3. On Wednesday, four customers bought 2 servings of milkshake each. How much milk was used? 4. On Thursday, two customers each bought a serving of milkshake. They placed the same order three more times for their friends that day. How much milk was used? 5. On Saturday, four friends each purchased a serving of milkshake for breakfast. They came back for the same after dinner. How much milk was used? $$4 \times (2\times \frac{1}{10})$$ $$4 \times \frac{2}{10}$$ $$8 \times \frac{1}{10}$$ $$2 \times (4 \times \frac{1}{10})$$ $$2 \times \frac{4}{10}$$ ### Student Response For access, consult one of our IM Certified Partners. If students do not see that each factor in the expressions can be interpreted in different ways, or that different expressions could represent the same quantity, invite them to use diagrams or record their thinking using diagrams to illustrate various groupings of the same quantity. Consider asking: “Where do you see _____ (whole number) groups of _____ (fraction)?” ### Activity Synthesis • See lesson synthesis. ## Lesson Synthesis ### Lesson Synthesis “Today, we matched expressions to situations. We learned that several expressions can represent the same situation.” Invite 1–2 students who chose different expressions for the same problem (one of the last two problems in the milkshake activity) to share. Record their ideas for all to see. “Who can explain how each expression matches the problem?” (On Thursday, there were 4 separate orders of 1 serving each, or $$4 \times \frac{1}{10}$$, that were made by 2 people, or $$2 \times (4 \times \frac{1}{10})$$. This is also the same as $$2 \times \frac{4}{10}$$.) “Did you notice something about the answers to the problems?” (They are all the same. They are all $$\frac{8}{10}$$.) “Why do you think they are all the same?” (They all involve 8 groups of $$\frac{1}{10}$$.) ## Cool-down: The Same or Not the Same? (5 minutes) ### Cool-Down For access, consult one of our IM Certified Partners. ## Student Section Summary ### Student Facing In this section, we learned to multiply a whole number and a fraction by thinking about equal-size groups, just as we did when multiplying two whole numbers. For instance, we can think of $$6 \times 4$$ as 6 groups of 4. A diagram like this can help to show that the product is 24: Likewise, we can think of $$6 \times \frac{1}{4}$$ as 6 groups of $$\frac{1}{4}$$. Diagrams can help us see that the product is $$\frac{6}{4}$$: After studying patterns, we saw that when we multiply a whole number and a fraction, the whole number is multiplied only by the numerator of the fraction and the denominator stays the same. For example: $$6 \times \frac{1}{2} = \frac{6}{2}$$ $$2 \times \frac{4}{5} = \frac{8}{5}$$ We also learned that: • Every fraction can be written as a product of a whole number and a unit fraction. For example, $$\frac{5}{4}$$ can be written as $$5 \times \frac{1}{4}$$. • We can write different multiplication expressions for the same fraction. For example, $$\frac{8}{3}$$ can be written as: $$8 \times \frac{1}{3}$$ $$4 \times 2 \times \frac{1}{3}$$ $$4 \times \frac{2}{3}$$ $$2 \times \frac{4}{3}$$
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# Quick Answer: How Do You Find 20% Of A Number? ## How do you find 70 percent of a number? Example 1. Find 70% of 80. Following the shortcut, we write this as 0.7 × 80. Remember that in decimal multiplication, you multiply as if there were no decimal points, and the answer will have as many “decimal digits” to the right of the decimal point as the total number of decimal digits of all of the factors.. ## How can I calculate average? How to Calculate Average. The average of a set of numbers is simply the sum of the numbers divided by the total number of values in the set. For example, suppose we want the average of 24 , 55 , 17 , 87 and 100 . Simply find the sum of the numbers: 24 + 55 + 17 + 87 + 100 = 283 and divide by 5 to get 56.6 . ## How do you find the difference between two numbers? How to Find the Difference between Two Numbers. To find the difference between two numbers, subtract the number with the smallest value from the number with the largest value. The product of this sum is the difference between the two numbers. Therefore the difference between 45 and 100 is 55. ## How do you find 20 percent of a number on a calculator? Example: 20% of what is 7?Written using the formula: X = 7 ÷ 20%Convert the percent to a decimal.20% ÷ 100 = 0.2.X = 7 ÷ 0.2.X = 35.So 20% of 35 is 7. ## How do you find 30% of a number? Once you have the decimal figure, multiply it by the number for which you seek to calculate the percentage; i.e., if you need to know 30 percent of 100, you convert 30 percent to a decimal (0.30) and multiply it by 100 (0.30 x 100, which equals 30). ## How can you find 75% of any number? Calculate the percent value:75 ÷ 100 =0.75 =0.75 × 100/100 =75/100 = ## What is the easiest way to find a percentage of a number? To calculate 10 percent of a number, simply divide it by 10 or move the decimal point one place to the left. For example, 10 percent of 230 is 230 divided by 10, or 23. ​5 percent​ is ​one half of 10 percent​. To calculate 5 percent of a number, simply divide 10 percent of the number by 2. ## How do I calculate percentage on calculator? How to Calculate Percentages with a CalculatorIf your calculator has a “%” button. Let’s say you wanted to find 19 percent of 20. Press these buttons: 1 9 % * 2 0 = … If your calculator does not have a “%” button. Step 1: Remove the percent sign and add a couple of zeros after the decimal point. 19% becomes 19.00.Nov 29, 2014 ## What is percentage of a number? In mathematics, a percentage is a number or ratio that represents a fraction of 100. It is often denoted by the symbol “%” or simply as “percent” or “pct.” For example, 35% is equivalent to the decimal 0.35, or the fraction. ## How do you find 15% of a number? 15% is 10% + 5% (or 0.15 = 0.1 + 0.05, dividing each percent by 100). Thinking about it this way is useful for two reasons. First, it’s easy to multiply any number by 0.1; just move the decimal point left one digit. For example, 75.00 x 0.1 = 7.50, or 346.43 x 0.1 = 34.64 (close enough). ## How do you find the reverse percentage? Step 1) Get the percentage of the original number. If the percentage is an increase then add it to 100, if it is a decrease then subtract it from 100. Step 2) Divide the percentage by 100 to convert it to a decimal. Step 3) Divide the final number by the decimal to get back to the original number. ## How do you find original price? This calculation helps you to find the original price after a percentage decrease.Subtract the discount from 100 to get the percentage of the original price.Multiply the final price by 100.Divide by the percentage in Step One.Nov 20, 2020 ## How do you find the selling price? Calculated by adding together all your costs, then adding a mark-up percentage that creates your profit margin. If a product costs \$50 to produce, and you want to apply a mark-up of 25% you multiply 50 by 1.25. The selling price would be \$62.50. This combines your cost per unit with projected output for your business. ## What is the formula for calculating percentage? To calculate the percentage, multiply this fraction by 100 and add a percent sign. 100 * numerator / denominator = percentage . In our example it’s 100 * 2/5 = 100 * 0.4 = 40 . Forty percent of the group are girls. ## How do I calculate a percentage between two numbers? First: work out the difference (increase) between the two numbers you are comparing. Then: divide the increase by the original number and multiply the answer by 100. ## How do I find the percentage of two numbers without a calculator? If you need to find a percentage of a number, here’s what you do – for example, to find 35% of 240: Divide the number by 10 to find 10%. In this case, 10% is 24. Multiply this number by how many tens are in the percentage you’re looking for – in this case, that’s 3, so you work out 30% to be 24 x 3 = 72. ## What is the difference between two numbers? The difference between two numbers on a number line is the distance between them and it varies whether you are going from left to right or right to left. The commonly used way is to go from right to left, which gives us a positive number. ## How do I calculate a percentage of a number? To find the percentage of a number when it is in decimal form, you just need to multiply the decimal number by 100. For example, to convert 0.5 to a percentage, 0.5 x 100 = 25% The second case involves a fraction. If the given number is in fractional form, first convert it to a decimal value and multiply by 100. ## How do you find the original price of 20 off? First consider the unknown original price as ‘x’. Then consider the rate of discount. To find the actual discount, multiply the discount rate by the original amount ‘x’. To find the sale price, subtract the actual discount from the original amount ‘x’ and equate this to given sale price.
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# Find the increasing OR decreasing point in an array Objec­tive:  Given an array of integer which is first increasing then decreasing. Find the element in array from which point it starts decreasing. OR Given an array of integer which is first increasing then decreasing. Find the maximum element in that array. Similar Problem: Given an array of integer which is first decreasing then increasing. Find the element in array from which point it starts increasing. OR Given an array of integer which is first decreasing then increasing. Find the minimum element in that array. Example: ```int [] a = {1,2,4,6,11,15,19,20,21,31,41,51,55,46,35,24,18,14,13,12,11,4,2,1}; output: 55 int [] a = {1,2,4};  //no deceasing element, so last element will be answer output: 4 int [] a = {4,2,1};  //no deceasing element, so last element will be answer output: 4 ``` Approach 1: Linear Search: • Navigate the array and track the element from where array starts decreasing. • Handle the edge cases. Time Complexity: O(n) Code: This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. public class FirstDecreasingBruteForce { public static void find(int [] a){ for (int i = 1; i Approach 2: Binary Search • Modify the binary search. • If mid element is greater than both its left and right neighbors then we have found our element. • If mid element is greater than its left neighbor and less than its right neighbor then array is still increasing. Do a recursive call on the right half of the array (mid+1,end). • If mid element is less than its left neighbor and greater than its right neighbor then array is now decreasing. Do a recursive call on the left half of the array (start, mid). • Handle the base cases (see the code). Time Complexity: O(logn) Code: This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. `(Binary Search) First Element from where elements starts decreasing: (index: 12) : 55`
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# Using definition of Laplace transform in determining Laplace of a step function #### shorty ##### New member I have a question that has stumped me a bit, i am not sure how to use the definition to calculate it, i can use the tables, but i don't think that's what is needed. Using the definition of the Laplace transform, determine the Laplace transform of I can do it with the table but i am not sure how to to this using the definition. Help please? #### Chris L T521 ##### Well-known member Staff member I have a question that has stumped me a bit, i am not sure how to use the definition to calculate it, i can use the tables, but i don't think that's what is needed. Using the definition of the Laplace transform, View attachment 153 determine the Laplace transform of View attachment 154 I can do it with the table but i am not sure how to to this using the definition. Help please? We can break up the integral into two parts since $f(t)$ is a piecewise function: $\mathcal{L}[f(t)] = \int_0^{\infty}e^{-st}f(t)\,dt=\int_0^2 e^{-st}0\,dt + \int_2^{\infty}e^{-st}t\,dt = \int_2^{\infty}te^{-st}\,dt.$ This should now be a relatively simple improper integral to compute. Can you take it from here? #### shorty ##### New member Thank you, but I got that far into the separation, but I wasn't sure how to proceed from there, my integrals kept repeating when I tried it by parts, and I wasn't getting anything to substitute to use that wasn't still leaving me with multiple variables to integrate. ... #### Chris L T521 ##### Well-known member Staff member Thank you, but I got that far into the separation, but I wasn't sure how to proceed from there, my integrals kept repeating when I tried it by parts, and I wasn't getting anything to substitute to use that wasn't still leaving me with multiple variables to integrate. ... In this case, you only need to apply integration by parts once. Let $u=t$, $dv=e^{-st}dt$; thus $du=dt$ and $v=-\dfrac{e^{-st}}{s}$. Plugging this into the integration by parts formula, we have $\int_2^{\infty}te^{-st}\,dt = \lim\limits_{b\to\infty}\left.\left[-\frac{te^{-st}}{s}\right]\right|_2^b + \frac{1}{s}\int_2^{\infty}e^{-st}\,dt=\ldots$ Can you take it from here?
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# How many dimensions does a square have...? formulae for cube? formulae for square? 1 by leon 2014-11-12T16:55:43+05:30 ### This Is a Certified Answer Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest. A square has 2 dimensions.  Length and breadth. Formulae for  cube : length of a side = a         number of sides/edges = 12 number of vertices or corners = 8 number of diagonals  on the surface = 12 , their length = a √2 number of diagonals inside cube = 4, their length = a √3 total number of faces = 6 Total surface area = 6 * a² total volume = a³ ================================= Formulae for square length of a side = a            area = a²    perimeter = 4 a number of diagonals = 2 length of diagonal = a √2
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# Find the lady probability trick 1. Feb 25, 2013 really basic trick, i'm quite tired and can't wrap my head around how it works out to 2/3 chances 2. Feb 25, 2013 ### trollcast This is just a rehash of the Monty Hall problem: http://en.wikipedia.org/wiki/Monty_Hall_problem Another hint would be to consider drawing a probability tree diagram for scenario. 3. Feb 25, 2013 ### nowonda It's a classic problem (http://en.wikipedia.org/wiki/Monty_Hall_problem), the easiest answer is that when you choose a door there's a 1/3 chance you get the woman, so it's more likely (twice as likely) that the woman is behind one of the other two doors. Of course, at first you can't know which one of the two, but after the host reveals the doll behind one of the doors, you're basically left with this choice: Should I stick with my original choice (with a 1/3 chance of finding the woman behind it) or should I choose "one of the other two doors" (which in total have a 2/3 chance of delivering the desired lady)? In reality you're not choosing "one of the other two doors" (since the host already exposed one), but actually the only door left, which now, after the host revealed the bogus one of the two, has a 2/3 chance of having the woman behind it. 4. Feb 25, 2013 It is not hard understanding through a tree diagram, it is the semantics that i'm struggling with I'm slightly convinced about the solution, but not entirely can someone explain with unconditional and conditional probability? explain it entirely? just saw nowonda's reply. Yeah, i've understood just as much but what i'm confused with is that why the odds change? even in the wikipedia page: I find this part to be quite critical. the wikipedia page doesn't clearly examine why many people wrongly believe the odds to be 1/2 and 1/2 can anybody elaborate? in the 1,000,000 goat example what I don't quite understand is how we can assume the door that the door he did NOT choose has the probability 999999/1000000 of winning while the door he chose is almost LIKELY to have a goat. Last edited: Feb 25, 2013 5. Feb 25, 2013 ### trollcast http://www.math.cornell.edu/~mec/2008-2009/TianyiZheng/Conditional.html That page should give a good explaination of the conditional probabilities in the Monty Hall Problem 6. Feb 25, 2013 thanks for that trollcast! here are some trivial but cool tricks, if anyone wants to take the time and breakdown each one, i'd appreciate it! i've figured most out by now http://nrich.maths.org/1441 7. Feb 25, 2013 ### nowonda I may be talking out of my ars, but I'll try an explanation that doesn't require extensive reading, just common sense. As a hint, it's not that "the odds change", magically surging from 33% to 50%, but the problem itself changes during the process, because there's more info than we had in the beginning. An example (stupid as it may be) is better, I suppose: I'm playing a game with you, where I put both my hands in my pockets or I keep them both out (randomly), and you have to guess my choice, without looking in any way. Obviously, you have a 50/50 chance of guessing what I did. Now suppose you don't look intentionally, but somehow you catch a glimpse of my left hand tucked in my pocket. And although you didn't have access to all the information (i.e. you only saw my left hand, not both), that info is enough for you to make a re-assessment of the probabilities you initially assigned for the two cases (hands in pockets - 50% and hands out of the pockets - 50%). Obviously, now the probability for "hands in pockets" magically increased to 100%, because you took into account the new available information. The example is quite retarded, but the principle is similar - new relevant info alters the problem, so you're not actually playing the same game. The game where you choose a door from three, the host reveals a bogus door (new relevant info) and then asks you if you want to change your choice is a completely different game compared to a game where you choose a door from three and the host asks you if you want to change, without showing you a bogus door (no new relevant info). 8. Feb 25, 2013 ### ssd Let the doors be marked A,B,C. You select A. One of the remaining is opened to show empty. You go for the other. Why? Because you win if the prize is behind any of B or C. So, by switching, you are allowed to bet for two doors and by sticking back to first choice you bet for one door. Resulting wining probabilities are 2/3 and 1/3 respectively. 9. Feb 28, 2013 ### Bacle2 Yet another approach: If you select doors randomly, you will choose a door with a goat behind it 2/3 of the time over the long run. In this case switching increases your odds , and 1/3 of the time you will select the car, in which case switching lowers your odds. So 2/3 of the times you play over the long-run it makes sense to change. IOW, you're twice as likely to have selected a door with a goat behind it than one with the car behind it, so switching will increase your odds of winning 2/3 of the time. My best understanding is that Marylin Idot-Savant did not pose the problem clearly-enough to eliminate ambiguity, i.e., the problem was not well-posed. 10. Mar 1, 2013
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https://www.queryhome.com/puzzle/30159/during-lucky-night-casino-maya-kayra-7740-made-2130-much-money
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crawl-data/CC-MAIN-2018-51/segments/1544376830479.82/warc/CC-MAIN-20181219025453-20181219051453-00252.warc.gz
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# During a lucky night out at the casino, Jay, Maya and Kayra won \$ 7740..........made \$ 2130, how much money did Jay win? 19 views During a lucky night out at the casino, Jay, Maya and Kayra won \$ 7740 between them. The difference between the sums of money won by Jay and Maya is twice as much as the difference between the amounts of cash won by Jay and Kayra. If Maya made \$ 2130, how much money did Jay win? posted Dec 6 Jay + Maya + Kayra = \$7740 ------ 1 (Jay - Maya) = 2(Jay - kayra) ------- 2 Maya = \$2130 ------ 3 Sub 3 in 2 (Jay - \$2130) = 2(Jay - kayra) ------- 4 Sub 3 in 1 Kayra = \$7740 - Jay - \$2130 Kayra = \$5610 - Jay --------- 5 Sub 5 in 4 Jay - \$2130 = 2(Jay) - 2(\$5610 - Jay) Jay = \$3030. Similar Puzzles +1 vote Mark is throwing his son Stewart a surprise birthday party but he has limited funds. He spent half of his money plus \$2.00 on the cake. Half of what was left plus \$2.00 was spent on balloons and decorations. Then he spent half of what he had left plus \$1.00 on candy. Now he is out of money, how much did Mark start with? +1 vote A man took a cab from his home to reach office. On the way back, he took a cab again. When he had reached his office, he paid 1/6th of what he had in his wallet and when he returned home, he paid 1/5th of what he had left in his wallet. In the first trip the fare was \$60. How much money did he initially have and how much is left with him now?
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https://www.callicoder.com/find-smallest-letter-greater-than-target/
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## Smallest Letter Greater than Target Given an array of lowercase letters sorted in ascending order, and a `target` letter, find the smallest letter in the array that is greater than the `target`. Note that, Letters also wrap around. For example, if `target = 'z'` and `letters = ['a', 'b']`, the answer is `'a'`. Example 1: ``````Input: letters = ["d", "h", "l"], target = "a" Output: "d"`````` Example 2: ``````Input: letters = ["d", "h", "l"], target = "d" Output: "h"`````` Example 3: ``````Input: letters = ["d", "h", "l"], target = "f" Output: "h"`````` Example 4: ``````Input: letters = ["d", "h", "l"], target = "j" Output: "l"`````` Example 4: ``````Input: letters = ["d", "h", "l"], target = "n" Output: "d"`````` ## Binary search to find the smallest letter greater than target This problem is a variation of the problem Find the ceiling of an element in a sorted array. The implementation approach is also similar. We apply binary search to find the smallest letter greater than the target. Here is the full implementation: ``````import java.util.Scanner; class NextLetter { private static int findNextLetter(char[] letters, char target) { int n = letters.length; int start = 0, end = n - 1; char nextLetter = letters[start]; while (start <= end) { int mid = (start + end) / 2; if (target < letters[mid]) { /* * letters[mid] is the smallest letter found so far that is greater than target. * So update nextLetter to this value and keep checking in the left side of the * array to find an even smaller letter that is greater than target */ nextLetter = letters[mid]; end = mid - 1; } else { start = mid + 1; } } return nextLetter; } public static void main(String[] args) { Scanner keyboard = new Scanner(System.in); int n = keyboard.nextInt(); char[] letters = new char[n]; for (int i = 0; i < n; i++) { letters[i] = keyboard.next().charAt(0); } char target = keyboard.next().charAt(0); keyboard.close(); System.out.printf("NextLetter(%c) = %c%n", target, findNextLetter(letters, target)); } }`````` ``````# Output \$ javac NextLetter.java \$ java NextLetter 3 d h l a NextLetter(a) = d \$ java NextLetter 3 d h l d NextLetter(d) = h \$ java NextLetter 3 d h l f NextLetter(f) = h \$ java NextLetter 3 d h l j NextLetter(j) = l \$ java NextLetter 3 d h l l NextLetter(l) = d \$ java NextLetter 3 d h l n NextLetter(n) = d``````
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1. ## Eulers Totient Hi Guys, Any information would be great. φ $(30)$ 2. ## Re: Eulers Totient Originally Posted by extatic Hi Guys, Any information would be great. φ $(30)$ The general expression of Euler's totiens function is... $\varphi(n)= n\ \prod_{p|n} (1-\frac{1}{p})$ (1) ... where $p|n$ means 'prime deviding n'... Kind regards $\chi$ $\sigma$ 3. ## Re: Eulers Totient Thank you mate, Looking at this example, can you tell me how they came to use 2 & 3 $\varphi(36)=\varphi\left(2^2 3^2\right)=36\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=36\cdot\frac{1}{2}\cdot\frac{2} {3}=12.$ 4. ## Re: Eulers Totient Originally Posted by chisigma The general expression of Euler's totiens function is... $\varphi(n)= n\ \prod_{p|n} (1-\frac{1}{p})$ (1) ... where $p|n$ means 'prime deviding n'... Kind regards $\chi$ $\sigma$ Can you check if im correct please, Since 30 is not prime, we list all of the positive integers less than 30 that are relatively prime to it: 1, 4, 7, 8, 9, 11 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29. $\varphi(n)= (30) = 23$ 5. ## Re: Eulers Totient Originally Posted by extatic Can you check if im correct please, Since 30 is not prime, we list all of the positive integers less than 30 that are relatively prime to it: 1, 4, 7, 8, 9, 11 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29. $\varphi(n)= (30) = 23$ Because 30= 2 x 3 x 5, it is more easy to count the number 1 and all primes greater than 5 and less than 30: 1 , 7 , 11 , 13 , 17 , 19 , 23 , 29... total 8 numbers!... Kind regards $\chi$ $\sigma$ 6. ## Re: Eulers Totient So i attempted $\varphi(n)= (10)$ 10 = (2 * 5) = 10 * (1 - 1/2)(1 - 1/5) = 10 * (1/2 * 4/5) = 4/10 * 10/1 = 4 So there should be 4 prime numbers less than 10 (not including 2 and 5) 1, 3, 7 (only three) What am i doing wrong? 7. ## Re: Eulers Totient Originally Posted by extatic So i attempted $\varphi(n)= (10)$ 10 = (2 * 5) = 10 * (1 - 1/2)(1 - 1/5) = 10 * (1/2 * 4/5) = 4/10 * 10/1 = 4 So there should be 4 prime numbers less than 10 (not including 2 and 5) 1, 3, 7 (only three) What am i doing wrong? In my previous post I have been a little 'approximative'... if n is not prime, i.e. is... $n= p_{1}^{k_{1}}\ p_{2}^{k_{2}}\ ...\ p_{i}^{k_{i}}$ (1) ... then the totiens function is 1 plus the number of primes p and their powers less than n. In Your case You have 'forgotten' 3 x 3=9... Kind regards $\chi$ $\sigma$ 8. ## Re: Eulers Totient by the chinese remainder theorem, if p,q are distinct primes, $\phi(p^rq^s) = \phi(p^r)\phi(q^s)$ thus it suffices to find $\phi(p^r)$, where r is the highest power of p that divides n, for each prime p in the factorization of n. by counting (never underestimate the power of counting), $\phi(p^r) = (p-1)(p^{r-1})$ so for n = 10, for example, 10 = (2)(5). φ(2) = 1, φ(5) = 4, so φ(10) = φ(2)φ(5) = (1)(4) = 4.
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https://movies.stackexchange.com/questions/63545/are-helicopters-capable-of-carrying-this-type-of-giants/63548
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# Are helicopters capable of carrying this type of giants? In The BFG, UK helicopters carry giants and leave them on an island: Are helicopters capable of carrying this type of giants? • Heck, I would have just gone with the Mi-26 or the Mi-v12 Which, if you think about how much muscle weighs, I am willing to bet that those giants weigh just a tad over the max lift capacity of the sky crane and chinook. Commented Nov 21, 2016 at 3:33 ## The Helicopters There are basically two models of helicopter in this image. The Sikorsky CH-54 is capable of carrying a payload of 20,000 pounds i.e 9000 Kg while a CH-47 Chinook can carry up to 10,000 pounds i.e 4500 Kg. ## The Giants Now guessing from size of those giants they might vary from 3000 to 6000 kg. Calculating weight of giants. The height of each man-eating giant is around 50 feet (15 metres), according to Wikipedia. And from the their body mass I will consider them obese which means they have a BMI ranging from 25 to 30. So BMI = Weight in KG / (Height in Metres)^2 Hence, Weight = BMI * (Height)^2 Assuming BMI = 25, Weight = 25 x 15 x 15 = 5625 Kg which is close to 6000 Kg. Assuming BMI = 30, Weight = 30 x 15 x 15 = 6750 Kg which is close to 7000 Kg. So the calculated weight ranges from 6000 to 7000 Kg. Right below the payload limit. ## Conclusion So yes, those helicopters can carry those giants. • BMI assumes body mass is proportional to height squared. We live in three dimensions. If giants were ten times our height (and width and depth), they'd be one thousand times our weight, ie. around 70,000 kg, much too heavy for the choppers. Commented Nov 21, 2016 at 11:13 • The computation must be bogus. I am (under) 2m tall and weigh 100 kg; overweight but not fat. These giants are fat. One of them is as long as the sky crane, i.e. 27m. That makes him linearly 13 times as large as me, making him have 13*13*13* my mass, i.e. 100 * 13^3 = 100 * 1728 * 172800 kg, or 172 metric tons, about 17 times as much as the sky crane could lift. That fits with the usual size of water loads for fire fighting (much smaller). Commented Nov 21, 2016 at 11:35 • @PeterA.Schneider: Definitely the CGI guys missed to keep proportion of helo's length and giants height. My computation is based on the Wiki info of their height to be around 50 feet. Commented Nov 21, 2016 at 11:43 • Ah, scaling strikes again. There are many problems with this. E.g., as you noted, the mass goes up with the cube of the scale, but the contact area of their feet with the ground only goes up with the square. Which means they put 8-15 times as much pressure on the ground as a human, and will probably sink in quite a bit. Also, the diameter of their bones will only go up with the square, but the weight they have to support goes up with the cube. Likewise, muscle strength is related to the diameter, i.e. goes up with the square, but the mass they have to move goes up with the cube. Commented Nov 21, 2016 at 13:04 • BMI is wildly unsuited to this kind of calculation. The power term used (square) is a stonking simplification of reality (somewhere up around 2.7 for humans) that doesn't even work that well for people near the edges of the normal range of human heights yet alone so far beyond the range. Accordingly your numbers are a huge under-estimate of the "likely" weight of the giants. Commented Nov 21, 2016 at 13:35 Transport helicopters are capable of carrying quite heavy and cumbersome loads externally as sling-load or underslung cargo, as long as the external load is properly attached and balanced and its weight doesn't exceed the max. permitted weight limit. One of the helicopters in the picture is a Sikorsky CH-54 Tarhe (S-64 Skycrane) which is purposefully built to transport heavy, extreme loads. It can carry a jet fighter, or a house as underslung cargo, as pictured below: And here is a Tarhe carrying a Chinook helicopter as underslung cargo: So the way they carry the giants is realistic, this method of external load transport by helicopters is not uncommon in real life. If the giant's weight is within the max. premitted weight limits, then it is OK to transport them this way. But, after all, this is a movie, and in the world of movies everything is possible. The way the helicopters transport the giants in The BFG is certainly much more realistic than the way the jaegers are transported in Pacific Rim: This will never happen in real life. • +1 for that last one - a bit of thought shows that the cables will tend to pull the helicopters together until they crash with each other, unless they expend considerable lift pushing away from the cable. – E.P. Commented Nov 20, 2016 at 18:05 • Which they may do - helicptters are capable of having sideways thrust. THey would not be that vertical then, though (CGI failure) and it would be RIDICULOUSLY dangerous. I Can not imagine it even working without way too many accidents. In the above example you must coordinate 8 pilots perfectly for a no crash. Not realistic. Commented Nov 20, 2016 at 18:43 • Forget the logistics of synchronized flying with a common load - those jaegers are big enough to use trains as weapons. They must weigh eleventy bajillion pounds each. Only Yoda could lift a jaeger. – user9311 Commented Nov 20, 2016 at 19:28 • @Tom maybe all the helicopter pilots drift tllt Commented Nov 21, 2016 at 5:28 • Note that the objects in your examples are as big as the giants in the movie, but essentially hollow shells; the two aircraft are actually designed for minimum weight. The giants are essentially water tanks. Considering what i have seen in fire fighting the giants are way too large to be carried. Commented Nov 21, 2016 at 11:27 I took easier approach than calculating mass. I took some info about similar sized animals. See info about Humpback Whale: Adult males measure up to a maximum length of 15 – 18m and weight of 40 tonne. Adult females measure up to 15m and weigh from 22 to 35 tonne. This is way over the limit as indicated by other posts. As I said in a comment, this is completely unrealistic (which is not surprising and not bad; it's a fantasy movie!). As an approximation, the giants roughly look the size of the helicopters. For an estimate let's assume block shape of the helicopter's dimensions, 27x7x7m, or 1323 m3, meaning >1000 metric tons of mass, assuming the usual body density of about 1. That's roughly 100 times the payload of a sky crane. If we mis-estimated by a linear factor of 2 (which would be a little less than the size the giants have in the book, 50 ft) the weight would still be 1323/2^3 = 165 tons, 16 times the payload. (But actually the guy in the front looks significantly larger than the copter carrying him.)
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### HS.G-CO: Congruence #### 1.1: Experiment with transformations in the plane HS.G-CO.1: Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, and plane. HS.G-CO.2.i: Represent transformations in the plane. HS.G-CO.2.ii: Describe transformations as functions that take points in the plane as inputs and give other points as outputs. HS.G-CO.2.iii: Compare transformations that preserve distance and angle to those that do not (e.g., translation versus horizontal stretch). HS.G-CO.3: Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry it onto itself. HS.G-CO.4: Develop or verify experimentally the characteristics of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments. HS.G-CO.5.i: Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, e.g., graph paper, tracing paper, or geometry software. HS.G-CO.5.ii: Specify a sequence of transformations that will carry a given figure onto another. #### 1.2: Understand congruence in terms of rigid motions HS.G-CO.6.i: Use geometric descriptions of rigid motions to predict the effect of a given rigid motion on a given figure. HS.G-CO.6.ii: Use the definition of congruence in terms of rigid motions to decide if two figures are congruent. HS.G-CO.8: Prove two triangles are congruent using the congruence theorems such as ASA, SAS, and SSS. #### 1.3: Prove and apply geometric theorems HS.G-CO.9: Prove and apply theorems about lines and angles. HS.G-CO.10: Prove and apply theorems about triangle properties. HS.G-CO.11: Prove and apply theorems about parallelograms. #### 1.4: Make geometric constructions HS.G-CO.12: Make basic geometric constructions with a variety of tools and methods. 1.4.1.2: Tools may include compass and straightedge, string, reflective devices, paper folding or dynamic geometric software. HS.G-CO.13: Apply basic constructions to create polygons such as equilateral triangles, squares, and regular hexagons inscribed in circles. ### HS.G-SRT: Similarity, Right Triangles, and Trigonometry #### 2.1: Understand similarity HS.G-SRT.1: Verify experimentally the properties of dilations given by a center and a scale factor. HS.G-SRT.2.i: Given two figures, use transformations to decide if they are similar. HS.G-SRT.2.ii: Apply the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides. HS.G-SRT.3: Use the properties of similarity transformations to establish the AA criterion for two triangles to be similar. #### 2.2: Prove theorems involving similarity HS.G-SRT.4: Prove similarity theorems about triangles. HS.G-SRT.5: Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. #### 2.3: Define trigonometric ratios and solve problems involving right triangles HS.G-SRT.6: Understand how the properties of similar right triangles allow the trigonometric ratios to be defined, and determine the sine, cosine, and tangent of an acute angle in a right triangle. HS.G-SRT.8: Use special right triangles (30°-60°-90° and 45°-45°-90°), trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. ### HS.G-C: Circles #### 3.1: Understand and apply theorems about circles HS.G-C.1: Understand and apply theorems about relationships with line segments and circles including radii, diameter, secants, tangents, and chords. HS.G-C.2.i: Understand and apply theorems about relationships with angles formed by radii, diameter, secants, tangents, and chords. HS.G-C.2.ii: Understand and apply properties of angles for a quadrilateral inscribed in a circle. HS.G-C.3: Construct the incenter and circumcenter of a triangle. Relate the incenter and circumcenter to the inscribed and circumscribed circles. #### 3.2: Find arc lengths and areas of sectors of circles HS.G-C.5: Explain and use the formulas for arc length and area of sectors of circles. ### HS.G-GPE: Expressing Geometric Properties with Equations #### 4.1: Understand and use conic sections HS.G-GPE.1.i: Derive the equation of a circle of given center and radius. HS.G-GPE.1.ii: Derive the equation of a parabola given a focus and directrix. HS.G-GPE.1.iii: Derive the equations of ellipses and hyperbolas given foci, using the fact that the sum or difference of distances from the foci is constant. HS.G-GPE.3.i: Identify key features of conic sections given their equations. HS.G-GPE.3.ii: Apply properties of conic sections in real world situations. #### 4.2: Use coordinates to verify simple geometric theorems algebraically HS.G-GPE.5.i: Develop and verify the slope criteria for parallel and perpendicular lines. HS.G-GPE.7: Use coordinates to compute perimeters of polygons and areas of triangles, parallelograms, trapezoids and kites. ### HS.G-GMD: Geometric Measurement and Dimension #### 5.1: Explain surface area and volume formulas and use them to solve problems HS.G-GMD.1: Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. 5.1.1.1: May use dissection arguments. Cavalieri’s Principle or informal limit arguments. 5.1.1.2: Cavalieri’s Principle: 2D: Suppose two regions in a plane are included between two parallel lines in that plane. If every line parallel to these two lines intersects both regions in line segments of equal length, then the two regions have equal areas. HS.G-GMD.2: Calculate the surface area for prisms, cylinders, pyramids, cones, and spheres to solve problems. HS.G-GMD.3: Know and apply volume formulas for prisms, cylinders, pyramids, cones, and spheres to solve problems. Correlation last revised: 9/22/2020 This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information.
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https://physics.stackexchange.com/questions/485603/what-is-the-reason-for-moon-and-satellite-in-free-fall-condition/485713
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# What is the reason for Moon and satellite in free fall condition? [duplicate] I am a high school student. I read in my book that moon and satellite are in free fall condition with Earth. I asked my teachers about this, but I'm not able to understand. Can anyone explain to me why this happens? Edit- if something is orbiting around the planet is he also in a free fall condition,. • Place a satellite high up and just leave it there. It falls and crashes. • Repeat, but this time give it a small push sideways. It still falls and crashes. But this time it didn't fall straight down. Because, it had some sideways speed that moved it sideways. It crashes slightly more to the side. • Repeat, but push much harder this time. It still falls and crashes, but it crashes maybe several hundred kilometers away from the point it is hovering above. This is because you gave it a large sideways speed; while falling it still moved sideways as well. • Now, finally, repeat, but push much, much, harder! As always, it still falls. But this time it misses Earth. It does fall towards Earth, but the sideways speed is so large, that before it lands, it has moved sideways away from Earth. It falls but misses. When it falls but misses, it will fall past Earth. Now it is on the other side, moving away from Earth. Gravity still pulls in it, so it slows down until it stops and starts falling back towards Earth, this time from the other side. The same thing happens: it falls but misses. And everything repeats itself. This will repeat itself forever. The path it takes in this way is an ellipse. Give it a bit more sideways speed to start with, and the ellipse becomes slightly wider. With some specific sideways speed, the elliptic path is just wide enough to be just as wide as it is long - the path is now circular (which is just a "special-case" of an ellipse). With an even larger sideways speed, the path will be wider than it is long, and we have an ellipse again, just a "fat" one instead of a "thin" one. This is how orbits work for any celestial object, including our own satellites, moons, suns, stars and planets. If something orbits in a circular path, there is nothing special about it; it just happened to have the fitting initial sideways speed. Free fall means the only force acting on the object is gravity. For objects orbiting the Earth this is the case, so we say they are in free fall. Indeed, you can think of objects that are orbiting the Earth as objects that are falling but never hit the ground. objects in orbit are falling, but they are moving forward fast enough so they never meet the Earth, they just keep falling around it. Picture throwing a ball, it is going forward, and it is falling, so it moves in a curve. If it were thrown in space where there is no air to slow it, and it were thrown at the right speed and the right height above Earth it would curve around Earth forever. A body is said to be in free fall when only gravitational force acts on it.So,Moon and satellites are also in free fall. if something is orbiting around the planet is he also in a free fall condition. Yes because gravitational force acts on it Any object moving in a circle is falling towards circle's center at every moment. A centripetal force is always acting on the object towards the center of circle. Because of this force, object velocity is keep on changing. Centripetal force always try to pull object towards center. In case of moon or satellite, speed is always constant but velocity keeps on changing. Because object is being pulled towards earth at every moment, it's said to be falling towards earth. • But the orbit is not perfectly circular. Nov 1, 2019 at 6:55 An object in orbit has two vectors acting on it simultaneously. The first is the gravitational vector wanting to pull the object directly to the center of the earth. The second vector acts on the object at 90 degrees to the first and we call this "angular momentum". When the second vector is set to zero we call this "free fall". When the second vector has a velocity greater than zero the object will still be in free fall but it will be displaced down range a little bit before returning to earth. The faster it goes the farther down range. Finally it reaches a point where it never does return to earth. Now it is circling the earth in a state of free fall. The gravitational vector keeps pushing it towards earth but the earth keeps curving away. An even faster velocity will take the object completely out of orbit and escape velocity will be achieved. Set a course for Rigel three Zulu. • I think your answer is rather unclear. What do you mean by angular momentum is a vector that acts on the object? Your answer could be severely improved if you mentioned explicitly energy considerations instead of just talking about "velocities". Jun 14, 2019 at 22:15 • It is assumed that the energy has already been applied per Newton's second law. The object is now drifting freely as per Newton's first law. The only thing that is different is that a large gravitational mass point has been introduced into the consideration. Now Newtons linear momentum is bent around the mass point converting it to angular momentum. It is now the objects velocity that prevents it crashing to earth, staying in orbit, or escaping to space. Jun 15, 2019 at 2:51
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# Vectors and 3D Coordinate Geometry 1. Oct 1, 2008 ### kehler 1. The problem statement, all variables and given/known data Suppose that II c R3 is a plane, and that P is a point not on II. Assume that Q is a point in II whose distance to P is minimal; in other words, the distance from P to Q is less than or equal to the distance from P to any other point in II. Show that the vector PQ is orthogonal to II. Hint given: Define a differentiable vector function r(t) with r(t) a subset of II, and r(0) = Q. Let p be the vector with components given by the coordinates of P. Let f(t) = |r(t)-p|2 = (r(t)-p) . (r(t)-p) df(t)/dt |t=0 ? 3. The attempt at a solution Not quite sure how to do it I let r(t) = ro + vt, where v is any vector in II and r(0) = Q So r(t) = Q + vt Then, f(t) = (Q + vt -p) . (Q + vt -p) df(t)/dt = (Q + vt - p) . (v) + (Q + vt - p) . (v) (wasn't quite sure how to differentiate dot products but i used the product rule) So, df(t)/dt |t=0 = 2(Q-p) . (v) I don't know where to go from here. I guess I must show that equation is zero to prove that it's ortogonal but I don't know how to.... Is this method even right? Any help would be much appreciated :) Last edited: Oct 1, 2008 2. Oct 1, 2008 ### Dick You should make it clear the Q=r(0) is the closest point in the plane to P. That makes r(t).r(t) a minimum at t=0. What's the derivative of a function at a minimum? 3. Oct 1, 2008 ### tiny-tim Hi kehler! I think you're missing the point … r(t) is a vector in II. In other words, r defines a perfectly general curve in II, with a parameter t … r(t) is the vector (from the origin) to the point on the curve with parameter t. Then you get a relationship between r'(t) and r(t). (and you're right, you can use the product rule on a dot-product!) 4. Oct 1, 2008 ### kehler I still don't really get it, tiny-tim :S. So did I do it correctly? Does this mean r(t) is not ro + vt because this is the equation of a line not a curve? r'(t) is the slope of r(t) isn't it? Thd derivative should be zero but I don't really know how to get it to be zero :S Thanks guys for your help :). Sorry I'm a bit slow to get stuff :( 5. Oct 1, 2008 ### tiny-tim Yes. It's a line, and therefore it's not in II. r(t) is in II. Hold it! r(t) is a vector. So r'(t) is a vector also (not a number, such as a slope). And in fact r'(t) is the vector which is … ? 6. Oct 1, 2008 ### kehler How then do I get an equation of a curve, tiny-tim? :S r(t) = At2 + Bt + C? Would it be the vector tangential to the point on the curve then? 7. Oct 1, 2008 ### tiny-tim Hi kehler! No, r(t) is the equation of the curve!! r is an "unknown curve", just as x can be an "unknown number". You don't need to know what r(t) is!! Yes. r'(t) is tangential to the (unknown, general) curve at that point. Now combine that with the differential of that dot-product. 8. Oct 1, 2008 ### kehler Hmm ok. So df(t)/dt = (r(t) - p) . (r'(t)) + (r(t) - p) . (r'(t)) = 2 (r(t) - p) . (r'(t)) So, df(t)/dt |t=0 = 2(Q-p) .(r'(0)) I guess this should be zero..... But I don't know how to show it without already knowing that they are ortogonal :S. 9. Oct 1, 2008 ### tiny-tim Are you forgetting how this started? When the distance is a minimum, df(t)/dt = … ? (going to sleep now … :zzz:) 10. Oct 1, 2008 ### kehler Hmm ok. I'll think about it some more and come back if I still don't get it. Thanks tiny-tim for your patience! Goodnight :D 11. Oct 1, 2008 ### HallsofIvy Staff Emeritus A very simple way to do this is purely geometric. Suppose PQ is NOT perpendicular to the plane. Draw the perpendicular from P to the plane and call the point at which the perependicular crosses the plane R. Then PQR is a right triangle with right angle at R. PR is a leg of that right triangle and PQ is the hypotenuse. By the Pythagorean theorem, $|PQ|= \sqrt{|PR|^2+ |QR|^2}< |PR|$ contradicting the fact that |PQ| is minimal. (| | here indicates the length of the line.) 12. Oct 2, 2008 ### kehler Oh wow, that method's a lot simpler :). Thanks HallsOfIvy. Tiny-tim, is my solution correct now: Let r(t) be a differentiable vector function r(t) with r(t) a subset of II, and r(0) = Q. Let p be the vector with components given by the coordinates of P. Let f(t) = |r(t)-p|2 = (r(t)-p) . (r(t)-p) So, f'(t) = (r(t)-p).r'(t) + (r(t)-p).r'(t) = 2(r(t)-p).r'(t) Since the distance from P to Q is less than or equal to the distance from P to any other point in II, f'(t) = 0 when r(t) = Q. This occurs when t=0 by definition So f'(0) = 0 But f'(0) = 2(r(0)-p).r'(0) = 2(Q-p).r'(0) Therefore, 2(Q-p).r'(0) = 2(PQ).r'(0) = 0 Since r'(0) lies in II, PQ is orthogonal to II 13. Oct 2, 2008 ### tiny-tim Hi kehler! Yes, that's fine, except for the last line. All you've proved is that PQ is perpendicular to the tangent of that particular curve, r(t). So you should add "but this applies for any curve through Q, and so PQ is perpendicular to the tangent of every curve in II through Q, ad so is perpendicular to II." btw, HallsofIvy's proof works fine in this case, where II is a plane, but the proof above will work for any surface, to prove that pQ is perpendicular to the tangent plane at Q. 14. Oct 2, 2008 ### kehler Thanks tiny-tim! :D
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Home | Math | Difference between Set and Subset # Difference between Set and Subset June 14, 2023 written by Azhar Ejaz The key difference between a set and a subset is that a set is a collection of distinct elements, while a subset is a subset of a larger set and contains elements that are also present in the larger set Set Vs Subset Difference between Set and Subset Set: • A collection of distinct elements. • Can contain any number of elements. • Examples: {1, 2, 3}, {a, b, c, d}, {red, blue, green}. Subset: • A subset is a subset of a larger set. • Contains elements that are also present in the larger set. • Can have fewer or the same number of elements as the larger set. • Examples: {1, 2} is a subset of {1, 2, 3}, {red, green} is a subset of {red, blue, green}. File Under:
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# Thread: Show that n^5 + n^4 + 1 is not prime for n>1 1. ## Show that n^5 + n^4 + 1 is not prime for n>1 No idea at all! 2. Originally Posted by fardeen_gen No idea at all! n^5+n^4+1=(x²+x+1)(x^3-x+1) But I cheated and asked it to Maxima. 3. Hello, fardeen_gen! Please don't hide the problem in the heading . . . Show that n^5 + n^4 + 1 is not prime for n > 1. Brilliant move, moo! I found a way to factor it. .(It helped that I already knew the factors!) Let: .P .= .n^5 + n^4 + 1 Multiply by (n - 1): . . (n - 1)P . = . (n - 1)(n^5 + n^4 + 1) - - . . . . . . .= . n^6 - n^4 + n - 1 - - . . . . . . .= . (n^6 - 1) - (n^4 - n) - - . . . . . . .= . (n³ - 1)(n³ + 1) - n(n³ - 1) - - . . . . . . .= . (n³ - 1)(n³ - n + 1) . . (n - 1)P . = . (n - 1)(n² + n + 1)(n³ - n + 1) Divide by (n - 1): . P . = . (n² + n + 1)(n³ - n + 1) 4. Thank you. Is expressing the given expression in factors sufficient to prove that it is not a prime? 5. Originally Posted by fardeen_gen Thank you. Is expressing the given expression in factors sufficient to prove that it is not a prime? Not really, because it still can be 1 multiplied by a prime number. But you can see that n²+n+1 > 1 for any n>1. For any n>1, n^3>n, and than n^3-n>0 --> n^3-n+1>1. So both factors are strictly > 1, which proves that it's a product of 2 numbers > 1, thus not prime. 6. You can note that $x^5+x^4+1$ can be factored using primitive root of unities. Let $\zeta$ be a primitive third root of unity then $\zeta^5+\zeta^4+1 = \zeta^2 + \zeta + 1 = 0$. Now the minimal polynomial for $\zeta$ is $\Phi_3(x)$ which is $x^2+x+1$. Thus, $x^2+x+1$ a factor of $x^5+x^4+1$. 7. ## Just for those, who got here from google 2nd solution: We have n5 + n4 + 1 = n5 + n4 + n3 - n3 - n2 - n +n2 + n + 1 = n3(n2 + n + 1) - n(n2 + n + 1) + (n2 + n + 1) = (n2 + n + 1)(n3 - n + 1) , , , , , , ### factorize n^5 n^4 1 Click on a term to search for related topics.
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posted by . brian cut an extra large round pizza into 12 pieces. 7 of the pieces were eaten. what angle measure of pizza is left? Do you know how many degrees are in a circle before it it cut? 360 If you divide that by 12, how many degrees in one of the 12 pieces? 30 So, each of the 12 pieces has 30 degrees. Now, if 7 pieces were eaten, how many pieces were left? so it would be 150? If each piece has 30 degrees, then how many degrees does 5 pieces have? 150? You are right! thank you very much! My pleasure! Thx soo much I hate school I'm in 4th grade and this helped me thanks :3 Thanks guys! It helped me alot. 150 ## Similar Questions 1. ### math the pizza was sliced into 6 equal pieces. Matrin ate 2 pieces. What fraction of the pizza did he eat. what percent did he eat? 2. ### math Whats the difference between an arc [of a circle]'s length and measure? 3. ### Math At Shannons birthday Kira ate 3 pieces, Brazdley ate 2,Shauna ate 4, and Sammi ate 3 pieces of pizza. Now 1/4 of the pizza is left. How many pieces was the pizza originally cut into? 4. ### math Brian cut an extra large round pizza into 12 pieces. Seven of the pieces were eaten. What angle measure of pizza is left? 5. ### math brain cut an extra large round pizza into 12 pieceas .Seven of then were eaten. WQhay angle measure of pizza is left 6. ### marh Martina and Charlotte are sharing a pizza. The pizza is cut into eight pieces. Martina ate a quarter of the pizza. Charlotte ate three pieces. How many pieces are left? 7. ### math Brian cut an extra large round pizza into 8 pieces. Seven of th pieces were eaten. What angle measures of pizza is left? 8. ### math My three friends and i ordered 2 pizzas. Each pizza was cut into 8 pieces.Each of us ate 3 pieces.What fractions of a pizza did each of us eat? 9. ### Math Mary had 3/4 of a pizza left over. She cut it into pieces, each of which is 3/8 of the whole pizza. How many pieces did Mary cut? 10. ### Math Nick pulled pizza out of the oven and cut it into four equivalent pieces he ate two of them.His sister came in and cut one of the remaining pieces into three smaller equivalent pieces she ate two of the small pieces .what fraction … More Similar Questions
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https://mathematica.stackexchange.com/questions/13226/how-can-i-get-exactly-5-logarithmic-divisions-of-an-interval?noredirect=1
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# How can I get exactly 5 logarithmic divisions of an interval? I'd like to get exactly 5 divisions from x to y on a log scale. Can FindDivisions do this? EDIT: After 3 years, it has been discovered that this oft-linked-to answer doesn't truly space logarithmically. It's close, which was all I was going for at the time (and handles zeros), but it's not quite right. Anyway, thanks to @Pickett's careful moderation, here's a better version... logspace[increments_, start_?Positive, end_?Positive] := Exp@Range[Log@start, Log@end, Log[end/start]/increments] This one, by the stodgy nature of Logs, won't handle non-positive numbers, so I'll leave the old answer. Sorry to all that lost millions in the stock market using the old function. :) OLD FUNCTION I built a function that calculates log spaced increments for a job at work. I've added a catch where it will handle log spacing from 0 to a number. logspace [increments_, start_, end_] := Module[{a}, ( a = Range[0, increments]; Exp[a/increments*Log[(end - start) + 1]] - 1 + start )] To try it out: N@logspace[5,1,1000] (*{1., 3.98107, 15.8489, 63.0957, 251.189, 1000.}*) To view it on a number line: a = N@logspace[10, 0, 10]; Graphics[Point@Transpose[{a, ConstantArray[.5, 11]}], Axes -> {True, False}, And if you want to find the distances between divisions, use Differences: Differences[a] (*{0.270982, 0.344413, 0.437742, 0.556362, 0.707126, 0.898744, 1.14229, 1.45183, 1.84524, 2.34527}*) • A problem (I think?) with this function was discovered, please see the comments here. Sep 14, 2015 at 21:53 • @Pickett Oh dear. I wrote that 3 years ago (about the amount of time I've used MMA).... eek. Yeah, it log-ish spaces, but isn't a "true" logspace. Let me work on that... – kale Sep 15, 2015 at 0:13 • Thanks for the quick response! :) Sep 15, 2015 at 8:41 • So how to get the logspace from -1 to 2? – yode Mar 7, 2016 at 8:23 Writing one wouldn't be that hard. You can convert it to Log10 and then let FindDivisions do all the work in log space before converting it back. For example: findLogDivisions[{xmin_, xmax_}, n_Integer] := 10^FindDivisions[Log10@{xmin, xmax}, n] Then, to find 4 "nice" divisions in log space between 1 and 1000, you simply need to do: findLogDivisions[{1, 1000}, 5] (* {1, 10, 100, 1000} *) Here's a mathematically simple approach, assuming that exactly n divisions are sought, no matter how nice or not. This produces exactly n intervals: Clear[logDiv]; logDiv[{x_?Positive, y_?Positive}, n_Integer /; n > 0] := x (y/x)^Range[0, 1, 1/n] logDiv[{10, 10000}, 3] (* {10, 100, 1000, 10000} *) If you want n "fence posts", use Clear[logDiv]; logDiv[{x_?Positive, y_?Positive}, n_Integer /; n > 1] := x (y/x)^Range[0, 1, 1/(n-1)] If you want n interior division points, change n to n+1 in the first version. Comparisons. Kuba's method produces the same output as logDiv mutatis mutandis​*. Below I use the first version of logDiv and omit Kuba's output. Comparison 1: N @ logDiv[{1/10, 100000}, 3] (* same output as Kuba *) N @ logspace[3, 1/10, 100000] (* kale *) N @ findLogDivisions[{1/10, 100000}, 3] (* rm -rf *) (* {0.1, 10., 1000., 100000.} {0.1, 45.516, 2153.55, 100000.} {0.01, 1., 100., 10000., 1.*10^6} *) Comparison 2: N @ logDiv[{10, 100000}, 5] N @ logspace[5, 10, 100000] N @ findLogDivisions[{10, 100000}, 5] (* {10., 63.0957, 398.107, 2511.89, 15848.9, 100000.} {10., 18.9998, 108.996, 1008.95, 10008.3, 100000.} {10., 100., 1000., 10000., 100000.} *) Note the outputs vary; in particular logspace does something quite different than the others when start is different than 1. Depending on the application, one or the other might be desired. *In honor of the language survey. • +1, what do you think about editing the question. Now the answer RandomReal[{start, end}, 5] fits quite well. – Kuba Mar 18, 2014 at 13:10 • +1 to you, too. I'm assuming RandomReal is a joke :) However the question could more clearly state what exactly counts as a division. It seems to me the OP should decide, but it's not a big deal. I wish the OP had accepted or commented on the other answers, though. Mar 18, 2014 at 13:17 • Yes it is a joke but it fits :) Maybe it's better with it's vague form, more answers are valid. – Kuba Mar 18, 2014 at 13:20 I do a lot of work where I need to have function evaluations at equal logarithmic spacings. The code I use is GeometricRange[imin_, imax_, perRange[n_]] := GeometricRange[imin, imax, (imax/imin)^(1/(n - 1))]; GeometricRange[imin_, imax_, r_] := Exp[Range @@ Log[N[{imin, imax, r}]]]; perOctave[n_] := 2^(1/(n - 1)); GeometricRange has the same calling syntax as Range, where the increment (in this case, the ratio r) is given as an argument, but also has the option of specifying the total number of samples in the range with perRange or commonly used log resolutions with perDecade or perOctave. The calling syntax is In[10]:= GeometricRange[10, 1000, 10] Out[10]= {10., 100., 1000.} or In[9]:=GeometricRange[10, 1000, 10 // perRange] Out[9]= {10., 16.681, 27.8256, 46.4159, 77.4264, 129.155, 215.443, 359.381, 599.484, 1000.} or In[6]:= GeometricRange[10, 1000, 10 // perDecade] Out[6]= {10., 12.9155, 16.681, 21.5443, 27.8256, 35.9381, 46.4159, 59.9484, 77.4264, 100., 129.155, 166.81, 215.443, 278.256, 359.381, 464.159, 599.484, 774.264, 1000.} Here is the complete code for Version 10 and up, using PowerRange
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55457 55,457 (fifty-five thousand four hundred fifty-seven) is an odd five-digits prime number following 55456 and preceding 55458. In scientific notation, it is written as 5.5457 × 104. The sum of its digits is 26. It has a total of 1 prime factor and 2 positive divisors. There are 55,456 positive integers (up to 55457) that are relatively prime to 55457. Basic properties • Is Prime? Yes • Number parity Odd • Number length 5 • Sum of Digits 26 • Digital Root 8 Name Short name 55 thousand 457 fifty-five thousand four hundred fifty-seven Notation Scientific notation 5.5457 × 104 55.457 × 103 Prime Factorization of 55457 Prime Factorization 55457 Prime number Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 55457 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.9234 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0 The prime factorization of 55,457 is 55457. Since it has a total of 1 prime factor, 55,457 is a prime number. Divisors of 55457 2 divisors Even divisors 0 2 2 0 Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 55458 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 27729 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 235.493 Returns the nth root of the product of n divisors H(n) 1.99996 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors The number 55,457 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 55,457) is 55,458, the average is 27,729. Other Arithmetic Functions (n = 55457) 1 φ(n) n Euler Totient Carmichael Lambda Prime Pi φ(n) 55456 Total number of positive integers not greater than n that are coprime to n λ(n) 55456 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5630 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares There are 55,456 positive integers (less than 55,457) that are coprime with 55,457. And there are approximately 5,630 prime numbers less than or equal to 55,457. Divisibility of 55457 m n mod m 2 3 4 5 6 7 8 9 1 2 1 2 5 3 1 8 55,457 is not divisible by any number less than or equal to 9. • Arithmetic • Prime • Deficient • Polite • Prime Power • Square Free Base conversion (55457) Base System Value 2 Binary 1101100010100001 3 Ternary 2211001222 4 Quaternary 31202201 5 Quinary 3233312 6 Senary 1104425 8 Octal 154241 10 Decimal 55457 12 Duodecimal 28115 20 Vigesimal 6ich 36 Base36 16sh Basic calculations (n = 55457) Multiplication n×i n×2 110914 166371 221828 277285 Division ni n⁄2 27728.5 18485.7 13864.2 11091.4 Exponentiation ni n2 3075478849 170556830528993 9458570150646364801 524543924844395452769057 Nth Root i√n 2√n 235.493 38.1346 15.3458 8.88774 55457 as geometric shapes Circle Diameter 110914 348447 9.6619e+09 Sphere Volume 7.14427e+14 3.86476e+10 348447 Square Length = n Perimeter 221828 3.07548e+09 78428 Cube Length = n Surface area 1.84529e+10 1.70557e+14 96054.3 Equilateral Triangle Length = n Perimeter 166371 1.33172e+09 48027.2 Triangular Pyramid Length = n Surface area 5.32689e+09 2.01003e+13 45280.5 Cryptographic Hash Functions md5 7cef7f0406190a1213922c4a9243638a c46d7c645751b5249bf7951a57f4732de8f27dd4 34fc31a0bc7c37f79026e8a405aaee0a2078cfbd87a002812d9fcbaf660234ff e860d92fe370f62a4f81560ac0262eeb5d835b0a174913c13f297cf3023f176a4ce9306452a6fc3351702848ef5c3e63f7854e2076334a3a79308933a94988fa 42c3ba05b924b68362dd78cf08e1c561eb69d0cd
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# Quick Answer: What Is The Face Value Of 7 In 4709606? ## What is the face value of 2 in 624? The face value of 2 in 93207 is 2.. ## What is the value of the 9 in 89? Answer: THE PLACE VALUE OF 9 IN 89 IS 9 ONES. ## What is the place value and face value of 7? Thus, for the place value of a digit, the digit is multiplied by the place value of 1 it has to be that place. the place value of 7 is 7 × 100 = 700. ## What is the place value of 0 in 109? tensThe place value of 0 in 109 is tens. ## How do you write place value? Finding and Writing the Place ValueIn the number 2965: (i) There is the digit 2 in the number 2965. This 2 is the fourth digit for the right in the number. It is at thousand’s place. So, its place value is 2000. We write 2 and put three zeros to its right as 2000. … In the number 9721: (i) There is the digit 9 in the number 9721. ## What is the difference of the place value and face value of 2 in 829? In the number 829. the place value of 2 is in ten’s place. Therefore the place value of 2 is 20 while the face value of 2 is 2. ## What is the place value of 7 in 478? Answer. the place value of 7 is 70. ## What is the place value of 2 in the number 123 456? tenthouandAnswer. the place value of 2 is tenthouand. ## What is the value of 2 in 23? Let’s look at an example. What does the 2 in the number 23 represent? Student: It represents twenty. ## What is the face value of 2 in 93207? What is the face value of 2 in 93207? The face value of 2 in 93207 is 2. ## How do you find the face value of a number? Face value of digit = numerical value of the digit itself. The place value of digit 0 in a given number is always 0. The place value of digit 0 is 0. Example: The place value of digit 8 in 5,831 = 8 × 100 = 800. ## What is the place value of 8 in 1278? The place value of 8 in 1278 is 8. ## What does at face value mean? 1 : for the price that is printed on something We bought the tickets at face value. 2 : as true or genuine without being questioned or doubted After all his lying, nothing he says now should be taken/accepted at face value. ## What is the face value of 7? 7 is in ones place, and its place value is 7. Place value and face value are not the same. The face value of a number is the value of the digit or numeral itself. For instance, the face value of 2 in 12783 is 2.
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# Chapter 3: Data ## Abstractions • Bits are grouped to represent abstractions. • These abstractions include but are not limited to numbers, characters, and colors. • Abstractions find common features to generalize the program. ## Analog vs. Digital Data • An analog signal has values that change smoothly over time, rather than in discrete intervals. • Analog signals are continuous signals, while digital signals are discrete time signals. • A digital signal is an analog signal that has been broken up into steps. ## Consequences of Using Bits to Represent Data • A variable is an abstraction inside a program that can hold a value. • Each variable has associated data storage that represents one value at a time. • However, value can be a list or other collection that, in turn, contains multiple values. • Some data types include integers, real numbers, Boolean, string, and list. ## Number Systems • Number bases, including binary, decimal, and hexadecimal, are used to represent and investigate digital data. DECIMAL BINARY 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001 10 1010 # Converting Numbers into Different Bases ### 11011BIN = ?DEC STEPS: Step 1. A five-column table is needed because 11011 has five digits. Start byputting a 1 into the upper-right box of the five-column table. Step 2. Fill in the remaining first row by continually multiplying by the base. Because the original number is in binary, fill the columns by continually multiplying the product by 2. Step 3. Place the numbers to be converted into the second row. Step 4. Add the result of multiplying row 1 by row 2. ### Convert a decimal (DEC) number to a binary (BIN) number. 30DEC = ?BIN Step 1. Create a flexible table with enough columns until the number in the upper row is just bigger than the number you are converting. Step 2. Start with the largest number that is still smaller than the target number. Subtract the number in the upper row of the table from the original number. Step 3. 14 − 8 = 6 Step 4. 6 − 4 = 2 Step 5. 2 − 2 = 0 Step 6: 0 # Various Errors ## Overflow Errors • An overflow error occurs when the result of a computation is too large for the available storage space. • This results in data loss, as some information gets cut off due to lack of memory. • Overflow errors can occur in almost any programming language and can be very difficult to debug. ## Roundoff Errors • A roundoff error occurs when decimals (real numbers) are rounded. • One computer might calculate 1/3 as 0.333333. Another computer might calculate ⅓ as 0.3333333333. • In this case, 1/3 on one computer is not equal to 1/3 on a second computer. ## Lossy and Lossless Data Compression • Data compression is reducing the size (number of bits) of transmitted or stored data. • Digital data compression often involves trade-offs in quality versus storage requirements. • Lossy compression can significantly reduce the file size while decreasing resolution. • Traditionally, lossy compression is used to reduce file size for storage and transmission (email). • Lossless data compression, no data are lost. • After compression, the original file can be reproduced without any lost data. ## Information Extracted From Data • People can use computer programs to process information as well as to gain insight and knowledge. • Information is the collection of facts and patterns extracted from data. • Depending on how the data were collected, the information may not be uniform. • For example, if users entered data into an open field, the way they chose to abbreviate, spell, or capitalize something may vary from user to user. • Cleaning data is a process that makes the data uniform without changing their meaning. ## Predicting Algorithms • Predicting algorithms use information collected from big data to influence our daily lives. • For example: • A credit card company can use purchasing patterns to identify when to extend credit or flag a purchase for possible fraud. • Social media sites can use patterns to target advertising based on viewing habits. ## Visualization of Data • Using appropriate visualizations when presenting digitally processed data can help one gain insight and knowledge. • Although big data is a powerful tool, the data will lose their value if they cannot be presented in a way that can be interpreted. • Visualization tools can communicate information about data. • Column charts, line graphs, pie charts, bar charts, XY charts, radar charts, histograms, and waterfall charts can make complex data easier to interpret. ## Privacy Concerns • Privacy concerns arise through the mass collection of data. • The content of the data may contain personal information and can affect the choice in storage and transmitting. • Geolocation, when used within a program, helps you find the approximate geographic location of an IP address along with some other useful information, including ISP, time zone, area code, state, and so on.
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