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## Calculus 8th Edition
Published by Cengage
# Chapter 1 - Functions and Limits - 1.1 Four Ways to Represent a Function - 1.1 Exercises - Page 22: 59
#### Answer
$A=\frac{\sqrt{3}}{4}x^2$, $x>0$
#### Work Step by Step
We have an equilateral triangle with sides $x$ and we need to find the area. We know that: $A=1/2*base*height=\frac{1}{2}xh$ We need to eliminate $h$. We construct a right triangle in the middle of the equilateral triangle with sides $h$, $x$, and $1/2x$. We use the Pythagorean Theorem with these three sides: $(\frac{1}{2}x)^2+h^2=x^2$ $h^2=x^2-(\frac{1}{2}x)^2$ $h=\pm\sqrt{x^2-(\frac{1}{2}x)^2}$ $h=+\sqrt{\frac{3}{4}x^2}=\frac{\sqrt{3}}{2}x$ (We eliminate the negative because lengths must be positive.) We plug in $h$ in the area formula: $A=\frac{1}{2}x\frac{\sqrt{3}}{2}x=\frac{\sqrt{3}}{4}x^2$ The domain is $x>0$ because lengths must be positive.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Financial Mathematics
1. Nov 9, 2005
### playboy
How do you find the annual effective rate of interest?
The question reads: You lend a freind $15 000 to be amortized by semiannual payments for 8 years, with interest at j2 = 9%. You deposit each payment in an account paying J12 = 7%. What annual effective rate of interest have you earned over the entire 8-year period? Ans = 8.17% Hmmm... i have absolutly no idea how to get the annuale effective rate of interest. My TA showed, (in another question) that its something like (1 + i)^n = 1 + r and solve for r? Please help somebody Thanks 2. Nov 10, 2005 ### hotvette Conceptually, it works like this. There is initial outlay of$15,000. The payments that come in annually are immediately invested. At the end of 8 years there is a total value of all investments. The 'effective' interest rate is the equivalent rate at which the initial outlay would compound at to achieve the same final result after 8 years. It might help to draw out a time line and treat each pmt and ensuing investment as a separate problem. Find out how much each is worth after the 8 years is up, sum the totals together, and then it's a straightforward back solution for a std compound interest problem.
By the way, you have 2 identical posts. If this was intentional, pls avoid that in the future.
P.S. One of the most useful classes (in terms of constantly using the material learned) I took in graduate school was called "Engineering Economy".
Last edited: Nov 10, 2005
3. Nov 10, 2005
### playboy
No, that was not intentional, i didn't know i did that :S... I will avoid that in the future!
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https://quizanswered.com/module-8-ac-amplifiers-based-on-jfets-multisim/
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Module 8: ac amplifiers based on jfets (multisim)
MULTISIM
Introduction
This experiment explores the uses of JFET as AC signal amplifiers. It also describes techniques for measuring the input and output impedance of circuits.
Remember that your lab report will need to include your measurements, calculations, screenshots, etc. as indicated at the end of this outline.
Procedure
1. Common Source Amplifier
1.1 Build the common source amplifier shown in Figure 8.1
Figure 8. 1: Common source amplifier
1.2 Using the oscilloscope, measure the voltage gain of the amplifier defined as Av = Vout/Vin.
2. Measurement of Input impedance
In this section we will learn a very useful technique to measure the input impedance of any circuit. This technique is based on placing a known resistor in series with the input of the circuit and measuring the voltage drop across the new resistor. This technique can also be used in live circuits and not just
simulations.
Figure 8. 2: Circuit to measure input impedance
2.1 Build the circuit shown in Figure 8.2 You will notice that this is the same circuit used in Figure 8.2 with the extra R4 resistor added to the input.
2.2 Measure with the oscilloscope the voltage at node V1 and the voltage at node V4.
2.3 Calculate input impedance as follows:
3. Measurement of output impedance
The following technique can be used to measure the output impedance of a circuit. In practical circuits, the best value of the load resistor must be selected by trial and error.
3.1 Measure Vout as shown in the circuit from Figure 8.1. We will name this voltage Vout
3.2 Connect a load resistor of 1 kΩ at the output of the same circuit. Measure the voltage across the load. We will call this voltage Vload
3.3 Calculate output impedance as:
(in the case of using a different value for the load resistor, change the 10 kΩ value in the equation to the appropriate value of resistor used).
Laboratory Report
Create a laboratory report using Word or another word processing software that contains at least these elements: – Introduction: what is the purpose of this laboratory experiment? – Results for each section : Measured and calculated values, calculations, etc. following the outline. Include screenshots for the circuits and waveforms as necessary — You can press Alt + Print_Screen inside Multisim or if using Windows 7, you can use the “Snipping tool”. Either way, you can paste these figures into your Word processor.
– Conclusion : What area(s) you had difficulties with in the lab; what did you lean in this experiment; how it applies to your coursework and any other comments.
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https://workforce.libretexts.org/Bookshelves/Electronics_Technology/Book%3A_Trigonometry_and_Single_Phase_AC_Generation_for_Electricians_(Flinn)/03%3A_AC_Generation/03.5%3A_Frequency_and_Alternators
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# 3.5: Frequency and Alternators
In the last chapter, we learned the term cycle means from the point in a waveform to where the waveform starts to repeat itself. When we discuss the term frequency, we are referring to how many cycles can occur in one second. Frequency is measured in hertz (shout out to Heinrich Hertz) or CPS (cycles per second). Two factors affect the frequency in an alternator: rotation speed and the number of poles.
Figure 52. Sine wave cycle
## Rotation speed
As the armature rotates through the field, it starts to create a waveform (as we saw in the last chapter). One full mechanical rotation of the armature creates one full sine wave on a two-pole alternator. If the two-pole alternator spins three complete revolutions in one second, it will create three full sine waves in that one second. We would say that the frequency is at three cycles per second or three hertz (as the cool kids say).
A machine’s rotational speed is measured in rotations per minute or RPM. However, we are not concerned with minutes, but rather, with seconds when dealing with frequency. Therefore, RPM must be converted to rotations per second (RPS). As there are 60 seconds in a minute, all we have to do is to divide the RPM by 60 to convert it to RPS.
For example, if the armature is spinning at a rate of 1800 RPM on a two-pole alternator, we can say that it is spinning at 30 rotations per second. If this alternator has two poles, then in one second it will generate 30 cycles of voltage. It then could be said to have a frequency of 30 cycles per second or 30 Hertz. The frequency of an alternator is directly proportional to the rotational speed of the alternator.
## Number of Poles
If we add poles to the alternator, we can change the frequency. In a two-pole alternator, Side A of the armature (Figure 53) passes from north to south, and then south to north, to create one complete sine wave. I f we add two more poles, as in Figure 54, then Side A of the armature will move past two north poles and two south poles in one full mechanical revolution.
Figure 53. Two pole alternator
Two full sine waves are created in one complete mechanical revolution. If a two-pole alternator creates one cycle of voltage in one second (or one hertz of frequency), a four pole alternator will create two cycles of voltage in one second (or two hertz).
The frequency of an alternator is directly proportional to the number of poles in the alternator.
Figure 54. Four pole alternator
## Formula time!
Knowing that rotation speed is directly proportional to frequency and that the number of poles is directly proportional to frequency, we can use a formula. The formula looks like this:
$f= \dfrac{P}{2} \times \dfrac{N}{60} \tag{Frequency formula}$
where…
• $$f$$ = frequency in hertz
• $$P$$ = number of poles
• $$N$$ = rotational speed in RPM
We divide the number of poles by two because there will always be a set of two poles. You can’t have a north pole without a south. We divide the RPM by 60 because we are concerned with rotations per second, not rotations per minute. The formula in Figure 56 can be combined to look like this:
$f = \dfrac{PN}{120} \tag{Combined frequency formula}$
Video! This video will walk you through how frequency is related to the RPM and the number of poles of an alternator.
A YouTube element has been excluded from this version of the text. You can view it online here: https://pressbooks.bccampus.ca/trigf...ricians/?p=278
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https://www.physicsforums.com/threads/inequalities-math-homework-problem.190125/
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# Inequalities math homework problem
1. Oct 9, 2007
### xCanx
A rectangular solid is to be constructed with a special kind of wire along all
the edges. The length of the base is to be twice the width of the base. The
height of the rectangular solid is such that the total amount of wire used (for
the whole figure) is 40 cm. Find the range of possible values for the width of
the base so that the volume of the figure will lie between 2 cm3 and 4 cm3.
Can someone show me how to start off?
2. Oct 9, 2007
### symbolipoint
w=width, L=length, h=height;
w+L+h=40, L=2w
wLh>2 and wLh<4
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https://17calculus.com/integrals/volume/washer-disc/
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## 17Calculus Integrals - Volume of Revolution Using The Washer-Disc Method
##### 17Calculus
This page covers single volume integrals when an area is rotated about a vertical or horizontal line. The area is defined by equations in the form $$y=f(x)$$ or $$x=f(y)$$ and we use the washer (disc) method. For other ways to calculate volume, see the links in the related topics panel.
If you want a full video lecture on this topic, we recommend this video and this instructor.
### Prof Leonard - Volume of Solids By Disks and Washers Method [2hr-47mins-48secs]
video by Prof Leonard
Alternate Names Disc Method Disk Method Ring Method
We choose to use the term washer-disc method to refer to this technique. We think it covers the two most commonly used and most descriptive names. This, of course, is a personal preference for this site and you need to check with your instructor to see what they require.
What Is A Washer?
If you are not familiar with a washer (other than to wash clothes), this wiki page has pictures and explains what a washer is. In short, it is a disc with a circular hole in it whose center is the same as the full disc.
Overview
When calculating the volume of rotation, there are 3 factors that determine how to set up the integral.
1. method (washer-disc or cylinder-shell)
2. axis of rotation
3. function (graph and form of the equations)
On this page, we discuss the washer-disc method where the axis of rotation will always be either an axis or a straight line that is parallel to one of the axes. However, before we discuss the rotation of an area, we need to know how to describe an area in the plane. This is a critical step to setting up your integral correctly. If you didn't completely understand this from the main volume integrals page, you can go over it again here.
### Describing A Region In The xy-Plane
To describe an area in the xy-plane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot (used to plot these graphs; we used gimp to add labels and other graphics). However, graphing by hand is usually the best and quickest way.
We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by $$f(x)$$ (red line), $$g(x)$$ (blue line) and $$x=a$$ (black line).
[Remember that an equation like $$x=a$$ can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]
Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).
Vertically
Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.
Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing $$g(x)$$, no matter where we draw it. Similarly, the arrow always exits the area by crossing $$f(x)$$, no matter where we draw it. Do you see that?
But wait, how far to the right does it go? We are not given that information. What we need to do is find the x-value where the functions $$f(x)$$ and $$g(x)$$ intersect. You should be able to do that. We will call that point $$(b,f(b))$$. Also, we will call the left boundary $$x=a$$. So now we have everything we need to describe this area. We give the final results below.
Vertical Arrow
$$g(x) \leq y \leq f(x)$$
arrow leaves through $$f(x)$$ and enters through $$g(x)$$
$$a \leq x \leq b$$
arrow sweeps from left ($$x=a$$) to right ($$x=b$$)
Horizontally
We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of $$f(x)$$ and $$g(x)$$ in terms of $$y$$. ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations $$f(x) \to F(y)$$ and $$g(x) \to G(y)$$.
Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line $$x=a$$. However, there is something strange going on at the point $$(b,f(b))$$. Notice that when the arrow is below $$f(b)$$, the arrow exits through $$g(x)$$ but when the arrow is above $$f(b)$$, the arrow exits through $$f(x)$$. This is a problem. To overcome this, we need to break the area into two parts at $$f(b)$$.
Lower Section - - This section is described by the arrow leaving through $$g(x)$$. So the arrow sweeps from $$g(a)$$ to $$g(b)$$.
Upper Section - - This section is described by the arrow leaving through $$f(x)$$. The arrow sweeps from $$f(b)$$ to $$f(a)$$.
The total area is the combination of these two areas. The results are summarized below.
Horizontal Arrow
lower section
$$a \leq x \leq G(y)$$
arrow leaves through $$G(y)$$ and enters through $$x=a$$
$$g(a) \leq y \leq g(b)$$
arrow sweeps from bottom ($$y=g(a)$$) to top ($$y=g(b)$$)
upper section
$$a \leq x \leq F(y)$$
arrow leaves through $$F(y)$$ and enters through $$x=a$$
$$f(b) \leq y \leq f(a)$$
arrow sweeps from bottom ($$y=f(b)$$) to top ($$y=f(a)$$)
Type 1 and Type 2 Regions
Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.
Here is a quick video clip going into more detail on Type 1 and Type 2 regions.
### Krista King Math - type I and type 2 regions [1min-39secs]
video by Krista King Math
washer-disc method
x-axis rotation
y-axis rotation
Now we will discuss each of these plots separately and explain each part of the plots.
Getting Started
Here are some key things that you need to do and know to get started.
1. Draw a rough plot of the area that is being rotated. This is usually best done by hand since you will need to label it.
2. Decide what method you will use, washer-disc or cylinder-shell.
3. On the rough plot from point 1, label the axis of rotation and draw a representative rectangle somewhere in the area.
4. Label R and r.
Once those steps are done, you are ready to set up your integral.
The volume integral using the washer-disc method is based on the volume of a washer or disc. Let's think a bit about the volume of a washer-disc. If we start with a full disc (no hole in the middle), the volume is the surface area times the thickness. Since the disc is a circle, the area of a circle is $$\pi R^2$$ where $$R$$ is the radius of the circle. The volume is $$\pi R^2 t$$ where $$t$$ is the thickness. We choose to use a capital R here as the radius of the disc.
Now, with a washer, we take the disc we just discussed and put a circular hole in it with it's center the same as the full disc. (Think of a CD or DVD disc.) Now the volume is reduced by what we have taken out of the center. This empty space has volume $$\pi r^2 t$$, where $$r$$ is the radius of the small hole. The thickness, $$t$$, is the same as the full disc.
So now we have what we need to put together an equation for the washer-disc with a hole in the middle. We take the volume of the full disc and subtract the volume of the hole to get $$V = \pi R^2 t - \pi r^2 t = \pi t(R^2-r^2)$$. [Notice the special case when there is no hole in the middle, can be thought of as $$r=0$$ giving the volume of the disk as just $$V=\pi R^2 t$$.]
summary of the washer-disc method
the representative rectangle is perpendicular to axis of revolution
$$R$$ is the distance from the axis of rotation to the far end of the representative rectangle
$$r$$ is the distance from the axis of rotation to the closest end of the representative rectangle
x-axis rotation equation
$$\displaystyle{ V = \pi \int_{a}^{b}{R^2-r^2~dx} }$$
y-axis rotation equation
$$\displaystyle{ V = \pi \int_{c}^{d}{R^2-r^2~dy} }$$
Note - Notice that $$R$$ and $$r$$ are distances, so they are always positive (although since we square them, the sign doesn't make any difference in the equations).
washer-disc method with x-axis rotation
If you feel like you need further explanation of this, here is a video that tries to explain this method by drawing in three dimensions. In this video, notice that the axis of rotation runs along one side of the figure and, consequently, $$r=0$$.
### Khan Academy - Disc Method [10min-4secs]
Okay, now let's work some problems using the washer-disc method revolving an area about the x-axis.
x-axis rotation practice
area: $$y=1/x, x=1, x=3$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, x=1, x=3$$ revolved about the x-axis. Give your answer in exact terms.
$$2\pi/3$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, x=1, x=3$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3487 video solution
$$2\pi/3$$
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area: $$y=x^2, y=4x-x^2$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=4x-x^2$$ revolved about the x-axis. Give your answer in exact terms.
$$V = 32\pi/3$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=4x-x^2$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3494 video solution
$$V = 32\pi/3$$
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area: $$y=\sqrt{x-1}, y=0, x=5$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x-1}, y=0, x=5$$ revolved about the x-axis. Give your answer in exact terms.
$$V=8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x-1}, y=0, x=5$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Krista King Math - 324 video solution
video by Krista King Math
$$V=8\pi$$
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area: $$y=x^2, x=y^2$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about the x-axis. Give your answer in exact terms.
$$\displaystyle{V=\frac{3\pi}{10}}$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about the x-axis. Give your answer in exact terms.
Solution
This problem is solved by two different instructors.
### Krista King Math - 877 video solution
video by Krista King Math
$$\displaystyle{V=\frac{3\pi}{10}}$$
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area: $$y=x^3, y=x, x \geq 0$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x \geq 0$$ revolved about the x-axis. Give your answer in exact terms.
$$V=4\pi/21$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x \geq 0$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### PatrickJMT - 1174 video solution
video by PatrickJMT
$$V=4\pi/21$$
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area: $$y=\sqrt{x}, y \geq 0, x=4$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y \geq 0, x=4$$ revolved about the x-axis. Give your answer in exact terms.
$$8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y \geq 0, x=4$$ revolved about the x-axis. Give your answer in exact terms.
Solution
This problem is solved by two different instructors.
In the second video, he doesn't finish the integration, so here are the details.
$$\displaystyle{ \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi [(4^2)/2 - (0^2)/2] = 8\pi}$$
### PatrickJMT - 1359 video solution
video by PatrickJMT
$$8\pi$$
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area: $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ revolved about the x-axis. Give your answer in exact terms.
$$V = 9\pi^2$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Dr Chris Tisdell - 2003 video solution
video by Dr Chris Tisdell
$$V = 9\pi^2$$
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area: $$y^2=x-2, x=5$$ in the first quadrant axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2=x-2, x=5$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
$$4.5\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2=x-2, x=5$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2272 video solution
video by Michel vanBiezen
$$4.5\pi$$
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area: $$y=x^2, y=x$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the x-axis. Give your answer in exact terms.
$$2\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2273 video solution
video by Michel vanBiezen
$$2\pi/15$$
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area: $$y=\sqrt{9-x^2}$$ in the first quadrant axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{9-x^2}$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
$$V=18\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{9-x^2}$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
Solution
### MIP4U - 2276 video solution
video by MIP4U
$$V=18\pi$$
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area: $$y=\sqrt{x}, y=x^2$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=x^2$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Khan Academy - 1184 video solution
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area: $$y=\sqrt{x}, y\geq0, x=1$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y\geq0, x=1$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Khan Academy - 1182 video solution
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Derive the equation for the volume of a sphere of radius r using the washer-disc method.
Problem Statement
Derive the equation for the volume of a sphere of radius r using the washer-disc method.
Solution
### Khan Academy - 1183 video solution
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Looking back at the plots, you will notice that the axis of revolution is always a coordinate axis, either the x-axis or the y-axis. A twist you will see is when the axis of revolution is another line. On this site, we will discuss only axes that are parallel to one of the coordinate axes.
In this case, the equations that will change are the ones that describe the distance from the axes of rotation. We suggest that you set up a sum from the parallel coordinate axis to the axis of rotation and then solve for whatever variable you need. This concept, especially, requires you to think over in your mind several times and look at examples.
parallel to x-axis rotation practice
area: $$y=x^2, x=0, y=4$$ axis of rotation: $$y=4$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the $$y=4$$. Give your answer in exact terms.
$$V = 256\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the $$y=4$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3490 video solution
$$V = 256\pi/15$$
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area: $$y=2-x^2, y=1$$ axis of rotation: $$y=1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2-x^2, y=1$$ revolved about the $$y=1$$. Give your answer in exact terms.
$$V = 16\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2-x^2, y=1$$ revolved about the $$y=1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3491 video solution
$$V = 16\pi/15$$
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area: $$y=x, y=x^2$$ axis of rotation: $$y=2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x, y=x^2$$ revolved about $$y=2$$. Give your answer in exact terms.
$$V = 8\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x, y=x^2$$ revolved about $$y=2$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3497 video solution
$$V = 8\pi/15$$
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area: $$y=1/x, y=0, x=1, x=3$$ axis of rotation: $$y=-1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, y=0, x=1, x=3$$ revolved about $$y=-1$$. Give your answer in exact terms.
$$V = 2\pi/3 + 2\pi\ln(3) = 2\pi(\ln(3)+1/3)$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, y=0, x=1, x=3$$ revolved about $$y=-1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3498 video solution
$$V = 2\pi/3 + 2\pi\ln(3) = 2\pi(\ln(3)+1/3)$$
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area: $$f(x)=x, g(x)=x^2-3x$$ axis of rotation: $$y=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=x, g(x)=x^2-3x$$ revolved about $$y=5$$. Give your answer in exact terms.
Hint
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=x, g(x)=x^2-3x$$ revolved about $$y=5$$. Give your answer in exact terms.
$$\displaystyle{ V = \frac{1472\pi}{15} }$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=x, g(x)=x^2-3x$$ revolved about $$y=5$$. Give your answer in exact terms.
Hint
Solution
1. Draw the plot and the example rectangle and label r and R. See the hint or the animation. We drew the example rectangle perpendicular to the axis of revolution since we were told to use the washer-disc method. The distance r is the distance from the axis of revolution to closest end of the rectangle. The distance R is from the axis of revolution to the farthest end of the example rectangle.
2. Choose The Integral - The volume integral we need is $$V = \pi\int_a^b{R^2-r^2~dx}$$. We chose this integral because we are told to use the washer-disc method, so we need an integral with r and R. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.
3. Determine r and R - From the plot, let's start on the axis of revolution and move down to the x-axis of revolution. This distance is 5 units. Moving back up to the end of the rectangle that lands on $$y=x$$, we move y units. What we are left with is r, so $$r=5-y$$. However, we need to replace y with expression for y in terms of x. Since $$y=x$$ is the line that we are working with, we have $$r=5-x$$.
Determining the expression for R is a bit trickier. Starting on the axis of revolution, we move down to the x-axis which is 5 units. However, when $$x < 3$$ we need to go a little further in the same direction to get the full distance R. Let's put that aside for a minute and think about the part of the graph for $$x > 3$$. In this case, we need to go back up y units, so $$R=5-y$$. This looks the same as r but in this case, we are landing on the plot $$y=x^2-3x$$. So $$R=5-(x^2-3x) = -x^2+3x+5$$. Let's plug in a few values and compare the values to graph to see if they match.
$$x=3$$ $$R=-3^2+3(3)+5=5$$ 𞀄 $$x=4$$ $$R=-4^2+3(4)+5=1$$ 𞀄
So far, so good. Let's plug in a few values $$x < 3$$ and see what we get.
$$x=0$$ $$R=-0^2+3(0)+5=5$$ 𞀄 $$x=2$$ $$R=-2^2+3(2)+5=7$$ 𞀄
So it looks like we have the correct equation for R. We can do the same with r to check our equation. This does not guarantee that we have the right equations but it may give an indication if they are incorrect. I usually check both endpoints and at least one other point, two other points is even better.
4. Set up and evaluate the integral - If we look at the animation above, we can tell that the rectangle sweeps across the area from $$x=0$$ to $$x=4$$. So our integral is $$\displaystyle{ V = \pi \int_0^4{ (-x^2+3x+5)^2 - } }$$ $$\displaystyle{ (5-x)^2 ~ dx }$$. Let's evaluate it.
$$\displaystyle{ V = \pi\int_0^4{ (-x^2+3x+5)^2 - (5-x)^2 ~ dx } }$$ $$\displaystyle{ V = \pi\int_0^4{ (x^4-3x^3-5x^2-3x^3+ } }$$ $$9x^2+15x-5x^2+15x+25) -$$ $$(25-10x+x^2) ~dx$$ $$\displaystyle{ V = \pi\int_0^4{ x^4-6x^3-2x^2+40x ~dx } }$$ $$\displaystyle{ V = \pi\left[ \frac{x^5}{5} - \frac{6x^4}{4} -\frac{2x^3}{3}+\frac{40x^2}{2} \right]_0^4 }$$ $$\displaystyle{ V = \pi\left[ \frac{4^5}{5} - \frac{6(4)^4}{4} -\frac{2(4)^3}{3}+\frac{40(4)^2}{2} \right] - 0 }$$ $$\displaystyle{ V = \pi\left[ \frac{1024}{5} - \frac{1536}{4} -\frac{128}{3}+\frac{640}{2} \right] }$$ $$\displaystyle{ V = \pi \frac{1472}{15} }$$
$$\displaystyle{ V = \frac{1472\pi}{15} }$$
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area: $$y=x^2, x=y^2$$ axis of rotation: $$y=1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$y=1$$. Give your answer in exact terms.
$$\displaystyle{V=11\pi/30}$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$y=1$$. Give your answer in exact terms.
Solution
### Krista King Math - 1172 video solution
video by Krista King Math
$$\displaystyle{V=11\pi/30}$$
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area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=-2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=-2$$. Give your answer in exact terms.
$$V=25\pi/21$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=-2$$. Give your answer in exact terms.
Solution
In the video, he set up the integral but did not evaluate it. His integral was $$\displaystyle{ \int_{0}^{1}{\pi[(2+x)^2 - \pi(2+x^3)^2]dx} }$$. This evaluates to $$\displaystyle{ \pi \left[ x^2 + \frac{x^3}{3} - x^4 - \frac{x^7}{7} \right]_{0}^{1} }$$
### PatrickJMT - 1175 video solution
video by PatrickJMT
$$V=25\pi/21$$
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area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=5$$. Give your answer in exact terms.
$$V=97\pi/42$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=5$$. Give your answer in exact terms.
Solution
### PatrickJMT - 1176 video solution
video by PatrickJMT
$$V=97\pi/42$$
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area: $$y=x^2, y=x$$ axis of rotation: $$y=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about $$y=5$$. Give your answer in exact terms.
$$23\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about $$y=5$$. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2275 video solution
video by Michel vanBiezen
$$23\pi/15$$
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Now, let's work some problems with the y-axis as the axis of rotation. Here is the plot that contains all the information you need to work these problems.
washer-disc method with y-axis rotation
y-axis rotation practice
area: $$y=x^2, x=0, y=4$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the y-axis. Give your answer in exact terms.
$$V = 8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3488 video solution
$$V = 8\pi$$
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area: $$y=x^{2/3}, x=0, y=1$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^{2/3}, x=0, y=1$$ revolved about the y-axis. Give your answer in exact terms.
$$V = \pi/4$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^{2/3}, x=0, y=1$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3489 video solution
$$V = \pi/4$$
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area: $$y^2 = x, x=2y$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2 = x, x=2y$$ revolved about the y-axis. Give your answer in exact terms.
$$V = 64\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2 = x, x=2y$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3496 video solution
$$V = 64\pi/15$$
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area: $$y=\sqrt{x}, y=0, x=3$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the y-axis. Give your answer in exact terms.
$$V = 36\pi\sqrt{3}/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3495 video solution
$$V = 36\pi\sqrt{3}/5$$
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area: $$x=2\sqrt{y}, x=0, y=9$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=2\sqrt{y}, x=0, y=9$$ revolved about the y-axis. Give your answer in exact terms.
$$V=162\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=2\sqrt{y}, x=0, y=9$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Krista King Math - 343 video solution
video by Krista King Math
$$V=162\pi$$
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area: $$y=1-x^2, y=0$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1-x^2, y=0$$ revolved about the y-axis. Give your answer in exact terms.
$$\displaystyle{V=\frac{\pi}{2}}$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1-x^2, y=0$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Krista King Math - 878 video solution
video by Krista King Math
$$\displaystyle{V=\frac{\pi}{2}}$$
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area: $$y=x^2, y=5, x=0$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=5, x=0$$ revolved about the y-axis. Give your answer in exact terms.
$$25\pi/2$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=5, x=0$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2271 video solution
video by Michel vanBiezen
$$25\pi/2$$
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area: $$y=x^2, y=x$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the y-axis. Give your answer in exact terms.
$$\pi/6$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2274 video solution
video by Michel vanBiezen
$$\pi/6$$
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area: $$y=\sqrt{x}$$ on $$[0,4]$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Unless otherwise instructed, use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}$$ on $$[0,4]$$ revolved about the y-axis. Give your answer in exact terms.
$$V=32\pi/5$$
Problem Statement
Unless otherwise instructed, use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}$$ on $$[0,4]$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### MIP4U - 2277 video solution
video by MIP4U
$$V=32\pi/5$$
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area: $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ revolved about the y-axis. Give your answer in exact terms.
$$V=5\pi/24$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2280 video solution
video by Michel vanBiezen
$$V=5\pi/24$$
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area: $$y=-3x+6$$, x-axis, y-axis axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=-3x+6$$, the x-axis and the y-axis revolved about the y-axis. Give your answer in exact terms.
$$V=8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=-3x+6$$, the x-axis and the y-axis revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2281 video solution
video by Michel vanBiezen
$$V=8\pi$$
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area: $$y=x^2, y=0, x=1$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=0, x=1$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Khan Academy - 1185 video solution
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parallel to y-axis rotation practice
area: $$y=\sqrt{x}, y=0, x=3$$ axis of rotation: $$x=3$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the $$x=3$$. Give your answer in exact terms.
$$V = 24\pi\sqrt{3}/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the $$x=3$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3492 video solution
$$V = 24\pi\sqrt{3}/5$$
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area: $$x=y^2, x=1$$ axis of rotation: $$x=1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=y^2, x=1$$ revolved about $$x=1$$. Give your answer in exact terms.
$$V = 16\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=y^2, x=1$$ revolved about $$x=1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3493 video solution
$$V = 16\pi/15$$
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area: $$y=\sqrt{x}, y=0, x=3$$ axis of rotation: $$x=6$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about $$x=6$$. Give your answer in exact terms.
$$V = 84\pi\sqrt{3}/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about $$x=6$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3499 video solution
$$V = 84\pi\sqrt{3}/5$$
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area: $$y=x^2, x=y^2$$ axis of rotation: $$x=-1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$x=-1$$. Give your answer in exact terms.
$$V = 29\pi\/30$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$x=-1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3500 video solution
$$V = 29\pi\/30$$
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area: $$y=x^3, y=0, x=1$$ axis of rotation: $$x=2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=0, x=1$$ revolved about $$x=2$$. Give your answer in exact terms.
$$V = 3\pi/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=0, x=1$$ revolved about $$x=2$$. Give your answer in exact terms.
Solution
### Krista King Math - 1173 video solution
video by Krista King Math
$$V = 3\pi/5$$
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area: $$y=2\sqrt{x}, y=0, x=4$$ axis of rotation: $$x=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2\sqrt{x}, y=0, x=4$$ revolved about $$x=5$$. Give your answer in exact terms.
$$V=832\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2\sqrt{x}, y=0, x=4$$ revolved about $$x=5$$. Give your answer in exact terms.
Solution
### MIP4U - 1916 video solution
video by MIP4U
$$V=832\pi/15$$
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area: $$x^2+y^2=1$$, x-axis, y-axis axis of rotation: $$x=2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1$$, the x-axis and the y-axis, revolved about $$x=2$$. Give your answer in exact terms.
$$V=(\pi/12)[3\pi-4]$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1$$, the x-axis and the y-axis, revolved about $$x=2$$. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2279 video solution
video by Michel vanBiezen
$$V=(\pi/12)[3\pi-4]$$
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https://www.clutchprep.com/physics/practice-problems/139925/explain-the-distinction-between-an-ohmic-and-non-ohmic-material-in-terms-of-how-
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crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00525.warc.gz
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|
Resistors and Ohm's Law Video Lessons
Concept
# Problem: a) Explain the distinction between an ohmic and non-ohmic material, in terms of how the current and resistance behave as the voltage difference across the material is changed.b) Now, imagine a single-loop circuit with a battery, two wires, and a 10 Ohm resistor. The wires are also ohmic, but with a resistance much smaller than the 10 Ohm resistor. Despite the disparity in resistance, the current in the wire is the same as that through the resistor since they are in series. Using Ohm's Law, explain how this uniformity in current relates to (or arises from) the individual potential differences across the wire and resistor.
###### FREE Expert Solution
a)
Ohmic materials are materials that obey Ohm's law: show linear relationship between the current and the voltage, and their resistances do not change with the variation in voltage.
83% (26 ratings)
###### Problem Details
a) Explain the distinction between an ohmic and non-ohmic material, in terms of how the current and resistance behave as the voltage difference across the material is changed.
b) Now, imagine a single-loop circuit with a battery, two wires, and a 10 Ohm resistor. The wires are also ohmic, but with a resistance much smaller than the 10 Ohm resistor. Despite the disparity in resistance, the current in the wire is the same as that through the resistor since they are in series. Using Ohm's Law, explain how this uniformity in current relates to (or arises from) the individual potential differences across the wire and resistor.
| 338
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http://www.fmxfeeds.com/2018/09/quick-algorithm-get-ideal-size-square-like-for-a-board-game-having-an-arbitrary-but-even-number-of-fields/
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# Quick Algorithm: Get Ideal Size (Square like) For a Board Game Having an Arbitrary (but Even) Number of Fields
Say you are developing a game like Chess, Go, Checkers, Tic-Tac-Toe or Memory. In each of those games the game board is a rectangle looking playfield of different size (rows x columns). Tic-Tac-Toe is 3×3, Checkers is 8×8, while Go can be 19×19 or 13×13 and similar.
In a game with an arbitrary number of game fields you might want to have the board look as closely to square as possible (rectangle where height and width are the same). Think of Memory. Let’s say we have 24 cards, that is 12 pairs. If you want to place them in a rectangle looking grid, most similar to square, you would go for 4 x 6 (or 6 x 4) board size (as it would look more square like than 3 x 8 and 2 x 12 or 1 x 24 would be too wide).
Therefore, the question: having an arbitrary number of game field pairs, what is the ideal, most square looking, grid size?
And the answer is an algorithm (some math knowledge required) like this one:
TGridSize = record
Rows, Columns : integer;
end;
function CalcGridSize(const numberOfPairs : Cardinal) : TGridSize;
//look for ideal rectangle dimensions (square is ideal)
//to present fields, number of fields = 2 * numberOfPairs
var
fieldCount : integer;
i : integer;
begin
fieldCount := 2 * numberOfPairs;
result.Rows := 1;
result.Columns := fieldCount;
if Sqrt(fieldCount) = Trunc(Sqrt(fieldCount)) then
begin
result.Rows := Trunc(Sqrt(fieldCount));
result.Columns := Trunc(Sqrt(fieldCount));
Exit;
end;
for i := Trunc(Sqrt(fieldCount)) downto 2 do
if (fieldCount mod i) = 0 then
begin
result.Rows := i;
result.Columns := fieldCount div i;
Exit;
end;
end; (*CalcGridSize*)
And here are some results:
var
gridSize : TGridSize;
i : integer;
begin
for i := 1 to 20 do
begin
gridSize := CalcGridSize(i);
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https://metanumbers.com/6011
| 1,708,554,300,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00537.warc.gz
| 409,762,550
| 7,370
|
# 6011 (number)
6011 is an odd four-digits prime number following 6010 and preceding 6012. In scientific notation, it is written as 6.011 × 103. The sum of its digits is 8. It has a total of one prime factor and 2 positive divisors. There are 6,010 positive integers (up to 6011) that are relatively prime to 6011.
## Basic properties
• Is Prime? yes
• Number parity odd
• Number length 4
• Sum of Digits 8
• Digital Root 8
## Name
Name six thousand eleven
## Notation
Scientific notation 6.011 × 103 6.011 × 103
## Prime Factorization of 6011
Prime Factorization 6011
Prime number
Distinct Factors Total Factors Radical ω 1 Total number of distinct prime factors Ω 1 Total number of prime factors rad 6011 Product of the distinct prime numbers λ -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ -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 8.70135 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 6011 is 6011. Since it has only one prime factor, 6011 is a prime number.
## Divisors of 6011
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ 2 Total number of the positive divisors of n σ 6012 Sum of all the positive divisors of n s 1 Sum of the proper positive divisors of n A 3006 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 77.5306 Returns the nth root of the product of n divisors H 1.99967 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 6011 can be divided by 2 positive divisors (out of which none is even, and 2 are odd). The sum of these divisors (counting 6011) is 6012, the average is 3006.
## Other Arithmetic Functions (n = 6011)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 6010 Total number of positive integers not greater than n that are coprime to n λ 6010 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 789 Total number of primes less than or equal to n r2 0 The number of ways n can be represented as the sum of 2 squares
There are 6,010 positive integers (less than 6011) that are coprime with 6011. And there are approximately 789 prime numbers less than or equal to 6011.
## Divisibility of 6011
m n mod m
2 1
3 2
4 3
5 1
6 5
7 5
8 3
9 8
6011 is not divisible by any number less than or equal to 9.
## Classification of 6011
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non hypotenuse
• Prime Power
• Square Free
## Base conversion 6011
Base System Value
2 Binary 1011101111011
3 Ternary 22020122
4 Quaternary 1131323
5 Quinary 143021
6 Senary 43455
8 Octal 13573
10 Decimal 6011
12 Duodecimal 358b
16 Hexadecimal 177b
20 Vigesimal f0b
36 Base36 4mz
## Basic calculations (n = 6011)
### Multiplication
n×y
n×2 12022 18033 24044 30055
### Division
n÷y
n÷2 3005.5 2003.67 1502.75 1202.2
### Exponentiation
ny
n2 36132121 217190179331 1305530167958641 7847541839599391051
### Nth Root
y√n
2√n 77.5306 18.1823 8.80515 5.69888
## 6011 as geometric shapes
### Circle
Radius = n
Diameter 12022 37768.2 1.13512e+08
### Sphere
Radius = n
Volume 9.09764e+11 4.5405e+08 37768.2
### Square
Length = n
Perimeter 24044 3.61321e+07 8500.84
### Cube
Length = n
Surface area 2.16793e+08 2.1719e+11 10411.4
### Equilateral Triangle
Length = n
Perimeter 18033 1.56457e+07 5205.68
### Triangular Pyramid
Length = n
Surface area 6.25827e+07 2.55961e+10 4907.96
## Cryptographic Hash Functions
md5 e3b80d30a727c738f3cff0941f6bc55a 83a08a7fdc4059c7ac2e647e546add90679b9ac4 c37fd5582393747ef03b83ad095a5650d2f5335acc65eaa7db54c4b2a21d1092 3dc2b7303e1342405a887f123ed4e5615ff7f0a3a805f9e157d23fb8fedee7bd4eb3302d39c67b8b66304f3309bb8e0720dcac9649bfa295eb8cc8e7a4451918 ac22fadd6e29870db1b5de10b3d7f60f5317d167
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| 3.78125
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CC-MAIN-2024-10
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latest
|
en
| 0.814264
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https://origin.geeksforgeeks.org/plot-multiple-data-sets-on-the-same-chart-in-excel/
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Skip to content
# Plot Multiple Data Sets on the Same Chart in Excel
• Last Updated : 29 Jun, 2021
Sometimes while dealing with hierarchical data we need to combine two or more various chart types into a single chart for better visualization and analysis. This type of chart having multiple data sets is known as “Combination charts”.
In this article, we are going to see how to make combination charts from a set of two different charts in Excel using the example shown below.
Example: Consider a famous coaching institute that deals with both free content in their YouTube channel and also have their own paid online courses. There are broadly two categories of students in this institute :
1. The students who enrolled in the coaching but are learning from YouTube free video content.
2. The students who enrolled as well as bought paid online courses.
So, the institute asked their Sales Department to make a statistical chart about how many paid courses from a pool of courses which the institute deals with were sold from the year 2014 to the last year 2020 and also show the percentage of students who have enrolled in these paid courses.
### Table :
Here, the first data is “Number of Paid courses sold” and the second one is “Percentage of Students enrolled”. Now our aim is to plot these two data in the same chart with different y-axis.
### Implementation :
Follow the below steps to implement the same:
Step 1: Insert the data in the cells. After insertion, select the rows and columns by dragging the cursor.
Step 2: Now click on Insert Tab from the top of the Excel window and then select Insert Line or Area Chart. From the pop-down menu select the first “2-D Line”.
From the above chart we can observe that the second data line is almost invisible because of scaling. The present y-axis line is having much higher values and the percentage line will be having values lesser than 1 i.e. in decimal values. Hence, we need a secondary axis in order to plot the two lines in the same chart. In Excel, it is also known as clustering of two charts.
The steps to add a secondary axis are as follows :
1. Open the Chart Type dialog box
`Select the Chart -> Design -> Change Chart Type`
Another way is :
`Select the Chart -> Right Click on it -> Change Chart Type`
2. The Chart Type dialog box opens. Now go to the “Combo” option and check the “Secondary Axis” box for the “Percentage of Students Enrolled” column. This will add the secondary axis in the original chart and will separate the two charts. This will result in better visualization for analysis purposes.
The combination chart with two data sets is now ready. The secondary axis is for the “Percentage of Students Enrolled” column in the data set as discussed above.
Now various formatting can be carried out in this secondary axis using the Format Axis window on the right corner of Excel.
`Select the secondary Axis -> Right Click -> Format Axis -> Format Axis Dialog Box`
Changing the Bounds of Secondary Axis
You can further format the above chart by making it more interactive by changing the “Chart Styles”, adding suitable “Axis Titles”, “Chart Title”, “Data Labels”, changing the “Chart Type” etc. It can be done using the “+” button in the top right corner of the Excel chart.
Finally, after all the modification, the chart with multiple data sets looks like :
We can infer from the above chart that in the year 2019, the percentage of students who enrolled in the online paid courses are relatively less but in 2020 more students have enrolled in paid courses than free content on YouTube.
My Personal Notes arrow_drop_up
Recommended Articles
Page :
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https://homework.cpm.org/category/CC/textbook/cca2/chapter/4/lesson/4.2.3/problem/4-95
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crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00183.warc.gz
| 255,003,461
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### Home > CCA2 > Chapter 4 > Lesson 4.2.3 > Problem4-95
4-95.
Solve the equations below.
1. $\sqrt { x + 15 } = 5 + \sqrt { x }$
Square both sides.
$\left(\sqrt{x+15}\right)^2=\left(5+\sqrt{x}\right)^2$
$x+15=\left(5+\sqrt{x}\right)\left(5+\sqrt{x}\right)$
$x+15=25+10\sqrt{x}+x$
Isolate the square root of $x$ on one side of the equation.
$-10=10\sqrt{x}$
Divide both sides by $10$.
$-1=\sqrt{x}$
What can you take the square root of and get $−1$?
No real solutions.
1. $(y−6)^2+10=3y$
Expand the $(y−6)^2$.
Rearrange the equation so that it equals zero.
Solve by factoring and using the Zero Product Property, or use the Quadratic Formula.
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Purchase Solution
# Functions: K-T Condition
Not what you're looking for?
Consider the following program:
Maximize f(x,y)=x^2+4xy+y^2
subject to g(x,y)=x^2+y^2-1=0
##### Solution Summary
A function is maximized using Kuhn-Tucker condition. The maximized function results are determined.
##### Solution Preview
Solution. Let us denote the gradient vector of the function f(x,y) by Df(x,y). We rewrite the original program as follows.
Minimize F(x,y)=-f(x,y)=-x^2-4xy-y^2
. g(x,y)=x^2+y^2-1=0.
Since DF(x,y)=(-2x-4y,-4x-2y)', Dg(x,y)=(2x,2y)', by K-T ...
Solution provided by:
###### Education
• BSc , Wuhan Univ. China
• MA, Shandong Univ.
###### Recent Feedback
• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
• "excellent work"
• "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
• "Thank you"
• "Thank you very much for your valuable time and assistance!"
##### Multiplying Complex Numbers
This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.
##### Geometry - Real Life Application Problems
Understanding of how geometry applies to in real-world contexts
##### Exponential Expressions
In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.
##### Graphs and Functions
This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
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### A: Number and Operations
#### A.1: Students will understand numerical concepts and mathematical operations
A.1.1: Understand numbers, ways of representing numbers, relationships among numbers, and number systems.
A.1.1.1: Exhibit an understanding of the place-value structure of the base-ten number system by:
A.1.1.1.a: reading, modeling, writing, and interpreting whole numbers up to 10,000
A.1.1.1.b: comparing and ordering numbers up to 1,000
A.1.1.1.c: recognizing the position of a given number in the base-ten number system and its relationship to benchmark numbers such as 10, 50, 100, 500
A.1.1.2: Use whole numbers by using a variety of contexts and models (e.g., exploring the size of 1,000 by skip-counting to 1,000 using hundred charts or strips 10 or 100 centimeters long).
A.1.1.4: Identify the relationship among commonly encountered factors and multiples (e.g., factor pairs of 12 are 1 x 12, 2 x 6, 3 x 4; multiples of 12 are 12, 24, 36).
A.1.1.5: Use visual models and other strategies to recognize and generate equivalents of commonly used fractions and mixed numbers (e.g., halves, thirds, fourths, sixths, eighths, and tenths).
A.1.1.6: Demonstrate an understanding of fractions as parts of unit wholes, parts of a collection or set, and as locations on a number line.
A.1.1.7: Use common fractions for measuring and money (e.g., using fractions and decimals as representations of the same concept, such as half of a dollar = 50 cents)
A.1.2: Understand the meaning of operations and how they relate to one another.
A.1.2.1: Use a variety of models to show an understanding of multiplication and division of whole numbers (e.g., charts, arrays, diagrams, and physical models [i.e., modeling multiplication with a variety of pictures, diagrams, and concrete tools to help students learn what the factors and products represent in various contexts]).
A.1.2.2: Find the sum or difference of two whole numbers between 0 and 10,000.
A.1.2.3: Solve simple multiplication and division problems (e.g., 135 รท.(ٱ= 5
A.1.2.4: Identify how the number of groups and the number of items in each group equals a product.
A.1.2.5: Demonstrate the effects of multiplying and dividing on whole numbers (e.g., to find the total number of legs on 12 cats, 4 represents the number of each [cat] unit, so 12 x 4 = 48 [leg] units).
A.1.3: Compute fluently and make reasonable estimates.
A.1.3.1: Choose computational methods based on understanding the base-ten number system, properties of multiplication and division, and number relationships.
A.1.3.2: Use strategies (e.g., 6 x 8 is double 3 x 8) to become fluent with the multiplication pairs up to 10 x 10.
A.1.3.4: Demonstrate reasonable estimation strategies for measurement, computation, and problem solving.
### B: Algebra
#### B.1: Students will understand algebraic concepts and applications.
B.1.1: Understand patterns, relations, and functions.
B.1.1.5: Recognize and use the commutative property of multiplication (e.g., if 5 x 7 = 35, then what is 7 x 5?).
B.1.1.6: Create, describe, and extend numeric and geometric patterns including multiplication patterns.
B.1.1.7: Represent simple functional relationships:
B.1.1.7.a: solve simple problems involving a functional relationship between two quantities (e.g., find the total cost of multiple items given the cost per unit)
B.1.2: Represent and analyze mathematical situations and structures using algebraic symbols.
B.1.2.2: Recognize and use the commutative and associative properties of addition and multiplication (e.g., "If 5 x 7 = 35, then what is 7 x 5? And if 5 x 7 x 3 = 105, then what is 7 x 3 x 5?").
B.1.2.3: Explore the ways that commutative, distributive, identity, and zero properties are useful in computing with numbers.
B.1.3: Use mathematical models to represent and understand quantitative relationships.
B.1.3.1: Model problem situations with objects and use representations such as pictures, graphs, tables, and equations to draw conclusions.
### C: Geometry
#### C.1: Students will understand geometric concepts and applications.
C.1.1: Analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships.
C.1.1.1: Describe and compare the attributes of plane and solid geometric figures to show relationships and solve problems:
C.1.1.1.a: identify, describe, and classify polygons (e.g., pentagons, hexagons, and octagons)
C.1.1.1.b: identify lines of symmetry in two-dimensional shapes
C.1.1.1.c: explore attributes of quadrilaterals (e.g., parallel and perpendicular sides for the parallelogram, right angles for the rectangle, equal sides and right angles for the square)
C.1.1.1.d: identify right angles
C.1.2: Specify locations and describe spatial relationships using coordinate geometry and other representational systems.
C.1.2.2: Use ordered pairs to graph, locate specific points, create paths, and measure distances within a coordinate grid system.
C.1.3: Apply transformations and use symmetry to analyze mathematical situations.
C.1.3.1: Predict and describe the results of sliding, flipping, and turning two-dimensional shapes.
C.1.3.2: Identify and describe the line of symmetry in two- and three-dimensional shapes.
### D: Measurement
#### D.1: Students will understand measurement systems and applications.
D.1.1: Understand measurable attributes of objects and the units, systems, and process of measurement.
D.1.1.3: Identify time to the nearest minute (elapsed time) and relate time to everyday events.
D.1.2: Apply appropriate techniques, tools, and formulas to determine measurements.
D.1.2.2: Estimate measurements.
### E: Data Analysis and Probability
#### E.1: Students will understand how to formulate questions, analyze data, and determine probabilities.
E.1.1: Formulate questions that can be addressed with data and collect, organize, and display relevant data to answer them.
E.1.1.1: Collect and organize data using observations, measurements, surveys, or experiments.
E.1.1.2: Represent data using tables and graphs (e.g., line plots, bar graphs, and line graphs).
E.1.1.3: Conduct simple experiments by determining the number of possible outcomes and make simple predictions:
E.1.1.3.a: identify whether events are certain, likely, unlikely, or impossible
E.1.1.3.b: record the outcomes for a simple event and keep track of repetitions
E.1.1.3.d: use the results to predict future events
E.1.2: Select and use appropriate statistical methods to analyze data.
E.1.2.1: Apply and explain the uses of sampling techniques (e.g., observations, polls, tally marks) for gathering data.
E.1.4: Understand and apply basic concepts of probability.
E.1.4.1: Discuss the degree of likelihood of events and use terminology such as "certain," "likely," "unlikely".
Correlation last revised: 1/20/2017
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://plainmath.org/other/51721-which-operation-could-we-perform-in-order-to-find-the
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Annette Sabin
## Answered question
2022-01-10
Which operation could we perform in order to find the number of milliseconds in a year?
### Answer & Explanation
Hector Roberts
Beginner2022-01-11Added 31 answers
The amount of milliseconds in a year must be determined.
First of all, milli means thousandth and thus one second contains 1000 milliseconds.
Then, there are 60 seconds in one minute and thus the number of milliseconds in a minute is then obtain by multiplying the number of milliseconds in a second by the number of seconds in a minute.
Then, there are 60 minutes in one hour and thus the number of milliseconds in a minute by the number of milliseconds in a minute by the number of minutes in a hour.
Then, there are 24 hours in a day and thus the number of milliseconds in a day is then obtain by multiplying the number of milliseconds in an hour by the number of hours in a day.
Finally, we assume that there are 365 days in a year. The number of milliseconds in a year is then by multiplying the number of milliseconds in a day by the number of days in a year.
Then mark the correct action: $1000\cdot 60\cdot 60\cdot 24\cdot 365$
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The Fano tetrahedron
The "Fano tetrahedron" is built from Fano planes. We start with a description of the Fano plane, which you can skip if you're already familiar with it.
The Fano plane
Diagram 1. The Fano plane
A Fano plane is is a finite projective plane of seven points and seven lines, with three points on every line and three lines through every point. It is a balanced block design, with seven blocks (the lines) and seven elements (the points). It can be specified as the balanced block design 2-(7,7,3,3,1); and as the last value of this balanced block design is 1, it is a Steiner system, and as such its descriptor can be written S(2,3,7). Any two points lie on one and only one line, and any two lines pass through one and only one point.
A Fano plane is shown to the right. Three of the lines are shown in dark gray, the in red, and one in green – this has no mathematical significance, but it may make the diagram easier to understand. Its vertices are labelled with numbers, which should be regarded, not as integers, but as bit-strings (001, 010, 011 etc.). Wherever two points on a line are labelled p and q, the third point on that line is labelled p xor y.
Building the Fano tetrahedron
Diagram 2a. A tetrahedral net built from four Fano planes Diagram 2b. The tetrahedral net with six more lines added.
Just as a tetrahedron is built from four triangles, a Fano tetrahedron is built from four Fano planes. First we build the "net" of the tetrahedron, as shown in diagram 2a. A tetrahedron can be built from this net by folding it along the lines labelled 3, 5 and 6, and uniting the pairs and triplet of vertices with the same label.
But what we have now is not a balanced block design or Steiner system, for two reasons. We need to add one more point, labelled 15, in the centre of the tetrahedron, with seven lines through it, joining each vertex (1,2,4,8) to its antipodal face (14,13,11,7) and each edge (3,5,6) to its antipodal edge (12,10,9). Also we need to add six more lines, shown in diagram 2b in blue and purple.
The diagrams do not show the central vertex with its label "15". You'll have to imagine it, and the lines that pass through it.
As with the Fano plane in diagram 1, each pair of points lies on one line. It is therefore a Steiner syetem, S(2,3,15). The labels on the points are again such that the xor of any two points is the label on the third point on their line.
Catalogue of lines
Below is a table cataloging the 35 lines (blocks) of the Fano tetrahedron, and listing the elements (Vertices, Edges, Faces, and the Centre) of the tetrahedron that lie on them. The others are easy to understand, but the blue and magenta lines may seem irregular. In fact they join a pair of faces and the edge antipodal to their mutual edge. The blue edges and the magenta ones are equivalent, but look different when drawn on the net in the diagram.
Colour of lines
in diagram
Elements
on line
Number of
such lines
Description
GreyV E V6edges
RedV F E12medians
GreenE E E4in-circles
BlueF E F6lie on great circles
of the tetrahedron
Magenta
(not shown)V C F4pass through the centre,
joining antipodes
E C E3
Total35
Another view
Diagram 3. No longer a net, but the tetrahedron drawn with one vertex (its apex, if you like) at infinity in all directions.
Diagram 3 shows another view of the Fano tetrahedron. Each edge is now shown only once. The apex vertex, labeled "8", is at infinity.
The magenta lines, which lie along great circles, ought to be straight, but would then coincide with red lines. So instead, they have been drawn with slight S-shaped curves.
Further simplexes
Just as we can use the Fano plane to build the Fano tetrahedron, we can use the Fano tetrahedron to build the Fano 5-cell, with 5 vertices, 31 numbered points, and 155 lines, the Steiner system S(2,3,31); and so on.
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https://jmservera.com/simplify-264-960-577-0-0924-b10-2592-b/
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|
text/html
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crawl-data/CC-MAIN-2022-49/segments/1669446711111.35/warc/CC-MAIN-20221206161009-20221206191009-00751.warc.gz
| 352,958,827
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# Simplify 264.96*(0.577-0.0924/b)*(1+0.2592/b)
264.96⋅(0.577-0.0924b)⋅(1+0.2592b)
Apply the distributive property.
(264.96⋅0.577+264.96(-0.0924b))⋅(1+0.2592b)
Multiply 264.96 by 0.577.
(152.88192+264.96(-0.0924b))⋅(1+0.2592b)
Multiply 264.96(-0.0924b).
Multiply -1 by 264.96.
(152.88192-264.960.0924b)⋅(1+0.2592b)
Combine -264.96 and 0.0924b.
(152.88192+-264.96⋅0.0924b)⋅(1+0.2592b)
Multiply -264.96 by 0.0924.
(152.88192+-24.482304b)⋅(1+0.2592b)
(152.88192+-24.482304b)⋅(1+0.2592b)
Move the negative in front of the fraction.
(152.88192-24.482304b)⋅(1+0.2592b)
Expand (152.88192-24.482304b)(1+0.2592b) using the FOIL Method.
Apply the distributive property.
152.88192(1+0.2592b)-24.482304b(1+0.2592b)
Apply the distributive property.
152.88192⋅1+152.881920.2592b-24.482304b(1+0.2592b)
Apply the distributive property.
152.88192⋅1+152.881920.2592b-24.482304b⋅1-24.482304b⋅0.2592b
152.88192⋅1+152.881920.2592b-24.482304b⋅1-24.482304b⋅0.2592b
Simplify and combine like terms.
Simplify each term.
Multiply 152.88192 by 1.
152.88192+152.881920.2592b-24.482304b⋅1-24.482304b⋅0.2592b
Multiply 152.881920.2592b.
Combine 152.88192 and 0.2592b.
152.88192+152.88192⋅0.2592b-24.482304b⋅1-24.482304b⋅0.2592b
Multiply 152.88192 by 0.2592.
152.88192+39.62699366b-24.482304b⋅1-24.482304b⋅0.2592b
152.88192+39.62699366b-24.482304b⋅1-24.482304b⋅0.2592b
Multiply -1 by 1.
152.88192+39.62699366b-24.482304b-24.482304b⋅0.2592b
Multiply -24.482304b⋅0.2592b.
Multiply 0.2592b and 24.482304b.
152.88192+39.62699366b-24.482304b-0.2592⋅24.482304b⋅b
Multiply 0.2592 by 24.482304.
152.88192+39.62699366b-24.482304b-6.34581319b⋅b
Raise b to the power of 1.
152.88192+39.62699366b-24.482304b-6.34581319b1b
Raise b to the power of 1.
152.88192+39.62699366b-24.482304b-6.34581319b1b1
Use the power rule aman=am+n to combine exponents.
152.88192+39.62699366b-24.482304b-6.34581319b1+1
152.88192+39.62699366b-24.482304b-6.34581319b2
152.88192+39.62699366b-24.482304b-6.34581319b2
152.88192+39.62699366b-24.482304b-6.34581319b2
Combine the numerators over the common denominator.
152.88192+39.62699366-24.482304b-6.34581319b2
152.88192+39.62699366-24.482304b-6.34581319b2
Subtract 24.482304 from 39.62699366.
152.88192+15.14468966b-6.34581319b2
Simplify 264.96*(0.577-0.0924/b)*(1+0.2592/b)
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| 3.640625
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CC-MAIN-2022-49
|
latest
|
en
| 0.529327
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# How do you write H(x) = 2/3|x-5| as a piecewise function?
Aug 25, 2017
$H \left(x\right) = - \frac{2}{3} \left(x - 5\right) , x < 5$
$H \left(x\right) = \frac{2}{3} \left(x - 5\right) , x \ge 5$
#### Explanation:
If x is less than 5, the number within the absolute value will be negative. Thus, if we wish to rid ourselves of the absolute value symbol, which as one recalls changes any negative expression within it positive, we must find a way to make $\frac{2}{3} \left(x - 5\right)$ positive when $x < 5$.
This is most easily accomplished by multiplying the expression by -1 when $x < 5$ ; in this case, since it will only be applied when $\frac{2}{3} \left(x - 5\right) < 0$, it will turn the expression positive. The breakpoint is a at x=5, because at x=5 the expression is equal to 0, and as x increases the expression becomes more positive.
Thus, we are left with
$H \left(x\right) = - \frac{2}{3} \left(x - 5\right) , x < 5$
$H \left(x\right) = \frac{2}{3} \left(x - 5\right) , x \ge 5$
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# what is a probability????????????????????????????????????????
Gowri sankar
292 Points
8 years ago
HAI GURAVAIAH,
THE PROBABILITY IS DIFINED BY THE QUALITY OF BEING PROBABLE IS CALLED THE PROBABILITY.........................................................................
317 Points
8 years ago
Hello Guravaiah
Probability is the chance that something will happen - how likely it is that some event will happen. Sometimes you can measure a probability with a number like "10% chance of rain", or you can use words such as impossible, unlikely, possible, even chance, likely and certain. Example: "It is unlikely to rain tomorrow".
Prabhakar ch
577 Points
8 years ago
Dear Guravaiah
Probability is the measure of the likelihood that an event will occur. Probability is quantified as a number between 0 and 1 (where 0 indicates impossibility and 1 indicates certainty).
mohan
216 Points
8 years ago
dear sivagiri......
probility is a quantified as a number between 0 and 1.the higher the probolity of an event,the more certain we are that the event will occur.....
N JYOTHEESWAR
342 Points
8 years ago
Probability is the chance that something will happen how likely it is that some event will happen. Sometimes you can measure a probability or you can use words such as impossible, unlikely, possible, even chance, likely and certain.
SAI SARDAR
1700 Points
8 years ago
Guravaiah, Probability is the measure of likliness.And the number of ways for reaching some thing.all the best.
sudarshan
174 Points
8 years ago
Probability is the chance that something will happen - how likely it is that some event will happen. Sometimes you can measure a probability with a number like "10% chance of rain", or you can use words such as impossible, unlikely, possible, even chance, likely and certain. Example: "It is unlikely to rain tomorrow".
raj
383 Points
8 years ago
hello guravaProbability is the chance that something will happen - how likely it is that some event will happen. Sometimes you can measure a probability with a number like "10% chance of rain", or you can use words such as impossible, unlikely, possible, even chance, likely and certain. Example: "It is unlikely to rain tomorrow".
T Dileep
100 Points
8 years ago
the extent to which an event is likely to occur ,measured by the ratio of the favourable cases to the whole number of casespossible.
T.kumar
281 Points
8 years ago
Probability is the measure of the likelihood that an event will occur. Probability is quantified as a number between 0 and 1 (where 0 indicates impossibility and 1 indicates certainty). The higher the probability of an event, the more certain we are that the event will occur.
Dheeru chowdary
100 Points
8 years ago
dear guravaih,
probabilty is the nmeasure of likehood i.e., how probable it will happens.................k
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https://www.jobilize.com/course/section/points-in-phase-introduction-and-key-concepts-by-openstax?qcr=www.quizover.com
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# Introduction and key concepts
Page 2 / 3
Amplitude
The amplitude is the maximum displacement of a particle from its equilibrium position.
## Investigation : amplitude
Fill in the table below by measuring the distance between the equilibrium and each peak and troughs in the wave above. Use your ruler to measure the distances.
Peak/Trough Measurement (cm) a b c d e f
2. Are the distances between the equilibrium position and each peak equal?
3. Are the distances between the equilibrium position and each trough equal?
4. Is the distance between the equilibrium position and peak equal to the distance between equilibrium and trough?
As we have seen in the activity on amplitude, the distance between the peak and the equilibrium position is equal to the distance between the trough and the equilibrium position. This distance is known as the amplitude of the wave, and is the characteristic height of wave, above or below the equilibrium position. Normally the symbol $A$ is used to represent the amplitude of a wave. The SI unit of amplitude is the metre (m).
If the peak of a wave measures $2\phantom{\rule{2pt}{0ex}}m$ above the still water mark in the harbour, what is the amplitude of the wave?
1. The definition of the amplitude is the height of a peak above the equilibrium position. The still water mark is the height of the water at equilibrium and the peak is $2\phantom{\rule{2pt}{0ex}}m$ above this, so the amplitude is $2\phantom{\rule{2pt}{0ex}}m$ .
## Investigation : wavelength
Fill in the table below by measuring the distance between peaks and troughs in the wave above.
Distance(cm) a b c d
2. Are the distances between peaks equal?
3. Are the distances between troughs equal?
4. Is the distance between peaks equal to the distance between troughs?
As we have seen in the activity on wavelength, the distance between two adjacent peaks is the same no matter which two adjacent peaks you choose. There is a fixed distance between the peaks. Similarly, we have seen that there is a fixed distance between the troughs, no matter which two troughs you look at. More importantly, the distance between two adjacent peaks is the same as the distance between two adjacent troughs. This distance is called the wavelength of the wave.
The symbol for the wavelength is $\lambda$ (the Greek letter lambda ) and wavelength is measured in metres ( $m$ ).
The total distance between $4$ consecutive peaks of a transverse wave is $6\phantom{\rule{2pt}{0ex}}m$ . What is the wavelength of the wave?
1. From the sketch we see that 4 consecutive peaks is equivalent to 3 wavelengths.
2. Therefore, the wavelength of the wave is:
$\begin{array}{ccc}\hfill 3\lambda & =& 6\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \\ \hfill \lambda & =& \frac{6\phantom{\rule{0.166667em}{0ex}}\mathrm{m}}{3}\hfill \\ & =& 2\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \end{array}$
## Investigation : points in phase
Fill in the table by measuring the distance between the indicated points.
Points Distance (cm) A to F B to G C to H D to I E to J
What do you find?
In the activity the distance between the indicated points was the same. These points are then said to be in phase . Two points in phase are separate by an integer (0,1,2,3,...) number of complete wave cycles. They do not have to be peaks or troughs, but they must be separated by a complete number of wavelengths.
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
Got questions? Join the online conversation and get instant answers!
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# 3.7: Practice SD Formula and Interpretation
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You may or may not understand the importance of calculating and understanding the variation of your data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean.
##### The Standard Deviation
• provides a numerical measure of the overall amount of variation in a data set, and
• can be used to determine whether a particular data value is close to or far from the mean.
There are a couple common kinds of questions that standard deviations can answer, in addition being foundational for later statistical analyses. First, a standard deviation helps understand the shape of a distribution. Second, a standard deviation can show if a score is extreme.
### Describing the Shape of a Distribution
The standard deviation provides a measure of the overall variation in a data set.
The standard deviation is always positive or zero. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. Distributions with small standard deviations have a tall and narrow line graph. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation. Distributions with large standard deviations may have a wide and flat lin graph, or they may be skewed (with the outlier(s) making the standard deviation bigger).
Suppose that we are studying the amount of time customers wait in line at the checkout at supermarket A and supermarket B. the average wait time at both supermarkets is five minutes. At supermarket A, the standard deviation for the wait time is two minutes; at supermarket B the standard deviation for the wait time is four minutes.
Because supermarket B has a higher standard deviation, we know that there is more variation in the wait times at supermarket B. Overall, wait times at supermarket B are more spread out from the average; wait times at supermarket A are more concentrated near the average.
### Identifying Extreme Scores
The standard deviation can be used to determine whether a data value is close to or far from the mean.
Suppose that Rosa and Binh both shop at supermarket A. Rosa waits at the checkout counter for seven minutes and Binh waits for one minute. At supermarket A, the mean waiting time is five minutes and the standard deviation is two minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean.
Rosa waits for seven minutes:
• Seven is two minutes longer than the average of five; two minutes is equal to one standard deviation.
• Rosa's wait time of seven minutes is two minutes longer than the average of five minutes.
• Rosa's wait time of seven minutes is one standard deviation above the average of five minutes.
Binh waits for one minute.
• One is four minutes less than the average of five; four minutes is equal to two standard deviations.
• Binh's wait time of one minute is four minutes less than the average of five minutes.
• Binh's wait time of one minute is two standard deviations below the average of five minutes.
• A data value that is two standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if it is more than two standard deviations away is more of an approximate "rule of thumb" than a rigid rule. In general, the shape of the distribution of the data affects how much of the data is further away than two standard deviations. (You will learn more about this in later chapters.)
The number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is one standard deviation to the right of five because $$5 + (1)(2) = 7$$.
If one were also part of the data set, then one is two standard deviations to the left of five because $$5 + (-2)(2) = 1$$.
• In general, a value = mean + (#ofSTDEV)*(standard deviation)
• where #ofSTDEVs = the number of standard deviations
• #ofSTDEV does not need to be an integer
• One is two standard deviations less than the mean of five because: $$1 = 5 + (-2)(2)$$. (The numbers in parentheses that touch should be multiplied)
The equation value = mean + (#ofSTDEVs)*(standard deviation) can be expressed for a sample and for a population.
• sample: $$x = \bar{x} + \text{(#ofSTDEV) \times (s)}$$
• Population: $$x = \mu + \text{(#ofSTDEV) \times (s)}$$
The lower case letter s represents the sample standard deviation and the Greek letter $$\sigma$$ (sigma, lower case) represents the population standard deviation.
The symbol $$\bar{x}$$ is the sample mean and the Greek symbol $$\mu$$ is the population mean.
### Calculating the Standard Deviation
If $$x$$ is a number, then the difference "$$x$$ – mean" is called its deviation. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is $$x - \mu$$. For sample data, in symbols a deviation is $$x - \bar{x}$$.
The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter s represents the sample standard deviation and the Greek letter $$\sigma$$ (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of $$\sigma$$.
To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the $$x - \bar{x}$$ values for a sample, or the $$x - \mu$$ values for a population). The symbol $$\sigma^{2}$$ represents the population variance; the population standard deviation $$\sigma$$ is the square root of the population variance. The symbol $$s^{2}$$ represents the sample variance; the sample standard deviation s is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations.
If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by $$N$$, the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n – 1, one less than the number of items in the sample.
##### Formulas for the Sample Standard Deviation
$s = \sqrt{\dfrac{\sum(X-\bar{X})^{2}}{n-1}} \nonumber$
For the sample standard deviation, the denominator is $$n - 1$$, that is the sample size MINUS 1.
## Practice!
##### Example $$\PageIndex{1}$$
In a fifth grade class at a private school, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a sample of n = 20 fifth grade students. The ages are rounded to the nearest half year in Table $$\PageIndex{1}$$, but first let's talk about the context.
1. Who was the sample? Who could this sample represent (population)?
The sample is the 20 fifth graders from a private school. The population could be all fifth graders from private schools?
1. What was measured?
Age, in years, was measured. This is the DV, the outcome variable.
Table $$\PageIndex{1}$$- Ages of a sample of 20 fifth graders
9
9.5
9.5
10
10
10
10
10.5
10.5
10.5
10.5
11
11
11
11
11
11
11.5
11.5
11.5
1. What is the mean?
$\bar{x} = \dfrac{(9+9.5+9.5+10+10+10+10+10.5+10.5+10.5+10.5+11+11+11+11+11+11+11.5+11.5+11.5)}{20} = 10.525 = 10.53 \nonumber$
The average age is 10.53 years, rounded to two places.
1. What is the standard deviation?
The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating s.
Table $$\PageIndex{1}$$- Ages of One Fifth Grade Class
Data Deviations Deviations2
x (X – $$\bar{X}$$) (X– $$\bar{X})^2$$
9 $$9 – 10.525 = –1.525$$ $$(–1.525)^2 = (-1.525 \times -1.525) = 2.325625$$
9.5 $$9.5 – 10.525 = –1.025$$ $$(–1.025)^2 = (–1.025 \times –1.025) = 1.050625$$
9.5 $$9.5 – 10.525 = –1.0.25$$ $$(–1.025)^2 = 1.050625$$
10 $$10 – 10.525 = –0.525$$ $$(–0.525)^2 = (–0.525 \times –0.525)= 0.275625$$
10 $$10 – 10.525 = –0.525$$ $$(–0.525)^2 = 0.275625$$
10 $$10 – 10.525 = –0.525$$ $$(–0.525)^2 = 0.275625$$
10 $$10 – 10.525 = –0.525$$ $$(–0.525)^2 = 0.275625$$
10.5 $$10.5 – 10.525 = –0.025$$ $$(–0.025)^2 = (–0.025 \times –0.025)= 0.000625$$
10.5 $$10.5 – 10.525 = –0.025$$ $$(–0.025)^2 = 0.000625$$
10.5 $$10.5 – 10.525 = –0.025$$ $$(–0.025)^2 = 0.000625$$
10.5 $$10.5 – 10.525 = –0.025$$ $$(–0.025)^2 = 0.000625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = (0.475 \times 0.475)= 0.225625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = 0.225625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = 0.225625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = 0.225625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = 0.225625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = 0.225625$$
11.5 $$11.5 – 10.525 = 0.975$$ $$(0.975)^2 = (0.975 \times 0.975)= 0.950625$$
11.5 $$11.5 – 10.525 = 0.975$$ $$(0.975)^2 = 0.950625$$
11.5 $$11.5 – 10.525 = 0.975$$ $$(0.975)^2 = 0.950625$$
$$\displaystyle\sum X$$ 0 (basically) $$\sum = 9.7375$$
The first column in Table $$\PageIndex{1}$$ has the data, the second column has has deviations (each score minus the mean), the third column has deviations squared. The first row is the row's title, the second row is the symbols for that column, the rest of the rows are the scores until the bottom row, which is the sum of each of the rows.
Take the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1):
$\dfrac{9.7375}{20-1} = 0.5125 \nonumber$
The sample standard deviation s is the square root of $$\dfrac{SS}{df} \nonumber$$:
$s = \sqrt{0.5125} = 0.715891 \nonumber$
and this is rounded to two decimal places, $$s = 0.72$$. The standard deviation of the sample fo 20 fifth graders is 0.72 years.
Typically, you do the calculation for the standard deviation on your calculator or computer. When calculations are completed on devices, the intermediate results are not rounded so the results are more accurate. It's also darned easier. So why are spending time learning this outdated formula? So that you can see what's happening. We are finding the difference between each score and the mean to see how varied the distribution of data is around the center, dividing it by the sample size minus one to make it like an average, then square rooting it to get the final answer back into the units that we started with ( age in years).
• For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation).
• For a sample: $$x$$ = $$\bar{x}$$ + (#ofSTDEVs)(s)
• For a population: $$x$$ = $$\mu$$ + (#ofSTDEVs)$$\sigma$$
• For this example, use x = $$\bar{x}$$ + (#ofSTDEVs)(s) because the data is from a sample
1. Verify the mean and standard deviation on your own.
2. Find the value that is one standard deviation above the mean. Find ($$\bar{x}$$ + 1s).
3. Find the value that is two standard deviations below the mean. Find ($$\bar{x}$$ – 2s).
4. Find the values that are 1.5 standard deviations from (below and above) the mean.
Solution
1. You should get something close to 0.72 years, but anything from 0.70 to 0.74 shows that you have the general idea.
2. ($$\bar{x} + 1s) = 10.53 + (1)(0.72) = 11.25$$
3. $$(\bar{x} - 2s) = 10.53 – (2)(0.72) = 9.09$$
• $$(\bar{x} - 1.5s) = 10.53 – (1.5)(0.72) = 9.45$$
• $$(\bar{x} + 1.5s) = 10.53 + (1.5)(0.72) = 11.61$$
Notice that instead of dividing by $$n = 20$$, the calculation divided by $$n - 1 = 20 - 1 = 19$$ because the data is a sample. For the sample, we divide by the sample size minus one ($$n - 1$$). The sample variance is an estimate of the population variance. After countless replications, it turns out that when the formula division by only N (the size of the sample) is used on a sample to infer the population’s variance, it always under-estimates the variance of the population.
Which one has the bigger solution, the one with the smaller denominator or the larger denominator?
• $$\dfrac{10}{2}=$$
• $$\dfrac{10}{5}=$$
Smaller denominators make the resulting product larger. To solve our problem of using the population’s variance formula on a sample under-estimating the variance, we make the denominator of our equation smaller when calculating variance for a sample. In other words, based on the mathematics that lies behind these calculations, dividing by ($$n - 1$$) gives a better estimate of the population.
## What does it mean?
The deviations show how spread out the data are about the mean. From Table $$\PageIndex{1}$$, The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value (age, in this case) is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean (that particular student is younger than the average age of the class) . The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero, so you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation. But the variance is a squared measure and does not have the same units as the data. No one knows what 9.7375 years squared means. Taking the square root solves the problem! The standard deviation measures the spread in the same units as the data.
The standard deviation, $$s$$ or $$\sigma$$, is either zero or larger than zero. When the standard deviation is zero, there is no spread; that is, all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make $$s$$ or $$\sigma$$ very large.
##### Exercise $$\PageIndex{1}$$
Scenario: Using one baseball professional team as a sample for all professional baseball teams, the ages of each of the players are shown in Table $$\PageIndex{2}$$.
Table $$\PageIndex{2}$$- One Baseball Team's Ages
Data Deviations Deviations2
x (x – $$\bar{x}$$) (x – $$\bar{x})^2$$
21
21
22
23
24
24
25
25
28
29
29
31
32
33
33
34
35
36
36
36
36
38
38
38
40
$$\displaystyle\sum X$$ = 767 $$\displaystyle\sum X$$ should be 0 (basically) $$\displaystyle\sum X$$ = ?
If you get stuck after the table, don't forget that: $$s=\sqrt{\dfrac{\sum(X-\overline {X})^{2}}{N-1}}$$
All of your answers should be complete sentences, not just one word or one number. Behavioral statistics is about research, not math.
1. Who was the sample? Who could this sample represent (population)?
2. What was measured?
3. What is the mean? (Get in the practice of including the units of measurement when answering questions; a number is usually not a complete answer).
4. What is the standard deviation?
$s=\sqrt{\dfrac{\sum(X-\overline {X})^{2}}{N-1}}=\sqrt{\dfrac{S S}{d f}} \nonumber$
1. Find the value that is two standard deviations above the mean, and determine if there are any players that are more than two standard deviations above the mean.
1. The sample is 25 players from a professional baseball team. They were chosen to represent all professional baseball players (it says so in the scenario description!).
2. Age, in years, was measured.
3. The mean of the sample ($$\bar{X}$$ was 30.68 years.
4. The standard deviation was 6.09 years ( $$s = 6.09$$ ), although due to rounding differences you could get something from about 6.05 to 6.12. Don't worry too much if you don't get exactly 6.09; if you are close, then you did the formula correctly!
5. The age that is two standard deviations above the mean is 42.86 years, and none of the players are older than that.
$(\bar{x} + 2s = 30.68 + (2)(6.09) = 42.86 \nonumber$.
What standard deviation show us can seem unclear at first. Especially when you are unfamiliar (and maybe nervous) about using the formula. By graphing your data, you can get a better "feel" for what a standard deviation can show you. You will find that in symmetrical distributions, the standard deviation can be very helpful. Because numbers can be confusing, always graph your data.
## Summary
• $$s = \sqrt{\dfrac{\sum(x-\bar{x})^{2}}{n-1}}$$ is the formula for calculating the standard deviation of a sample. f(xμ)2N
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http://mathforum.org/kb/plaintext.jspa?messageID=7942838
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crawl-data/CC-MAIN-2018-17/segments/1524125948285.62/warc/CC-MAIN-20180426144615-20180426164615-00411.warc.gz
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```Date: Dec 23, 2012 6:20 AM
Author: quasi
Subject: Re: convex polyhedra with all faces regular
achille wrote:>quasi wrote:>>>> Prove or disprove:>> >> For each positive integer n, there are only finitely many >> convex polyhedra, up to similarity, such that all faces are >> regular polygons (not necessarily of the same type) with at >> most n edges.>>Yes, it is finite.>>It is known that the strictly convex regular-faced polyhedra>comprises >> 2 infinite families (the prisms and antiprisms)> 5 Platonic solids,> 13 Archimedian solids>and 92 Johnson solids> >Let N(n) be the number of convex polyhedra with regular polygons>up to n sides as faces. One has:>> N(n) <= 2n+104>>Actually, it is pretty simple to prove N(n) < oo directly.>WOLOG, let us fix the sides of the regular polygons to has >length 1.>>Let's pick any convex polyhedron and one of its vertex v. >Let say's v is connected to k edges >e_0, e_1, e_2, ... e_k = e_0 >and a_i ( i = 1..k ) is the angle between e_(i-1) and e_i.>For this v, let >> A(v) := 2 pi - sum_{i=1..k} a_i>>Being a convex polyhedron, we have A(v) > 0. It is also easy>to see if we sum over all vertices of the convex polyhedron, >we get:> > sum_v A(v) = 4 pi >>If one build a convex polyhedron using regular polygons up to>n sides, it is easy to see 3 <= k <= 5 and there are only>finitely many possible choices of a_i:>> (1 - 2/3) pi, (1 - 2/4) pi, ... ( 1 - 2/n) pi>>This mean there are finitely many possible choices of >a_1,.., a_k which satisfy:>>(*) 2 pi - sum_{i=1..k} a_i > 0 >>Let M(n) be the smallest possible value of L.H.S of (*) for >given n. On any vertex v of any convex polyhedron build from >regular polygons up to n sides, A(v) >= M(n) and hence the >convex polyhedron has at most 4 pi / M(n) vertices.>>Since the number of vertices is bounded, there are finitely >many ways to connect them to build a polyhedron. Using Cauchy >theorem of convex polytopes, each way of connecting the >vertices to from a polyhedron corresponds to at most 1 convex>polyhedron in Euclidean space. (since the length of all edges >has been fixed to 1).>>As a result, there are only finitely many convex polyhedra one>can build using regular polygons up to n sides.When I get a chance, I'll try to digest the details, but at first glance, it looks very good. Thanks.quasi
```
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EULER'S DAY OFF
# Pi is the key to a beautiful, impossible formula that shows math is scarily perfect
Mathematics is beautiful.
Image: Quartz
By
In high school, I studied advanced mathematics. This has two uses: Firstly, I can tell people that I once studied advanced mathematics. Secondly, I know about Euler’s equation.
Euler’s equation should not exist. A science fiction author would never imagine a formula so mind-blowingly perfect; it’s far too neat to seem real in an imaginary world. And yet it’s true, and absolutely true in the way only mathematics can be. I may have forgotten how to derive Euler’s equation from first principles (along with nearly every other detail from my high school classes), but I will always remember the resulting formula. It is irrefutable proof of mathematical perfection, and a reminder that we will never be able to fully grasp the meaning and reasons behind math.
Lo, here is the equation:
𝑒𝑖𝜋 = -1
If it’s been a while since you studied math, perhaps this doesn’t look like much. But breaking it down reveals its beauty.
Firstly, in honor of pi day, take pi, or 𝜋: An irrational number, so-called because it goes on infinitely without repeating, 𝜋 was first discovered as the number that describes the relationship between a circle’s circumference and its diameter (circumference = 𝜋 x diameter.) In the centuries since, scientists have determined that 𝜋 also describes the way rivers wind and ripples of light in physics. Originally, though, 𝜋 belonged to the realm of mathematics that deals with shapes and sizes: geometry.
Another famous irrational number, 𝑒, comes from logarithms, which are a part of calculus—a totally different branch of mathematics. The full significance of 𝑒 takes a while to explain, but one key detail, forming the basis of its role in logarithms, is that the rate at which 𝑒x grows is 𝑒x. Like pi, e is the basis of many different formulas. Numerically, it equals 2.71828… going on continually, without repeating.
Then there’s 𝑖, which is an imaginary number. It’s a theoretical concept that can never practically exist. 𝑖 signifies the square root of -1, which is impossible. No two identical numbers can be multiplied together to get a negative, meaning there is no numerical square root of -1. The square root of 4 is 2, and you can have 2 apples. But you can never have √-1 apples.
And yet, take these three utterly different, complicated numbers and bring them together in Euler’s equation and you get a magically neat result: e to the power of (𝑖 multiplied by pi) equals -1. Or:
𝑒𝑖𝜋 = -1
You can also write this as:
𝑒𝑖𝜋 + 1 = 0
The number one, of course, is the first natural number, the first positive integer, and the most common lead in sets of data: Quite simply, it’s how we start counting. And the number zero is the only non-positive natural number, the smallest non-negative quantity, signifying nothing.
In other words, one short equation includes five of the most important numbers in all of mathematics.
It’s almost unnerving. Taken alone, numbers like e, 𝑖, and 𝜋 seem like the results of an imperfect human effort to understand the complexity of the world through mathematical relationships. Euler’s equation, though, shows there’s a unity behind these numbers. The sum total of human mathematic knowledge is no more than a tiny fraction of the complete, perfect system. And every number or equation we discover is a reflection of this abstract, inherent truth, rather than a human invention. Humans can hope to uncover mathematical truths, but we cannot create them.
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Compound angles help
Watch
Announcements
#1
I can't figure this compound angles question out for the life of me.
Question:
Determine for what range of values of θ between 0 and 2π, Sinθ > Sin3θ
A walkthrough of how to do it would be much appreciated. Thanks in advance.
0
3 years ago
#2
Have u tried drawing the graphs of both of these functions on the same axis?
0
3 years ago
#3
(Original post by TheGibbsy)
I can't figure this compound angles question out for the life of me.
Question:
Determine for what range of values of θ between 0 and 2π, Sinθ > Sin3θ.
Start by sketching y = sin θ and y = sin 3θ on the same set of axes. The former should be very easy, and the latter only requires you to know the effect of transformations of thie kind y = f(x) to y = f(3x).
You should now be able to see where these intervals are, from where the first curve is higher up than the second curve.
Now the tricky part! You need to find the points of intersection of the two curves, that is, you need to solve sin θ = sin 3θ. Any ideas?
0
#4
(Original post by Pangol)
Start by sketching y = sin θ and y = sin 3θ on the same set of axes. The former should be very easy, and the latter only requires you to know the effect of transformations of thie kind y = f(x) to y = f(3x).
You should now be able to see where these intervals are, from where the first curve is higher up than the second curve.
Now the tricky part! You need to find the points of intersection of the two curves, that is, you need to solve sin θ = sin 3θ. Any ideas?
Thanks for the pointer. Using Sin (A+B) i've changed the equation to:
Sin(2θ+θ) = Sinθ and from there using the rule,
Sin2θcosθ + Cos2θsinθ = sinθ
However, I've hit a wall again as to where to go from there.
0
3 years ago
#5
(Original post by TheGibbsy)
Thanks for the pointer. Using Sin (A+B) i've changed the equation to:
Sin(2θ+θ) = Sinθ and from there using the rule,
Sin2θcosθ + Cos2θsinθ = sinθ
However, I've hit a wall again as to where to go from there.
You need to express the LHS purely in terms of sin θ. Start by using the identites for sin 2θ and cos 2θ, and see if you can then get rid of any remaining cosines that you have.
1
#6
(Original post by Pangol)
You need to express the LHS purely in terms of sin θ. Start by using the identites for sin 2θ and cos 2θ, and see if you can then get rid of any remaining cosines that you have.
Thank you! I never even considered the Identities.
Substituting the identities in:
(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get
Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:
0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π
Thank you! You're a life saver!
0
3 years ago
#7
(Original post by TheGibbsy)
Thank you! I never even considered the Identities.
Substituting the identities in:
(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get
Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:
0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π
Thank you! You're a life saver!
For solving , you can save a lot of time if you consider when two sine functions are equal instead of using identities to change the equation. This method confuses students in my experience but it's very useful.
So clearly one solution is and there are other solutions if you consider the graph / CAST etc.
Let us know if you want to have a go at this method and if you need help with it.
0
3 years ago
#8
(Original post by TheGibbsy)
Thank you! I never even considered the Identities.
Substituting the identities in:
(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get
Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:
0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π
Thank you! You're a life saver!
Following on from Notnek's suggestion you could look at:
sin 3θ - sin θ = 0 then use the formula to turn this into a product
1
3 years ago
#9
(Original post by Muttley79)
Following on from Notnek's suggestion you could look at:
sin 3θ - sin θ = 0 then use the formula to turn this into a product
Yes this is also a nice quick way to do it.
0
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# Quants Questions : Time and Distance Set 4
Hello Aspirants. Welcome to Online Quantitative Aptitude in AffairsCloud.com. Here we are creating question sample in Time and Distance, which is common for all the IBPS,SBI exam and other competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!
1. A Bus covers a distance of 36km at a uniform speed.Had the speed been 6 kmps less, it would have taken one hour more for the journey.The original speed of the bus is
A)30kmps
B)12kmps
C)18mmps
D)20kmps
Explanation :
X^2 – 6x – 216 = 0
x^2-18x+12x-216 = 0
(x+12)(x-18) = 0
X = 18kmph
2. Two buses of same length are running in in the same direction with speed 50km/hr and 80km/ph.The latter completely cross the man in 18sec. The length of the each bus is
A)70m
B)75m
C)80m
D)150m
Explanation :
80 – 50 = 30
30×(5/18) = 25/3 m/s
Distance covered in 20sec = 18 ×(25/3) = 150
Length of each train = 150/2 = 75
3. Rahul started his journey on bike at 7.30pm at a speed of 8km/ph. After 30m, Lenin started his journey from the same place with the speed of 10km/ph. At what time did Lenin overtake Rahul ?
A)10 a.m
B)9 a.m
C)8.30 am
D)8 a.m
Explanation :
Relative speed = 10 – 8 = 2km/ph
Distance covered in 30 min = 8×(30/60) = 4km
Lenin take time to overtake Rahul = 4/2 = 2hr
8+2 = 10 a.m
4. A man can reach a certain place in 30hrs.If he reduces his speed by 1/10 th, he goes 9km less in that time. Find his speed ?
A)8kmph
B)7kmph
C)3kmph
D)1kmph
Explanation :
Let x be the speed
30x – (30*9/10)x = 9
30x – 27x = 9
3x = 9
X= 3 kmph
5. If Anil persons walks at 12kmph instead of 10kmph, he would have walked 18km more.The actual distance travelled by him is
A)100km
B)90km
C)80km
D)60km
Explanation :
x/10 = (x+18)/12
12x = 10x + 180
2x = 180
X = 90km
6. A lady travels equal distances with speed 4kmph,6kmph and 8kmph and takes a total time of 52min.The distance she travelled is
A)4km
B)4.2km
C)4.4km
D)4.8km
Explanation :
(x/4)+(x/6)+(x/8) = 52/60
(6x+4x+3x)/24 = 52/60
13x/24 = 52/60
X = (52×24)/(13×60) = 1.6km
Total distance = 3×1.6 = 4.8km
7. A man crosses a 500m long street in 8min.What is his speed in kmph
A)3.25kmph
B)3.5kmph
C)3.75kmph
D)4kmph
Explanation :
500/(8×60) = 1.04m/sec
1.04×(18/5) = 3.75kmph
8. A bus reached Mumbai from Hyderabad in 30min with an average speed of 50kmph. If the average speed is increased by 35kmph ,How long will it take to cover the same distance ?
A)17min
B)18min
C)19min
D)20min
Explanation :
Time = [ 50 × (30/60) ] / (50 +35)
= [150/60] / 85
= 17.6 / 60 hr = 17.6 min = 18min
9. An aeroplane covers a certain distance at a speed of 240kmph in 5hours. To cover the same distance in 1(2/3) hours it must travel at a speed of
A)700kmph
B)730kmph
C)720kmph
D)710kmph
Explanation :
D = 240×5 = 1200km
New speed = 1200/[5/3] = (1200×3) /5 = 720kmph
10. An athlete runs 240m race in 24 sec.Find his speed
A)30kmph
B)36kmph
C)40kmph
D)63kmph
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https://math.stackexchange.com/questions/3808159/are-the-ideals-of-a-ring-with-cyclic-additive-group-always-principal
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# Are the ideals of a ring with cyclic additive group always principal?
Note for me rings need not be unital or commutative.
Let $$R$$ be a ring with cyclic additive group $$(R, +, 0)$$ and let $$I$$ be an ideal in $$R$$. Is $$I$$ principal?
Here's my attempt, assuming $$R$$ has a $$1$$ and $$1$$ generates the additive group $$(R,+,0)$$:
Since $$(R,+,0)$$ is cyclic and $$(I,+,0)$$ is an additive subgroup of $$(R,+,0)$$, it is also cyclic and generated by some $$a \in R$$. Best guess is $$I = (a)$$.
By definition, as sets $$(I, +, 0 ) = (\langle a \rangle , +, 0) \subseteq (a)$$ . Also if $$x \in (a)$$ then $$x = \sum _i r_i a s_i$$ for some $$r_i, s_i$$. Hence ( using poor notation)
$$x = \sum_i r_i a (1+...+1) = \sum_i r_i (a+...+a) \\ = \sum_i (1+...+1) (a+...+a) = \sum_i ((a+...+a) +... +(a+...+a)) \in (\langle a \rangle, +, 0)$$.
By double inclusion we have the desired equality. $$\blacksquare$$
Firstly is this correct and also what about the case where $$R$$ is not unital or the case where $$R$$ is unital but $$1$$ doesn't generate the additive group?
Many thanks!
EDIT:
For future reference. It is argued here Does the unit generate the additive group in a unital ring with cyclic additive group? that the condition that $$1$$ generates the additive group is infact implied by $$R$$ being unital and is therefore not needed.
Not necessarily. Consider the ideal $$8\mathbb Z$$ within the ring $$4\mathbb Z$$.
Edit: Or, maybe this one is clearer: consider the ideal $$6\mathbb Z$$ within the ring $$2\mathbb Z$$.
• I'm a bit confused. $8\mathbb{Z}$ is principal in $4\mathbb{Z}$ isn't it? Aug 30, 2020 at 9:54
• What ring element would generate it? Aug 30, 2020 at 9:58
• (just to cross word limit) 8? Aug 30, 2020 at 9:59
• Within $4\mathbb Z$, the ideal $(8)=32\mathbb Z\neq 8\mathbb Z$. Aug 30, 2020 at 10:00
• Oh! Nice! I see it now, thanks! Aug 30, 2020 at 10:02
Well, an ideal is an additive subgroup of the given ring. If the additive structure of the ideal is cyclic, then each element of the ideal can be written as $$rg$$, where $$r\in R$$ and $$g$$ is a generator of the additive cyclic structure. Hence, the ideal is principal.
• How do you prove the claim "If the additive structure of the ideal is cyclic, then each element of the ideal can be written as $rg$..."? This is equivalent to saying if $I$ is cyclic then it is principal, which is my question. Many thanks! Aug 30, 2020 at 9:43
• @Bellem Hmmm... it seems to me that you are mixing notation. You are using concatenation both as addition and multiplication, no? Aug 30, 2020 at 9:47
• I have been sloppy, I will answer better. Aug 30, 2020 at 9:55
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https://math.stackexchange.com/questions/4769025/simplify-asymptotic-notation
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# Simplify asymptotic notation
Premise for context: Reading a paper I stumbled upon the following expression $$n(1+p)^{O(\log_{\sigma}n)} + 2$$ which is claimed to be $$O(n)$$ when $$p \in O(1/\log_{\sigma}n)$$, under the (reasonable) assumption that $$n > \sigma$$. Substituting we get $$n\left(1+O\left(\frac{1}{\log_{\sigma}n}\right)\right)^{O(\log_{\sigma}n)} + 2$$
As we are dealing with asymptotic notation I would look only at the higher order terms (this passage actually turned out to be completely wrong, have a look at the comments for details) getting to
$$n\left(1+O\left(\frac{1}{\log_{\sigma}n}\right)^{O(\log_{\sigma}n)}\right) + 2$$
Let now $$x=\log_{\sigma}n$$ and discard the constant to get to $$O(n)+n \cdot O\left(\frac{1}{x}\right)^{O(x)}$$
end of premise, if you like more details feel free to ask them.
To verify the claim we need to check the value of $$O\left(\frac{1}{x}\right)^{O(x)}$$, which should turn out to be $$O(1)$$ to satisfy the initial claim.
Looking at the value of $$y^{-y}$$ which tends to 1 for growing values of $$y$$ in $$\mathbb{N}$$ (e.g. considering $$y^{-y} = e^{-y\cdot \log y }$$ and seeing that the value $$y\cdot \log y$$ goes to $$\infty$$).
This seems correct to me but I was wondering if my (not so formal) reasoning holds or not and whether there is a better line of reasoning to verify this.
• It's not true that $(1+f(n))^{g(n)} = O(1+f(n)^{g(n)})$. (For one thing, increasing $g(n)$ makes the left side increase, but makes the right side decrease when $f(n)<1$.) In this case (and often when dealing with functions in exponents), I recommend writing $(1+p)^{O(g(n))} = \exp( g(n)\log(1+p) )$ and finding an upper bound for $g(n)\log(1+p)$ as an intermediate step; here you should be able to prove that it's $O(1)$. Commented Sep 14, 2023 at 21:39
• Start by $1 + p \le {\rm e}^p$ for $p\ge 0$.
– Gary
Commented Sep 15, 2023 at 4:05
As $$n\to\infty$$, $$n(1+p)^{O(\log_{\sigma}n)}=\exp\left(\ln\left( n\right)+O(\log_{\sigma}n)\ln(1+p)\right)=n\exp(O(\log_{\sigma}n)\ln(1+p))$$ substitue as you did, but note that by Taylor's formula, $$\ln(1+O(1/\log_\sigma n))\sim O(1/\log_\sigma n)$$ as $$n\to\infty$$, then using the fact that $$e^{O(1)}=O(1)$$, i.e., it is bounded, $$n\exp(O(1))=O(n)$$ as $$n\to\infty$$. As you have noticed, discarding the constant $$2$$ which is $$O(1)$$ doesn't matter here.
• Writing $\exp(O(1))\sim 1+O(1)$ does not make much sense. Just note that $\exp(O(1))$ is a bounded quantity, i.e., it is $O(1)$.
• @Gary. You are right, $e^{O(1)}$ is just a number. I will edit that.
• Thanks a lot. In place of Taylor's formula I think we can also use the simpler obesrvation $log(1+x) \le x$ s.t. $x > -1$ (which holds since $x = O(1/\log_{\sigma}n) \ge 0$ in our case) in this case as we are looking for an upper bound right? Commented Sep 15, 2023 at 7:48
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https://hackaday.com/2021/07/21/what-exactly-is-a-gaussian-blur/?replytocom=6371740
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# What Exactly Is A Gaussian Blur?
Blurring is a commonly used visual effect when digitally editing photos and videos. One of the most common blurs used in these fields is the Gaussian blur. You may have used this tool thousands of times without ever giving it greater thought. After all, it does a nice job and does indeed make things blurrier.
Of course, we often like to dig deeper here at Hackaday, so here’s our crash course on what’s going on when you run a Gaussian blur operation.
## It’s Math! It’s All Math.
Digital images are really just lots of numbers, so we can work with them mathematically. Each pixel that makes up a typical digital color image has three values- its intensity in red, green and blue. Of course, greyscale images consist of just a single value per pixel, representing its intensity on a scale from black to white, with greys in between.
Regardless of the image, whether color or greyscale, the basic principle of a Gaussian blur remains the same. Each pixel in the image we wish to blur is considered independently, and its value changed depending on its own value, and those of its surroundings, based on a filter matrix called a kernel.
The kernel consists of a rectangular array of numbers that follow a Gaussian distribution, AKA a normal distribution, or a bell curve.
Our rectangular kernel consists of values that are higher in the middle and drop off towards the outer edges of the square array, like the height of a bell curve in two dimensions. The kernel corresponds to the number of pixels we consider when blurring each individual pixel. Larger kernels spread the blur around a wider region, as each pixel is modified by more of its surrounding pixels.
For each pixel to be subject to the blur operation, a rectangular section equal to the size of the kernel is taken around the pixel of interest itself. These surrounding pixel values are used to calculate a weighted average for the original pixel’s new value based on the Gaussian distribution in the kernel itself.
Thanks to the distribution, the central pixel’s original value has the highest weight, so it doesn’t obliterate the image entirely. Immediately neighboring pixels having the next highest influence on the new pixel, and so on. This local averaging smoothes out the pixel values, and that’s the blur.
Edge cases are straightforward too. Where an edge pixel is sampled, the otherwise non-existent surrounding pixels are either given the same value of their nearest neighbor, or given a value matching up with their mirror opposite pixel in the sampled area.
The same calculation is run for each pixel in the original image to be blurred, with the final output image made up of the pixel values calculated through the process. For grayscale images, it’s that simple. Color images can be done the same way, with the blur calculated separately for the red, green, and blue values of each pixel. Alternatively, you can specify the pixel values in some other color space and smooth them there.
Here we see an original image, and a version filtered with a Gaussian blur of kernel size three and kernel size ten. Note the increased blur as the kernel size increases. More pixels incorporated in the averaging results in more smoothing.
Of course, larger images require more calculations to deal with the greater number of pixels, and larger kernel sizes sample more surrounding pixels for each pixel of interest, and can thus take much longer to calculate. However, on modern computers, even blurring high-resolution images with huge kernel sizes can be done in the blink of an eye. Typically, however, it’s uncommon to use a kernel size larger than around 50 or so as things are usually already pretty blurry by that point.
The Gaussian blur is a great example of simple mathematics put to a powerful use in image processing. Now you know how it works on a fundamental level!
## 29 thoughts on “What Exactly Is A Gaussian Blur?”
1. quinor says:
That’s a great paper. I’ve been using a hillbilly version of that algorithm for my hanging plotter project but this one is significantly nicer! How did you stumble upon it?
1. I don’t remember … It was years ago where I implemented an BW filter for laser.im-pro.at ….
1. Steven Clark says:
Also the classic Sharpen filter is a RREF-style reversal of a gaussian blur kernel. Meanwhile Unsharp Mask is a gaussian blur applied in negative which is why it’s easier to parametrize: there’s no solving step.
2. snarkysparky says:
now how about inverting an applied gaussian blur to recover the prior image.
1. RustyHydrogen says:
Ah, the fun worlds of deconvolution, and blind deconvolution.
3. grounded says:
And if you apply the same blur sequentially? Are two 3×3 blurs the same as a single 5×5?
I guess I should do some math…
1. Marcus says:
Yep, doing the math helps here, to some degree:
the convolution of a Gaussian with a Gaussian is still a Gaussian (without that, there wouldn’t be the central limit theorem, i.e. the reason why sums of independent identically distributed random variables add up to Gaussian distribution for large numbers of summands). So, yes, if we take a Gaussian with some sigma² and convolve it with itself, we get a new Gaussian with twice the sigma² (Variance).
This means that if we decided the original discrete 3×3 kernel was “large” enough to cover the “interesting parts of the Gaussian because the stuff that was “cut off” from the (infinitely extending) Gaussian function was small enough for our tastes to not matter.
(I’m ***very*** much at odds with the article: “larger kernels spread further”: No; you can put the same 3×3 in a 11×11 matrix and fill in with the smaller values from the actual Gaussian function. What the author means is that a higher-variance Gaussian spreads further.)
But 5 is not at least twice 3… so, you’d get a worse approximation to an actual Gaussian filter with this finite kernel.
However 5×5 is indeed the size you get when you, in the discrete domain, convolve two 3×3 kernels. So, by doing it that way, you get a different Gaussian kernel, but it’s a worse approximation to the Gaussian of twice the variance of the Gaussian in the 3×3 kernel.
1. Lasse says:
Also the Gaussian is separable, so you can do the 2D blur by doing a 1D horizontally and then vertically.
4. Marcus says:
Is it just me that’s bothered by the logo in the post header *not* being the result of the sharp hackaday logo being filtered with a symmetric kernel,i.e. especially *not* with a Gaussian? That blur isn’t symmetrical at all!
1. Chris Cox says:
Nope, not just you. That kinda jumped out at me as well.
5. Marcus says:
Honestly, I like when HaD does such articles!
I must still voice my disagreement on a few things: the first two are fundamental, the third I think is a bit of an overgeneralization but not a fundamentally wrong statement; my apologies.
> The kernel consists of a rectangular array of numbers that follow a Gaussian distribution,
These numbers don’t follow a Gaussian distribution. They are deterministic; they are samples taken from the *Gaussian function*, which happens to be the probability density of the Gaussian distribution.
> Larger kernels spread the blur around a wider region,
This is wrong – the size of the kernel doesn’t define the spread, just the number of pixels taken into account.
The spread is defined by the variance of the Gaussian function that gets sampled for the kernel. You can have a very narrow blur “spread” in a large filter, if you need the accuracy, or a very wide (=high variance) blur in a small filter.
The question is really just how much of the actual continuous function you cut off when you decide on your filter size.
But these are two different properties of the filter, and you’re conflating them!
> as each pixel is modified by more of its surrounding pixels.
That’s right – a larger Kernel takes more pixels into account.
> Edge cases are straightforward too. Where an edge pixel is sampled, the otherwise non-existent surrounding pixels are either given the same value of their nearest neighbor, or given a value matching up with their mirror opposite pixel in the sampled area.
These are **two** options, but they are not necessarily the right ones for any particular application. In fact, that choice can become very awkward for the pixels that are not at the very image border – the actual border pixel get their “weight” increased very much, so that their values overshadow that of their neighbors.
So, not a great generalization to make!
6. echodelta says:
Now for a math routine to put shake and jiggle into stable video and stills for enhanced viewing.
1. Nick says:
Oh god, don’t give them ideas! We already have fake film lines and blemishes, and fake artefacts and glitches. Overuse of those is bad enough.
7. Ian Dobbie says:
I would question the large kernels taking significant processing time. Applying the blur is a convolution and almost any sensible algorithm will actually compute this as a multiplication in Fourier space. Take your kernel and image, FFT both, multiple them together and then do in the inverse FFT. Almost certainly massively faster than a direct real space convolution where you generate a new image, multiple every pixel by the kernel and sum the the relevant pixels into the results image.
1. ROFL. Nope. Gaussian kernels are separable, and can use the central limit theorem, plus some linear and constant time algorithms, to do the work much more quickly than you could do a 2D FFT (much less the multiply and reverse FFT). Plus the separability makes the operation much more cache friendly than an FFT. (see also: bicubic and sinc based resampling kernels)
If you are doing a convolution that isn’t separable, then an FFT (or other frequency space transform) might be a good idea. But please talk to algorithms and performance experts first, and measure, measure, measure.
1. intheoryonly says:
The source you linked to only considers theoretical number of operations. In practice, memory access order and caches are far more important. The crossover point where FFT-based convolution is faster than separable convolution will only happen for much larger inputs.
2. Yeah, that completely ignores better algorithms and cache effects. It’s using the number of operations for a direct convolution – not separable, not box filters (certainly not the constant time algorithms for box filters).
Next time, you probably should consult with someone who actually knows algorithms and performance.
1. snarkysparky says:
These people at Analog Devices are idiots. Shouldn’t consult them at all.
“”This chapter presents two important DSP techniques, the overlap-add method, and FFT
convolution. The overlap-add method is used to break long signals into smaller segments for
easier processing. FFT convolution uses the overlap-add method together with the Fast Fourier
Transform, allowing signals to be convolved by multiplying their frequency spectra. For filter
kernels longer than about 64 points, FFT convolution is faster than standard convolution, while
producing exactly the same result. “”
https://www.analog.com/media/en/technical-documentation/dsp-book/dsp_book_Ch18.pdf
Mr Algorithm guru would you be so kind as to point me to where I can gain understanding about this from the REAL experts?
2. Yes sparky, you are in way over your head. FFTs are good in some places, and not in many others. In real world image processing, an FFT for gaussian blur would be among the slowest algorithms (only exceeded by brute force convolution). You are confusing theory (written by people who don’t know the practice and who are talking entirely in generalities) with actual practice. I’m talking about actual code in major applications that has been highly optimized, measured, copied by other applications, and hits the limits for throughput on a variety of architectures and inputs (image size, kernel size, etc.).
How can you gain experience? Listen to the experts and stop trying to claim you know more than they do by quoting other people who know even less.
Here’s a hint: click on the link in my name above.
3. snarkysparky says:
Ok smartie Chris. You win. All i did was google fft vs convolution and the first 63455 articles all went with fft as faster once past a very minimal kernel size. I looked for something agreeing with you but couldn’t find it. One thing i have always noticed is that when someone makes the boneheaded claim about the difference between theory and practice they don’t actually know the theory. What they have in brain is a conglomeration of anecdotes that support their view.
You thought you could “one up” the experts by looking things up on Google. But you got the theory (and ignored all the warning signs that it was just theory), with zero practice and zero experience.
Again, you probably should click on the link in my name here to learn who it is that you are addressing. Yes, I know the theory, and I lecture on it a few times a year at the larger universities here in the SF Bay Area. I also spent many years on the practice: measuring everything, improving the algorithms until they were limited by the speed of RAM instead of the CPU, squeezing out a bit more performance with significant cache optimization, then re-optimizing for each new major chip architecture. There is a reason why the chip and motherboard makers consult with me about their latest designs and how their changes will affect real world throughput…
5. Hallo Chris,
I really appreciate your input to this topic. I’m now myself working in a research institute and my experience is not really good. We have an high incentive to write publications and there the quality is often not optimal. Furthermore when doing research I takes a lot of effort to filter out bad papers and find the practical problems.
I hob if somebody searches about FFT for convolution this will show up to save them a lot of time to go throw with the implementation and find in the end that it will not be faster for real applications.
Thanks Chris!
@sparky here is one link about chis you should open : http://www.photoshophalloffame.com/chris-cox
8. Chris Maple says:
The article would be greatly improved if the actual formula for pixel weighting was provided.
9. brandon says:
you guys are gaussian blurring and im still reticulating splines :(
10. Alan says:
If you have any interest in stitching images together (to make a panorama) then it helps to know about Gaussian blur.
Blurring to different kernel sizes, then subtracting (Difference of Gaussians), is one of the steps in keypoint detection algorithms like SIFT and SURF.
https://en.wikipedia.org/wiki/Image_stitching#Keypoint_detection
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ex2-sol10
# ex2-sol10 - AMS 341 (Spring, 2010) Exam 2 - Solution notes...
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Unformatted text preview: AMS 341 (Spring, 2010) Exam 2 - Solution notes Estie Arkin Mean 73.925, median 76, top quartile 87, high 99, low 12. 1. (10 points) I am planning a vacation once the spring semester is over. There are many tasks that have to be completed before I can go. Fortunately, I can get help from family and friends. The following are the tasks that would have to be undertaken before the vacation: Task Predecessors Time (hours) A- 2 B- 3 C A 1 D B 8 E B,C 7 F D,E 5 (a). Draw a project network. 1 2 3 4 5 6 A-2 B-3 C-1 dummy-0 D-8 E-7 F-5 Common mistakes: several start nodes, undirected edges, wrong predecessors. (b). What is my critical path? You may find the path either by computing the total float for each node, or by inspection. (Your answer should be a list all critical activities.) Tasks B,D,F (or nodes 1,3,5,6). 2. (10 points) Consider the following (minimum) Balanced Transportation problem: Find an initial BFS for the problem using the min cost method: 1 100 100 100 75 75 125 25 1 2 3 4 5 6 7 8 9 1 1 25 75 100 25 75 Common mistake: not enough basic variables. You should have 4 + 3 − 1 = 6. 3. (15 points) Plans are being made for the energy systems for a new building. The three possible sources of energy are electricity, natural gas and a solar heating unit. The energy requirements are: 20 units of electricity, 10 units for water heating and 30 units for space heating. The size of the roof limits the solar heater to 30 units. There is no limit on the amount of electricity or natural gas bought. Electricity needs can only be met by buying electricity at a cost of \$ 50 per unit. Other energy needs can be met by any sources with the following unit costs: Electricity Natural Gas Solar heater Water heating \$ 90 \$ 60 \$ 30 Space heating \$ 80 \$ 50 \$ 40 Formulate a Balanced Transportation Problem to minimize the...
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## This note was uploaded on 05/28/2011 for the course AMS 341 taught by Professor Arkin,e during the Spring '08 term at SUNY Stony Brook.
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https://www.bartleby.com/questions-and-answers/ou-have-been-given-the-following-information-for-moores-honeybee-corp.-net-sales-dollar44000000.-gro/4f4e8e93-02ec-40ca-b0dc-f2060c6fb2b1
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|
text/html
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crawl-data/CC-MAIN-2020-16/segments/1585370528224.61/warc/CC-MAIN-20200405022138-20200405052138-00323.warc.gz
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# ou have been given the following information for Moore’s HoneyBee Corp.: Net sales = \$44,000,000.Gross profit = \$19,400,000.Other operating expenses = \$3,400,000.Addition to retained earnings = \$8,328,000.Dividends paid to preferred and common stockholders = \$2,100,000.Depreciation expense = \$2,000,000. The firm’s tax rate is 21 percent. The firm's interest expense is all tax deductible.Calculate the cost of goods sold and the interest expense for Moore’s HoneyBee Corp. (Round your answers to the nearest dollar amount.) Cost of goods sold Interest expense
Question
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ou have been given the following information for Moore’s HoneyBee Corp.:
1. Net sales = \$44,000,000.
2. Gross profit = \$19,400,000.
3. Other operating expenses = \$3,400,000.
4. Addition to retained earnings = \$8,328,000.
5. Dividends paid to preferred and common stockholders = \$2,100,000.
6. Depreciation expense = \$2,000,000.
The firm’s tax rate is 21 percent. The firm's interest expense is all tax deductible.
Calculate the cost of goods sold and the interest expense for Moore’s HoneyBee Corp. (Round your answers to the nearest dollar amount.)
Cost of goods sold Interest expense
check_circle
Step 1
Recall the mathematical equation:
Gross Profit = Net Sales - Cost of Goods sold
Hence, 19,400,000 = 44,000,000 - Cost of Goods sold
Hence, Cost of Goods sold = 44,000,000 - 19,400,000 = \$ 24,600,000
Step 2
Recall the fundamental equation of accounting of net income:
Net income = Addition to retained earnings + Dividends paid to preferred and common stockholders =(Gross Profit - Other Operating expenses - Depreciaton - Interest) x (1 - Tax rat...
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# Category Archives: Maths MAQ
The margin of error is an important measure in statistics. The degree of error in random sampling surveys is known as the margin of error.… Read More
Cubic Inches to Cubic Centimeter conversion is widely used in mathematics. A cubic centimeter and cubic inches are the units of measurement of volume. Volume… Read More
Volume is defined as the measure of space occupied by a three-dimensional object. The volume of a 3D geometric object is expressed quantitatively using SI-derived… Read More
A rectangular pyramid is a three-dimensional object that has a rectangular base upon which are erected four triangular faces that meet at a common point… Read More
A cubic meter and a cubic yard are the units of measurement of volume. Volume is a mathematical quantity that is used to measure the… Read More
Cubic centimeters to liters conversion is very useful for converting different units of volumes in our daily life. A cubic centimeter and a liter are… Read More
In mathematics, a numeral or number system is a writing system for expressing numbers in various forms. It is a mathematical notation for expressing numbers… Read More
A cubic foot and an imperial gallon are the units of measurement of a unit volume. Volume is defined as the measure of space occupied… Read More
A cubic yard and a cubic inch are the units of measurement of a unit volume. Volume is defined as the measure of space occupied… Read More
A function is defined as the relation between a set of inputs and their outputs, where the input can have only one output. It depicts… Read More
A rectangular pyramid is a three-dimensional object that has a rectangular base upon which are erected four triangular faces that meet at a common point… Read More
Volume is a mathematical quantity that is the measure of space occupied by a three-dimensional object. The volume of a three-dimensional object varies with the… Read More
Concurrent lines are line segments, two or more, crossing through a single point of intersection. The point is called the point of concurrency. The point… Read More
In geometry, a pentagonal pyramid is a three-dimensional figure with a pentagonal base upon which five triangular faces are erected and meet at a meeting… Read More
Volume is a mathematical quantity that is the measure of space occupied by a three-dimensional object. The volume of a three-dimensional object varies with the… Read More
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This site is being phased out.
# Calculus III -- Spring 2014 -- final exam
Name:_________________________
10 problems, 10 points each
• Justify every step you make with as thorough explanation as possible.
• Unless requested, no decimal representation of the answers is necessary.
• Start every problem at the top of the page.
$\bullet$ 1. (1) Sketch the parametric curve $x=\cos t,y=\sin 2t$. (2) The curve intersects itself. Find the angle of this intersection.
$\bullet$ 2. Sketch the graph of a function of two variables $z=f(x,y)$ the derivatives of which have the following signs: $f_x>0, f_{xx}>0, f_y<0, f_{yy}<0$.
$\bullet$ 3. The graph of a function of two variables $z=f(x,y)$ is given below along with four points on the graph. Sketch the gradient for each on a separate $xy$-plane:
$\bullet$ 4. Give the definition of the curvature. Give examples of curves with various curvatures.
$\bullet$ 5. A vector field is sketched below. Suppose $C$ is the clockwise oriented square centered at the origin. Is $\int _{C}\mathbf{F}\cdot d\mathbf{r}$ positive, negative or $0$? Explain.
$\bullet$ 6. This is the formula of Green's Theorem: $$\oint_{C} (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, dA.$$ Explain its parts and the conditions that have to be satisfied.
$\bullet$ 7. Prove that the vector field $F(x,y,z)=z\mathbf{j} - y\mathbf{k}$ is not conservative.
$\bullet$ 8. Make a sketch of contour (level) curves for the function below:
$\bullet$ 9. Sketch the vector field $$F(x,y)=\dfrac{1}{x^{2}+y^{2}}(y\mathbf{i}-x\mathbf{j}).$$
$\bullet$ 10. Find the work done by force field $$F(x,y)=< xy,y^{2}>.$$ in moving an object along the parabola $x=t,y=t^{2},0\leq t\leq1.$
$\bullet$ 11. Use a Riemann sum with $8$ terms to estimate the value of the integral $$\iiint_{D}(x+y+z)dV,$$ over the cube $D=[0,1]\times[0,1]\times[0,1]$. Choose your own sample points.
$\bullet$ 12. (1) Represent the cylinder of radius $1$ and height $1$ centered on the $z$-axis as a parametric surface. (2) Find the tangent plane to the cylinder at the point $(\sqrt{2}/2,\sqrt{2}/2,1/2)$. (3) Compute the flux of the vector field $F=< 2,1,1 >$ across the part of the cylinder that lies in the first octant.
$\bullet$ Extra credit problem. (5 points, no partial credit) Suppose you are towing a trailer-home. During the first few minutes, every time you look at the rear view mirror you can see only the lower part of the home. Later, every time you look you can see only the top part. Discuss the profile of the road.
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https://www.esaral.com/q/find-the-coordinates-of-the-foci-the-vertices-the-length-of-major-axis-the-minor-axis-the-eccentricity-and-the-length-81740
| 1,725,882,353,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651098.19/warc/CC-MAIN-20240909103148-20240909133148-00624.warc.gz
| 704,819,564
| 11,920
|
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length
Question:
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $16 x^{2}+y^{2}=16$
Solution:
The given equation is $16 x^{2}+y^{2}=16$.
It can be written as
$16 x^{2}+y^{2}=16$
Or, $\frac{x^{2}}{1}+\frac{y^{2}}{16}=1$
Or, $\frac{x^{2}}{1^{2}}+\frac{y^{2}}{4^{2}}=1$ $\ldots(1)$
Here, the denominator of $\frac{y^{2}}{4^{2}}$ is greater than the denominator of $\frac{x^{2}}{1^{2}}$.
Therefore, the major axis is along the $y$-axis, while the minor axis is along the $x$-axis.
On comparing equation (1) with $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, we obtain = 1 and a = 4.
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{16-1}=\sqrt{15}$
Therefore,
The coordinates of the foci are $(0, \pm \sqrt{15})$.
The coordinates of the vertices are $(0, \pm 4)$.
Length of major axis = 2a = 8
Length of minor axis = 2b = 2
Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{15}}{4}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 1}{4}=\frac{1}{2}$
| 441
| 1,179
|
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| 4.625
| 5
|
CC-MAIN-2024-38
|
latest
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en
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https://math.stackexchange.com/questions/15620/questions-about-well-order
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# questions about well-order
1. In Wikipedia, well-order is defined as a strict total order on a set $S$ with the property that every non-empty subset of $S$ has a least element in this ordering.
But then later, well-order is defined as a total order on $S$ with the property that every non-empty subset of $S$ has a least element in this ordering.
As far as I know, a total order and a strict total order are different. One is not the other. So I was wondering if well-order is defined for total order or strict total order or both? If for both, are they equivalent in the sense that if a total order is well-order, then its corresponding strict total order is also well-order? Vice versa?
2. At the same Wikipedia page, it also says "a well-ordering is a well-founded strict total order". As I clicked into the definition of Well-founded_relation, it says "a binary relation, $R$, is well-founded (or wellfounded) on a class $X$ if and only if every non-empty subset of $X$ has a minimal element with respect to $R$". As minimal element is defined for partial order not for strict total order, is it true that well-founded order is a partial order and not a strict total order? So the aforementioned "a well-ordering is a well-founded strict total order" is not well-stated?
Thanks and regards!
If $\leq$ is a total order on a set $S$, then the new relation $<$ defined by $x < y$ iff ($x \leq y$ and $x \neq y$) is a strict total order on $S$.
If $<$ is a strict total order on a set $S$, then the new relation $\leq$ defined by $x \leq y$ iff ($x < y$ or $x = y$) is a total order on $S$.
In other words, in a fairly evident way one can always exchange a total order for a strict total order and conversely. So it doesn't really matter which definition is taken. One can easily check that the definition of a well-order in one setting carries over to the definition of a well-order in the other setting.
• This is one of the simplest examples of the clarifying power of category: although set-theoretically a total order and a strict total order on a set are not the same thing, categorically the category of total orders and strict total orders are isomorphic (at least if one defines morphisms correctly). Dec 27 '10 at 8:10
• Qiaochu: Clarity is for chumps ;-) Dec 27 '10 at 8:12
• @Qiaochu [why is this considered a power?] ops I mis-read the English sentence. Never mind!
– user2468
Dec 27 '10 at 8:51
For the first question, taking strict and non-strict orders to be well-ordering is up to you and the way you use it.
However, in set theory it is generally easier to use strict ordering in the definition because it saves the trouble with $x\le y\wedge y\le x$, furthermore we want $\in$ to define some relations and it has to be strict by the axiom of foundation (also known as axiom of regularity in some parts of the globe).
As for minimal elements, they are not only for partial orders - but for any order. If $R$ is some relation on $A$ then $x$ is $R$-minimal if $\forall y (yRx \rightarrow y=x)$, note by the way that if $R$ is not reflexive then this is still true, but you could phrase it as $\forall y\neg(yRx)$ instead, which is clearer.
The best way, in my opinion to understand deeply these choices of definitions is to study some set theory theorems about recursive definitions, transfinite inductions and the needed theorems for those. In these proofs it becomes very clear why one prefer strict relations over non-strict ones.
One final remark, although strict and non-strict total orders are "very different", they only differ by reflexivity which is some vacuous condition that you want to add when it's easier to have it - and you remove it when you find it easier to handle without it.
1. If a set is totally ordered, the strict total order is implied.
Definition of total order from Wikipedia:
$\forall a, b, c \in X$
If $a ≤ b$ and $b ≤ a$ then $a = b$ (antisymmetry);
If $a ≤ b$ and $b ≤ c$ then $a ≤ c$ (transitivity);
$a ≤ b$ or $b ≤ a$ (totality).
Definition of strict total order from Wikipedia:
$\forall a, b \in X$
$a < b$ if and only if $a ≤ b$ and $a ≠ b$
$a < b$ if and only if not $b ≤ a$
If you compare the definitions, you will see that there are no conditions under which the second set of rules is invalid, assuming the first set of rules. So, saying that a set is well-ordered if it is totally ordered makes sense, because total order implies existence of strict total order (and hence, well-ordering) on the given set.
2. I am fairly sure that one can use the term 'minimal element' with respect to totally ordered sets also.
• "So, saying that a set is totally ordered if it is well-ordered makes sense" you mean the other way around. The integers have a strict order which is not a well-order. Dec 27 '10 at 8:10
• Thank you, made a typo. Asaf, could you please explain why well-order (strict total order) does not work on integers? I am blanking here. :) Dec 27 '10 at 8:21
• I didn't say that there is none. Just that there is one which is strict and not well-ordered, namely the usual $<$ relation on integers... It has no minimal element so it is not well founded, and therefore not well ordered :) Dec 27 '10 at 8:31
• Oh I see what you are saying, I was blanking in a different direction. Thanks. :) Dec 27 '10 at 8:32
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Physics Problem
Discus: SAT/ACT Tests and Test Preparation: October 2003 Archive: Physics Problem
By Mo222 (Mo222) on Wednesday, October 01, 2003 - 04:04 pm: Edit
How fast must a ball be thrown upward to reach a height of 12 m?
By Euphoria (Euphoria) on Wednesday, October 01, 2003 - 04:14 pm: Edit
15.3 m/s
By Geniusash (Geniusash) on Wednesday, October 01, 2003 - 04:17 pm: Edit
iv=x
fv=0
h=12
use equation, fv^2=iv^2+2a(h) (fv=final velocity, iv=initial velocity, a=acceleration, h=height)
0=x^2+2(-9.8)(12)
235.2=x^2
x=sqr235.2 (no calc, sorry)
By Perry2006 (Perry2006) on Wednesday, October 01, 2003 - 04:21 pm: Edit
Hope this helps.
Method 1 (Comprehensive)
0=vi + -9.81t
t=vi/9.81
12=vi (vi/9.81) + .5(-9.81)(vi/9.81)^2
solve for vi.
Method 2 (Easy plug-in - Recommended)
Vf^2 = Vi^2 + 2aX
0^2 = vi^2 + 2(-9.81)(12)
solve for vi.
By Mo222 (Mo222) on Wednesday, October 01, 2003 - 04:26 pm: Edit
Thanks --
The thing I don't get is why you are using -9.81 m/s^2 for the acceleration. How can all objects that are thrown up decellerate at a constant rate? Dosent it matter how strong the push upwards is?
By Fairyofwind (Fairyofwind) on Wednesday, October 01, 2003 - 04:31 pm: Edit
The standard way to do it is: the derivative of position is velocity. Ther derivative of velocity is acceleration. So integrate. a=-9.8, v=-9.8t+v0, s=-4.9t^2+v0t+s0.
Solve -9.8t+v0=0, t=v0/9.8, -4.9(v0/9.8)^2+(v0)(v0/9.8)+s0=12, s0=0, so v0=15.3.
15.3 m/s.
Alternatively, you can use the cheesy little kinematic equation that doesn't work when acceleration isn't constant: V_f^2=V_i^2+2a(delta S)
By Geniusash (Geniusash) on Wednesday, October 01, 2003 - 04:33 pm: Edit
No, because the pull of gravity is the same on everything, no matter how fast it's going.
By Mo222 (Mo222) on Wednesday, October 01, 2003 - 04:38 pm: Edit
Yea but it's going up so its acceleration will be made less by gravity, but woulden't the rate at which it goes upward still depend on how strong the push up is? If there is a really strong person who throws it then it will be affected by gravity the same as if a really weak person throws it, but the acceleration will be different. What am I thinking wrong?
By Fairyofwind (Fairyofwind) on Wednesday, October 01, 2003 - 04:41 pm: Edit
It's not acceleration. You're thinking initial velocity. When the ball leaves the person's hand, it starts off at a nonzero velocity. There is no acceleration involved there. But remember acceleration has a direction. So say it starts off at 60 m/s UPWARDS. Gravity will accelerate it downwards at 9.8 m/s^2, so that after one second, it'll be moving at 50.2 m/s (STILL UPWARDS), etc. until it becomes negative, then the ball starts moving downwards. This is why you are looking for the time the velocity = 0.
By Geniusash (Geniusash) on Wednesday, October 01, 2003 - 04:42 pm: Edit
gravity=the acceleration
By Geniusash (Geniusash) on Wednesday, October 01, 2003 - 04:44 pm: Edit
Fairy, have you looked at that Probability problem that's posted on this board? Am I just getting REALLY stupid?
By Mo222 (Mo222) on Wednesday, October 01, 2003 - 04:45 pm: Edit
OK I think I get it. Thanks fairy.
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https://assignmentgrade.com/question/663914/
| 1,590,962,839,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00527.warc.gz
| 261,702,157
| 8,469
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If the sales tax rate is 7 percent, what is the sales tax on a \$50 mp3 player? - AssignmentGrade.com
# If the sales tax rate is 7 percent, what is the sales tax on a \$50 mp3 player?
QUESTION POSTED AT 23/05/2020 - 04:30 PM
The mp3 player costs \$53.50
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## Sunday, July 12, 2015
### Playing with Math: Can you write a review?
Playing with Math: Stories from Math Circles, Homeschoolers, and Passionate Teachers is on Amazon now! But we don't yet have any reviews. If you've gotten a copy of the book, can you write a review on Amazon? We would be so grateful.
Warmly,
Sue
## Friday, July 10, 2015
I'll be leading a Math Jam for eight days just before Fall semester starts, helping students prepare to succeed in Beginning Algebra. My eight topics:
1. Number Sense
2. Fractions
3. Negatives
4. Algebra
5. Percents
6. Graphing
7. Slopes
8. Problem-Solving
For fractions, I plan to do a bit with Egyptian Fractions. Here's a site that looks good for that. I looked at the Beast Academy site to see if they had anything good. I found 5 things I liked: one game and two puzzles using the area meaning of multiplication, one puzzle on ordering of decimals, and one game like Taboo for communicating about shapes.
## Thursday, July 2, 2015
### Playing with Math: Inspiring Online Conversations
First sighting of a comment on a mathematical blog post that was inspired by seeing the content in my book...
Jonathan Halabi writes jd2718. His post, Puzzle: Who am I?, became one of the puzzles in Playing with Math: Stories from Math Circles, Homeschoolers, and Passionate Teachers.
Today Lara H replied to his post:
I came across this puzzle in the book “Playing with Math.” I found a different solution based on a wrong assumption I made at the beginning of solving the puzzle. I was thinking that a number with 3 digits also has 2 digits so I made both of those statements true and came up with 4097, which works for all the other conditions.
I responded with:
I’d say ‘different interpretation’ instead of ‘wrong assumption’. I wonder how many solutions the puzzle has using your interpretation. (Pretty exciting to see my book has inspired new discussion on Jonathan’s blog post!)
We are hoping that the book will inspire online conversations. This is the first drop of what we hope will eventually become a deluge.
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Math 3150, Section 1, Fall 2017
# Analysis I
Instructor's Office Hours:
Office hours are in Monteith 326
(most) Wednesdays 11:00 - 12:00
(most) Thursdays 2:00 - 3:00
Important Dates
In-Class Midterm Exam: Friday, October 27
Covers Chapters 1 and 2
## Portfolio Problems
Some Remarks on Writing Mathematical Proofs
by Jack Lee
1st Problem Set: Due September 29
(Now includes correction to Problem 1)
2nd Problem Set: Due October 20
3rd Problem Set: Due November 17
Update: For the final revisions, you only need to turn in revisions to 5 problems. The choice of problems is up to you.
## Suggested Exercises
It is a good idea to work out exercises from each section of the book that we cover. These will not be collected.
Section 1.1: Exercises 1.1.1, 1.1.2, 1.1.4, 1.1.6, 1.1.8, 1.1.9, 1.1.10, 1.1.11 (lookup definition of equivalence relation), 1.1.12
Section 1.2: Exercises 1.2.2, 1.2.4, 1.2.5, 1.2.6, 1.2.7, 1.2.8
Section 1.3: Exercises 1.3.2, 1.3.3, 1.3.4, 1.3.5, 1.3.6, 1.3.7, 1.3.8, 1.3.9, 1.3.10, 1.3.16
Section 1.4: Exercises 1.4.1, 1.4.2 (can you give two proofs?), 1.4.4, 1.4.6, 1.4.7, 1.4.10, 1.4.12, 1.4.13
Section 2.1: Exercises 2.1.2, 2.1.3, 2.1.4, 2.1.7
Section 2.2: Exercises 2.2.6, 2.2.3, 2.2.7, 2.2.8, 2.2.11, 2.2.14
Section 2.3: Exercises 2.3.1, 2.3.4, 2.3.5, 2.3.12
Section 2.4: Exercises 2.4.1, 2.4.2, 2.4.3, 2.4.4, 2.4.5 (use a lemma we proved in class), 2.4.7, 2.4.8
Section 3.1: Exercises 3.1.1, 3.1.2, 3.1.3, 3.1.4, 3.1.7, 3.1.11
Section 3.2: Exercises 3.2.1, 3.2.2, 3.2.3, 3.2.4, 3.2.5, 3.2.7, 3.2.10
Section 4.1: Exercises 4.1.1, 4.1.2, 4.1.3, 4.1.5
Section 4.2: Exercises 4.2.1, 4.2.2, 4.2.4, 4.2.10, 4.2.13
Last updated: November 17, 2017
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http://physics.stackexchange.com/questions/78323/motion-in-a-straight-line
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# Motion in a straight line? [closed]
I am trying to solve the following problem that involves motion in a straight line. I am not quite sure how to go about it though. I know it will involve formulas and derivatives but which specific ones would I use?
An electron has a constant acceleration of 2.6 m/s^2. At a certain instant its velocity is 12.1 m/s.
What was its velocity 2.5 s earlier? What was its velocity 2.5 s later?
-
## closed as off-topic by John Rennie, akhmeteli, Qmechanic♦Sep 23 '13 at 20:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, akhmeteli, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.
For constant acceleration, $\Delta v = a \Delta t$. – Alfred Centauri Sep 23 '13 at 17:27
Unless the particle being an electron is a catch, standard equations of motion can be used. $v= u + at$ with symbols having usual meanings. for 2.5 sec earlier: $v = 12.1 \mathrm{\ m/s}$, $a = 2.6 \mathrm{\ m/s^2}$, $u = ?$, $t = 2.5 \mathrm{\ s}$
Similarly for 2.5 sec later $v = ?$, $a = 2.6 \mathrm{\ m/s}^2$, $u = 12.1\mathrm{\ m/s}$, $t = 2.5 \mathrm{\ s}$.
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crawl-data/CC-MAIN-2018-43/segments/1539583516480.46/warc/CC-MAIN-20181023153446-20181023174946-00129.warc.gz
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# Intro
(Throughout this post I will be using the convention $\log \equiv \log_e$ and $\lg \equiv \log_2$).
In my post on Shannon entropy I introduced the notion of information entropy with respect to discrete random variables. There we derived, albeit in a hand-wavy fashion, the entropy formula:
$H(X) = H(p_1,\ldots,p_n) = -\sum p_i \cdot \lg p_i$
I hinted at the fact that $H$ is maximized when the underlying random variable $X$ is distributed uniformly. We saw this in the coin-toss example by seeing how any deviation from a fair-coin reduced the entropy.
My aim in this post is to prove (at least in the discrete case) that $H$ is indeed maximized when $X$ is distributed uniformly.
As an aside, one thing I neglected to mention in my original post on entropy is how $H$ is computed when $p = 0$. By convention, $-p\cdot \lg p \equiv 0$ when $p = 0$.
This agrees with our intuition since the case $p = 0$ is analogous to the case $p = 1$ in that the event will almost-surely not occur. And recall that $-1\cdot \lg 1 = 0$. So $p = 0$ can be seen as the symmetric case to $p = 1$, making the above convention not far-fetched.
You can actually show using methods of indeterminate forms that $p\cdot \lg p \to 0 \text{ as } p \to 0$. So that if we interpret the product as a limit $p\to 0$ the above equality ceases to be a convention but instead a fact.
$\log x \leq x - 1 \enspace \forall x >0$
Where equality holds if and only if $x = 1$.
# Step 1
The proof follows directly from the definition,
$\log x = \int_1^x \frac{1}{t} \enspace .$
If $x\geq 1$ then
$\log x = \int_1^x \frac{1}{t} \leq \int_1^x 1 = x - 1 \enspace .$
If $0 < x < 1$ then
$\log x = \int_1^x \frac{1}{t} = -\int_x^1 \frac{1}{t} \leq -\int_x^1 1 = -(1-x) = x - 1 \enspace .$
Where the inequality follows from the fact that
$t\in [x,1] \Rightarrow 1/t \geq 1 \Rightarrow -1/t \leq -1 \enspace .$
If $x = 1$ it follows trivially that $\log x = x - 1$. We want to show, however, that $x = 1$ is the only root.
For this we define the auxiliary function
$f(x) = \log x - x + 1$
Suppose that $f$ has two roots
\begin{aligned} x &= 1 \\ x_0 &\neq 1 \end{aligned}
Assume, without loss of generality, that $x_0 < 1$.
According to Rolle’s theorem, since $f$ is continuous on $[x_0,1]$ and differentiable on $(x_0,x)$, there must exist a point $\xi \in (x_0,1)$ such that
$f'(\xi) = \frac{1}{\xi} - 1 = 0 \enspace .$
However, the only root of $1/\xi -1 = 0$ is $\xi = 1$.
But $1 \notin (x_0,1)$.
Therefore, $x = 1$ is the only root of $f$.
Next we want to show that $\lg n$ is an upper bound for $H$.
# Step 2
By the change of base formula I can write
$\lg(n) = \frac{\log(n)}{\log(2)}$
Suppose for now that $p_j > 0$ for each $j = 1,\ldots, n$.
Then,
\begin{aligned} H(X) - \lg n &= H(p_1,\ldots,p_n) - \lg n \\ &= -\sum p_j \cdot \lg p_j - \lg n \\ &= -\sum p_j \cdot \frac{\log p_j }{\log 2} - \frac{\log p_j }{\log 2} \\ &= -\frac{1}{\log 2}\cdot \sum p_j \cdot\left(\log p_j + \log n \right) \qquad \small{\text{since} \sum p_j = 1} \\ &= -\frac{1}{\log 2}\cdot \sum p_j \cdot\left(\log p_j \cdot n \right) \qquad \star \\ &= \frac{1}{\log 2} \cdot \sum p_j \cdot\left(\log \frac{1}{p_j \cdot n} \right) \\ &\leq \frac{1}{\log 2} \cdot \sum p_j \cdot\left(\frac{1}{p_j \cdot n} - 1 \right) \qquad \small{\text{from the inequality in step 1}} \\ &= \frac{1}{\log 2} \cdot \sum \left(\frac{1}{n} - p_j \right) \\ &= \frac{1}{\log 2}\cdot\left(n\cdot\frac{1}{n} - 1\right) \\ &= \frac{1}{\log 2}\cdot\left(1 - 1\right) \\ &= 0 \\ &\Rightarrow H \leq \lg n \\ &\square \end{aligned}
Suppose $p_j = 0$ for some $j$. Recall from the preliminary discussion above that $-p\cdot\lg p \equiv 0$ by convention. Then from step $\star$ above simply note that
$p_j = 0 \Rightarrow -p_j\cdot \lg p = 0 < \frac{1}{n} - 0 = \frac{1}{n} - p_j \enspace$
which is precisely the inequality that is required to be shown in the last step. The inequality therefore still holds.
Finally, as discussed earlier, equality is reached if and only if
$\frac{1}{p_j\cdot n} = 1 \quad \forall j \enspace .$
That is, when $p_j = \frac{1}{n}$ for each $j$.
That is, when $X$ is uniformly distributed.
$\square$
Published 12 Jul 2018
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Define the Term 'Focal Length of a Mirror'. with the Help of a Ray Diagram, Obtain the Relation Between Its Focal Length and Radius of Curvature. - Physics
Diagram
Short Note
With the help of a ray diagram, obtain the relation between its focal length and radius of curvature.
Solution
The distance between the centre of a lens or curved mirror and its focus.
The relationship between the focal length f and the radius of curvature R = 2f.
Consider a ray of light AB, parallel to the principal axis and incident on a spherical mirror at point B. The normal to the surface at point B is CB and CP = CB = R is the radius of curvature. The ray AB, after reflection from a mirror, will pass through F (concave mirror) or will appear to diverge from F (convex mirror) and obeys the law of reflection i.e. i = r.
From the geometry of the figure,
∠BCP = θ = i
In D CBF, θ = r
∴BF = FC (because i = r)
If the aperture of the mirror is small, B lies close to P, and therefore BF = PF
Or FC = FP = PF
Or PC = PF + FC = PF + PF
Or R = 2 PF = 2f
Or f = R/2
Similar relation holds for convex mirror also. In deriving this relation, we have assumed that the aperture of the mirror is small.
Concept: Reflection of Light by Spherical Mirrors
Is there an error in this question or solution?
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# 4510 in Words
The word form of the number 4510 is Four Thousand Five Hundred Ten. For instance, Rahul purchased a weighing machine for Rs. 4510, then you can say, “Rahul purchased a weighing machine for Rupees Four Thousand Five Hundred Ten”. 4510 is a cardinal number as it indicates a certain amount. In this article, we will learn the simple tricks of writing the numerical name for the number 4510.
4510 in Words Four Thousand Five Hundred Ten Four Thousand Five Hundred Ten in numerical form 4510
## 4510 in English Words
We usually, write the numbers in words using the letters of the English alphabet. Hence, we can write and spell the number 4510 in English as Four Thousand Five Hundred Ten.
## How to Write 4510 in Words?
The table mentioned below depicts the place value chart for the number 4510 with 4 columns as it is a four-digit number.
Thousands Hundreds Tens Ones 4 5 1 0
Hence, we can write the expanded form as:
4 x Thousand + 5 x Hundred + 1 x Ten + 0 x One
= 4 x 1000 + 5 x 100 + 1 x 10 + 0 x 1
= 4000 + 500 + 10 + 0
= 4000 + 500 + 10
= 4510
= Four Thousand Five Hundred Ten
Therefore, 4510 in words is written as Four Thousand Five Hundred Ten
Interesting way of writing 4510 in words
4 = Four
45 = Forty-Five
451 = Four Hundred Fifty-One
4510 = Four Thousand Five Hundred Ten
Thus, the word form of the number 4510 is Four Thousand Five Hundred Ten
4510 is a natural number that precedes 4511 and succeeds 4509
• 4510 in words – Four Thousand Five Hundred Ten
• Is 4510 an odd number? – No
• Is 4510 an even number? – Yes
• Is 4510 a perfect square number? – No
• Is 4510 a perfect cube number? – No
• Is 4510 a prime number? – No
• Is 4510 a composite number? – Yes
## Frequently Asked Questions on 4510 in Words
### How do you write 4510 in words?
We write 4510 in words as Four Thousand Five Hundred Ten.
### Simplify 3000 + 1510, and express in words.
Simplifying 3000 + 1510, we get 4510. Hence, the number 4510 in words is Four Thousand Five Hundred Ten.
### 4510 is a composite number. True or False.
True, the number 4510 is a composite number.
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# The curriculum for maths in Year 4
Years 3 and 4 (lower Key Stage 2) share similar curriculum targets.
In lower Key Stage 2, the principal focus of maths teaching is to ensure that pupils become increasingly fluent with whole numbers and the four operations, including number facts and the concept of place value. This should ensure that pupils develop efficient written and mental methods and perform calculations accurately with increasingly large whole numbers.
At this stage, pupils should develop their ability to solve a range of problems, including with simple fractions and decimal place value. Pupils will also draw with increasing accuracy and develop mathematical reasoning so they can analyse shapes and their properties, and confidently describe the relationships between them. It should ensure that they can accurately use measuring instruments and make connections between measure and number.
By the end of Year 4, pupils should have memorised their times tables up to and including the 12 times table, and they will show precision and fluency in their work. Pupils should read and spell mathematical vocabulary correctly and confidently, using their growing word reading knowledge and their knowledge of spelling.
We’ve outlined England’s curriculum for maths in Year 4 below. Follow the links for help and activities:
### Children will learn to:
• count in multiples of 6, 7, 9, 25 and 1000
• find 1000 more or less than a given number
• count backwards through zero to include negative numbers
• recognise the place value of each digit in a four-digit number (thousands, hundreds, tens, and ones)
• order and compare numbers beyond 1000
• identify, represent and estimate numbers using different representations
• round any number to the nearest 10, 100 or 1000
• solve number and practical problems that involve all of the above and with increasingly large positive numbers
• read Roman numerals to 100 (I to C) and know that over time, the numeral system changed to include the concept of zero and place value.
### More information and fun activities
You’ll find a more detailed guide to spelling, advice on how to help your child at home, and free activities on our Number & place value in Year 4 page.
### Children will learn to:
• add and subtract numbers with up to 4 digits using the formal written methods of columnar addition and subtraction where appropriate
• estimate and use inverse operations to check answers to a calculation
• solve addition and subtraction two-step problems in contexts, deciding which operations and methods to use and why.
### More information and fun activities
You’ll find a more detailed guide to handwriting, advice on how to help your child at home, and free activities on our Addition & subtraction in Year 4 page.
### Children will learn to:
• recall multiplication and division facts for multiplication tables up to 12 × 12
• use place value, known and derived facts to multiply and divide mentally, including: multiplying by 0 and 1; dividing by 1; multiplying together three numbers
• recognise and use factor pairs and commutativity in mental calculations
• multiply two-digit and three-digit numbers by a one-digit number using formal written layout
• solve problems involving multiplying and adding, including using the distributive law to multiply two digit numbers by one digit, integer scaling problems and harder correspondence problems such as n objects are connected to m objects.
More information and fun activities
You’ll find a more detailed guide to grammar and punctuation, advice on how to help your child at home, and free activities on our Multiplication & division in Year 4 page.
### Children will learn to:
• recognise and show, using diagrams, families of common equivalent fractions
• count up and down in hundredths; recognise that hundredths arise when dividing an object by one hundred and dividing tenths by ten.
• solve problems involving increasingly harder fractions to calculate quantities, and fractions to divide quantities, including non-unit fractions where the answer is a whole number
• add and subtract fractions with the same denominator
• recognise and write decimal equivalents of any number of tenths or hundredths
• recognise and write decimal equivalents to $\huge&space;\frac{1}{4}$, $\huge&space;\frac{1}{2}$, $\huge&space;\frac{3}{4}$
• find the effect of dividing a one- or two-digit number by 10 and 100, identifying the value of the digits in the answer as ones, tenths and hundredths
• round decimals with one decimal place to the nearest whole number
• compare numbers with the same number of decimal places up to two decimal places
• solve simple measure and money problems involving fractions and decimals to two decimal places.
### More information and fun activities
You’ll find a more detailed guide to fractions, advice on how to help your child at home, and free activities on our Fractions in Year 4 page.
### Children will learn to:
• compare and classify geometric shapes, including quadrilaterals and triangles, based on their properties and sizes
• identify acute and obtuse angles and compare and order angles up to two right angles by size
• identify lines of symmetry in 2-D shapes presented in different orientations
• complete a simple symmetric figure with respect to a specific line of symmetry.
### More information and fun activities
You’ll find a more detailed guide to geometry, advice on how to help your child at home, and free activities on our Geometry in Year 4 page.
### Children will learn to:
• describe positions on a 2D grid as coordinates in the first quadrant
• describe movements between positions as translations of a given unit to the left/right and up/down
• plot specified points and draw sides to complete a given polygon.
### More information and fun activities
You’ll find a more detailed guide to geometry, advice on how to help your child at home, and free activities on our Geometry in Year 4 page.
### Children will learn to:
• Convert between different units of measure (for example, kilometre to metre; hour to minute)
• measure and calculate the perimeter of a rectilinear figure (including squares) in centimetres and metres
• find the area of rectilinear shapes by counting squares
• estimate, compare and calculate different measures, including money in pounds and pence
### More information and fun activities
You’ll find a more detailed guide to measurement, advice on how to help your child at home, and free activities on our Measurement in Year 4 page.
### Children will learn to:
• interpret and present discrete and continuous data using appropriate graphical methods, including bar charts and time graphs.
• solve comparison, sum and difference problems using information presented in bar charts, pictograms, tables and other graphs.
### More information and fun activities
You’ll find a more detailed guide to statistics, advice on how to help your child at home, and free activities on our Statistics in Year 4 page.
Copyright Oxford University Press 2022
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# Simplification of Decimal
Use the identities to solve the problems based on simplification of decimal. We will make use of these identities in some of the questions related to division of decimals.
Learn the following identities to apply these in simplifying decimal.
(a) (a + b)2 = a2 + b2 + 2ab
(b) (a - b)2 = a2 + b2 - 2ab
(c) a2 - b2 = (a + b) (a - b)
(d) a3 + b3 = (a + b) (a2 - ab + b2)
(e) a3 - b3 = (a - b) (a2 + ab + b2)
Worked-out examples on simplification of decimal:
Let us observe how to simplify decimals using identities with detailed step-by-step explanation.
Simplify the following:
(a) {(0.9 - 0.6)2}/{(0.9)2 - 2(0.9)(0.6) + (0.6)2}
Solution:
Let, a = 0.9 and b = 0.6
So, [(a - b)3]/[a2 - 2(a)(b) + b2]
= (a - b)3/(a - b)2
= (a - b)
Now putting the value of a and b we get,
= 0.9 - 0.6
= 0.3
(b) [(5.8)3 - (2.6)3]/[(5.8)2 + (2.6)2 - 2(5.8) + (2.6)2]
Solution:
Let a = 5.8 and b = 2.6
So, we have
= [a3 - b3]/[a2 - 2ab + b2]
= [(a - b) (a2 + ab + b2)]/[(a - b)2]
= (a2 + ab + b2)/(a - b)
Now putting the value of a and b we get,
= [(5.8)2 + (5.8)(2.6) + (2.6)2]/(5.8 - 2.6)
= 55.48/3.2
= (55.48 × 10)/(3.2 × 10), Multiply both numerator and denominator by 10
= 554.8/32
= 17.3375
(c) [(8.65)2 - (4.35)2]/(8.65 - 4.35)
Solution:
Let a = 8.65 and b = 4.35
So, we have
= [a2 - b2]/(a - b)
= [(a + b)(a-b)]/(a - b)
= a + b
Now putting the value of a and b
= 8.65 + 4.35
= 13
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https://www.studyrankersonline.com/74267/solve-15-2-2x-1-15-x-%E2%88%88-z
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# Solve: 15 – 2(2x – 1) < 15, x ∈ Z.
1 view
Solve: 15 – 2(2x – 1) < 15, x ∈ Z.
answered May 25 by (-3,230 points)
15 – 2(2x – 1) < 15, x ∈ Z.
⇒ 15 – 4x + 2 < 15
⇒ 17 – 4x < 15
⇒ -4x < 15 – 17
⇒ -4x < -2
⇒ (-4/-4)x > -2/-4 = 1/2 (Dividing by -4)
∴ x 1, 2, 3, 4, 5,….
∴ Solution set = {1, 2, 3, 4, 5,….}
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# algebra
posted by .
the units digit of a 2-digit number exceeds twice the tens digit by 1. find the number if the sum of its digits is 10.
• algebra -
x y
y = 2x+1
y = 10-x
so
10-x = 2x+1
3 x = 9
etc
• algebra -
im still pretty confused, but i think u r suppossed to solve the 2 equations u gave and got (3,7) but i think im way off
• algebra -
Check it by using 37 in the problem statement.
the units digit (7) of a 2-digit number (37)exceeds twice the tens (2*3=6) digit by 1.(6+1=7 sure enough)) find the number if the sum of its digits is 10. (3+7=10 sure enough)
• algebra -
thank u
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# MTH101 -Calculus and Analytical Geometry Assignment NO #1 Discussion and Solutions Spring Fall 2013 Due date:27 nov 2013
Assignment: # 01 (Fall-2013)
Mth101 (Calculus & Analytical Geometry)
Lecture: 01 – 10
Total Marks = 20
Due date: 27-11-2013
INSTRUCTIONS:-
1. In order to attempt this assignment you should have full command on Lecture# 01 to Lecture # 10
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Q.1 Marks 5
Solve the inequality and write the solution set in interval form.
Q.2 Marks 5
Find the equation of a circle whose diameter has endpoints (4, -1) and (-6, 7).
Q.3 Marks 5
Evaluate the following limit:
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If and, then find the composite functions .
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Our main purpose here discussion not just Solution
We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.
If and, then find the composite functions .
Please friends exmpalin me how this steps come out...
Thanks
true
bur = is not use . >is use
MTH101 Calculus And Analytical Geometry Assignment No. 01 Question No.01 See this method to solve
but it shoul be in iterval form
3 hours ago
MTH101 - Calculus And Analytical Geometry Assignment No. 01 Solution Question No.02 See this example and change values
for solving q no 3
Can anyone confirm my answer plz.....
Q no. 3.
you have to use Cube Formula and the answer is
-1
Yes.. I know ... My answer Also -1 ...
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https://homeschoolmath.blogspot.com/2012/01/
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## Posts
Showing posts from January, 2012
### Free ebook "All Children Can Be Great Listeners"
... this free e-book is meant for preschool-kindergarten age kids, and you can download it here. It's not my book, but courtesy of Renee from SchoolSparks.com
### Triangle problem with equal areas - solution
This is the solution for the triangle problem with equal areas that I posted earlier.
We are given that the areas of the three right triangles are equal, that is the area of the triangle DAE = area of the triangle EBF = area of the triangle FCD.
We will make an equation based on that fact. For that, I like to use x as my variable, so I denote the longer side of the rectangle with a, the other side with b, the distance AE with x, and the distance BF with y. We are asked the ratio AE:EB, which is the same as x : (a − x) using my notation, and the ratio BF:FC, which is the same as y : (b − y) using my notation.
The area of triangle ADE is its base times altitude divided by 2, or bx/2.
The area of triangle EBF is its base times altitude divided by 2, or y(a − x)/2.
The area of triangle CDF is its base times altitude divided by 2, or a(b − y)/2.
And these three are equal. Basically you just make two equations from the above information, and manipulate your equations until you get the ratio…
Bon from Math is not a four-letter word made this little counting song, sung to the tune of "Twinkle Twinkle Little Star". Hope your little ones enjoy it!
### Three-day sale for Math Mammoth
SORRY I forgot to post it here (I just sent this to my email list).
For January 23-25, get 23% off of all Math Mammoth downloads & CDs
at Kagi store.
Use the coupon code THREEDAYS.
Go to http://www.mathmammoth.com first, then find the links to
Kagi's order pages. OR, use these direct links:
~ Light Blue series (complete curriculum)
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=Math_Mammoth_LightBlue_Series
~ Blue series
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE
~ Golden and Green Series
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=MathMammoth_Workbooks
~ Make It Real Learning workbooks
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=Make_It_Real_Learning
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=Math_Mammoth_Pack…
### Triangle puzzle - equal areas
I hope Pat doesn't mind that I copied the image from his blog... He posted this triangle puzzle on his blog and I thought you might enjoy it, too!
Basically, we have a triangle DFE inside a rectangle, dividing the rectangle into various triangles. And, the three areas of fainter color are equal. That is, the area of the triangle DAE = area of the triangle EBF = area of the triangle FCD. (Notice the image is not drawn to scale at all.)
And, we're asked to solve the RATIOS AE : EB and BF : FC.
The solution is here.
### Versa Ruler
Well, this IS something different! A physical ruler that can draw shapes with any angle measure you want.
Unfortunately it's not yet in production. In fact, the project is needing funding... but very interesting! Please read more at
Rule Like Never Before! NEW Shape-making Versa Ruler
This ruler give you accurate measurements for every side and angle. You can connect sides to form angles, which scale and skew smoothly, and you can lock angles and sides. Cool!
### Vi Hart and mathematical doodling
I just learned about Vi Hart's "doodling in math class" videos (hat tip goes to Fawn Nguyen).
Vi calls herself mathemusician - and definitely, she's a talented and smart gal! I'm sure you'll enjoy her videos (as long as you can follow her super fast speeaking). Here are some that I enjoyed:
Doodling in Math: Spirals, Fibonacci, and Being a Plant [1 of 3]
Binary Hand Dance was pretty cool too!
Her series of "mathematical doodling" videos have become somewhat of a viral success. Here's one more:
Doodling in Math Class: Infinity Elephants
### Math teachers are at play again
Denise's done a beautiful job with the current Math Teachers at Play carnival number 46, lots to read and explore and see... head on over!
### Compound interest
Photo courtesy of Mooster Someone sent me in a question concerning compound interest...
Please send me the formula for compound interest and explain line by line.
p(1 + rate)3
What does the 1 stand for and must you add it to the rate of say 10%?
100(1+10)3 years = ???
Obviously I am looking for a basic course?
The formula that this person is using is correct... the formula for compound interest is
A = p(1 + r)t
but this formula doesn't give us the amount of interest -- it gives us the amount of money you would withdraw after t years. In the formula, p is the original principal, r is the interest rate, and t is the time in years.
However, we cannot put the interest rate in as he did. If r = 10%, then r = 0.1 must be used in this formula. In other words, FIRST convert your percentage into a decimal.
For example, if the principal is \$5000 and r = 10% = 0.1, then we get
A = \$5000 × 1.1t
The number 1 in the formula p(1 + r)t doesn't stand for anything by itself. It come…
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# Objective Recognize and graph periodic and trigonometric sine and cosine functions.
## Presentation on theme: "Objective Recognize and graph periodic and trigonometric sine and cosine functions."— Presentation transcript:
Objective Recognize and graph periodic and trigonometric sine and cosine functions.
Vocabulary periodic function cycle period amplitude frequency
phase shift
Periodic functions are functions that repeat exactly in regular intervals called cycles. The length of the cycle is called its period. A cycle may begin at any point on the graph of a function.
Example 1A: Identifying Periodic Functions
Identify whether each function is periodic. If the function is periodic, give the period. The pattern repeats exactly, so the function is periodic. Identify the period by using the start and finish of one cycle. This function is periodic with a period of .
Example 1B: Identifying Periodic Functions
Identify whether each function is periodic. If the function is periodic, give the period. Although there is some symmetry, the pattern does not repeat exactly. This function is not periodic.
Example Identify whether each function is periodic. If the function is periodic, give the period. a. b.
The trigonometric functions that you studied in Chapter 10 are periodic. You can graph the function f(x) = sin x on the coordinate plane by using y-values from points on the unit circle where the independent variable x represents the angle θ in standard position.
Similarly, the function f(x) = cos x can be graphed on the coordinate plane by using x-values from points on the unit circle. The amplitude of sine and cosine functions is half of the difference between the maximum and minimum values of the function. The amplitude is always positive.
Key Values for Functions
Sine Function Cosine Function
You can use the parent functions to graph transformations y = a sin bx and y = a cos bx.
Recall that a indicates a vertical stretch (|a|>1) or compression (0 < |a| < 1), which changes the amplitude. If a is less than 0, the graph is reflected across the x-axis. The value of b indicates a horizontal stretch or compression, which changes the period.
Example 2: Stretching or Compressing Functions Sine and Cosine Functions
Using f(x) = sin x as a guide, graph the function g(x) = Identify the amplitude and period. Step 1 Identify the amplitude and period. Because the amplitude is Because the period is
Example 2 Continued Step 2 Graph. The curve is vertically compressed by a factor of horizontally stretched by a factor of 2. The maximum value of g is , and the minimum value is .
Example Using f(x) = cos x as a guide, graph the function h(x) = Identify the amplitude and period. Step 1 Identify the amplitude and period.
Example Continued Step 2 Graph. The maximum value of h is , and the minimum value is .
You Try Pg. 759 # 2, 3, 4, 5
Sine and cosine functions can be used to model real-world phenomena, such as sound waves. Different sounds create different waves. One way to distinguish sounds is to measure frequency. Frequency is the number of cycles in a given unit of time, so it is the reciprocal of the period of a function. Hertz (Hz) is the standard measure of frequency and represents one cycle per second. For example, the sound wave made by a tuning fork for middle A has a frequency of 440 Hz. This means that the wave repeats 440 times in 1 second.
Example 3: Sound Application
Use a sine function to graph a sound wave with a period of s and an amplitude of 3 cm. Find the frequency in hertz for this sound wave. period amplitude Use a horizontal scale where one unit represents s to complete one full cycle. The maximum and minimum values are given by the amplitude. The frequency of the sound wave is 500 Hz.
The frequency of the sound wave is 250 Hz.
Example Use a sine function to graph a sound wave with a period of s and an amplitude of 3 cm. Find the frequency in hertz for this sound wave. Use a horizontal scale where one unit represents s to complete one full cycle. The maximum and minimum values are given by the amplitude. period amplitude The frequency of the sound wave is 250 Hz.
Sine and cosine can also be translated as y = sin(x – h) + k and y = cos(x – h) + k.
Recall that a vertical translation by k units moves the graph up (k > 0) or down (k < 0). A phase shift is a horizontal translation of a periodic function. A phase shift of h units moves the graph left (h < 0) or right (h > 0).
Example 4: Identifying Phase Shifts for Sine and Cosine Functions
Using f(x) = sin x as a guide, graph g(x) = g(x) = sin Identify the x-intercepts and phase shift. Step 1 Identify the amplitude and period.
Example 4 Continued Step 2 Identify the phase shift. Identify h. Because h = the phase shift is radians to the right. All x-intercepts, maxima, and minima of f(x) are shifted units to the right.
Example 4 Continued Step 3 Identify the x-intercepts. The first x-intercept occurs at Because sin x has two x-intercepts in each period of 2, the x-intercepts occur at + n, where n is an integer.
Example 4 Continued Step 4 Identify the maximum and minimum values. The maximum and minimum values occur between the x-intercepts. The maxima occur at n and have a value of 1. The minima occur at n and have a value of –1.
Step 5 Graph using all the information about the function.
Example 4 Continued Step 5 Graph using all the information about the function. sin x sin
Example Using f(x) = cos x as a guide, graph g(x) = cos(x – ). Identify the x-intercepts and phase shift. Step 1 Identify the amplitude and period.
Example Continued Step 2 Identify the phase shift.
Example Continued Step 3 Identify the x-intercepts.
Example Continued Step 4 Identify the maximum and minimum values.
Step 5 Graph using all the information about the function.
Example Continued Step 5 Graph using all the information about the function. y cos x – x cos (x–)
You can combine the transformations of trigonometric functions
You can combine the transformations of trigonometric functions. Use the values of a, b, h, and k to identify the important features of a sine or cosine function. Amplitude Phase shift y = asinb(x – h) + k Vertical shift Period
Example
Example
You Try Pg. 759 # 7, 8, 9
Homework Pgs #12 – 22 # 25 – 28 # 40 – 42
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http://www.docstoc.com/docs/80155570/lesson
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# lesson by nuhman10
VIEWS: 6 PAGES: 7
• pg 1
``` Lesson - Atwood's Pulley - Massive
Name: _________________________
The applet Atwood simulates the motion of two masses connected by a massless, ideal string
which passes over a massive pulley.
Prerequisites
Students should be familiar with the concepts of potential and kinetic energy.
Learning Outcomes
Students will become familiar with energy conservation and energy transformation.
Instructions
Students should know how the applet functions, as described in Help and ShowMe. Many of the step-by-
step instructions in the following lesson are to be done in the applet.
Contents
Calculating Potential Energy and setting a zero-point for Potential Energy
Calculating Potential Rotational and Kinetic Energy
Total Mechanical Energy of a System and Conservation of Energy
Potential Energy and defining a zero-point for Potential Energy
In order to assign a numerical value to the potential energy of a body
it is first necessary to define a "zero point" or reference level at which
the potential energy is considered to be 0 J. You do this on the applet
by repositioning the Ep Reference line (moving it up or down)
On the applet, position the Ep Reference line at the
same height as the center of the pulley as illustrated in Figure 1.
Using the mass sliders, set
m1 to 250 g
m2 to 750 g
pulley mass to 400 g
Figure 1
Lesson – Atwood’s Pulley – Massive 1 of 7
Are the initial potential energies for mass 1 and mass 2 positive, negative or zero in this case? Hint: are
the masses above or below the Ep Reference line? Explain your answer.
Play the applet ( ), and produce two graphs showing the potential energy for each
mass as a function of time (Do not reset the applet - if you do you will need to reset the Ep Reference
line and masses). To view the graphs:
1. select the graph option ( ) after the applet is finished playing
2. select "time" for the horizontal axis
3. select "potential energy - 1" for the vertical axis (this graphs the energy of the mass on the left of
the pulley)
4. fit the graph ( ) on the display and sketch it below
5. to display the energy graph for the second mass, change the vertical axis to "potential energy - 2"
6. fit the graph ( ) on the display and sketch it below
Graph 1: Potential Energy - 1 vs. Time Graph 2: Potential Energy - 2 vs. Time
Explain why the graphs look the way they do.
Lesson – Atwood’s Pulley – Massive 2 of 7
Mass 1 climbs by 1.119 m and mass 2 drops by 1.119 m. Calculate the change in
potential energy for each mass. Recall that , where Ep is the change in potential energy, m
is the mass, g is the acceleration of gravity and h is the change in height.
Can you determine the change in potential energy of each mass using only your graphs
from Exercise 2? If so, how is this done? (Tip: use the "drag-and-zoom" button ( ) to zoom-in on points of interest on the
graph. Alternately, generate a data table using the option menu ( ) and look up the potential energy at the beginning and end of the
motion.)
Calculate the initial potential energy of the pulley-mass system? Using your graphs
from Exercise 2, add the initial potential energy of each mass (Ep1 + Ep2). This represents the initial
potential energy of the pulley - mass system.
Calculate the final potential energy of the pulley-mass system? Using your graphs
from Exercise 2, add the final potential energy of each mass (Ep1 + Ep2). This represents the final
potential energy of the pulley - mass system.
Is the final potential energy of the system equal to the initial potential energy of the
system?
Lesson – Atwood’s Pulley – Massive 3 of 7
Calculating Potential, Rotational and Kinetic Energy
In the previous example the initial potential energy of the system was greater than the final potential
energy (by about 5.56 J). What happened to this energy?
Answer: When you press play, the masses begin to accelerate. Mass 1 moves up, mass 2 moves
down. A new energy form, kinetic energy, is now being created. Recall, kinetic energy is given by the
expression , where m is the mass and v is the velocity.
Re-run the applet using the same mass values used in the previous section. Produce
kinetic energy - time graphs for each mass and sketch them below.
Graph 3: Kinetic Energy - 1 vs. Time Graph 4: Kinetic Energy - 2 vs. Time
Using the graphs above, determine the final kinetic energy of each mass. (Tip: use the
"drag-and-zoom" button ( ) to zoom-in on points of interest on the graph. Alternately, generate a data table using the option menu ( )
and look up the final kinetic energy.)
Ek1final = _________________ Ek2final = ___________________
Add the final kinetic energy of each mass. Has the missing 5.56 J been accounted for?
Does the kinetic energy you just measured equal the missing potential energy?
Lesson – Atwood’s Pulley – Massive 4 of 7
If you did all of your calculations carefully you will have discovered that there is still
some energy missing! We forgot to account for the fact that the pulley has mass and that this mass is
spinning. We have encountered a new form of energy - Rotational Energy. On the basis of your
calculations, how much rotational energy must be stored in the pulley itself?
A spinning mass has rotational energy. The rotational energy of the pulley equal to one half the
product of the pulley's moment of inertia and the square of its angular speed.
Expressed as an equation:
Quantity Symbol SI Unit
rotational energy E J
moment of inertia I kg.m2
angular speed radian/s
The moment of inertia (I) is a measure both of how much mass is spinning and how this mass is
distributed around its rotational axis.
It is assumed in this applet that the pulley is a uniform density cylinder, ( ),
where Mp is the pulley mass and R is the radius of the pulley. Further, it is assumed that the string does
not slip on the pulley and therefore, . Show by algebraic manipulation that the rotational energy
can be re-written as .
Using the applet, produce a rotational energy - time graph for the motion set-up in
Exercise 1. Based on the graph or data tables produced by the applet what is the final rotational energy
of the pulley? Does this verify your answer to Exercise 11?
Lesson – Atwood’s Pulley – Massive 5 of 7
The total energy of the system is the sum of all the kinetic, rotational and potential energy terms at any
instant. In Exercise 5, you calculated the initial energy of the system (it is all in the form of potential
energy before the masses begin to move). Now, calculate the final energy of the system by adding the
final potential energy (calculated in Exercise 6) and the final kinetic energy of each mass.
final potential energy of mass 1 and 2
rotational energy of the pulley
final kinetic energy of mass 1
final kinetic energy of mass 2
+ _______________________
Final total energy
Compare the initial and final total energies. What do you notice about these numbers?
Total Mechanical Energy of a System and Conservation of Energy
In the previous example, you should have noticed that the total energy of the system is constant. This is
to say that the total energy before the masses move is identical to the total energy at any point during
the movement. The applet Atwood assumes that there is no loss of energy from the system. This
means that there is no frictional loss in the pulley and that air resistance on the moving masses can be
ignored. It is assumed that energy is conserved. When this happens we can conclude that the total
energy of the system is constant. This can be expressed in the following ways:
The net change in energy in the system is zero. Energy is
total energy = zero
neither lost or created
The total energy of the system before is equal to total
Etotal initial = Etotal final
energy of the system after any motion or change.
The individual expressions for the energy can change but
their sum must be zero. Increases in one term will be offset
by decreases in other terms.
These are just three ways of stating the Principle of Conservation of Mechanical Energy.
Lesson – Atwood’s Pulley – Massive 6 of 7
Set up the applet with the following quantities.
m1 = 200 g
m2 = 800 g
pulley mass = 600 g
Place the reference line at the center of the pulley. Play the motion and use the graphing tool and data
table option to collect the necessary data required to complete the following table and verify the Principle
of Conservation of Mechanical Energy.
Before Masses
System When t = 0.500 s
Released
Ep1(J)
Ep2(J)
Ek1(J)
Ek2(J)
Erot (J)
E total
(J)
Lesson – Atwood’s Pulley – Massive 7 of 7
```
To top
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http://mathoverflow.net/feeds/question/118417
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Isomorphic simple groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:41:49Z http://mathoverflow.net/feeds/question/118417 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118417/isomorphic-simple-groups Isomorphic simple groups Yu 2013-01-09T03:58:53Z 2013-01-10T18:29:54Z <p>It is known that \$SL_{4}(\mathbb{F}_2)\cong A_8\$. Obviously, this is equivalent to the existence of a subgroup of \$Sl_4(\mathbb{F}_2)\$ of index \$8\$. How to find such a subgroup? </p> http://mathoverflow.net/questions/118417/isomorphic-simple-groups/118559#118559 Answer by Quang Hoang for Isomorphic simple groups Quang Hoang 2013-01-10T18:14:53Z 2013-01-10T18:29:54Z <p>An elementary answer in terms of symmetric groups. </p> <p>Let \$V\cong\mathbb F^3\$ be a \$3\$-dim \$\mathbb F_2\$ space. Consider \$V\$ as a subgroup of \$S(V)\$. It is well-known that \$N_{S(V)}(V) \cong V\rtimes GL(4)\$. Then the isomorphism \$A_8\cong GL_4\$ because they are both even subgroups of \$S_8=S(V)\$. </p> <p>Now the subgroup of index \$8\$ is the group \$A_7\$ which fixes the origin of \$V\$. </p> <p>Edit: Apparently, I made a silly mistake along the way that \$N_{S(V)}(V) \cong V\rtimes GL(4)\$ (Should be \$GL(3)\$). Yet somehow I think it could be fixed and that the argument is somewhat equivalent to that of Elkies.</p>
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| 1,366
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| 3.5625
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CC-MAIN-2013-20
|
latest
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en
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http://blog.csdn.net/kakulukia/article/details/46816607
| 1,503,200,033,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-34/segments/1502886105961.34/warc/CC-MAIN-20170820015021-20170820035021-00488.warc.gz
| 61,000,826
| 15,350
|
# 八皇后递归求解
170人阅读 评论(0)
bool isConfilctQueen(int **martix, int row, int column, int count) {
bool result = false;
for (int i = 0; i < count; i++) { //行检测
if (i == column) continue;
if (martix[row][i] == 1)
return true;
}
for (int i = 0; i < count; i++) { //列检测
if (i == row) continue;
if (martix[i][column] == 1)
return true;
}
for (int i = 1; i < count; i++) { //对角线检测，以目标点为中心,对角线相对长度一次检测四个角
if (row - i >= 0 && column - i >= 0) { //左上角
if (martix[row - i][column - i] == 1)
return true;
}
if (row - i >= 0 && column + i < count) { //右上角
if (martix[row - i][column + i] == 1)
return true;
}
if (row + i < count && column - i >= 0) { //左下角
if (martix[row + i][column - i] == 1)
return true;
}
if (row + i < count && column + i < count) { //右下角
if (martix[row + i][column + i] == 1)
return true;
}
}
return result;
}
void backtrackingQueen(int **martix, int n, int count, int *num) {
if (n == count) {
*num += 1;
printf("第%d组解：\n", *num);
for (int i = 0; i < count; i++) {
for (int j = 0; j < count; j++)
printf("[%d]", martix[i][j]);
printf("\n");
}
printf("\n");
return;
}
for (int i = 0; i < count; i++) { //此循环里2个memset的位置有玄妙
// memset(martix[n], 0, sizeof(int) * count);
martix[n][i] = 1;
if (!isConfilctQueen(martix, n, i, count))
backtrackingQueen(martix, n + 1, count, num);
memset(martix[n], 0, sizeof(int) * count);
}
}
int nQuenn(int count) {
if (count < 4) {
printf("Error! n < 4\n");
return 0;
}
int **martixQuenn = (int **)malloc(sizeof(int *) * count);
for (int i = 0; i < count; i++) {
martixQuenn[i] = (int *)malloc(sizeof(int) * count);
memset(martixQuenn[i], 0, sizeof(int) * count);
}
int n = 0;
int num = 0;
backtrackingQueen(martixQuenn, n, count, &num);
for (int i = 0; i < count; i++)
free(martixQuenn[i]);
free(martixQuenn);
return num;
}
void printNQueens(int **matrix, int n, int row, int column, int *count) {
if (row >= n || column >= n)
return;
bool flag = true;
for (int i = 0; i < n && flag; i++) {
for (int j = 0; j < n && flag; j++) {
if (matrix[i][j] == 1) {
if (row == i || abs(row - i) == abs(column - j)) {
flag = false;
}
}
}
}
if (flag) {
matrix[row][column] = 1;
if (column == n - 1) { //最后一列到达，开始打印
*count += 1;
printf("第%d组解：\n", *count);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf("%d ", matrix[i][j]);
}
printf("\n");
}
printf("\n");
matrix[row][column] = 0;
return;
}
}
if (flag) {
printNQueens(matrix, n, 0, column + 1, count);
matrix[row][column] = 0;
}
printNQueens(matrix, n, row + 1, column, count);
}
bool nQueens(int n) {
bool result = true;
int **martrix = (int **)calloc(n, sizeof(int *));
for (int i = 0; i < n; i++) {
martrix[i] = (int *)calloc(n, sizeof(int));
for (int j = 0; j < n; j++)
martrix[i][j] = 0;
}
int *count = (int *)calloc(1, sizeof(int));
int row = 0, column = 0;
printNQueens(martrix, n, row, column, count);
printf("all nQueens are %d.\n\n", *count);
for (int i = 0; i < n; i++) {
free(martrix[i]);
martrix[i] = NULL;
}
free(martrix);
martrix = NULL;
free(count);
count = NULL;
return result;
}
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| 1,276
| 3,151
|
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| 3.515625
| 4
|
CC-MAIN-2017-34
|
longest
|
en
| 0.128254
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https://brainmass.com/chemistry/stoichiometry/molality-molarity-questions-542318
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Explore BrainMass
# Molality and Molarity Questions
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Question #1: For the reaction:
2 Al + 6 HBr (aq) ---> 3H2 + 2 AlBr3
What volume of 0.300 M HBr(aq) is needed to react with 22.6 g of AL?
Question #2: The density of a 3.92 M solution of NaCl in water is 1.148 g/cm^3 at 20 degrees celcius. Find the molality (not molarity) of this solution.
https://brainmass.com/chemistry/stoichiometry/molality-molarity-questions-542318
#### Solution Preview
Q 1: 1 mole of HBr == 79.9+1 = 80.9 gm
1 mole of Al == 26.98 gm
2 moles of Al == 6 moles of HBr
=> 2*26.98 gm Al == 6 moles of HBr
=> 22.6 gm Al == 6*22.6/(2*26.98) = 2.513 mole of HBr
0.3M ...
#### Solution Summary
A question is solved to find volume of a solution of given molarity while reacting with the other element of given quantity. The other problem solves for estimating molality of given solution.
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# Class 7 Maths Chapter 13 Exponents and Powers
Posted on: October 27, 2021 Posted by: user Comments: 0
Chapter 13 Exponents and Powers
## NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1
Question 1.
Find the value of:
1. 26
2. 93
3. 112
4. 54.
Solution:
1. 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64
2. 93 = 9 × 9 × 9 = 729
3. 112 = 11 × 11 = 121
4. 54 = 5 × 5 × 5 × 5 = 625.
Question 2.
Express the following in exponential form:
1. 6 × 6 × 6 × 6
2. t × t
3. b × b × b × b
4. 5 × 5 × 7 × 7 × 7
5. 2 × 2 × a × a
6. a × a × a × c × c × c × c × d.
Solution:
1. 6 × 6 × 6 × 6 = 64
2. t × t = t2
3. b × b × b × b = b4
4. 5 × 5 × 7 × 7 × 7 = 52 × 73
5. 2 × 2 × a × a = 22 × a2
6. a × a × a × c × c × c × c × d = a3 × c4 × d.
Question 3.
Express each of the following numbers using the exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125.
Solution:
Question 4.
Identify wherever possible, in each of the following?
Solution:
Question 5.
Express each of the following as a product of powers of their prime factors:
(i) 648
(ii) 405
(iii) 540
(iv) 3600
Solution:
Question 6.
Simplify:
Solution:
Question 7.
Simplify:
Solution:
Question 8.
Compare the following numbers:
Solution:
## NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2
Question 1.
Using laws of exponents, simplify and write the answer in exponential form :
Solution:
Question 2.
Simplify and express each of the following in exponential form:
Solution:
Question 3.
Solution:
Question 4.
Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768.
Solution:
Question 5.
Simplify
Solution:
## NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3
Question 1.
Write the following numbers in the expand forms :
(i) 279404
(ii) 3006194
(iii) 2806196
(iv) 120719
(v) 20068.
Solution:
Question 2.
Find the number from each of the following expanded forms :
Solution:
Question 3.
Express the following numbers in standard form:
1. 5,00,00,000
2. 70,00,000
3. 3,18,65,00,000
4. 3,90,878
5. 39087.8
6. 3908.78
Solution:
1. 5,00,00,000 = 5 × 107
2. 70,00,000 = 7 × 106
3. 3,18,65,00,000 = 3.1865 × 109
4. 3,90,878 = 3.90878 × 105
5. 39087.8 = 3.90878 × 104
6. 3908.78 = 3.90878 × 103
Question 4.
Express the number appearing in the following statements in standard form :
1. The distance between Earth and Moon is 384,000,000 m.
2. The speed of light in a vacuum is 300,000,000 miles.
3. The diameter of the Earth is 1,27,56,000 m.
4. Diameter of the Sun is 1,400,000,000 m.
5. In a galaxy, there are on average 100,000,000,000 stars.
6. The universe is estimated to be about 12,000,000,000 years old.
7. The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m.
8. 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.
9. The earth has 1,353,000,000 cubic km of seawater.
10. The population of India was about 1,027,000,000 in March 2001.
Solution:
1. The mean distance between Earth and Moon is 3.84 × 108 m.
2. The speed of light in a vacuum is 3 × 108 m/s.
3. The diameter of the Earth is 1.2756 × 107 m.
4. The diameter of the Sun is 1.4 × 109 m.
5. In a galaxy, there are on average 1 × 1011 stars.
6. The universe is estimated to be about 1.2 × 1010 years old.
7. The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 3 × 1020 m.
8. 6.023 × 1022 molecules are contained in a drop of water weighing 1.8 gm.
9. The earth has 1.353 × 109 cubic km of seawater.
10. The population of India was about 1.027 × 109 in March 2001.
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https://www.cram.com/flashcards/math-87-test-1-review-202111
| 1,532,312,803,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676594790.48/warc/CC-MAIN-20180723012644-20180723032644-00166.warc.gz
| 843,447,126
| 18,244
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B - 4782 = 2084 B = 6866 What are the next three numbers in this sequence? 63, 60, 57, 54, 51, ___, ___, ___ 48, 45, 42 Use the numbers 4 and 5 to illustrate the commutative property of multiplication. 4 x 5 = 5 x 4 Show this subtraction problem on a number line: 6 - 3 Start at zero, draw a ray to the right, 6 units long. Then from 6, draw a ray going left from 6 back 3 units. Ending at point 3 6048 - Y = 2532 Y = 3516 Compare: -6 _______ -8 < > = > \$41.30 / 10 = \$4.13 T + \$5.50 = \$12.00 T = \$6.50 Use words to write 14328735. Fourteen million, three hundred twenty-eight thousand, seven hundred thirty-five 38,154 / 6 = 6359 If the product of 12 and 60 is divided by the sum of 12 and 36, what is the quotient? ( ) ( ) 15 1587 + C = 2950 C = 1363 100 - (720 - 38) = 318 Use digits to write three million, forty thousand, seven hundred. 3,040,700 Write 75,000 in expanded notation. (7 x 10,000) + (5 x 1000) F x 7 = \$51.80 F = \$7.40 9 * 22 * 25 = 4950 150(18) = 2700 Use digits and symbols to write "Negative five is less than positive five." -5 < 5 15 x P = 270 P = 18
| 538
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https://socratic.org/questions/how-do-you-solve-0-12x-2-1-4x-3-5-0-5-using-the-quadratic-formula
| 1,596,561,170,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439735881.90/warc/CC-MAIN-20200804161521-20200804191521-00340.warc.gz
| 505,541,540
| 6,070
|
How do you solve -0.12x^2 + 1.4x - 3.5 = 0.5 using the quadratic formula?
Jul 9, 2015
I found:
${x}_{1} = 5$
${x}_{2} = \frac{20}{3}$
Explanation:
Ok...I`ll try to "simplify" it a bit to avoid decimals...(I do not like them!).
I can write it as:
$- \frac{12}{100} {x}^{2} + \frac{14}{10} x - \frac{35}{10} = \frac{5}{10}$
getting rid of the denominators (after taking $100$ as common) I got:
$- 12 {x}^{2} + 140 x - 400 = 0$
that is in the form:
$a {x}^{2} + b x + c = 0$
With:
$a = - 12$
$b = 140$
$c = - 400$
I now use the Quadratic Formula:
${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 140 \pm \sqrt{{140}^{2} - 4 \left(- 12 \cdot - 400\right)}}{2 \cdot - 12} =$
$= \frac{- 140 \pm \sqrt{400}}{-} 24 = \frac{- 140 \pm 20}{-} 24 =$
You get two solutions:
${x}_{1} = \frac{- 140 + 20}{-} 24 = 5$
${x}_{2} = \frac{- 140 - 20}{-} 24 = \frac{160}{24} = \frac{20}{3}$
| 402
| 891
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https://articles.abilogic.com/371043/examples-add-extra-mark-your.html
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crawl-data/CC-MAIN-2021-04/segments/1610703581888.64/warc/CC-MAIN-20210125123120-20210125153120-00203.warc.gz
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This website uses cookies to improve user experience. By using our website you consent to all cookies in accordance with our Privacy Policy.
by Assignment Desk
Posted: Jun 27, 2019
@@@@@@
Thermodynamics is a branch of physics which deals with heat and temperature and their relationship with work and energy. Thermodynamics has four laws, namely, first law, second law, third law and zeroth law. These laws help to drive the things happening in the universe. A thermodynamics assignment could make you learn new, unknown potentials about the subject. A thermodynamics assignment is formed by the combination of ideas and conclusion. Many students opt for @@@@thermodynamics assignment help as the subject requires use of theory as well as practical application of physics laws.
Here are the few examples you can use in your thermodynamic assignment to get an extra mark. These examples are categorized according to the laws.
## Real life Examples of Thermodynamics
1. First Law of Thermodynamics: The first law states that total energy in any system cannot be generated nor destroyed but can only be transferred from one form to another. Example:
• Diesel Engine: When a diesel engine burns fuel, it converts the energy stored in fuel’s chemical bonds into productive work and energy. Whatever the amount of energy you will input(fuel), the same amount of energy will be generated.
Some more examples of first law of thermodynamics are loudspeakers, microphones, motors, etc.
2. Second Law of Thermodynamics: Simplest explanation of second law of thermodynamics is that heat will naturally flow from a hotter body to a colder body. Example:
• Melting Ice: When an ice cube is kept in the room temperature, the ice cube absorbs the heat or thermal energy from the temperature.
Some more examples are hot fluid becomes cold, on glasses in winters, etc.
3. Third Law of Thermodynamics: The third law of thermodynamics is actually easy to understand but has no applicability in day to day life. It states that if something becomes or reaches 0°K or -273.15°C or -459.67°F, then its atoms stops moving. This temperature is called zero temperature. Example:
• Experiments: This law is useful in observing the responses of various substances to temperature change.
4. Zeroth Law of Thermodynamics: This law simply states that if one system say A is in thermal equilibrium with another system say B, and B is also in thermal equilibrium with another third system say C, then A and C will also be in thermal equilibrium. Example:
• Thermometer: If we dip a thermometer into a cup of boiling water, then the thermometer warms up until it gets to the same temperature as water. This means when we dip the thermometer in water, the water(A) and glass(B) of thermometer attain the thermal equilibrium and then the glass(B) and the mercury in the thermometer (C) attain the thermal equilibrium. In this way water and mercury that is A and C also attain thermal equilibrium.
You Must Be Galvanized Now!!!
These thermodynamics examples you can use in your assignment as well as experiments, as these are, really simple day to day examples. If you use any above mentioned examples, surely, your professor will be interested throughout your whole assignment.
Summary: If you have got a thermodynamics assignment, here are few examples for your thermodynamics assignment. These examples will help you grab some extra marks in your assignment.
Author’s Bio: Marie Claire is associated with Assignment Desk for 2 years now and have been providing @@@@assignment assistance to students online. In her free time, she likes to go on a family vacation.
Marie Claire is the assignment writing expert at Assignment Desk. He love reading looks and playing outdoor games in his free time.
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+1-316-444-1378
I MCQSI11In standard normal distribution the means is always (Points : 3)2. The area under the standard normal curve is (Points : 3)3. 3.If Larry gets a 70 on a physics test where the mean is 65 and the standard deviation is 5.5.8 where does he stand in relation to his classmates? (Points : 3)4. In a normal distribution with mu = 25 and sigma = 3 what number corresponds to z = -3? (Points : 3)5. Let s assume you have taken 100 samples of size 49 each from a normally distributed population. Calculate the standard deviation of the sample means if the population s variance is 16. (Points : 3)6. The area to the left of z is 0.9976. What z-score corresponds to this area? (Points : 3)7. Find P(9 8. What is the critical z-value that corresponds to a confidence level of 88%? (Points : 3)9. Compute the population mean margin of error for a 95% confidence interval when sigma is 4 and the sample size is 36. (Points : 3)10. A standard IQ test has a mean of 98 and a standard deviation of 16. We want to be 90% certain that we are within 8 IQ points of the true mean. Determine the sample size. (Points : 3)11. A private medical clinic wants to estimate the true mean annual income of its patients. The clinic needs to be within \$200 of the true mean. The clinic estimates that the true population standard deviation is around \$1300. If the confidence level is 95% find the required sample size in order to meet the desired accuracy. (Points : 6)112. An auditor wants to estimate what proportion of a bank s commercial loan files are incomplete. The auditor wants to be within 7% of the true proportion when using a 95% confidence level. How many files must the auditor sample? No estimate of the proportion is available so use 0.5 for the population proportion. (Points : 6)QuestionsIInterpret a 90% confidence interval of (4.355 4.445) for a population mean.A nursing school wants to estimate the true mean annual income of its alumni. It randomly samples 200 of its alumni. The mean annual income was \$55200 with a standard deviation of \$1500. Find a 95% confidence interval for the true mean annual income of the nursing school alumni. Write a statement about the confidence level and the interval you find
Categories: Uncategorized
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http://www.simetric.co.uk/sudoku/tip3.htm
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| 3,818
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# sudoku
<home for links to the tutorials
explanations:-
tip 3 - the third column
This has an example of what I call 'the third column'.
The game is part done and we are looking at 3's here and from this information we can find the 3 in c5r4.
Looking at Box2, we can see that the 3 in row1 means that a 3 must be in column4 or column6 of Box2.
Looking at Box5, we can see that the 3 in row6 means a 3 must be in row4 of Box5.
Looking at Box8, the 3 in row9 means that a 3 must be in c4 or c6 of Box8.
Now there's that column4 and 6 again!
So logically, if a 3 must be in c4 or c6 of Box2 and Box8 there cannot be a 3 in those columns in Box5. A 3 must be in the third column of Box5 - column5.
As we have already proved that a 3 must be in row4 of Box5 and it cannot be in c4 or c6, therefore a 3 must be in c5r4.
This is the 'tough' puzzle from the Daily Telegraph on 15th. October 2009
(ok, in this instance, the 3 can also be found by the fact that there is only one cell left for it in c5 but this is not always the case with this tip)
c1
c2
c3
c4
c5
c6
c7
c8
c9
7
2 3 9
r1
box
1
2
&
3
box 1
3?
box 2
1
3?
box 3
r2
3?
4
3?
2
r3
7 9
3?
3?
3?
8
r4
box
4
5
&
6
2
box 4
8 4
box 5
7
6
box 6
5
r5
3 2
r6
2
3?
9
3?
6
r7
box
7
8
&
9
box 7
3?
box 8
6
3?
box 9
r8
3 1 9 8
r9
___________________________________ [ contact me, Roger Walker ] 15th january 2010 © 2012
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https://math.stackexchange.com/questions/2229703/can-my-consuption-of-coffee-be-modelled-using-markovian-master-equations
| 1,566,632,130,000,000,000
|
text/html
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crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00410.warc.gz
| 550,734,639
| 29,962
|
# ¿Can my consuption of coffee be modelled using Markovian master equations?
During a 3h shift, an employee estimates that the probability for him to drink a cup of coffee rises at a constant rate $\omega=0.8/hour$. Given that the rate is time-independent, he expects an average of 2.4 cups consumed in the whole shift. No cup has been consumed at the beginning of the shift ($P_n(t=0)=\delta_{n 0}$).
I want to model this problem by considering Markovian time evolution for the probabilities in order to keep track of fluctuactions.
This is my attemp:
Probability vector whose components $p_i$ give the probability that $i$ have been consumed after the 3h shift at a given time $t>0$.
$$P(t)=\begin{bmatrix} p_{0}(t) \\ p_{1}(t)\\ p_{2}(t)\\ p_{3}(t)\\ \end{bmatrix}$$
The transition matrix, in left stochastic notation, with elements given by:
$$(W)_{ij}=w_{ij}-\delta_{ij}\sum_{k=0}^{3}w_{kj}, w_{ii}=0$$
where $w_{ij}$ are the non-negative rates for transition from state $j$ to state $i$ per unit time. In this case, the transition rate from having $j$ cups of coffe to $i$ cups.
In this problem I have the following transition rates: $$w_{ij}=0 \hspace{1cm} j>i \\ w_{i+1,i}=0.8\\ w_{i+2,i}=0.8/2=0.4\\ w_{i+3,i}=0.8/3=0.27$$\
I'm considering that there is not possibility to "untake" a cup a coffe, $w_{ij}=0$ for $j>i$ and that if we have 1 cup of coffe, the probability to take 2 more cups of coffe in the next hour ($w_{i+2,i}\rightarrow w_{31}$) will be half the probability to take 1 cup. I don't know if this assumption makes sense, but I couldn't think of anything better with the information "the probability for him to drink a cup of coffee rises at a constant rate $\omega=0.8/hour$" I have been given.
I end up with this transition matrix: $$M= \left( {\begin{array}{cc} -1.47 & 0 & 0 & 0 \\ 0.8 & -1.2 & 0 & 0 \\ 0.4 & 0.8 & -0.8 & 0 \\ 0.27 & 0.4 & 0.8 & 0 \\ \end{array} } \right)$$
Now the next step would be to solve the master equation: $$\frac{d}{dt}P(t)=W·P(t)$$
Nevertheless, I'm not very confident with the transition matrix I got because I am not sure the transitions rates I wrote make sense. Any comments on this? Would you model it in a different way?
| 691
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https://www.physicsforums.com/threads/plotting-a-plane-in-mathematica.101840/
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# Plotting a plane in Mathematica (1 Viewer)
### Users Who Are Viewing This Thread (Users: 0, Guests: 1)
#### Pengwuino
Gold Member
Does anyone know how to plot the plane y=2 in 3d? I'm looken for a curve of intersection with a surface and I don't seem to know how to plot the y=2 plane :)
#### vaishakh
What do you mean by plotting? You can understand that it contains all points in 3d space where the co-ordinate of y is 2. For example (0,2,0), (2,2,0) and (0,2,2). You must know that three non-collinear points define a plane. So it is the plane that consists of the following points
#### Pengwuino
Gold Member
Well I needed to know how to enter that into Mathematica. I was trying parametric crap with t's and it just gave me a stupid line. I finally found out what i had to do though to get an actual plane to show up though.. thanks anyways.
#### loweaxerium
hey having the same problem what u did to get the plane?
#### wil3
Use the Plot3D command:
Plot3D[y = 2, {x, 0, 10}, {y, -2, 2}]
Nota bene: If you fail to express the equation as an equality, it automatically treats the first argument as equal to an unspecified z variable. So the function
Plot3D[y - 2, {x, 0, 10}, {y, -2, 2}] would *not* plot a y-plane, since it actually would end up plotting z = y - 2 (although the zero level set of that function would be your plane)
Last edited:
#### Bill Simpson
Use the Plot3D command: Plot3D[y = 2, {x, 0, 10}, {y, -2, 2}]
I believe that is plotting z=2 and not y=2. You can verify that with this
Plot3D[y = 2, {x,0,10}, {y,-2,2}, AxesLabel->{x,y,z}]
to label the axes so you can see if your plane is at 2 on the z axis or the y axis.
To plot in the y=2 plane I believe this might do what you want:
ParametricPlot3D[{x,2,z}, {x,-2,2}, {z,-2,2}, AxesLabel->{x,y,z}]
There is a nice tutorial I found using Google that you can use to learn this stuff.
http://www.math.uconn.edu/~hurley/math220/Mathematica_docs/Planes.pdf
#### wil3
I see what you mean... doesn't plotting y = 2 look the same as plotting z = 2, just with different axes labeling? And that's strange that even though y = 2 is explicitly stated, Mathematica interprets it as z = 2
#### Bill Simpson
And that's strange that even though y = 2 is explicitly stated, Mathematica interprets it as z = 2
Mathematica is very likely not the language you think it is or doing what you think it is doing.
When you put y=2 in there it created a definition that in the future if you ever used y it should be immediately replaced with 2, at least until you changed it, AND THEN the "value" of y=2 was also 2, so the "y" was gone by the time Plot3D got its hands on what you wrote. So what Plot3D "saw" was Plot3D[2, etc,etc,etc]. Now since Plot3D is expecting some sort of function giving the value of z for the various values of x and y it then plotted the plane z==2 and that was what you saw when you added axis labels.
Learning the strange way that Mathematica evaluates expressions is a serious undertaking. Even when you think you are finally getting the hang of it you are probably only getting close to ready to move up to the next level in the game.
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Latest Teaching jobs » Maths Quiz for KVS and CTET...
# Maths Quiz for KVS and CTET Exams
Q1. The maximum value of 81^sinx . 27^cosx is
(a) 81
(b) 243
(c) 27
(d) None of these
Q2. If 29 tanθ=31 than the value of (1+2 sinθ.cosθ)/(1-2 sinθ.cosθ)
(a) 810
(b) 900
(c) 300
(d) 540
Q3. Two circles with centre A and B and radius 2 units touch each other externally at ‘C’. A third circle with centre ‘C’ and radius ‘2’ units meets other two at D and E. Then the area of the quadrilateral ABDE is
(a) 2√2 sq. units
(b) 3√3 sq. units
(c) 3√2 sq. units
(d) 2√3 sq. units
Q4. A shopkeeper earns a profit of 12% on selling a book at 10% discount on printed price. The ratio of the cost price to printed price of the book is
(a) 45 : 56
(b) 50 : 61
(c) 90 : 974
(d) 99 : 125
Q5. A dealer purchased a washing machine for Rs. 7,660. After allowing a discount of 12% on its marked price, he still gains 10%. Find the marked price of the washing machine.
(a) Rs. 9,575
(b) Rs. 8,426
(c) Rs. 8,246
(d) Rs. 9,755
Q6. A saleable article passes successively in the hands of three traders. Each trader sold it further at a gain of 25% of the cost price. If the last trader sold it for Rs. 250 then what was the cost price for the first trader?
(a) Rs. 128
(b) Rs. 150
(c) Rs. 192
(d) Rs. 200
Q7. Rs. 6,000 becomes Rs. 7,200 in 4 years. If the rate becomes 1.5 times of itself, the amount of the same principal in 5 years will be
(a) Rs. 8,000
(b) Rs. 8,250
(c) Rs. 9,250
(d) Rs. 9,000
Q8. Two alloys contain tin and iron in the ratio of 1 : 2 and 2 : 3. If the two alloys are mixed in the proportion of 3 : 4 respectively (by weight), the ratio of tin and iron in the newly formed alloy is:-
(a) 14 : 25
(b) 10 : 21
(c) 12 : 23
(d) 13 : 22
Q9. Tom is chasing Jerry. In the same interval of time Tom jumps 8 times while Jerry jumps 6 times. But the distance covered by Tom in 7 jumps in equal to the distance covered by Jerry in 5 jumps. The ratio of speed of Tom and Jerry is
(a) 48 : 35
(b) 28 : 15
(c) 24 : 20
(d) 20 : 21
Q10. A labourer was appointed by a contractor on the condition he would be paid Rs. 75 for each day of his work but would be, fined at the rate of Rs. 15 per day for his absent. After 20 days, the contractor paid the labourer Rs. 1140. The number of days the labourer absented from work was
(a) 3 days
(b) 5 days
(c) 4 days
(d) 2 days
Solutions
S1. Ans. (b)
Sol. 81^sinx. 27^cosx
3^(4 sinx ) × 3^(3 cosx )
3^(4 sinx + 3 cosx)
=3^√(4)^2+(3)^2 )
=3^5
= 243
S4. Ans.(a)
Sol. (C.P.)/(M.P.)=(100 – D)/(100 + P)
(C.P.)/(M.P.)=90/112
(C.P.)/(M.P.)=45/56
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https://math.stackexchange.com/questions/2671885/integration-boundary-condition-dependent-on-integral-derivative
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text/html
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crawl-data/CC-MAIN-2019-22/segments/1558232256887.36/warc/CC-MAIN-20190522163302-20190522185302-00147.warc.gz
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# Integration boundary condition dependent on integral derivative?
I need to solve integral:
$\int_0^{r(z)} \dfrac{\mathrm{d}z}{r^4(z)-2r^2(z)}$,
where $r(z)=r_i-z(r_i-1)$, where $r_i$ is constant, $z$ is longitudinal coordinate. Boundary condition $r(z)$ is dependent on $z$, so $r$ is not constant and I am not sure how to solve this integral.
If I change boundary immediately in this inegral I will have this shape:
$\int_0^{r(z)} \dfrac{\mathrm{d}z}{a-bz+cz^2-dz^3+ez^4}$, should I solve this integral instead of upper integral and how?
• If $z$ is the integration variable, it is bad style, or even possibly hinting at an error in the surrounding computation, to also have it as outer independent variable in the upper end of the integration interval. Please edit to distinguish between these two variables. Or is the integral really $$\int_0^{r(z)}\frac{dx}{x^4-2x^2}?$$ – LutzL May 3 '18 at 11:46
First, do a change of variable $x = r(z)$. Then do a partial fraction decomposition: $$\frac{1}{x^4 - 2x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - \sqrt2} +\frac{D}{x + \sqrt2}.$$ Can you continue?
• Yes, but I have that $r=r(z)$ and my integral is in form $\int_0^{r(z)} \mathrm{d}z$, what is happening with this dependence $r$ on $z$: $r=r(z)$ in integral with $\mathrm{d}z$? – nick_name Mar 1 '18 at 15:08
• If I would use $x=r(z)$ then it would be $\mathrm{d}x=\mathrm{d}r$, but I have only $\mathrm{d}z$ in my equation, what would I do with it on your way? – nick_name Mar 1 '18 at 15:40
• @nick_name, wrong. $dx = dz$. – Martín-Blas Pérez Pinilla Mar 1 '18 at 15:41
• @nick_name, yes. $dx = −(r_i − 1)dz$. I wrote too fast. My point was that the original variable is $z$. – Martín-Blas Pérez Pinilla Mar 2 '18 at 11:54
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https://www.physicsforums.com/threads/proof-of-a-theorem-y-n-x.588126/
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# Proof of a theorem (y^n=x)
1. Mar 18, 2012
### jwqwerty
the theorem goes likes this:
For every x>0 and every integer n>0 there is one and only one real y>0 such that y^n=x
The book starts the proof by stating E as set consisting of all positive real numbers t such that t<x^n. Then it states that:
If t= x/(1+x) then 0<t<1. Hence t^n<t<x. Thus t exists in E and E is not empty
If t>1+x then t^n>t>x so that t does not exist in E. Thus 1+x is an upper bound of E.
My questions is this:
Why does it divide into two cases, t= x/(1+x) and t>1+x? And instead, why can't we divide into t> x/(1+x) and t=1+x? Don't they still have the same meaning as the previous one in the way that 0<t<1 and t>1?
Last edited: Mar 18, 2012
2. Mar 18, 2012
### mathman
The original statement is wrong. If n is even there are two values of y, one positive and the other negative.
3. Mar 18, 2012
### jwqwerty
oh sorry i changed the statement from all real y to all real y>0
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### Square root of 11
Is square root of 11 an irrational number? How do you know from using a calculator? Thank you.
Well, I happen to know that if the square root of a natural number is NOT a whole number, then it is an irrational number. There are no other possibilities.
Of course you can't tell by the calculator. The calculator will show you 8 or 10 decimals, but you won't know from that if it's going to continue or not, or if it is periodical or not.
But pure mathematics and established, proven theorems will tell you that! : )
A proof that the square root of 2 is irrational
Lots of proofs of the same... plus one proving that any root is irrational if it's not a whole number
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https://math.stackexchange.com/questions/1786249/can-this-sum-over-the-q-pochhammer-symbol-be-simplified
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crawl-data/CC-MAIN-2019-35/segments/1566027315695.36/warc/CC-MAIN-20190821001802-20190821023802-00430.warc.gz
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Can this sum over the q-Pochhammer symbol be simplified?
While considering the problem of the expected value of a dice fixing strategy on a two-sided die that comes up as $1$ with a probability of $\alpha$ and $0$ otherwise. I was studying the strategy where one fixes every die if they are all $1$'s and otherwise fixes only one die. I found that the expected value of this strategy satisfies the functional equation $$A(x)=C(x)-\frac{\alpha x}{1-x} A(\alpha x)$$ where $C$ is some particular rational function (although, I'd be interested in solutions for any or every rational $C$). By making infinitely many substitutions of this equation into itself, we can solve this as: $$A(x)=\sum_{n=0}^{\infty}\left(\prod_{k=0}^{n-1}\frac{-\alpha^{k+1} x}{1-\alpha^kx}\right)C(\alpha^n x)=\sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{n(n+1)/2}x^nC(\alpha^nx)}{(x;\alpha)_n}$$ where $(x;\alpha)_n$ is the q-Pochhammer symbol defined as $$(x;\alpha)_n = \prod_{k=0}^{n-1}(1-x\alpha^k).$$ Is there a simpler form for the sum for $A$?
I am hopeful that there is such a form since I the Wikipedia page on q-Pochhammer symbols lists the following identity: $$(x;q)_{\infty}=\sum_{n=0}^{\infty}\frac{(-1)^nq^{n(n-1)/2}x^n}{(q;q)_n}$$ which looks very similar to what I'm trying to get. In fact, this lets us calculate particular values of our generating function for certain $C$. For instance, if $C(x)=\frac{1}{1-x}$, we would find that at $x=\alpha$, we would have $$A(\alpha)=\sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{n(n+1)/2}\alpha^n}{(\alpha;\alpha)_{n+1}}=\frac{1-(\alpha;\alpha)_{\infty}}{\alpha}.$$ However, I don't see any other useful identites.
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https://socratic.org/questions/how-do-you-find-the-stationary-points-of-a-function
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text/html
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crawl-data/CC-MAIN-2020-05/segments/1579251778168.77/warc/CC-MAIN-20200128091916-20200128121916-00073.warc.gz
| 656,194,363
| 6,356
|
# How do you find the stationary points of a function?
Jun 26, 2018
Shown below
#### Explanation:
As we can see from this image, a stationary point is a point on a curve where the slop is zero
Hence the stationary points are when the derivative is zero
Hence to find the stationary point of $y = f \left(x\right)$, find $\frac{\mathrm{dy}}{\mathrm{dx}}$ and then set it equal to zero
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
Then solve this equation, to find the values of $x$ for what the function is stationary
For examples
$y = {x}^{2} + 3 x + 8$
To find the stationary find $\frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x + 3$
Set it to zero
$2 x + 3 = 0$
Solve
$x = - \frac{3}{2} \implies y = \frac{23}{4}$
Hence the stationary point of this function is at $\left(- \frac{3}{2} , \frac{23}{4}\right)$
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Practice Midterm8
# Practice Midterm8 - Solutions to Exam#1 ST 8533 Applied...
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Solutions to Exam #1 - ST 8533: Applied Probability Spring, 2001 February 29, 2001 Directions: Answer all questions precisely and completely. To receive full credit, you must de fi ne all notation, and show a reasonable amount of work. 1. (20 points) Giggleswick is a gift store on Main Street in downtown Starkville. Let N represent the number of gifts purchased at Giggleswick, and suppose N has a Poisson distribution with parameter λ . Further suppose that each item purchased will be gift wrapped with probability p . Let X denote the number of items that are wrapped. Use a conditional probability argument to prove that X has a Poisson distribution with parameter λ p . Solution: The proof is as follows: P { X = x } = X n =0 P { X = x | N = n } P { N = n } = X n = x μ n x p x (1 p ) n x e λ λ n n ! = e λ p x X n = x n ! x !( n x )! (1 p ) n x λ n n ! = e λ p x X n = x [ λ (1 p )] n x ( n x )! λ x x ! = ( λ p ) x x ! e λ h e λ (1 p ) i = ( λ p ) x x ! e λ p 2. (20 points) Joe Bob, a small town boy, goes on vacation, and visits “the big city.” He leaves his hotel and walks around to do a little site-seeing. He’s not paying much attention, and all of the sudden realizes he is lost. He comes to an intersection which will take him in three directions. If he chooses Main Street, he will be back at his hotel in 1 hour. If he choose University Avenue, he will walk for 1.5 hours and end up back in the same spot. If he chooses “Big City Circle” he will walk for 2 hours and end up in the same spot. Unfortunately, there are no street signs or distinguishing landmarks at this three way intersection, so at each time, Joe Bob is equally likely to choose any of the three streets.
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Transient Response of Resistor-Capacitor (RC) and Resistor-Inductor (RL) Circuits
Time Constants
Outline: The Transient Response of RC Circuits The Transient Response of RL Circuits The Complete Response Analysis Steps for finding the Complete Response of RC and RL Circuits Study Problems
The Transient Response of RC Circuits
The Transient Response (also known as the Natural Response) is the
way the circuit responds to energies stored in storage elements, such as
capacitors and inductors. If a capacitor has energy stored within it, then
that energy can be dissipated/absorbed by a resistor. How that energy is
dissipated is the Transient Response.
In this circuit, there is a pulse, a resistor, and a capacitor. Assume here that the pulse goes from 10V down to 0V at t=0. Assume also that the circuit is in Steady State at t=0-. This implies that the capacitor is 'open' at t=0-. In order for KVL to be true at t=0- then the capacitor voltage must be 10V at t=0-. This is because there is no current in the circuit, therefore the voltage across the resistor is zero. Vc(0-) = Vc(0+) = 10V Note that since the Transient Response is the circuit's response to energies stored in storage elements, we will 'kill' the pulse source. This leaves us with a simple Resitor-Capacitor circuit with an initial 10V on the capacitor at t=0+. Applying KCL to an RC circuit: Cdv/dt + V/R = 0 dv/dt + V/(RC) = 0 ∫dv/V = ∫-1/(RC) dt ln V = -t/(RC) + K ln V(t=0) = K ln Vo = K ←Vo is the voltage on the cap at t=0+. lnV - ln Vo = -t/(RC) ln (V/Vo) = -t/(RC) V/Vo = e-t/(RC) V(t) = Vo e-t/(RC) ←Vo = 10V in this example. Note that the speed at which the capacitor discharges from 10V to 0V is determined by the product R×C When t=RC, the voltage on the capacitor is Vo/e or 37% of it's initial value. We call RC the time constant and the symbol is τ For an RC circuit, τ=RC In this particular circuit τ = RC = 100Ω×1mF = 0.1 seconds This means it takes 0.1 seconds for the capacitor to discharge from 10V down to 3.7V. Looking at the response at the right, does this look about right? Here is the same circuit as that above, except that the resistor value is doubled. This means that τ is also doubled. τ = RC = 200Ω×1mF = 0.2 seconds This circuit is twice as slow as the last circuit. Compare this response to the last one. Does it appear that it takes twice as long for the capacitor to discharge? Finding the Time Constant τ: Using an oscilloscope or P-Spice, we can calculate τ by inspecting the response curve. Locate the place where V = 37% × Vo. Then find the time at which this occurs. This time is the time constant, τ.
The Time Constant τ = RC for a simple RC-circuit.
The bigger τ is the longer it takes for the circuit to discharge.
The smaller τ is the faster the response.
τ is the time needed for the Transient Response to decay by a factor of 1/e.
## Study Problems
After clicking on the following link enter 7-1 for the problem and 1 for the step:
Study Problem 7-1
Top of Page
The Transient Response of RL Circuits
The Transient Response (also known as the Natural Response) is the
way the circuit responds to energies stored in storage elements, such as
capacitors and inductors. If an inductor has energy stored within it, then
that energy can be dissipated/absorbed by a resistor. How that energy is
dissipated is the Transient Response.
In this circuit, there is a pulse, a resistor, and an inductor. Assume here that the pulse goes from -10V to 0V at t=0. Assume also that the circuit is in Steady State at t=0-. This implies that the inductor is a 'short' at t=0-. In order for KCL to be true at t=0- the inductor current must be -1A at t=0-. IL(0-) = IL(0+) = -1A Consider the circuit at t=0+, the voltage across the pulse is zero but since IL(0+) = -1A then VR = -10V. Therefore for KVL to be true VL = +10V. Therefore VL = +10V is the initial voltage across the inductor. Note that since the Transient Response is the circuit's response to energies stored in storage elements, we will 'kill' the pulse source. This leaves us with a simple Resitor-Inductor circuit with an initial -10A going through the inductor at t=0+. Applying KVL to an RL circuit: iR + Ldi/dt = 0 iR/L + di/dt = 0 -iR/L = di/dt -R/L dt = di/i ∫-R/L dt = ∫di/i -Rt/L + K = ln i K = ln i(t=0) K = ln io -Rt/L = ln i - K -Rt/L = ln i - ln io -Rt/L = ln(i/io) i/io = e-Rt/L i(t) = ioe-Rt/L ←io in this case is -1A Since the plot on the right is for voltage we will find VL using VL = Ldi/dt VL = (1H) d[ioe-Rt/L]/dt = (1H) (-10) ioe-Rt/L VL = -10 e-Rt/L When t=L/R, the voltage on the inductor is Vo/e or 37% of it's initial value. We call L/R the time constant and again the symbol is τ For an RL circuit, τ=L/R In this particular circuit τ = L/R = 1H/10Ω = 0.1 seconds This means it takes 0.1 seconds for the inductor to go from 10V down to 3.7V. Looking at the response at the right, does this look about right? Here is the same circuit as that above, except that the resistor value is halved. This means that τ is doubled. τ = L/R = 1H/5Ω = 0.2 seconds This circuit is twice as slow as the last circuit. Compare this response to the last one. Does it appear that it takes twice as long for the circuit to dissipate it's energy? Finding the Time Constant τ: Using an oscilloscope or P-Spice, we can calculate τ by inspecting the response curve. Locate the place where V = 37% × Vo. Then find the time at which this occurs. This time is the time constant, τ.
The Time Constant τ = L/R for a simple RL-circuit.
The bigger τ is the longer it takes for the circuit energy to discharge.
The smaller τ is the faster the response.
τ is the time needed for the Transient Response to decay by a factor of 1/e.
## Study Problems
After clicking on the following link enter 7-2 for the problem and 1 for the step:
Study Problem 7-2
Top of Page
The Complete Response
The Complete Response is the circuit's response to both
an independent source as well as energies stored in the circuit.
A circuit driven by an independent source is said to have
a forcing function.
Vcomplete response = Vnatural + Vforced
Here is an RC Circuit with a Forcing Function: Assume the source is a pulse which goes from 0V to 10V at t=0. If we assume steady state at t=0-, then there is no initial energy stored in the circuit. Intuitively we know that the capacitor is going to charge up to 10V. When the capacitor gets to 10V then the circuit is again at steady state. The pulse is forcing the capacitor to 10V, thus the 10V on the capacitor is called the forced response. The time it takes the capacitor to charge up to 10V is determined by the time constant. The response of getting to 10V is the transient response. Now we will find the Complete Response for V across the capacitor. This equation will match the curve shown at the right. From the last section we know that the transient response for an RC circuit is: V(t) = Vo e-t/(RC) = A e-t/(RC) Note that A is just some constant. We also know from inspection that eventually the capacitor will charge up to 10V. Now putting the transient and forced responses together we get: Vcomplete = A e-t/(RC) + Vforced Vcomplete = A e-t/(RC) + 10V Now we need to find A such that the equation equals Vo at t=0. In other words, the equation must satisfy the initial condition. V(t=0) = 0, therefore: 0 = A e0 + 10V = A + 10V A = -10V Vcomplete = -10e-t/(RC) + 10V Vcomplete = -10e-10t + 10V Note that when t>>0, Vcomplete = 10V. This intuitively means that when the transient response is gone the forced response still remains. In this circuit, the capacitor DOES NOT start at 0V. In other words the capacitor has a non-zero initial condition of 5V: Note that the left switch closes at the same time the right switch opens. Intuitively we can see that the capacitor is going to start at 5V and then charge up to 15V. For t<0 the 5V source is the forcing function and for t>0 the 15V source is the forcing function. Since this is an RC circuit with a forcing function, the response takes the following form: Vcomplete = A e-t/(RC) + Vforced By inspection we know that Vforced = 15V Vcomplete = A e-t/(RC) + 15V Now we need to find A such that the entire equation satisfies the value of V at t=0. V(t=0) = 5V = A e0 + 15V A = -10V Vcomplete = -10e-t/(RC) + 15 V Vcomplete = -10e-10t + 15 V Here is the Complete Response: Does this curve match the equation: Vcomplete = -10e-10t + 15 V Now let's find the voltage across the resistor for the RL circuit to the right. Note that the pulse goes from 5V to 15V at t=0. Assume that the circuit is in steady state at t=0-. At steady state inductors look like 'shorts' therefore the voltage across the resistor must be equal to the pulse voltage of 5V at t=0-. At t=0+ the voltage across the resistor is still 5V, but WHY????? Since the current in the inductor is continuous from 0- to 0+, and the current in the resistor is the same as the current in the inductor, and the voltage across the resistor is determined by its current, then we can say that if the resistor's current is continuous then the resistor's voltage must also be continuous. At t>>0 the voltage across the resistor is 15V. For t>0 we eventually reach steady state (as the transient response dies away), so we know that at t>>0 the inductor will look like a 'short'. Therefore the voltage across the resistor will equal the voltage of the pulse. Therefore we have both the initial condition and the forced response for the voltage across the resistor: Vo = 5V Vforced = 15V Now we will find the Complete Response: Vcomplete = A e-t/τ + Vforced Vcomplete = A e-Rt/L + Vforced Vcomplete = A e-Rt/L + 15V To find A we must let t=0 and assign the equation to Vo: V(t=0) = 5V = A e0 + 15V = A + 15 A = -10 Vcomplete = -10 e-Rt/L + 15V Vcomplete = -10 e-Rt/L + 15V R/L = 5 Vcomplete = -10 e-5t + 15V Assume the pulse oscillates between 15V and 5V every 1 s. Consider the waveforms to the right. One is the pulse and the other is the voltage across the inductor. Note that as the transient response for VL dies away, the voltage across the inductor falls to zero. This is because the inductor acts like a 'short' in steady state conditions. However immediately after the switch moves a voltage is induced across the inductor. This voltage opposes the change that is occuring to the current in the circuit. Apply KVL to the circuit at t=0+ and t=1+ and see if you can get the 10V and -10V initial conditions for VL.
The Complete Response has two parts: The Transient Response
and the Forced Response:
Vcomplete response = Vtransient + Vforced
Vtransient is found by 'killing' the forcing function. If the
circuit is RC then τ=RC and if the circuit is RL then τ=L/R.
Vforced is found by assuming Steady State. In other words, we
put the forcing function back into the circuit and assume that the
Transient Response has died out.
Any constants are found by setting Vcomplete(t=0) = Initial Conditions (at t=0)
Top of Page
Analysis Steps for finding the Complete Response of RC and RL Circuits
Use these Steps when finding the Complete Response for a 1st-order Circuit:
• Step 1: First examine the switch to see if it is opening or closing and at what time.
• Step 2: Next draw the circuit right before the switch moves. You will probably assume
steady state at this time but not always. The problem needs to tell you to assume steady state.
• Step 3: Find all voltages and currents that can not change instantaneously when the switch moves.
In other words, Find voltages across all capacitors and currents through all inductors!
• Step 4: Now draw the circuit right after the switch moves. Label the circuit with all the
capacitor voltages and inductor currents you found in step 3.
• Step 5: Now you are ready to find your initial condition(s). Analyze the circuit to find the
initial condition(s) of what it is your solving for.
• Step 6: Next you will find the transient/natural reponse, or τ. To do this 'kill' all forcing functions.
Make all voltage sources 'shorts' and all current sources 'opens'. Remember that the transient response is the circuit's
response to energies stored in storage elements, so we need to remove forcing functions to find this.
Recall that every voltage and current will have the same τ value.
You now have Ae-t/τ for what your solving for.
• Step 7: Now we need to find the forced response.
The forced response is the state of the circuit after the switch has moved AND after
the transient response has died-off. To find the forced response assume Steady State, i,e, t>>>0.
Find the final resting value (forced response - VF) of whatever it is you are solving for.
• Step 8: You should now have an equation which looks like v(t) = Ae-t/τ + VF
or i(t) = Ae-t/τ + IF. To find the unknown 'A' you will apply the initial condition to
this equation. Usually the initial condition is the value at t=0, so you will plug in t=0 to get the following:
I.C. = A + VF or I.C. = A + IF You can now solve for A.
• Step 9: Plugging the value of A into: v(t) = Ae-t/τ + VF, you now
know the voltage for all time greater than t=0 (assuming that the switch moved at t=0).
• Step 10: Using your equation for v(t) or i(t), you can find other things (voltages, currents,
power, etc.) using KVL, KCL, and Ohm's Law.
Top of Page
Study Problems
The following problems will follow the steps above to find the Complete Response of
first order circuits.
## Study Problems
After clicking on the following link enter 7-3 for the problem and 1 for the step:
Study Problem 7-3
After clicking on the following link enter 7-4 for the problem and 1 for the step:
Study Problem 7-4
Top of Page
Back To Index
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https://testbook.com/question-answer/the-expression-228-229-230-is-divi--608d39026c1f0213ce027d92
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text/html
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crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00628.warc.gz
| 646,908,087
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The expression (228 + 229 + 230) is divisible by which of the following number?
1. 7
2. 9
3. 5
4. 3
Option 1 : 7
Detailed Solution
Given:
The expression: (228 + 229 + 230)
Calculation:
(228 + 229 + 230)
When we take common out,
⇒ 228(20 + 21 + 22)
⇒ 228(1 + 2 + 4)
⇒ 228 × 7
The expression (228 + 229 + 230) is divisible by 7.
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# How to find the Surface Area of a Prism?
• Last Updated : 16 Jun, 2022
A prism is a 3-D shape having two identical polygons facing each other. The identical polygons are called the bases of the prism, where they can be triangles, squares, rectangles, pentagons, or any other n-sided polygon. The other faces of a prism are parallelograms or rectangles. A prism can be a regular prism or an irregular prism based on the shape of its base, i.e., whether the base is a regular polygon or an irregular polygon. Also, there are different types of prisms based on the shape of the base of a prism, such as
• Triangular prisms,
• Square prisms,
• Rectangular prisms,
• Pentagonal prisms,
• Hexagonal prisms, etc.
### Surface Area of Prism
The total surface area of a prism is the total area occupied by all its faces. To find out the surface area of a prism, we need to calculate the areas of all of its faces and then add all the areas obtained. The lateral surface area of a prism is the area occupied by the faces of a prism, excluding the identical faces (bases of a prism) facing each other.
Lateral surface area of a prism = Base perimeter × height
Now, the total surface area of a prism is equal to the sum of its lateral surface area and the areas of its two bases.
The general formula for calculating the surface area of any type of prism is:
Total surface area of a Prism = 2(Base Area)+ (Base perimeter × height)
### Triangular prism
A triangular prism is a prism whose bases are triangular. Let “H” be the height of the prism; “a, b, and c” are the lengths of the sides of triangular bases, and “h” be their height.
Now, lateral surface area of the triangular prism = Base perimeter × height
We know that perimeter of a triangle = Sum of its three sides = a + b + c
Lateral surface area of the triangular prism = (a + b + c) H
Total surface area of a triangular prism = 2(Base Area)+ (Base perimeter × height)
We know that area of a triangle = 1/2 base × height = 1/2 b × h
Total surface area of a triangular prism = 1/2bh + (a + b + c) H
### Square Prism
A square prism is a prism whose bases are squares. Let “a” be the length of the side of a square and “h” be the height of the prism.
Now, lateral surface area of the square prism = Base perimeter × height
We know that perimeter of a square = Sum of its four sides = 4a
Lateral surface area of the square prism = 4ah
Total surface area of a square prism = 2(Base Area)+ (Base perimeter × height)
We know that the area of a square = a2 square units.
Total surface area of a square prism = 2a2 + 4ah
### Rectangular prism
A rectangular prism is a prism whose bases are rectangles. Let “h” be the height of the prism, and “l and b” be the length and breadth of the rectangular bases.
Now, lateral surface area of the rectangular prism = Base perimeter × height
We know that perimeter of a rectangle = Sum of its four sides = 2 (l + b)
Lateral surface area of the rectangular prism = 2(l + b)h
Total surface area of a rectangular prism = 2(Base Area)+ (Base perimeter × height)
We know that the area of a rectangle = (l × b) square units.
Total surface area of a rectangular prism = 2lb + 2bh + 2lh
### Pentagonal prism
A pentagonal prism is a prism whose bases are pentagons. Let “h” be the height of the prism, “a” be the apothem length of the prism, and “b” be the base length of the prism.
Lateral surface area of the pentagonal prism = 5bh
Total surface area of a pentagonal prism = 5ab + 5bh
### Hexagonal Prism
A hexagonal prism is a prism whose bases are hexagons. Let “a” be the length of the side of a hexagon and “h” be the height of the prism.
Lateral surface area of the hexagonal prism = 6ah
Total surface area of a hexagonal prism = 3√3a2 + 6ah
### Sample Problems
Problem 1: Find the height of the square prism if its total surface area is 58 cm2, and the length of the side of the square base is 2 cm.
Solution:
Given data,
The total surface area of the square prism = 58 cm2
The length of the side of the square base = 2 cm
We know that,
The total surface area of a square prism = 2a2 + 4ah
⇒ 58 = 2 × (2)2 + 4 × 2 × h
⇒ 58 = 8 + 8h
⇒ 50 = 8h
⇒ h = 50/8 = 6.25 cm,
Hence, the height of the given prism is 6.25 cm.
Problem 2: What is the surface area of a prism whose base area is 15 square units, a base perimeter of 24 units, and its height is 8 units?
Solution:
Given data,
Base area = 15 square units
Base perimeter = 24 units
Height of the prism = 8 units
We have,
The total surface area of the prism = (2 × Base Area) + (Base perimeter × height)
= (2 × 15) + (24 × 8)
= 30 + 192 = 222 square units.
Hence, the surface area of the given prism = 222 square units.
Problem 3: What is the lateral surface area of a triangular prism whose base perimeter is 30 cm and the height of the prism is 12 cm?
Solution:
Given data,
The base perimeter of the prism = 30 cm2
The height of the prism = 12 cm
We know that,
The lateral surface area of the prism = Base perimeter × height
= 30 × 12= 360 sq. cm
Hence, the lateral surface area of the prism is 360 sq. cm.
Problem 4: Find the surface area of the regular hexagonal prism if the height of the prism is 10 in and the length of the side of the base is 7 in.
Solution:
Given data,
The height of the prism (h) = 10 in
The length of the side of the base (a) = 7 in
The surface area of a regular hexagonal prism = 6ah + 3√3a2
= 6 × 7 × 10 + 3√3(7)2
= (420 + 147√3) sq. in
Hence, the surface area of the given prism is (420 + 147√3) sq. in.
Problem 5: Determine the lateral surface area and the total surface of a rectangular prism if the length and breadth of the base are 11 cm and 8 cm, respectively, and the height of the prism is 14 cm.
Solution:
Given data,
The length of the rectangular base (l) = 11 cm
The breadth of the rectangular base (b) = 8 cm
The height of the prism (h) = 14 cm
We have,
The lateral surface area of the prism = Base perimeter × height
= 2 (l + b) h = 2 × (11 + 8) × 14 = 532 sq. cm
We know that,
The total surface area of the rectangular prism = 2 (lb + bh + lh)
= 2 × (11 × 8 + 8 × 14 + 11 × 14)
= 2 × (88 + 112 + 154) = 708 sq. cm
Hence, the lateral and total surface areas of the given rectangular prism are 532 sq. cm and 708 sq. cm, respectively.
Problem 6: If the total area of a pentagonal prism is 125 square units and its height and apothem length are 10 units and 6 units respectively, determine its base length.
Solution:
Given data,
The total surface area of the pentagonal prism = 125 square units
The height of the prism (h) = 10 units
Apothem length (a) = 6 units
We know that,
The total surface area of the pentagonal prism = 5ab + 5bh
⇒ 125 = 5b (a+ h)
⇒ 125/5 = b (6 + 10)
⇒ 25 = b × (16) ⇒ b = 25/16 = 1.5625 units
Hence, the base length is 1.5625 units
My Personal Notes arrow_drop_up
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https://justaaa.com/physics/331467-if-337-m3-of-a-gas-initially-at-stp-is-placed
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text/html
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Question
# If 3.37 m3 of a gas initially at STP is placed under a pressure of 2.30...
If 3.37 m3 of a gas initially at STP is placed under a pressure of 2.30 atm , the temperature of the gas rises to 33.0 ?C. What is the volume?
Using PV = nRT
The gas is going to retain the same number of molecules, and the constant R remains the same, so:
=
n and R cancel out for the previously explain reason; so you have P1V1/T1 = P2V2/T2
STP is defined as 273K (0 C) and 100kPa (or 10^5 Pa)
1 atm = 1.013 x 10^5 Pa
No conversion necessary for m^3
Substituting: (100,000)(3.37)/ 273 = (2.30 x 101,300)(V2) / (273 + 33)
(V2) = 1234.43 / 761.40
(V2) = 1.621 m3
Thankyou !!! Have fun
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CC-MAIN-2024-18
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https://www.emathhelp.net/en/calculators/calculus-1/derivative-calculator/?f=cot%28x%29&var=x&steps=on
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# Derivative of $\cot{\left(x \right)}$
The calculator will find the derivative of $\cot{\left(x \right)}$, with steps shown.
Related calculators: Logarithmic Differentiation Calculator, Implicit Differentiation Calculator with Steps
Leave empty for autodetection.
Leave empty, if you don't need the derivative at a specific point.
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.
Find $\frac{d}{dx} \left(\cot{\left(x \right)}\right)$.
### Solution
The derivative of the cotangent is $\frac{d}{dx} \left(\cot{\left(x \right)}\right) = - \csc^{2}{\left(x \right)}$:
$${\color{red}\left(\frac{d}{dx} \left(\cot{\left(x \right)}\right)\right)} = {\color{red}\left(- \csc^{2}{\left(x \right)}\right)}$$
Thus, $\frac{d}{dx} \left(\cot{\left(x \right)}\right) = - \csc^{2}{\left(x \right)}$.
$\frac{d}{dx} \left(\cot{\left(x \right)}\right) = - \csc^{2}{\left(x \right)}$A
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Free Essay
# Logarithms
In: Other Topics
Submitted By rmphillips1985
Words 408
Pages 2
This is an essay about nothing in order to qualify for this site it must contain at least 250 words. So On the left-hand side above is the exponential statement "y = bx". On the right-hand side above, "logb(y) = x" is the equivalent logarithmic statement, which is pronounced "log-base-b of y equals x"; The value of the subscripted "b" is "the base of the logarithm", just as b is the base in the exponential expression "bx". And, just as the base b in an exponential is always positive and not equal to 1, so also the base b for a logarithm is always positive and not equal to 1. Whatever is inside the logarithm is called the "argument" of the log. Note that the base in both the exponential equation and the log equation (above) is "b", but that the x and y switch sides when you switch between the two equations.PrintHidden<p><font face="Arial" size="2" color="#000000">Note: The graphic in the box below is animated in the original ("live") web lesson.</font></p> —The Relationship Animated— | |
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#### College Algebra
...Unit 7 Test 5/1/15, 4:09 PM MAT 120.17, Spring 2015 Assessments Unit 8 Test Results Unit 7 Test - Grade Report Score: 100% (19 of 19 pts) Submitted: Apr 19 at 12:43pm Question 1 Question Grade: 1.0 Weighted Grade: (1/1.0) If q and f are inverse functions and q(−2) = 8, what is f (9) ? Your Answer: cannot be determined Correct Answer: cannot be determined Comment: If q and f are inverse functions and q(a) = b, then f (b) = a. However, since 9 is not the given domain value for q, the answer cannot be determined. Question 2 Question Grade: 1.0 Weighted Grade: (1/1.0) Choose any false statements regarding the graph. Select all that apply. Choice Selected Points The graph is a function. Yes +1 The graph is a function that has an inverse function. Yes +1 The graph is a one-to-one function. Yes +1 The inverse of the graph is not a function. No The graph passes the horizontal line test. Yes +1 The graph passes the vertical line test. Yes +1 Number of available correct choices: 5 Comment: A vertical line can be drawn through the graph intersecting the graph in more than one place, so the graph fails the vertical line test. Therefore, the graph does not represent a function. A horizontal line can be drawn through the graph intersecting the graph in more than one place. So, the graph fails the horizontal line test. Therefore, the graph is not a......
Words: 2442 - Pages: 10
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#### Dflgvre
...Algebra 2 Quarter 4 Review Name: ________________________ Class: ____________ Date: _______________ Section 1: Logarithms and Exponential Relations Definitions to Know: * Natural Logarithm * Common Logarithm * Mathematical * Exponential Growth * Exponential Decay Question 1) Change the following from exponential form to logarithmic form (1 mark each): a) b) Question 2) Change the following from logarithmic form to exponential form (1 mark each): a) b) Question 3) Solve for WITHOUT using a calculator. Show all of your work. (Hint: Use the definition of a logarithm.) (2 marks each) a) b) c) d) Question 4) Apply the Change of Base Formula to rewrite the logarithms with the common logarithm. (1 mark each) a) b) Question 5) Solve for the variable. Show all of your work and all of your steps. (Hint: Use the properties of logarithms.) (4 marks each) a) b) c) d) Question 6) Solve for the variable. Show all of your work and all of your steps. Show the answer to 4 decimal places. (Hint: Use the common logarithm.) (4 marks each) a) b) c) Question 7) Solve for . Show all of your work and all of your steps. Show the answer to 4 decimal places. (Hint: Use the natural logarithm and the definition of a logarithm.) (4 marks each) a) b) c) Question 8) Ms. Mary bought a condo for \$145 000. Assuming that the value of the condo will appreciate at most 5% a year, how much will the condo be worth in 5......
Words: 612 - Pages: 3
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#### 12345
...UNIVERSITI TUNKU ABDUL RAHMAN ACADEMIC YEAR 2012/2013 APRIL EXAMINATION FHMM1014 MATHEMATICS I THURSDAY, 25 APRIL 2013 TIME: 5.00 PM – 7.00 PM (2 HOURS) FOUNDATION IN SCIENCE SOLUTIONS This is not an official document of UTAR. The University is not responsible for any errors found in the solutions. Solutions to FHMM1014 Mathematics I (April 2013) 2 FHMM1014 MATHEMATICS I Q1. (a) (i) a + bi = +i 1 (a − 2) + bi ⇒ a + bi = a − 2) + bi )(1 + i ) (( a + bi = (a − 2 − b) + (a − 2 + b)i a a = − 2 − b So b = a−2+b ⇒ ⇒ − b=2 a= 2 z= 2 − 2i (ii) (iii) z= 4+4 = 2 2 ⇒ −π θ= 4 −2 = tan θ =−1 2 = arg( z ) θ= −π 4 (b) (i) 3log x x log 3 = 3log x =81 ⇒ ⇒ 2 × 3log x 162 = 3log x =34 log x = 4 ⇒ x = 104 (ii) = log10 x + log(1 + 2 x ) log 5x + log 6 log10 x (1 + 2 x )= log 5 x × 6 10 x (1 + 2 x ) = x × 6 5 Let ⇒ 2 x (1 + 2 x ) = 6 y = 2 x then y 2 + y − 6 = 0 ( y − 2)( y + 3) = 0 y − 2 = 0 ⇒ y = 2 ⇒ 2x = 2 ⇒ x = 1 y + 3 = ⇒ y = 3 ⇒ 2 x = 3 impossible − − 0 Solutions to FHMM1014 Mathematics I (April 2013) 3 FHMM1014 MATHEMATICS I Q1. (Continued) (c) P ( x)= A( x − 2)( x − (2 − i ))( x − (2 + i )) P( x) = A( x − 2)( x 2 − 4 x + 5) P( x) = A( x3 − 6 x 2 + 13 x − 10) P(0) = −5 P( x) = ⇒ 1 A= 2 1 3 ( x − 6 x 2 + 13 x − 10) 2 (d) A(B ( A − ( B C ) )′ = C )′ ( )′ = A′ ( B C ) = ( A′ B ) ( A′ C ) = ( A B′ )′ ( A C ′ )′ =B )′ ( A − C )′......
Words: 991 - Pages: 4
...QUESTION PAPER CODE 65/1/1 EXPECTED ANSWERS/VALUE POINTS SECTION - A Q. No. 1-10. 1. x = 25 2. x = 6. 2x3/2 + 2 10. 1 5 Marks 3. 10 π 12 4. x = 2 5. x = + 6 2π 3 x +c 7. 8. 5 9. ˆ ˆ { r – (aˆi + bˆj + ck ) }⋅ (ˆi + ˆj + k ) = 0 or r⋅ ˆ+ˆ+k =a+b+c i j ˆ ( ) 1×10 =10 m SECTION - B 11. ∀ (a, b) ∈ A × A a + b = b + a ∴ (a, b) R (a, b) ∴ R is reflexive For (a, b), (c, d) ∈ A × A If (a, b) R (c, d) i.e. a + d = b + c ⇒ c + b = d + a then (c, d) R (a, b) ∴ R is symmetric For (a, b), (c, d), (e, f) ∈ A × A If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e Adding, a + d + c + f = b + c + d + e then (a, b) R (e, f) ∴ R is transitive ∴ R is reflexive, symmetric and transitive 1m 1m ⇒ a+f = b+e 1m hence R is an equivalance relation [(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)} ½m ½m 2 12. 1 + sin x + 1 – sin x cot–1 1 + sin x – 1 – sin x 2 2 x x x x cos + sin + cos − sin 2 2 2 2 2 2 x x x x cos + sin − cos − sin 2 2 2 2 = cot–1 2½ m x 2 cos 2 x x −1 = cot −1 = cot cot = 2 2 2 sin x 2 OR 5 2 1 1 LHS = 2 tan −1 + tan −1 + sec −1 7 5 8 1½ m 1 1 + −1 5 8 + tan −1 1 = 2 tan 7 1– 1 40 2⋅1 1 3 + tan −1 1 = tan −1 2 7 7 1– 1 3 1½+½ m = 2 tan −1 1 + tan −1......
Words: 1846 - Pages: 8
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#### Lab 9
...1. What is the primary place to store log files on a local Linux system and what are recommended procedures for that location? Almost all logfiles are located under /var/log directory It is very important that the information that comes from syslog not be compromised. Making the files in /var/log readable and writable by only a limited number of users is a good start. 2. Why remote logging to a central server is considered a best practice? To identify a baseline system state with the use of the logs & to keep the information from prying eyes. 3. What is the syntax and file you would edit with the necessary entries to send syslogs from your Linux system to a logging server at 172.130.1.254? su –c ’ vi/etc/rsyslog.conf ‘, then remove the # from in front of \$ModLoadimudp and \$UDPServerRun514 if it hasn’t already been done. Then add a line below remote host with the following syntax *.*@@172.130.1.254:541 4. Why is the “Tripwire” application considered a file integrity checker? File Integrity Monitoring is available as a standalone solution or as part of Tripwire’s Security Configuration Management suite, where you have continual assurance of the integrity of security configurations and complete visibility and control of all change for your continuous monitoring, change audit and compliance demands. 5. Could rkhunter be considered a file integrity checker? Why or why not? Rootkit Hunter is considered a file integrity checker because......
Words: 608 - Pages: 3
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#### Pros and Cons of Media
...For the benefit of the students, specially the aspiring ones, the question of JEE(advanced), 2013 are also given in this booklet. Keeping the interest of students studying in class XI, the questions based on topics from class XI have been marked with ‘*’, which can be attempted as a test. For this test the time allocated in Physics, Chemistry & Mathematics and Physics are 22 minutes, 21 minutes and 25 minutes respectively. FIITJEE SOLUTIONS TO JEE(ADVANCED)-2013 CODE PAPER 2 3 Time: 3 Hours Maximum Marks: 180ase read the instructions carefully. You are allotted 5 minutes specifically for this purpose. INSTRUCTIONS A. General: 1. This booklet is your Question Paper. Do not break the seals of this booklet before being instructed to do so by the invigilators. 2. Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadgets are NOT allowed inside the examination hall. 3. Write your name and roll number in the space provided on the back cover of this booklet. 4. Answers to the questions and personal details are to be filled on a two-part carbon-less paper, which is provided separately. These parts should only be separated at the end of the examination when instructed by the invigilator. The upper sheet is a machine-gradable Objective Response Sheet (ORS) which will be retained by the invigilator. You will be allowed to take away the bottom sheet at the end of the examination. 5. Using a black ball point pen......
Words: 328 - Pages: 2
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] |
https://socratic.org/questions/circle-a-has-a-radius-of-4-and-a-center-of-5-3-circle-b-has-a-radius-of-3-and-a-
| 1,642,391,962,000,000,000
|
text/html
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crawl-data/CC-MAIN-2022-05/segments/1642320300289.37/warc/CC-MAIN-20220117031001-20220117061001-00051.warc.gz
| 583,582,350
| 5,919
|
# Circle A has a radius of 4 and a center of (5 ,3 ). Circle B has a radius of 3 and a center of (1 ,4 ). If circle B is translated by <2 ,4 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?
Feb 9, 2018
Circles A and B overlap.
#### Explanation:
Circle A - $C e n t e r {O}_{A} \left(5 , 3\right) , {R}_{A} = 4$
Circle A - $C e n t e r {O}_{A} \left(1 , 4\right) , {R}_{A} = 3$
Circle B translated by (2,4)
New center of B ${O}_{B} \left(\begin{matrix}1 + 2 \\ 4 + 4\end{matrix}\right) \implies \left(\begin{matrix}3 \\ 8\end{matrix}\right)$
$\vec{{O}_{A} {O}_{B}} = \sqrt{{\left(5 - 3\right)}^{2} + {\left(3 - 8\right)}^{2}} = \sqrt{29} \approx 5.39$
Sum of radii ${R}_{A} + {R}_{B} = 4 + 3 = 7$
Since ${R}_{A} + {R}_{B} > \vec{{O}_{A} {O}_{B}}$, circles A and B overlap.
| 335
| 832
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CC-MAIN-2022-05
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https://www.instructables.com/How-to-Factor-Polynomials-on-a-Graphing-Calculator/
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Introduction: How to Factor Polynomials on a Graphing Calculator (TI-83 and TI-84)
These instructions will explain step-by-step on how to factor polynomials on a TI-83/TI-84 graphing calculator
Step 1:
Begin by selecting the PRGM button and scroll over to NEW, click ENTER and name the program and then click ENTER. Having the name relate to the formula is always a good idea. (Example: Factors)
Step 2:
Press the PRGM button, scroll once to the right to I/O, scroll down and select ClrHome. Press PRGM again, scroll once to the right to I/O and select Input, then hit 2ND ALPHA and type in “ENTER A:” (use the + to make quotations). After “ENTER A” put a comma followed by the variable A.
Step 3:
Follow the above instructions to create the same input but with “ENTER B” and “ENTER C” You should have 3 sets of inputs at the end of this step. ex - Input “ENTER B:” , B Press ENTER
Step 4:
Press MATH, scroll once to the right and select “gcd(“. Press MATH again, scroll right and select “abs(“. In the of the “abs(“ put your variable A and then close the parenthesis. Repeat these steps for the variable B. For variable C all that is needed is “abs” followed by three sets of parenthesis. After the parenthesis press STO,located above the ON button, which is the store button followed by the variable G. Press ENTER
Step 5:
Select PRGM and select the If statement. Skip a few lines and select a left parenthesis and put the variable G into it. Select 2nd MATH and then select the not equal sign followed by zero and the end of the parenthesis. Press ENTER
Step 6:
Press PRGM and select the Then statement. Press ENTER
Step 7:
Begin with a parenthesis on a new line followed by the variable A divided by variable G and end parenthesis. Then STO the answer with variable A. Press ENTER. Repeat these same steps but with variables B and C. ex - (C/G)->C (To store an answer press STO, it will be followed by an arrow then enter the variable as instructed)
Step 8:
Select PRGM and put End statement after the three above lines in step 7.
Step 9:
Begin with a parenthesis and put variable A multiplied by C followed by a parenthesis. After the parenthesis STO with variable D. Press ENTER.
Step 10:
Select 0 and STO it with variable L. Press ENTER.
Step 11:
Select 1 and STO it with variable J. Press ENTER.
Step 12:
Select PRGM and select the While statement. Begin a parenthesis followed by variable L does not equal(≠) variable B, end with a parenthesis. Press ENTER. ex - While (L x=B)
Step 13:
Begin with a parenthesis followed by variable D divided by variable J. End the parenthesis and STO to variable K. Press ENTER.
Step 14:
Select PRGM and select the If statement. Press the MATH button and scroll once to the right and select fPart( (#4 on the list) followed by a parenthesis then variable K followed by another parenthesis and set that equal to 0 followed by once more parenthesis. ex - If (fPart(K)=0) Press ENTER
Step 15:
Select PRGM and select Then. Press ENTER.
Step 16:
Select variable J and add K to STO variable L. Press ENTER. Select variable J and add 1 to STO variable J. Press ENTER.
Step 17:
Select PRGM and select statement Else. Press ENTER.
Step 18:
Select variable J and add 1 to it followed by STO variable J. Press ENTER.
Step 19:
Select PRGM and select statement END. Repeat this once on a new line. Press ENTER.
Step 20:
Select variable J and subtract by 1 and STO variable J. Press ENTER. Select variable A and STO it to O. Press ENTER.
Step 21:
Press MATH and scroll once to the right to select gcd( (located at the bottom of NUM) followed by abs( which is located at the same spot. After the parenthesis on abs( follow it with variable K. then put a comma followed by abs( with a 0 in the parenthesis. Add two more parenthesis to finish the statement followed by STO variable H. Press ENTER
Step 22:
Begin with a parenthesis followed by variable K divided by variable H followed by a parenthesis and STO with variable K. Press ENTER. Begin with a parenthesis followed by variable O divided by H and STO it to O. Press ENTER
Step 23:
Select gcd( and abs( followed by variable J and end parenthesis. Begin with a comma followed by abs( with variable A followed by two parenthesis and STO it to variable M. Press ENTER.
Step 24:
Begin with a parenthesis followed by variable J divided by M, end parenthesis and STO it to variable J. Press ENTER. Begin with a parenthesis followed by variable A divided by M, end parenthesis and STO it to variable A.
Step 25:
Select 2nd 0 which will bring up the CATALOG and select PlotsOff. Press ENTER. Select CATALOG again and select AxesOff. Press ENTER. CATALOG once again and select ClrDraw. Press ENTER.
Step 26:
This step is for the display of the formula. Select 2nd PRGM and then select Text followed by a parenthesis with the numbers -1,15,0,G,”(,O,”X+”,K,”)(“,A,”X+”,K,”)(“,A,”X+”,J,”)” (You can skip to a new line for each parenthesis to make it cleaner)
Step 27:
To test program press PRGM (make sure EXEC is highlighted) and scroll to your program name and press ENTER.
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Proving logarithm rules
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We're doing logarithms right now (grade 10 honours math, the US math curriculum goes incredibly slowly... but that's another topic), and one assignment was to prove logbx - logby = logbx/y This is what I did: logbx - logby = logbx/y let m = logbx <=> bm = x (by definition of logarithms) let n = logby <=> bn = y (by definition of logarithms) logbbm - logbbn ? logbbm/bn (by substitution) logbbm - logbbn ? logbbm - n (by law of exponents) m - n = m - n (by definition of logarithms, i.e. "b to what power equals bz, the answer is z") Q.E.D. Now, from my point of view, this is nice, simple, and logical. From my teacher's point of view, it's a very very poor proof because I lack experience with logarithms. Now, clearly, I haven't been using them for years or anything, but that doesn't sound like a very logical reason as to why it's a poor proof... Would someone please chime in with their point of view? I don't care so much if I'm wrong, but I care as to why I'm wrong so I understand things better. (I know I had another thread like this previously, but I'm always getting into debates with my math teacher (who is also my APCS teacher) and I can't find anyone else to ask as to who's right)
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I'm gonna move this to Math & Physics.
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What exactly do you mean when you write
logbbm - logbbn ? logbbm/bn (by substitution)?
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Quote:
Original post by RavuyaI'm gonna move this to Math & Physics.
Gotcha, just thought that this:
Quote:
Discussion of math and physics, primarily as they apply to games.
discouraged non-game-applying math discussion [grin]
Quote:
Original post by Gil GrissomWhat exactly do you mean when you writelogbbm - logbbn ? logbbm/bn (by substitution)?
I mean that I substitute x and y with bm and bn into the original equation and use a question mark instead of an equal sign since we don't know if it's true yet.
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Your reasoning is right, but your proof is wrong. You can't just start with the statement under dispute and massage it into a statement that is trivially true. Look, here's a proof that 0=1:
0 = 1
0*0 = 1*0
0 = 0
...which is trivially true. But not useful. To prove a statement, you must START with trivially true statements, and transform them into the statement under dispute.
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Quote:
Original post by SneftelTo prove a statement, you must START with trivially true statements, and transform them into the statement under dispute.
OK, gotcha. I can't understand why my teacher couldn't just have stated that.
Thanks [smile].
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Quick follow up - does that mean I can essentially reverse what I had? Like so:
prove: logbx - logby = logbx/y
proof:
let m = logbx <=> bm = x (by definition of logarithms)
let n = logby <=> bn = y (by definition of logarithms)
m - n = m - n
logbbm - logbbn = logbbm - n (by definition of logarithms, this is equivalent to the above equation)
logbbm - logbbn = logbbm/bn (exponent rules)
logbx - logby = logbx/y (substitution)
Q.E.D.
Would that be valid, since I started with a trivially true statement and worked up to the original?
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Quote:
Original post by nullsquaredI mean that I substitute x and y with bm and bn into the original equation and use a question mark instead of an equal sign since we don't know if it's true yet.
So my problem with this proof is that it goes in the wrong direction. You start with the premise, manipulate it in some ways and get a true statement (m-n = m-n). As I understand, you then assume that the premise itself is proved, which of course it is not since you can start with a false premise and still get a true conclusion. Your idea itself is OK, but you need to work it out a bit to get an actual proof. I wouldn't call it "lack of experience with logarithms", though -- maybe more "lack of experience with formal proofs".
I'm not sure what your teacher meant by lack of experience with logarithms. Did he give an example of what would be a good proof? Maybe, say, you already know that log x + log y = log xy, and he just wanted you to use that for a much shorter proof.
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Quote:
Original post by Gil Grissom
Quote:
Original post by nullsquaredI mean that I substitute x and y with bm and bn into the original equation and use a question mark instead of an equal sign since we don't know if it's true yet.
So my problem with this proof is that it goes in the wrong direction. You start with the premise, manipulate it in some ways and get a true statement (m-n = m-n). As I understand, you then assume that the premise itself is proved, which of course it is not since you can start with a false premise and still get a true conclusion.
Yup, as Sneftel mentioned. Now my question is, is my new version of the proof (which is essentially a reversal of what I originally had) better? See the post above yours.
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Quote:
Original post by nullsquaredWould that be valid, since I started with a trivially true statement and worked up to the original?
Yes, it looks valid, as long as by "substitution" you mean "the equation one before last is true for any m, n, and, therefore, for m = log x and n = log y in particular".
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Quote:
Original post by Gil Grissom
Quote:
Original post by nullsquaredWould that be valid, since I started with a trivially true statement and worked up to the original?
Yes, it looks valid, as long as by "substitution" you mean "the equation one before last is true for any m, n, and, therefore, for m = log x and n = log y in particular".
Thanks, appreciate it. If anyone else wants to chime in, feel free.
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Quote:
m - n = m - nlogbbm - logbbn = logbbm - n (by definition of logarithms, this is equivalent to the above equation)
You have not shown that this is true.
Namely, you have yet to demonstrate that log b^m - log b^n = log b^(m-n)
You're still assuming that which you have to prove.
Here's a better way:
log x - log y ? log x/y
Let m = log x <=> b^m = x
Let n = log y <=> b^n = y
m - n ? log x/y
b^(m-n) ? log x/y
log b^(m-n) ? log x/y
log b^m/b^n ? log x/y
log x/y ? log x/y
log x/y = log x/y
Thus: log x - log y = log x/y
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Quote:
Original post by SticksandStones
Quote:
m - n = m - nlogbbm - logbbn = logbbm - n (by definition of logarithms, this is equivalent to the above equation)
You have not shown that this is true.
Namely, you have yet to demonstrate that log b^m - log b^n = log b^(m-n)
You're still assuming that which you have to prove.
Well, by definition of logs, the question is b to the what power is bm? and the answer is clearly m because bm = bm. Therefore, logbbm is just m, and the same goes for n and (m - n). Perhaps if I explicitly write out bm = bm it would work?
<edit> Actually, my part might be better rewritten as "since logarithms and exponentiation are inverse operations, logbbm is equivalent to m" </edit>
Quote:
Here's a better way: log x - log y ? log x/yLet m = log x <=> b^m = xLet n = log y <=> b^n = ym - n ? log x/yb^(m-n) ? log x/ylog b^(m-n) ? log x/ylog b^m/b^n ? log x/ylog x/y ? log x/ylog x/y = log x/yThus: log x - log y = log x/y
I understand your proof, but don't your first and third steps do exactly what you told me I can't do? You change m - n into logbbm-n, which is essentially what I did before.
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If you want to use the same steps as your first proof, start by assuming that the statement is not true:
logbx - logby != logbx/y
Standard proof by contradiction.
However, assuming you already have some other rules for logarithms, there's a more direct proof. Here's a hint: u - v = u + (-v)
I think this leads to a better proof since it's more direct, leads to a better understanding of why the relation is true, and, to a teacher, shows a better understanding of the underlying concepts.
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My book does it neatly by first proving this property :
ln(ab) = ln(a) + ln(b)
Then letting a = 1/b :
ln(1/b * b) = ln(1/b) + ln(b)
since ln(1/b * b ) = ln(1) = 0;
then
0 = ln(1/b) + ln(b)
ln(1/b) = -ln(b)
and we know that ln(1/b) = -ln(b).
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Here's how I would do it. Let's forget about b for a second an use natural logarithms (it mostly just simplifies notation):
k := log(x)-log(y)
exp(k) = exp(log(x)-log(y))
exp(k) = exp(log(x))/exp(log(y))
exp(k) = x/y
k = log(x/y)
Therefore,
log(x)-log(y) = log(x/y)
Now you can divide everything by log(b) and get the same result with base-b logarithms.
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Quote:
Original post by alvaroHere's how I would do it. Let's forget about b for a second an use natural logarithms (it mostly just simplifies notation):k := log(x)-log(y)exp(k) = exp(log(x)-log(y))exp(k) = exp(log(x))/exp(log(y))exp(k) = x/yk = log(x/y)Therefore,log(x)-log(y) = log(x/y)Now you can divide everything by log(b) and get the same result with base-b logarithms.
Oh, that's incredibly simple and elegant. Thanks for pointing it out!
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it seems liek you are getting interested in mathematics. I have a large number of electronic resources i have collected over the years that go along with the classes at MIT OCW and stanford online etc etc.. I feel like getting to see some real mathematics would be great for you at this early stage. If you are interested email me at [email protected]
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https://www.khanacademy.org/math/precalculus-2018/conics-precalc/foci-of-a-hyperbola/v/foci-of-a-hyperbola
| 1,695,282,858,000,000,000
|
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## Precalculus (2018 edition)
### Course: Precalculus (2018 edition)>Unit 2
Lesson 9: Foci of a hyperbola
# Foci of a hyperbola from equation
Sal discusses the foci of hyperbolas and shows how they relate to hyperbola equations. Created by Sal Khan.
## Want to join the conversation?
• So a hyperbola is essentially an "inside out" ellipse?
• It is technically the opposite of a ellipse. One of the x/y is negative making it "inside out."
• I am confused. Where is the point 'b' on hyperbola?
• "b" is not a point on a hyperbola, but the number is important to the overall shape. The b comes in when finding the slope of asymptotes of the hyperbola.
"(b/a) x" will give the equations for those lines, in the event the it is centered on the origin.
"(b/a) (x-c) + d", where c is the change in x and d is the change in y, will solve for any hyperbola of the origin.
• I'm confused of the definition of ellipse and hyperbola.. Why is this video called 'Foci of hyperbola' while what being talked about is the foci of ellipse? Or is there any close relationship between ellipse and hyperbola? Many thanks!
• At the beginning of the video he shows you the ellipse because he wanted you to see that f = sqrt(a^2 - b^2) is an equation that applies to the ellipse and then then after that he starts to talk about the hyperbola and how the equation is f = sqrt(a^2 + b^2), which shows the relationship you are asking of between the ellipse and the hyperbola because you can see the only change is that the - became a +
• can the focal lengths of an ellipse be any distance away from the center along the major axis ??
• Focal length is the distance away from the center the 2 Foci are. Foci will always exist on the major radius so no.
(1 vote)
• Does anyone hear the alarm going off in the background?
• Shouldn't the focal length just be the square root of the absolute value of a squared minus b squared? This way it doesn't matter which is bigger.
f=sqrt(|a^2-b^2|)
• ??? sqrt(abs(3^2-4^2))=sqrt(5)
sqrt(abs(4^2+3^2))=sqrt(25)=5.
If you are just talking about an ellipse then yes that is right.
• Does Sal have any proof videos for d1 + d2 = 2a for ellipses or |d2 - d1| = 2a for hyperbolas?
• at didn't sal write out the Pythagorean theorem?
• Yes, you are correct. At , Sal did write out the Pythagorean Theorem.
• what does the ' a ' and ' b ' represent ? a radius of something ?
(1 vote)
• 'a' is the distance between the center of the hyperbola and the vertices. I am not sure if 'b' represents a specific distance related to the hyperbola, but as we see in this video, it is related to the focus, and it is also related to the slope of the asymptotes: b / a
My suspicion is that an equation that uses only distances defined by the hyperbola might substitute |a^2 - f^2| for b^2, but then you wouldn't have easy access to the slope of the asymptotes, which would be inconvenient.
• I don't get why the formula for a hyperbola is so similar to the formula for an ellipse. Can someone explain? Thanks.
Also, at wasn't sal using the formula of an ellipse and using the focal length formula of a hyperbola?
• They are similar because the equation for a hyperbola is the same as an ellipse except the equation for a hyperbola has a - instead of a + (in the graphical equation). As for your second question, Sal is using the foci formula of the hyperbola, not an ellipse. The foci formula for an ellipse is
c^2=|a^2-b^2|
And the foci formula for a hyperbola is:
c^2 = a^2 + b^2 (same as the pythagorean theorem)
does this help?
(1 vote)
| 992
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http://www.physicsforums.com/showthread.php?t=385188&page=7
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## Mechanics Problem HELP!
Quote by D44 So the ground reaction force is the hypotenuse of the triangle used to calculate the other forces? I'd previously written it as the adjacent. Just for future reference, how would you know which it was?
When you first look at the frame in its entirety, always find the reactions first, using sum of y forces =0, sum of x forces = 0, and sum of moments about any point = 0. When you look in the x direction, there are no applied forces in that direction, hence, no horizontal reaction at the support. There is only a vertical reaction P1 +P2, and a moment, P1x1 + P2x2. The vertical reaction is the resultant hypotenuse if you wish to break it up into comonents parallel and perpendicular to the sloped member.
The shear force, 64.5sin15, does this not act at the rhs (ground)?
yes...
If not, so far at the lhs I have a moment of 68.1Nm, a shear force of 64.5sin15N - at the rhs I have a horizontal force of 64.5cos15N and an moment of 128.4Nm? Am I also trying to find a vertical force at the rhs too?
yes, the shear force at the right hand side as you noted. Direction of force please at the left and right hand sides?
Quote by D44 Using the equillibrium equations, the vertical up force at the rhs is equal to the 64.5sin15N force?
yes....
What happens with the horizontal force at the rhs? Because the sum of Fx=0, but where does the other horizontal force come from to balance the rhs force? 64.5cos15 from lhs from splitting the 64.5N force up?
yes...the horizontal (axial) force in the member is 64.5 cos 15. You are doing very well so far...now just draw the shear and moment diagrams for this slanted member and you've got it nailed...
Shear forces at both ends then, lhs acting downwards at 64.5sin15 and rhs acting upwards at 64.5sin15? So the axial force of 64.5cos15 is what equals the ground reaction force to make the sum of Fx=0? When drawing the shear and moment diagrams, how is the axial force incorporated? What effect does it have?
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Quote by D44 Shear forces at both ends then, lhs acting downwards at 64.5sin15 and rhs acting upwards at 64.5sin15?
Yes! Since the sum of all forces perp to the member must be 0, if you have the shear force down at the left end, then it must act up at the right end,equal in mgnitude, because there are no other external forces applied on the beam in that direction.
So the axial force of 64.5cos15 is what equals the ground reaction force to make the sum of Fx=0?
The axial compressive force of 64.5cos15 at the lhs must be balanced by an equal and opposite axial component of the reaction force at the ground.
When drawing the shear and moment diagrams, how is the axial force incorporated? What effect does it have?
It has no effect when drawing the shear and moment diagrams. You could draw an axial force digram separately, which would be a constant force throughout. When determining the axial and bending memberstresses later, then you combine those stresses for the final results, per P/A +/- Mc/I.
Brill, thank you!! My shear force diagram is looking like a vertical line down to -16.69N, horizontal line straight along to the rhs then back up to 0? As for the BM diagram, I'm not so sure, but it has to go vertically staight down to -68.1Nm and then up to 128.4Nm at the rhs? Is this a curve crossing at the centre point of the 0 line, not a linear line? The induced stresses are a combination of which, sorry? The shear forces and the axial force? Although I know what the P/A +/- Mc/I means, how am I to apply this? How do you mean per?
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Quote by D44 Brill, thank you!! My shear force diagram is looking like a vertical line down to -16.69N, horizontal line straight along to the rhs then back up to 0?
Yes, you are quite correct.
As for the BM diagram, I'm not so sure, but it has to go vertically staight down to -68.1Nm
yes
and then up to 128.4Nm at the rhs? Is this a curve crossing at the centre point of the 0 line, not a linear line?
It doesn't cross the line. The moment at the lhs and the shear at the lhs, both produce ccw moments about the support, so the straight line (not a curved line) from left to right slants down, not up. You must remember that the slope of the moment diagram at a given point is equal to the shear at that point. Thus, its slope is -16.69, a negative slope, and the moment at the right end is thus -68.1 - 16.69(4) = about -134 ( there is a round off error somwhere, as it should agee with the 128.4 you calculated earlier).
The induced stresses are a combination of which, sorry? The shear forces and the axial force? Although I know what the P/A +/- Mc/I means, how am I to apply this? How do you mean per?
It looks like the problem as given on page 1 did not ask for stresses..that's probably the 2nd part of the question.. If you haven't got to stresses yet, don't go any further until you study that topic.
i am confused about the moment at the slanting bit... i have worked it out as 134 .. is that right
Quote by PhanthomJay Yes, you are quite correct. yes It doesn't cross the line. The moment at the lhs and the shear at the lhs, both produce ccw moments about the support, so the straight line (not a curved line) from left to right slants down, not up. You must remember that the slope of the moment diagram at a given point is equal to the shear at that point. Thus, its slope is -16.69, a negative slope, and the moment at the right end is thus -68.1 - 16.69(4) = about -134 ( there is a round off error somwhere, as it should agee with the 128.4 you calculated earlier). It looks like the problem as given on page 1 did not ask for stresses..that's probably the 2nd part of the question.. If you haven't got to stresses yet, don't go any further until you study that topic.
Recognitions: Gold Member Homework Help Science Advisor Ok let's crank it out. P1 is (20)(2.15) = 43 and P2 is (20/2)(2.15) = 21.5, so the vert load is 64.5 N. The moment about the support is 43[(2.15/2) +.1 + 4 sin15] + 21.5[(2.15/3) + .1 + 4 sin15] = 134.9 N-m. Or, if you look at the moment at the support in the free body of the slanted member, it's 64.5(4)(sin15) + 68.1 = 134.9 N-m.....checks out OK.
Using the bending equations, I have a value of approx 88Pa for the induced stresses. Am I right in using the the biggest moment in the equation, which is at the ground?
Sorry, 44.24MPa
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Quote by D44 Sorry, 44.24MPa
It's somewhere around there, I didn't calc out the numbers, but yes, max bending stress is Mc/I where M is the max moment (139 Nm) occurring at the support, which controls the overall frame structural design. The I term comes from the properties of the hollow circle, and c is its outside radius.
Hi, i am having the same trouble with this assignment, mainly drawing the free body diagram of each section. Not sure exactly whether to draw it as a full frame or two seperate parts. thanks
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# ECO406 Mathematical Economics assignment no 01 solution and discussion due date 08-12-2014
ECO406 Mathematical Economics assignment no 01 solution and discussion due date 08-12-2014
MATHEMATICAL ECONOMICS (ECO406)
ASSIGNMENT NO. 01
DUE DATE: 8DEC, 2014
MARKS: 20
ASSIGNMENT:
Question No. 1
Assume that total cost ‘C’ of a firm for daily output Q is given by the function,
C= 100+20Q and maximum daily productivity capacity of the firm is 300 units of
the output. Find the domain and the range of the function.
(Marks: 5)
Question No. 2
Obtain the linear equation of demand or supply from the following:
(Marks: 5)
Question No. 3
Which of the following are identities and which are equations, and why?
i. (2x+3y)2
=12(xy+3)
ii. (x+3)2
=x
2
+6x+9
(Marks: 2.5+2.5)
P 10 12
Q 12 10 Question No. 4
Solve the following equation and find the values of independent variables.
1. 5x + 10 – 7(15-x) =0
2. 2448
5 2 10 3
x x
x x
(Marks: 2.5+2.5)
Views: 718
### Replies to This Discussion
Our main purpose here discussion not just Solution
We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.
1st qustion solution
DOMAIN :
0<=Q<=300
RANGE :
Total cast at zero point will be :
C=100+20(0)
C=100+0
C=100
Total cast at 300 output will be :
C=100+20(300)
C=100+6000
C=6100
100<=C<=6100
1
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https://dsp.stackexchange.com/questions/66702/calculate-the-derivative-of-gradient-field-of-an-image
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# Calculate the Derivative of Gradient Field of an Image
I meet a confusing thing in image processing recently....
Assume the image $$x \in \mathbb{R}^n$$, with its derivative (difference) matrix: $$D^+ = \begin{bmatrix} D_h \\ Dv \end{bmatrix} \in \mathbb{R}^{2n\times n}$$ ($$+$$ means forward difference), also equal to $$\nabla$$. Therefore, it is natural to define the divergence: $$\triangle = \nabla \cdot \nabla$$.
I have seen some papers use $$div = \triangle = D_h^-D_h^+ + D_v^-D_v^+ \in \mathbb{R}^{n\times n}$$, where $$-$$ denotes the backward difference.
Here is my question: assume I want to calculate the $$\frac{\partial\|\nabla x -p\|_2^2}{\partial x}$$ where $$p\in \mathbb{R}^{2n\times 1}$$ is a vector not related to $$x$$, what is the result? I have seen some authors use $$\nabla\cdot (\nabla x -p)$$.
However, if writing the $$\nabla$$ as matrix form $$D$$ as I have introduced before, $$D^T$$ is exactly adjoint of gradient, not backward difference. Hence $$-\triangle x$$ would appear! So what is the right formula? Could anyone tell me?
• What is p, and please specify dimensions of all the matrices and vectors Apr 21, 2020 at 6:36
• @Dspguysam Thanks for your advise. I have edited the question. Apr 21, 2020 at 6:46
• $\nabla x$ will be a vector of dimension 2n and you are subtracting that to a matrix $p$ of dimension 2nxn? Did I get that right? Apr 21, 2020 at 6:59
• @Dspguysam Yes, you are right. I have already marked the dimension of $p$. Apr 21, 2020 at 8:15
• a matrix times a vector will result in a vector, and you can only subtract or add vectors of same dimension in a space, therefore $p$ should be a vector, not a matrix, and the dimension of "vector" $p$ should be the same as number of rows of $\nabla$, $p$ should be $\mathbb{R}^{2n}$ not $\mathbb{R}^{2n*n}$ Apr 21, 2020 at 8:29
Consider the expansion of the term below $$\|\nabla x -p\|_2^2 = (\nabla x -p)^T(\nabla x -p)$$ $$\|\nabla x -p\|_2^2 = (x^T\nabla^T -p^T)(\nabla x -p)$$ $$\|\nabla x -p\|_2^2 = (x^T\nabla^T\nabla x - x^T\nabla^T p - p^T\nabla x +p^Tp)$$
Now consider the following basic definition:
$$\frac{\partial(A x)}{\partial x} = A^T$$ $$\frac{\partial(x^TA)}{\partial x} = A$$
Now applying the above two definitions together with the expansion of the objective above to differentiate the objective we have
$$\frac{\partial\|\nabla x -p\|_2^2}{\partial x} = 2\nabla^T\nabla x - 2\nabla^Tp$$
$$\frac{\partial\|\nabla x -p\|_2^2}{\partial x} = 2\nabla^T(\nabla x - p)$$
since $$\nabla^T = \nabla$$, therefore we have
$$\frac{\partial\|\nabla x -p\|_2^2}{\partial x} = 2\nabla(\nabla x - p)$$ the constant 2 is just a constant, so the result we have is consistent with the ones that authors are using, its simply a consequence of vector differentiation
• The question is here: where is the conclusion of $\nabla^T = \nabla$? When we express it in matrix form, for example, let $D$ represents the forward difference, $D^T \neq D$ sinc $D_h^T \neq D_h$ Apr 21, 2020 at 8:37
• $D^T$ is the adjoint of gradient. If you implement it in programming language, $D_h^T = -D^-_h$ Apr 21, 2020 at 8:39
• For this to be a valid definition $\triangle = \nabla \cdot \nabla$, firstly $\nabla$should be a square matrix and is $\triangle$ should be positive semidefinite, in that case $\nabla^T = \nabla$ Apr 21, 2020 at 8:44
• I agree. In mathematics this is valid. But how about changing the $\nabla$ to $D$? The things goes different. Apr 21, 2020 at 8:51
• this is all maths. According to the definition you provide D is not a square matrix, then it can never be substituted as $\nabla$ , which has to be a square matrix, based on the definitions you provide as I explained above, If you want give me an objective function in terms of D that you want to differentiate with respect to x. We will have no problem doing it. Currently the objective is in terms of $\nabla$ for which the result is consistent. D and $\nabla$ have to have equal dimesnions first to be equated! Apr 21, 2020 at 8:56
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##### Help with geometry problem
Mathematics Tutor: None Selected Time limit: 1 Day
Nov 3rd, 2015
Here we are going to need to use one of the three trigonometric identities:
sin(x) = opposite / hypotenuse, cos(x) = adjacent / hypotenuse, tan(x) = opposite / adjacent
We are given the length of the hypotenuse (the longest side) and the angle and we are asked to find the length of the opposite side.
Therefore we will use the equation sin(x) = opposite / hypotenuse.
To find the value of the opposite side, we can rearrange this equation to leave opposite by itself on one side of the equals sign:
sin(x) * hypotenuse = opposite
Substituting our values for x and the hypotenuse we get:
opposite = sin(47) * 51
Using a calculator this gives:
opposite = 37.3
and so the answer is a = 37.3
Nov 3rd, 2015
...
Nov 3rd, 2015
...
Nov 3rd, 2015
Dec 6th, 2016
check_circle
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https://www.numbersaplenty.com/142413
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Search a number
142413 = 3371283
BaseRepresentation
bin100010110001001101
321020100120
4202301031
514024123
63015153
71132125
oct426115
9236316
10142413
1197aa7
126a4b9
134ca8b
1439c85
152c2e3
hex22c4d
142413 has 8 divisors (see below), whose sum is σ = 195168. Its totient is φ = 92304.
The previous prime is 142403. The next prime is 142421. The reversal of 142413 is 314241.
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 142413 - 25 = 142381 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 142392 and 142401.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (142403) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 531 + ... + 752.
It is an arithmetic number, because the mean of its divisors is an integer number (24396).
2142413 is an apocalyptic number.
It is an amenable number.
142413 is a deficient number, since it is larger than the sum of its proper divisors (52755).
142413 is a wasteful number, since it uses less digits than its factorization.
142413 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1323.
The product of its digits is 96, while the sum is 15.
The square root of 142413 is about 377.3764698547. The cubic root of 142413 is about 52.2215644670.
Adding to 142413 its reverse (314241), we get a palindrome (456654).
It can be divided in two parts, 142 and 413, that added together give a palindrome (555).
The spelling of 142413 in words is "one hundred forty-two thousand, four hundred thirteen".
Divisors: 1 3 37 111 1283 3849 47471 142413
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## Algebra 1: Common Core (15th Edition)
We have to consider the two areas where the circles for hot dogs and hamburgers intersect. We add 24 and 12 to obtain that 36 people like hot dogs and hamburgers. We now can add up all of the people in the survey. We find that there are 150 people that were surveyed. Thus, we find the percentage of people that like hot dogs and hamburgers: $(36/150) \times 100 =24$%. 24% of people like hot dogs and hamburgers, so we can multiply 850 by .24 to obtain that 204 people should be expected to like hot dogs and hamburgers.
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# Thread: Proof question: What's happening?
1. ## Proof question: What's happening?
Okay, so I did get the problem right the first try. However I am not sure what's going on.
I have the problem: Give an epsilon-delta proof of the limit(x^2-5x-3) = 3 as X approaches -1.
At a point in the proof I come to a point where I have ...|(x-6)(x+1)|<E
My notes and book have some really, really wild stuff going on. Basically "Agree that delta is less than or equal to 1", then it does something really wild. The problem ends with it being epsilon over 8, then the proof continues like normal.
I'm not sure what's happening mathematically here. The PDF on limits has only confused me further, as the notes I have ( which are a copy of the instructor's ) don't mention using a variable such as "M" in them.
Thanks!
2. Originally Posted by Wolvenmoon
Okay, so I did get the problem right the first try. However I am not sure what's going on.
I have the problem: Give an epsilon-delta proof of the limit(x^2-5x-3) = 3 as X approaches -1.
At a point in the proof I come to a point where I have ...|(x-6)(x+1)|<E
My notes and book have some really, really wild stuff going on. Basically "Agree that delta is less than or equal to 1", then it does something really wild. The problem ends with it being epsilon over 8, then the proof continues like normal.
I'm not sure what's happening mathematically here. The PDF on limits has only confused me further, as the notes I have ( which are a copy of the instructor's ) don't mention using a variable such as "M" in them.
Thanks!
You have $\displaystyle \lim_{x \rightarrow -1} x^2-5x-3 =3$
$\displaystyle |(x^2-5x-3)-3| < \epsilon$ If $\displaystyle 0<|x+1|<\delta$
That's just using the definition to prove the limit. The first part means that value of your function $\displaystyle f(x)=x^2-5x-3$ will fall between $\displaystyle 3- \epsilon$ and $\displaystyle 3+ \epsilon$. The second bit just means that x lies in the interval $\displaystyle (-1-\delta , -1+\delta)$. To complete your proof you only need to discover a value of $\displaystyle \delta$ for which this statement holds, and then you must prove that the statement holds for $\displaystyle \delta$.
By the way, $\displaystyle |(x^2-5x-3)-3|= |x^2-5x-6|$ $\displaystyle = |(x-6)(x+1)| < \epsilon$
It's just been factorized, if you were wondering where it came from.
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# 192 kg to lbs - 192 kilograms to pounds
Do you want to learn how much is 192 kg equal to lbs and how to convert 192 kg to lbs? Here it is. In this article you will find everything about kilogram to pound conversion - theoretical and practical too. It is also needed/We also want to emphasize that all this article is dedicated to a specific number of kilograms - exactly one kilogram. So if you want to know more about 192 kg to pound conversion - read on.
Before we get to the more practical part - that is 192 kg how much lbs calculation - we are going to tell you some theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 192 kg to lbs? 192 kilograms it is equal 423.28754304 pounds, so 192 kg is equal 423.28754304 lbs.
## 192 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, known also as International System of Units (in short form SI).
At times the kilogram could be written as kilogramme. The symbol of the kilogram is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. First definition was not complicated but totally impractical to use.
Then, in 1889 the kilogram was defined by the International Prototype of the Kilogram (in short form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was replaced by a new definition.
Today the definition of the kilogram is build on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It is also divided into 100 decagrams and 1000 grams.
## 192 kilogram to pounds
You learned something about kilogram, so now let's go to the pound. The pound is also a unit of mass. It is needed to underline that there are not only one kind of pound. What does it mean? For instance, there are also pound-force. In this article we are going to to focus only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. To be honest, this unit is used also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is exactly 0.45359237 kilograms. One avoirdupois pound can be divided into 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 192 kg?
192 kilogram is equal to 423.28754304 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 192 kg in lbs
The most theoretical section is already behind us. In next part we are going to tell you how much is 192 kg to lbs. Now you know that 192 kg = x lbs. So it is high time to know the answer. Let’s see:
192 kilogram = 423.28754304 pounds.
That is an accurate outcome of how much 192 kg to pound. You can also round it off. After it your outcome will be exactly: 192 kg = 422.4 lbs.
You know 192 kg is how many lbs, so look how many kg 192 lbs: 192 pound = 0.45359237 kilograms.
Obviously, in this case you may also round off the result. After rounding off your outcome is as following: 192 lb = 0.45 kgs.
We are also going to show you 192 kg to how many pounds and 192 pound how many kg results in charts. Let’s see:
We are going to start with a chart for how much is 192 kg equal to pound.
### 192 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
192 423.28754304 422.40
Now look at a chart for how many kilograms 192 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
192 0.45359237 0.45
Now you learned how many 192 kg to lbs and how many kilograms 192 pound, so we can move on to the 192 kg to lbs formula.
### 192 kg to pounds
To convert 192 kg to us lbs a formula is needed. We will show you a formula in two different versions. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 423.28754304 outcome in pounds
The first formula give you the most correct result. Sometimes even the smallest difference can be considerable. So if you need a correct result - this formula will be the best solution to convert how many pounds are equivalent to 192 kilogram.
So go to the second formula, which also enables calculations to know how much 192 kilogram in pounds.
The second formula is as following, have a look:
Number of kilograms * 2.2 = the outcome in pounds
As you see, the second version is simpler. It can be better solution if you want to make a conversion of 192 kilogram to pounds in easy way, for instance, during shopping. Just remember that your result will be not so accurate.
Now we want to show you these two formulas in practice. But before we are going to make a conversion of 192 kg to lbs we are going to show you another way to know 192 kg to how many lbs totally effortless.
### 192 kg to lbs converter
An easier way to know what is 192 kilogram equal to in pounds is to use 192 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Converter is based on longer version of a formula which we gave you in the previous part of this article. Thanks to 192 kg pound calculator you can quickly convert 192 kg to lbs. You only have to enter number of kilograms which you need to convert and click ‘calculate’ button. The result will be shown in a flash.
So let’s try to calculate 192 kg into lbs using 192 kg vs pound converter. We entered 192 as an amount of kilograms. This is the outcome: 192 kilogram = 423.28754304 pounds.
As you see, this 192 kg vs lbs calculator is intuitive.
Now we can move on to our main issue - how to convert 192 kilograms to pounds on your own.
#### 192 kg to lbs conversion
We will begin 192 kilogram equals to how many pounds calculation with the first version of a formula to get the most correct outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 423.28754304 the outcome in pounds
So what have you do to learn how many pounds equal to 192 kilogram? Just multiply number of kilograms, this time 192, by 2.20462262. It is 423.28754304. So 192 kilogram is exactly 423.28754304.
You can also round off this result, for instance, to two decimal places. It is equal 2.20. So 192 kilogram = 422.40 pounds.
It is time for an example from everyday life. Let’s calculate 192 kg gold in pounds. So 192 kg equal to how many lbs? As in the previous example - multiply 192 by 2.20462262. It is 423.28754304. So equivalent of 192 kilograms to pounds, when it comes to gold, is exactly 423.28754304.
In this example you can also round off the result. This is the result after rounding off, this time to one decimal place - 192 kilogram 422.4 pounds.
Now we can go to examples calculated using short formula.
#### How many 192 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 422.4 the outcome in pounds
So 192 kg equal to how much lbs? And again, you need to multiply number of kilogram, in this case 192, by 2.2. Let’s see: 192 * 2.2 = 422.4. So 192 kilogram is exactly 2.2 pounds.
Let’s make another calculation with use of this version of a formula. Now calculate something from everyday life, for example, 192 kg to lbs weight of strawberries.
So calculate - 192 kilogram of strawberries * 2.2 = 422.4 pounds of strawberries. So 192 kg to pound mass is 422.4.
If you learned how much is 192 kilogram weight in pounds and are able to convert it using two different formulas, let’s move on. Now we want to show you these results in tables.
#### Convert 192 kilogram to pounds
We know that outcomes shown in tables are so much clearer for most of you. We understand it, so we gathered all these results in charts for your convenience. Thanks to this you can quickly make a comparison 192 kg equivalent to lbs outcomes.
Let’s start with a 192 kg equals lbs table for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
192 423.28754304 422.40
And now see 192 kg equal pound chart for the second version of a formula:
Kilograms Pounds
192 422.4
As you see, after rounding off, if it comes to how much 192 kilogram equals pounds, the results are the same. The bigger number the more significant difference. Please note it when you want to make bigger number than 192 kilograms pounds conversion.
#### How many kilograms 192 pound
Now you know how to calculate 192 kilograms how much pounds but we want to show you something more. Are you interested what it is? What about 192 kilogram to pounds and ounces calculation?
We want to show you how you can convert it step by step. Begin. How much is 192 kg in lbs and oz?
First thing you need to do is multiply amount of kilograms, this time 192, by 2.20462262. So 192 * 2.20462262 = 423.28754304. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To convert how much 192 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So final outcome is exactly 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then final result is exactly 2 pounds and 33 ounces.
As you can see, calculation 192 kilogram in pounds and ounces easy.
The last calculation which we will show you is calculation of 192 foot pounds to kilograms meters. Both of them are units of work.
To convert foot pounds to kilogram meters you need another formula. Before we give you this formula, let’s see:
• 192 kilograms meters = 7.23301385 foot pounds,
• 192 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 192 foot pounds to kilograms meters you need to multiply 192 by 0.13825495. It is 0.13825495. So 192 foot pounds is 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 192 foot pounds will be exactly 0.14 kilogram meters.
We hope that this calculation was as easy as 192 kilogram into pounds conversions.
This article was a huge compendium about kilogram, pound and 192 kg to lbs in calculation. Thanks to this conversion you learned 192 kilogram is equivalent to how many pounds.
We showed you not only how to do a conversion 192 kilogram to metric pounds but also two other conversions - to check how many 192 kg in pounds and ounces and how many 192 foot pounds to kilograms meters.
We showed you also other solution to do 192 kilogram how many pounds conversions, that is with use of 192 kg en pound calculator. This is the best option for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you can do 192 kilogram equal to how many pounds conversion - on your own or with use of our 192 kgs to pounds converter.
So what are you waiting for? Convert 192 kilogram mass to pounds in the best way for you.
Do you want to make other than 192 kilogram as pounds calculation? For instance, for 5 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so simply as for 192 kilogram equal many pounds.
### How much is 192 kg in pounds
We want to sum up this topic, that is how much is 192 kg in pounds , we prepared for you an additional section. Here we have for you all you need to remember about how much is 192 kg equal to lbs and how to convert 192 kg to lbs . You can see it down below.
How does the kilogram to pound conversion look? The conversion kg to lb is just multiplying 2 numbers. Let’s see 192 kg to pound conversion formula . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 192 kilogram to pounds? The accurate result is 423.28754304 lb.
It is also possible to calculate how much 192 kilogram is equal to pounds with another, shortened type of the formula. Have a look.
The number of kilograms * 2.2 = the result in pounds
So this time, 192 kg equal to how much lbs ? The answer is 423.28754304 lbs.
How to convert 192 kg to lbs in just a moment? It is possible to use the 192 kg to lbs converter , which will make the rest for you and you will get an exact result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
Read more related articles:
193 kg to lbs = 425.492 194 kg to lbs = 427.697 195 kg to lbs = 429.901 196 kg to lbs = 432.106 197 kg to lbs = 434.311 198 kg to lbs = 436.515 199 kg to lbs = 438.72 200 kg to lbs = 440.925 201 kg to lbs = 443.129 202 kg to lbs = 445.334 203 kg to lbs = 447.538 204 kg to lbs = 449.743 205 kg to lbs = 451.948 206 kg to lbs = 454.152 207 kg to lbs = 456.357 208 kg to lbs = 458.562 209 kg to lbs = 460.766 210 kg to lbs = 462.971 211 kg to lbs = 465.175 212 kg to lbs = 467.38 213 kg to lbs = 469.585 214 kg to lbs = 471.789 215 kg to lbs = 473.994 216 kg to lbs = 476.198 217 kg to lbs = 478.403
218 kg to lbs = 480.608 219 kg to lbs = 482.812 220 kg to lbs = 485.017 221 kg to lbs = 487.222 222 kg to lbs = 489.426 223 kg to lbs = 491.631 224 kg to lbs = 493.835 225 kg to lbs = 496.04 226 kg to lbs = 498.245 227 kg to lbs = 500.449 228 kg to lbs = 502.654 229 kg to lbs = 504.859 230 kg to lbs = 507.063 231 kg to lbs = 509.268 232 kg to lbs = 511.472 233 kg to lbs = 513.677 234 kg to lbs = 515.882 235 kg to lbs = 518.086 236 kg to lbs = 520.291 237 kg to lbs = 522.496 238 kg to lbs = 524.7 239 kg to lbs = 526.905 240 kg to lbs = 529.109 241 kg to lbs = 531.314 242 kg to lbs = 533.519
243 kg to lbs = 535.723 244 kg to lbs = 537.928 245 kg to lbs = 540.133 246 kg to lbs = 542.337 247 kg to lbs = 544.542 248 kg to lbs = 546.746 249 kg to lbs = 548.951 250 kg to lbs = 551.156 251 kg to lbs = 553.36 252 kg to lbs = 555.565 253 kg to lbs = 557.77 254 kg to lbs = 559.974 255 kg to lbs = 562.179 256 kg to lbs = 564.383 257 kg to lbs = 566.588 258 kg to lbs = 568.793 259 kg to lbs = 570.997 260 kg to lbs = 573.202 261 kg to lbs = 575.407 262 kg to lbs = 577.611 263 kg to lbs = 579.816 264 kg to lbs = 582.02 265 kg to lbs = 584.225 266 kg to lbs = 586.43 267 kg to lbs = 588.634
268 kg to lbs = 590.839 269 kg to lbs = 593.043 270 kg to lbs = 595.248 271 kg to lbs = 597.453 272 kg to lbs = 599.657 273 kg to lbs = 601.862 274 kg to lbs = 604.067 275 kg to lbs = 606.271 276 kg to lbs = 608.476 277 kg to lbs = 610.68 278 kg to lbs = 612.885 279 kg to lbs = 615.09 280 kg to lbs = 617.294 281 kg to lbs = 619.499 282 kg to lbs = 621.704 283 kg to lbs = 623.908 284 kg to lbs = 626.113 285 kg to lbs = 628.317 286 kg to lbs = 630.522 287 kg to lbs = 632.727 288 kg to lbs = 634.931 289 kg to lbs = 637.136 290 kg to lbs = 639.341 291 kg to lbs = 641.545 292 kg to lbs = 643.75
| 4,376
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11,
311,
1403,
12395,
7634,
13,
1102,
374,
6273,
220,
17,
13,
508,
13,
2100,
220,
5926,
15395,
13255,
284,
220,
16460,
13,
1272,
16701,
382,
2181,
374,
892,
369,
459,
3187,
505,
18254,
2324,
13,
6914,
753,
11294,
220,
5926,
21647,
6761,
304,
16701,
13,
2100,
220,
5926,
21647,
6273,
311,
1268,
1690,
29160,
30,
1666,
304,
279,
3766,
3187,
482,
31370,
220,
5926,
555,
220,
17,
13,
7854,
19808,
5538,
13,
1102,
374,
220,
19711,
13,
17897,
19642,
2371,
13,
2100,
13890,
315,
220,
5926,
85402,
311,
16701,
11,
994,
433,
4131,
311,
6761,
11,
374,
7041,
220,
19711,
13,
17897,
19642,
2371,
382,
644,
420,
3187,
499,
649,
1101,
4883,
1022,
279,
1121,
13,
1115,
374,
279,
1121,
1306,
52662,
1022,
11,
420,
892,
311,
832,
12395,
2035,
482,
220,
5926,
15395,
13255,
220,
16460,
13,
19,
16701,
382,
7184,
584,
649,
733,
311,
10507,
16997,
1701,
2875,
15150,
382,
827,
2650,
1690,
220,
5926,
21647,
311,
29160,
271,
10438,
584,
1501,
499,
459,
3187,
482,
264,
4062,
27626,
315,
24210,
15150,
1473,
10301,
315,
85402,
353,
220,
17,
13,
17,
284,
220,
16460,
13,
19,
279,
15632,
304,
16701,
271,
4516,
220,
5926,
21647,
6273,
311,
1268,
1790,
29160,
30,
1628,
1578,
11,
499,
1205,
311,
31370,
1396,
315,
15395,
13255,
11,
304,
420,
1162,
220,
5926,
11,
555,
220,
17,
13,
17,
13,
6914,
753,
1518,
25,
220,
5926,
353,
220,
17,
13,
17,
284,
220,
16460,
13,
19,
13,
2100,
220,
5926,
15395,
13255,
374,
7041,
220,
17,
13,
17,
16701,
382,
10267,
753,
1304,
2500,
22702,
449,
1005,
315,
420,
2373,
315,
264,
15150,
13,
4800,
11294,
2555,
505,
18254,
2324,
11,
369,
3187,
11,
220,
5926,
21647,
311,
29160,
4785,
315,
76203,
382,
4516,
11294,
482,
220,
5926,
15395,
13255,
315,
76203,
353,
220,
17,
13,
17,
284,
220,
16460,
13,
19,
16701,
315,
76203,
13,
2100,
220,
5926,
21647,
311,
31123,
3148,
374,
220,
16460,
13,
19,
382,
2746,
499,
9687,
1268,
1790,
374,
220,
5926,
15395,
13255,
4785,
304,
16701,
323,
527,
3025,
311,
5625,
433,
1701,
1403,
2204,
55179,
11,
1095,
753,
3351,
389,
13,
4800,
584,
1390,
311,
1501,
499,
1521,
3135,
304,
12920,
382,
827,
7316,
220,
5926,
15395,
13255,
311,
16701,
271,
1687,
1440,
430,
20124,
6982,
304,
12920,
527,
779,
1790,
49479,
369,
1455,
315,
499,
13,
1226,
3619,
433,
11,
779,
584,
20802,
682,
1521,
3135,
304,
27223,
369,
701,
19679,
13,
11361,
311,
420,
499,
649,
6288,
1304,
264,
12593,
220,
5926,
21647,
13890,
311,
29160,
20124,
382,
10267,
753,
1212,
449,
264,
220,
5926,
21647,
17239,
29160,
2007,
369,
279,
1176,
15150,
1473,
42,
321,
56485,
393,
3171,
393,
3171,
320,
10924,
52662,
1022,
311,
1403,
12395,
7634,
340,
5926,
220,
19711,
13,
17897,
19642,
2371,
220,
16460,
13,
1272,
271,
3112,
1457,
1518,
220,
5926,
21647,
6273,
31123,
9676,
369,
279,
2132,
2373,
315,
264,
15150,
1473,
42,
321,
56485,
393,
3171,
198,
5926,
220,
16460,
13,
19,
271,
2170,
499,
1518,
11,
1306,
52662,
1022,
11,
422,
433,
4131,
311,
1268,
1790,
220,
5926,
15395,
13255,
17239,
16701,
11,
279,
3135,
527,
279,
1890,
13,
578,
11493,
1396,
279,
810,
5199,
6811,
13,
5321,
5296,
433,
994,
499,
1390,
311,
1304,
11493,
1396,
1109,
220,
5926,
85402,
16701,
14747,
382,
827,
2650,
1690,
85402,
220,
5926,
31123,
271,
7184,
499,
1440,
1268,
311,
11294,
220,
5926,
85402,
1268,
1790,
16701,
719,
584,
1390,
311,
1501,
499,
2555,
810,
13,
8886,
499,
8173,
1148,
433,
374,
30,
3639,
922,
220,
5926,
15395,
13255,
311,
16701,
323,
49138,
22702,
1980,
1687,
1390,
311,
1501,
499,
1268,
499,
649,
5625,
433,
3094,
555,
3094,
13,
19110,
13,
2650,
1790,
374,
220,
5926,
21647,
304,
29160,
323,
25616,
1980,
5451,
3245,
499,
1205,
311,
656,
374,
31370,
3392,
315,
85402,
11,
420,
892,
220,
5926,
11,
555,
220,
17,
13,
7854,
19808,
5538,
13,
2100,
220,
5926,
353,
220,
17,
13,
7854,
19808,
5538,
284,
220,
19711,
13,
17897,
19642,
2371,
13,
3861,
15395,
13255,
374,
220,
17,
13,
7854,
19808,
5538,
16701,
382,
791,
7698,
961,
374,
1396,
315,
16701,
13,
2100,
304,
420,
3187,
1070,
527,
220,
17,
16701,
382,
1271,
5625,
1268,
1790,
220,
5926,
15395,
13255,
374,
6273,
311,
16701,
323,
49138,
499,
1205,
311,
31370,
19983,
961,
555,
220,
845,
13,
2100,
31370,
220,
7854,
19808,
5538,
555,
220,
845,
13,
1102,
374,
220,
13817,
19615,
5926,
49138,
382,
4516,
1620,
15632,
374,
7041,
220,
17,
16701,
323,
220,
13817,
19615,
5926,
49138,
13,
1472,
649,
1101,
4883,
1022,
49138,
11,
369,
3187,
11,
311,
1403,
7634,
13,
5112,
1620,
1121,
374,
7041,
220,
17,
16701,
323,
220,
1644,
49138,
382,
2170,
499,
649,
1518,
11,
22702,
220,
5926,
15395,
13255,
304,
16701,
323,
49138,
4228,
382,
791,
1566,
22702,
902,
584,
690,
1501,
499,
374,
22702,
315,
220,
5926,
4579,
16701,
311,
85402,
20645,
13,
11995,
315,
1124,
527,
8316,
315,
990,
382,
1271,
5625,
4579,
16701,
311,
15395,
13255,
20645,
499,
1205,
2500,
15150,
13,
13538,
584,
3041,
499,
420,
15150,
11,
1095,
753,
1518,
1473,
6806,
220,
5926,
85402,
20645,
284,
220,
22,
13,
12994,
16368,
5313,
4579,
16701,
345,
6806,
220,
5926,
4579,
16701,
284,
220,
15,
13,
10350,
12375,
2721,
85402,
20645,
382,
7184,
1427,
520,
264,
15150,
1473,
2903,
27807,
1726,
2189,
315,
4579,
16701,
353,
220,
15,
13,
10350,
12375,
2721,
284,
279,
1121,
304,
85402,
20645,
271,
4516,
311,
5625,
220,
5926,
4579,
16701,
311,
85402,
20645,
499,
1205,
311,
31370,
220,
5926,
555,
220,
15,
13,
10350,
12375,
2721,
13,
1102,
374,
220,
15,
13,
10350,
12375,
2721,
13,
2100,
220,
5926,
4579,
16701,
374,
220,
15,
13,
10350,
12375,
2721,
15395,
13255,
20645,
382,
2181,
374,
1101,
3284,
311,
4883,
1022,
420,
1121,
11,
369,
3187,
11,
311,
1403,
12395,
7634,
13,
5112,
220,
5926,
4579,
16701,
690,
387,
7041,
220,
15,
13,
975,
15395,
13255,
20645,
382,
1687,
3987,
430,
420,
22702,
574,
439,
4228,
439,
220,
5926,
15395,
13255,
1139,
16701,
49822,
382,
2028,
4652,
574,
264,
6908,
1391,
408,
2411,
922,
15395,
13255,
11,
31123,
323,
220,
5926,
21647,
311,
29160,
304,
22702,
13,
11361,
311,
420,
14747,
499,
9687,
220,
5926,
15395,
13255,
374,
13890,
311,
1268,
1690,
16701,
382,
1687,
8710,
499,
539,
1193,
1268,
311,
656,
264,
14747,
220,
5926,
15395,
13255,
311,
18767,
16701,
719,
1101,
1403,
1023,
49822,
482,
311,
1817,
1268,
1690,
220,
5926,
21647,
304,
16701,
323,
49138,
323,
1268,
1690,
220,
5926,
4579,
16701,
311,
85402,
20645,
382,
1687,
8710,
499,
1101,
1023,
6425,
311,
656,
220,
5926,
15395,
13255,
1268,
1690,
16701,
49822,
11,
430,
374,
449,
1005,
315,
220,
5926,
21647,
665,
31123,
31052,
13,
1115,
374,
279,
1888,
3072,
369,
1884,
315,
499,
889,
656,
539,
1093,
38714,
389,
701,
1866,
520,
682,
477,
1205,
311,
1304,
571,
3231,
10301,
2645,
21647,
1268,
29160,
49822,
304,
39392,
1648,
382,
1687,
3987,
430,
1457,
682,
315,
499,
649,
656,
220,
5926,
15395,
13255,
6273,
311,
1268,
1690,
16701,
14747,
482,
389,
701,
1866,
477,
449,
1005,
315,
1057,
220,
5926,
597,
5981,
311,
16701,
28157,
382,
4516,
1148,
527,
499,
8748,
369,
30,
7316,
220,
5926,
15395,
13255,
3148,
311,
16701,
304,
279,
1888,
1648,
369,
499,
382,
5519,
499,
1390,
311,
1304,
1023,
1109,
220,
5926,
15395,
13255,
439,
16701,
22702,
30,
1789,
2937,
11,
369,
220,
20,
85402,
30,
4343,
1057,
1023,
9908,
0,
1226,
15803,
430,
49822,
369,
1023,
5219,
315,
85402,
527,
779,
5042,
439,
369,
220,
5926,
15395,
13255,
6273,
1690,
16701,
382,
14711,
2650,
1790,
374,
220,
5926,
21647,
304,
16701,
271,
1687,
1390,
311,
2694,
709,
420,
8712,
11,
430,
374,
1268,
1790,
374,
220,
5926,
21647,
304,
16701,
1174,
584,
10235,
369,
499,
459,
5217,
3857,
13,
5810,
584,
617,
369,
499,
682,
499,
1205,
311,
6227,
922,
1268,
1790,
374,
220,
5926,
21647,
6273,
311,
29160,
323,
1268,
311,
5625,
220,
5926,
21647,
311,
29160,
662,
1472,
649,
1518,
433,
1523,
3770,
382,
4438,
1587,
279,
15395,
13255,
311,
31123,
14747,
1427,
30,
578,
14747,
21647,
311,
19398,
374,
1120,
85292,
220,
17,
5219,
13,
6914,
753,
1518,
220,
5926,
21647,
311,
31123,
14747,
15150,
662,
12522,
264,
1427,
1473,
791,
1396,
315,
85402,
353,
220,
17,
13,
7854,
19808,
5538,
284,
279,
1121,
304,
16701,
271,
4516,
1148,
374,
279,
1121,
315,
279,
14747,
315,
220,
5926,
15395,
13255,
311,
16701,
30,
578,
13687,
1121,
374,
220,
19711,
13,
17897,
19642,
2371,
19398,
382,
2181,
374,
1101,
3284,
311,
11294,
1268,
1790,
220,
5926,
15395,
13255,
374,
6273,
311,
16701,
449,
2500,
11,
66663,
955,
315,
279,
15150,
13,
12522,
264,
1427,
382,
791,
1396,
315,
85402,
353,
220,
17,
13,
17,
284,
279,
1121,
304,
16701,
271,
4516,
420,
892,
11,
220,
5926,
21647,
6273,
311,
1268,
1790,
29160,
949,
578,
4320,
374,
220,
19711,
13,
17897,
19642,
2371,
29160,
382,
4438,
311,
5625,
220,
5926,
21647,
311,
29160,
304,
1120,
264,
4545,
30,
1102,
374,
3284,
311,
1005,
279,
220,
5926,
21647,
311,
29160,
28157,
1174,
902,
690,
1304,
279,
2800,
369,
499,
323,
499,
690,
636,
459,
4839,
1121,
6905,
827,
38988,
56485,
510,
7501,
2595,
791,
15395,
13255,
11,
477,
15395,
13255,
2727,
11,
374,
279,
2385,
5089,
315,
4785,
304,
279,
53558,
1887,
13,
1102,
374,
279,
45968,
4785,
315,
264,
24671,
315,
3090,
220,
605,
2960,
55336,
389,
264,
3185,
382,
827,
393,
3171,
510,
54044,
2595,
32,
31123,
374,
264,
5089,
315,
4785,
17037,
1511,
304,
279,
3723,
4273,
323,
279,
8013,
4279,
31633,
82,
13,
362,
31123,
374,
4613,
439,
7041,
220,
15,
13,
20235,
20128,
1806,
85402,
627,
4518,
810,
5552,
9908,
512,
220,
7285,
21647,
311,
29160,
284,
220,
17837,
13,
21776,
220,
6393,
21647,
311,
29160,
284,
220,
20465,
13,
25388,
220,
6280,
21647,
311,
29160,
284,
220,
16371,
13,
19319,
220,
5162,
21647,
311,
29160,
284,
220,
16739,
13,
7461,
220,
4468,
21647,
311,
29160,
284,
220,
20165,
13,
15134,
220,
3753,
21647,
311,
29160,
284,
220,
21299,
13,
19633,
220,
2550,
21647,
311,
29160,
284,
220,
20596,
13,
5332,
220,
1049,
21647,
311,
29160,
284,
220,
14868,
13,
21910,
220,
679,
21647,
311,
29160,
284,
220,
17147,
13,
9748,
220,
2366,
21647,
311,
29160,
284,
220,
19697,
13,
17153,
220,
9639,
21647,
311,
29160,
284,
220,
20800,
13,
22600,
220,
7854,
21647,
311,
29160,
284,
220,
21125,
13,
26260,
220,
10866,
21647,
311,
29160,
284,
220,
20360,
13,
26979,
220,
11056,
21647,
311,
29160,
284,
220,
20555,
13,
9756,
220,
12060,
21647,
311,
29160,
284,
220,
10961,
13,
18520,
220,
12171,
21647,
311,
29160,
284,
220,
21209,
13,
19242,
220,
12652,
21647,
311,
29160,
284,
220,
16551,
13,
25358,
220,
8848,
21647,
311,
29160,
284,
220,
20911,
13,
25693,
220,
11483,
21647,
311,
29160,
284,
220,
19988,
13,
10005,
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## 每週問題 January 26, 2015
A matrix satisfying $A^2=I$ is said to be an involutory matrix, and a matrix $B$ satisfying $B^2=B$ is said to be an idempotent matrix. Show that there is a one-to-one correspondence between the set of $n\times n$ involutory matrices and the set of $n\times n$ idempotent matrices.
$A^2=(I-2B)^2=I-4B+4B^2=I$
$A^2=I$，則
$\displaystyle B^2=\left(\frac{I-A}{2}\right)^2=\frac{I-2A+A^2}{4}=\frac{I-A}{2}=B$
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# Posts tagged math games for every day
## Teaching Number Concepts
4
I am trying to teach my son a concept of positive whole numbers being made up of other, smaller, positive whole numbers. This has been a tough going so far, full of unexpected obstacles. There was, for example, the part where I tried to explain and show that although a larger number can be made up of smaller numbers, it doesn’t work in reverse and a smaller number cannot be made up of larger numbers.
An even more formidable obstacle was (and still is) showing that a larger number can be made out of various combinations of smaller numbers. Say, 5=2+3, but also =4+1 and even 1+2+2. And by showing I mean proving. And by proving, I mean having my son test the rule and prove (or disprove) it to himself.
That’s why I was very happy when I got a hold of Oleg Gleizer’s book Modern Math for Elementary School. By the way, the book is free to download and use. We’ve been building and drawing multi-story buildings (mostly Jedi academies with x number of training rooms) ever since. If this sounds cryptic, I urge you to download the book and go straight to page 12, Addition, Subtraction and Young Diagrams.
And just yesterday I found this very simple activity on Mrs. T’s First Grade Class blog, via Love2Learn2Day‘s Pinterest board. All you need for it is a Ziploc bag, draw a line across the middle with a permanent marker, then add x number of manipulatives. Took me like 2 minutes to put it together, mostly because I had to hunt for my permanent marker.
The way we played with it was I gave the bag to my son and asked him how many items were in the bag. He counted 8. I showed him that the bag was closed tight, so nothing could fall out of it or be added to it. I also put a card with a large 8 on it in front of him as a reminder. At this point all 8 items were on one side of the line. I showed him how to move items across the line and let him play. As he was moving the manipulatives, I would simply provide the narrative:
Ok, so you took 2 of these and moved them across to the other side. Now you have 2 on the left and how many on the right? Yes, six (after him counting). Two here and six here. Two plus six. And how many items do we have in this bag? Good remembering, there are 8. So two plus six is 8. Want to move a few more over?
It went on like this for a few minutes until he got bored with it. Overall, I thought it was a good way of teaching, especially for children who do not like or can’t draw very well yet. Plus upping the complexity is really easy – draw more than one line on the bag and create opportunities for discovering that a number can be made of more than two smaller numbers.
## Playing Math Every Day – November 28 – Dec 4, 2011
2
Math games can be played any time anywhere. Here are some ideas for each day of the week. These games require very little, if any, advance prep. Give them and feel free to change them to make math more interesting for your children.
November 28 – Spots and Dots Day
This is a perfect day to play subitizing games, playing dominoes or any board games that including throwing dice. If you have simple dot stickers and 3×5 cards, you can create subitizing cards. To make the game easier, keep the number of dots small and/or arrange them in an easily recognizable pattern (i.e. like dots on dominoes). For a harder game, increase the number or dots, mix dots of different colors and sizes, or place them on the cards randomly.
Quickly show the card to your child. Your child should have just enough time to estimate the number of dots, but not enough time to allow your child to count them. Then, depending on the age of the child, you can either ask how many dots were on the card or ask to show the number of dots on the card using some other manipulative (i.e. bear counters, beads, etc). For very young children, you can show the first card briefly, then display two cards – the first one and another one and ask your child to point to the one she just saw.
November 29 – Louisa May Alcott’s Birthday
Louisa May Alcott was a big-time journal writer. Help your child start a math journal. You can make it a daily tradition of making an entry into the journal. The questions don’t have to be from worksheets (although they can be). You can ask your child to build a pyramid with 6 blocks, then sketch it out in the journal. I love searching Pinterest for great pre-K and K math journal ideas.
November 30 – Mark Twain’s Birthday
Do you remember The Great Jumping Frog of Calaveras County? Let’s make cute origami frogs today. Origami is surprisingly mathematical. On the surface, it’s a lesson in shapes and symmetry. But as you start folding, you’ll notice a lot more math opportunities. For example, do you have to start with a square? What if it’s a rectangle? Can I make a frog if I start with a Post-It note square? What words should I use to explain each fold?
If you start with a rectangle of paper, you can make a whole family of proportionally smaller frogs and a leftover rectangle of paper too small for frog making. Ask the “what if” question: “what if we could continue folding ever-smaller frogs”.
December 1 – Let’s Play Ball
And after all the running around, you can explore a type of fractal called Apollonian gasket. You can print it out or draw it (get inspired with this video). Depending on the age of your children, you can ask them to decorate, trace or draw the circles. If you have a young child, you probably have a collection of balls of various sizes, from basketballs to tennis balls to marbles to pompoms. See if you can arrange this collection into a gasket.
December 2 – Map and Measure
If you are planning a holiday road trip, then get the map out and see how long the drive will be… in origami frogs from November 30th. Measure it on the map, then measure distances to other interesting points just to compare. No road trip in the plans? No worries! You can measure a room in jumping frogs, then create a map using these measurements.
December 3 – The Rule of Three
Today’s game is noticing the number 3 in your daily activities and surroundings. Record the findings in the math journal. You can start at breakfast with figuring out how many meals (not counting snacks) we have every day.
December 4 – Reindeer Day
Explore odd and even numbers by talking about Santa Claus’s flying reindeer. Can we tell, just by looking at Santa’s sleigh, if Santa has an odd or even number of reindeer? How can we tell? What if Santa had more or fewer reindeer?
## Playing Math Every Day – November 14 – 20, 2011
0
Math games can be played any time anywhere. Here are some ideas for each day of the week. These games do not require any advance prep either. Give them a try this week and feel free to change them to make more interesting for your kids.
November 14 – Claude Monet’s Birthday
Monet would often paint the same subject at different times of the day as the light changed. Let’s create a color gradient collage today. All you need is a bunch of paint chips from your home improvement store. Suggest arranging different shades of the same color from lightest to darkest. Now try it with other colors. In case you don’t have time to run to a home improvement store, you can modify this game. Replace paint chips with liquid food coloring and give your child a dropper and several clear containers filled with water (glasses, clear jars or white ice-cube trays all work great).
November 15 – Children’s Book Day
There are quite a few wonderful children’s story books that go beyond basic counting and shapes. We are going to be reading Spaghetti and Meatballs for All by Marilyn Burns and Anno’s Magic Seeds by Mitsumasa Anno. If you or your child prefer to make up your own stories that include math, nothing beats another great book by Mitsumasa Anno, called Anno’s Counting Book.
November 16 – Talking Turkey Day
For this game you’ll need a marker, a piece of paper and a bag of bird seed. If you don’t have bird seed, a mix of 2 or more different pasta shapes or dried beans will do. First, trace your or your child’s hand on a piece of paper – that’s your turkey. Now, decide on a pattern, but don’t tell your child what it is. Let him guess which seed (or pasta shape) the turkey would like to eat next. Start with something simple, such as ABAB pattern. Then move to more complicated ones. Then let your child decide on a pattern and you’ll try to guess it.
November 17 – Bread Baking Day
Ah, kitchen is a perfect place for math! Let your children do all the measuring. Then let them experiment with estimating (i.e. how many tea spoons make a table spoon). The result is going to be some delicious math. And if you don’t have time to bake bread from scratch, there’s absolutely nothing wrong with picking up a muffins or cupcakes mix at the store.
November 18 – Mickey Mouse’s Birthday
Let’s watch a Disney cartoon today. How about this one – Donald Duck in Mathmagic Land (all three parts are available on YouTube). You can even try some of the math activities Donald tries during his adventure, starting with playing tic-tac-toe.
If you would rather stick with the Mickey Mouse’s theme, then how about revisiting November 14th idea of gradients, only using Disney Paint Chips.
November 19
Let’s start getting ready for the Pie Day!
November 20 – Pie Day
Nope, not the “pi day” which happens on March 14th (you know, 3.14). Instead, today is all about baking and enjoying pies! So why not do some more kitchen math. You can also cut a few pie shapes out of construction paper, let your child decorate them, then ask to share it with her toys (hello, fractions!).
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https://www.shaalaa.com/question-bank-solutions/find-shortest-distance-between-lines-x-1-7-y-1-6-z-1-1-x-3-1-y-5-2-z-7-1-shortest-distance-between-two-lines_12251
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# Find the Shortest Distance Between the Lines (X+1)/7 = (Y+1)/(-6) = (Z+1)/1 and (X-3)/1 = (Y-5)/(-2) = (Z-7)/1 - CBSE (Commerce) Class 12 - Mathematics
ConceptShortest Distance Between Two Lines
#### Question
Find the shortest distance between the lines (x+1)/7 = (y+1)/(-6) = (z+1)/1 and (x-3)/1 = (y-5)/(-2) = (z-7)/1
#### Solution
The given lines are (x+1)/7 = (y+1)/(-6) = (z+1)/1 and (x-3)/1 = (y-5)/(-2) = (z-7)/1
It is known that the shortest distance between the two lines,
Since distance is always non-negative, the distance between the given lines is 2sqrt29 units.
Is there an error in this question or solution?
#### APPEARS IN
NCERT Solution for Mathematics Textbook for Class 12 (2018 to Current)
Chapter 11: Three Dimensional Geometry
Q: 15 | Page no. 478
#### Video TutorialsVIEW ALL [5]
Solution Find the Shortest Distance Between the Lines (X+1)/7 = (Y+1)/(-6) = (Z+1)/1 and (X-3)/1 = (Y-5)/(-2) = (Z-7)/1 Concept: Shortest Distance Between Two Lines.
S
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# Addition Subtraction Fact Families Worksheets
Addition Subtraction Fact Families Worksheets - 5 + 3 = 8. Web in this free printable worksheet, kids can learn what a fact family is and then make their own by completing a series of addition and subtraction math problems located in each house. Fact family worksheets help your students begin to see the relationship between numbers and number bonds. 3 + 5 = 8. This fact family worksheet is great for testing children in their ability to build fact families for addition and subtraction. The three numbers will have two addition and two subtraction relationships in an addition and subtraction fact family.
The first worksheets only have single digit numbers; Web in this free printable worksheet, kids can learn what a fact family is and then make their own by completing a series of addition and subtraction math problems located in each house. Web these fact family worksheets are great for testing children in their ability to build fact families for addition and subtraction. Web addition and subtraction fact family. Write the numbers in the boxes.
The later worksheets use slightly larger numbers. This fact family worksheet is great for testing children in their ability to build fact families for addition and subtraction. 5 + 3 = 8. 12 + 4 = 16, 4 + 12 = 16, and 16. Web a few addition and subtraction family facts are included in this worksheet.
## Fact Family Complete each fact family 2 Worksheets / FREE Printable
Addition Subtraction Fact Families Worksheets - Fill in the numbers for each fact family house. Multiplication and division fact families. Web working with fact families addition and subtraction. Web addition and subtraction worksheets. These fact family worksheets demonstrate the relation between addition and subtraction. 12 + 4 = 16, 4 + 12 = 16, and 16. Web use addition and subtraction to write the fact family for each. This page includes fact family worksheets including addition and subtraction relationships, and multiplication and division relationships. Maths calculation addition and subtraction. The key concept here is that you can use these four related equations to understand the relationships between these three numbers.
Worksheet #1 worksheet #2 worksheet #3 worksheet #4. For example, numbers such as 12, 4, and 16 have the following addition/subtraction fact families: These fact family worksheets provide practice in simple addition and subtraction facts, and reinforce the relationships between the two operations. Web addition and subtraction fact family. When you download this resource, you’ll have instant access to a really handy worksheet to use in your classroom for maths!
In this fact family, you have two addition facts and two subtraction facts. Students write sets of related addition / subtraction facts. Web addition and subtraction fact families. Use addition and subtraction to fill in the missing fact in each family.
Fact family worksheets help your students begin to see the relationship between numbers and number bonds. 12 + 4 = 16, 4 + 12 = 16, and 16. 3 + 5 = 8.
Web in this free printable worksheet, kids can learn what a fact family is and then make their own by completing a series of addition and subtraction math problems located in each house. These worksheets ask students to write out the 4 different addition or subtraction equations representing each fact family. Web create addition and subtraction fact family worksheets with our worksheet generator, and give kids unlimited addition and subtraction facts practice.
## Fill In The Numbers For Each Fact Family House.
12 + 4 = 16, 4 + 12 = 16, and 16. These fact family worksheets demonstrate the relation between addition and subtraction. Whether you're teaching your students about addition facts to 100, 20 or even 10, these differentiated sheets will work for you. To create fact family worksheets to specific sum, change the total to be more than maximum values.
## Web Addition And Subtraction Fact Family.
Web addition and subtraction worksheets. Web fact families to 100 addition and subtraction worksheet. Students write sets of related addition / subtraction facts. These fact family worksheets provide practice in simple addition and subtraction facts, and reinforce the relationships between the two operations.
## Maths Calculation Addition And Subtraction.
Web this useful set of differentiated subtraction and addition number facts worksheets is a great way to give students experience of studying subtraction and addition facts families and inverse relationships. Web 96 fact family worksheets. Adding up to 10, addition and subtraction fact families, commutative property of addition, subtracting up to 10. For each number below, create your own fact family.
## These Math Worksheets Have 100 Addition And Subtraction Fact Family Problems And Make For A Challenging Two Minute Test.
In this fact family, you have two addition facts and two subtraction facts. For example, numbers such as 12, 4, and 16 have the following addition/subtraction fact families: The three numbers will have two addition and two subtraction relationships in an addition and subtraction fact family. Write the correct numbers in each addition/subtraction fact family triangle.
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https://www.coursehero.com/file/5777018/light-ray-given-by-the-vector-a-a-1-a-2-a-3-rst-strikes/
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# Light ray given by the vector a a 1 a 2 a 3 rst
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Unformatted text preview: each half of the clothesline. 34. The tension T at each end of the chain has magnitude 25 N. What is the weight of the chain? 37° light ray given by the vector a a 1, a 2 , a 3 first strikes the xz-plane, as shown in the figure. Use the fact that the angle of incidence equals the angle of reflection to show that the direction of the reflected ray is given by b a 1, a 2 , a 3 . Deduce that, after being reflected by all three mutually perpendicular mirrors, the resulting ray is parallel to the initial ray. (American space scientists used this principle, together with laser beams and an array of corner mirrors on the Moon, to calculate very precisely the distance from the Earth to the Moon.) z 37° 35. If A, B, and C are the vertices of a triangle, find l AB l BC l CA. b 36. Let C be the point on the line segment AB that is twice as far from B as it is from A. If a that c 2 a 1 b. 3 3 l OA, b l OB, and c l OC, show a x y 5E-13(pp 838-847) 1/18/06 11:15 AM Page 843 SECTION 13.3 THE DOT PRODUCT |||| 13.3 ❙❙❙❙ 843 The Dot Product So far we have added two vectors and multiplied a vector by a scalar. The question arises: Is it possible to multiply two vectors so that their product is a useful quantity? One such product is the dot product, whose definition follows. Another is the cross product, which is discussed in the next section. a 1, a 2 , a 3 and b 1 Definition If a and b is the number a b given by ab b1, b2 , b3 , then the dot product of a a 1 b1 a 2 b2 a 3 b3 Thus, to find the dot product of a and b we multiply corresponding components and add. The result is not a vector. It is a real number, that is, a scalar. For this reason, the dot product is sometimes called the scalar product (or inner product). Although Definition 1 is given for three-dimensional vectors, the dot product of two-dimensional vectors is defined in a similar fashion: a 1, a 2 b1, b2 a 1 b1 a 2 b2 EXAMPLE 1 2, 4 3, 6, 2, i 2j 1 2 2j 1, 7, 4 1 k 3k 23 4 1 16 10 2 4( 72 22 3 ) 6 1 7 1...
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## This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas at Austin.
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# Principles of mechanism - Jameson
PRINCIPLES OF MECHANISM
BY JOSEPH M. JAMESON
GIRARD COLLEGE; 1918
Principles of mechanism
PREFACE
This book is intended to present the elementary principles of mechanism in a way that will make it adapted for use in evening technical schools, trade schools, mechanic arts high schools, and other schools where it is desired to teach the subject thoroughly yet without going into the highly mathematical treatment. Typical problems are solved throughout the text and a large number of problems are included for solution by the student.
CONTENTS
- General Definitions
- Revolving and Oscillating Bodies
- Transmission of Motion by Means of Cylinders, Cones and Discs.
- Gears and Gear Teeth
- Belts, Ropes, and Chains
- Inclined Plane, Wedge, Screw, Worm and Wheel
- Cams
- Simple Wheel Trains
- Problems for Solution
CHAPTER I - GENERAL DEFINITIONS
1. A Machine. A machine consists of a number of pieces or groups of pieces so arranged that, when a driving force is applied to the proper place, the various pieces operate together to do some useful work. Each of the pieces in a machine either moves or helps to guide some of the other pieces in their motion. A familiar example is the common sewing machine. Power is applied at the treadle, causing motion of the rod connected with the treadle; this motion is passed on up through the various parts with the final result that the cloth is stitched. It is evident that the machine consists of a frame and moving pieces; the frame supporting and guiding the moving pieces, while the moving pieces pass the forces along.
2. A Mechanism. A mechanism is one of the groups of pieces in a machine, all of which pieces are so connected that, when a definite motion is given to one of them, the others are caused to move in definite ways. The treadle, driving rod, crank shaft, and that part of the frame which supports them in the sewing machine, constitute a mechanism. A machine, therefore, consists of a series of mechanisms, each of which is doing its own work, and all of which work together to accomplish the purpose for which the machine is intended.
3. The Study of Mechanism. The study of mechanism is the study of the laws which govern the motions of the various parts of a machine and the forces which exist in or are transmitted by those parts. That branch of the subject which deals with the motions only, regardless of the forces, is often given the name Kinematics of Machines, while the part dealing with forces is called Dynamics of Machines,
In this book only the kinematics of machines and a few of the simpler problems relating to forces will be treated.
4. Importance of a Study of Mechanism. The importance of this study to one who has to do with machinery is very great. Whether he is engaged in using or in designing machinery he ought to know how to analyze the motions which the various pieces have and to determine the speeds of the pieces. He also ought to know how to design and connect the mechanisms in order to obtain a desired motion for any particular piece. Furthermore, he needs to be able to determine the forces which may be expected to exist in the several pieces when a known force is applied at a known point. In short, in order intelligently to design or handle a machine, a man should thoroughly understand the natural laws which enter into the operation of the machine, and the various methods which have been devised to apply those laws to the performance of definite work. It is to this end that the science of mechanism is devoted.
5. Design. - The design of a machine consists in adapting known mechanical appliances to meet special conditions. That is, there are certain elementary mechanical units such as levers, revolving wheels, screws, cams, cranks, connecting rods, sliders, etc., which form the basis of all mechanisms. The, designer must so proportion and arrange these as to accomplish the desired end. It is purposed in the present work to enumerate some of the more common of these units or elements and to consider each one separately, discussing the natural laws governing its design and operation. Some of the ways in which these units are combined into mechanisms will be studied, also the motions and forces in these mechanisms. Sometimes it is better to carry on such studies by means of graphical work on the drawing board and sometimes by calculation. Whichever method seems best in any particular case will be adopted in the present text, and very frequently both will be used as a check on each other.
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https://documen.tv/question/1-willie-enlarged-the-size-of-a-painting-to-a-height-of-12-in-what-is-the-new-width-if-it-was-or-17653567-97/
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## 1.Willie enlarged the size of a painting to a height of 12 in. What is the new width if it was originally 2 in wide and 4 in tall. 2
Question
1.Willie enlarged the size of a painting to a height of 12 in. What is the new width if it was originally 2 in wide and 4 in tall.
2. A rectangle is 6 inches wide and 12 inches tall. If it is reduced to a width of 2 inches, then how tall will it be?
in progress 0
3 years 2021-09-04T14:25:36+00:00 1 Answers 3 views 0
1 6 inches
2 4 inches
Step-by-step explanation:
1. 2wide : 4tall
×3
6wide: 12tall
2. 6wide : 12tall
÷ 3
2wide to 4tall
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0
# Is 68 composite
Updated: 4/28/2022
Wiki User
7y ago
A Prime number has only 2 factors which are 1 and itself. Composite numbers are everything else except 1 and 0. 1 and 0 are neither prime, nor composite. 68 is composite.
Wiki User
11y ago
Wiki User
7y ago
Yes.
Earn +20 pts
Q: Is 68 composite
Submit
Still have questions?
Related questions
68 is composite.
### Is 68 a composite or prime?
68 is a Composite Number.
### What are the composite numbers of 68?
The composite factors of 68 are 4, 34 and 68
It is composite.
### Is 68 or 79 a prime number?
68 is composite and 79 is prime
yes
composite
### Is 68 a prime number or a composite number?
68 is not prime. 68 = 2 * 2 * 17
### Is 68 the sum a prime number?
No, it is composite.
### Is 68 a prime or compsite?
Composite (1 and 68, 2 and 34,, 4 and 17)
### What numbers can go in to 68?
It is composite...... factors:1,2,4,68
### Are 17 and 68 prime numbers?
17 is a prime number but 68 is Composite.
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+0
help
0
147
1
Janice bought 30 items each priced at 30 cents, 2 dollars or 3 dollars. If her total purchase price was \$30.00, how many 30-cent items did she purchase?
Nov 4, 2019
#1
+272
+1
we know that 10 3-dollars will worth 30 dollars, so items that cost 3 dollars have to be less than ten.
We also know that 10 30-cents will worth 3 dollars, so the number of cents has to come in 10s since the final cost is 30 which is a whole number.
There will be a lot of cents, so I will just say that there are 20 30-cents, which will cost 6 dollars. If I add 6 2-dollars, then I will get 12 dollars. 12 +6=18. 30-18=12. and 12/3=4.
4 items cost \$3
6 items cost \$2
20 items cost 30 cents
Nov 4, 2019
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3. Factor.
9x2 + 15x
(Points : 1) x(3x + 5)
3x(3x + 5)
9x(x + 5)
3x2(3x + 5)
Question 4.4. Factor.
20mn − 30m
(Points : 1) 2m − 3
10m(2n − 3)
10mn(2m + 3n)
10n(2m − 3) Question 5.5. Factor.
8x2y2 − 4x2y − 12xy
(Points : 1) 4(2xy − 4x2y − 12xy)
4(8x2y2 − x − 12xy)
4x2y2(2xyxy −3)
4xy(2xyx − 3)
It seems that some the work is already here, but I’d be glad to! So for #3 which is 9x^2+15x, we can factor out both a 3 and an x (3x) so we know that 3x * 3x =9x^2 and 3x * 5 = 15x so once we take the 3x out of the equation, we are left with 3x(3x+5) and that’s as far as you can factor.
For #4, we see that the common factor is 10m because 10m * 2n = 20mn and 10m * 3 = 30m so once we take 10m out of the original, it becomes 10m(2n-3)
For #5, this one the common factor is 4xy because 4xy * 2xy=8x^2y^2 and 4xy*x= 4x^2y and 4xy*3=12xy so once we take the 4xy out of the equation, it becomes 4xy(2xy-x-3)
!
RELATED:
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• anonymous
Match each set of measures of a and b with the measure of c that gives a volume of 24 cubic inches. a = 2in and b = 8 in a = 4in and b = 2 in a = 3in and b = 4 in a = 6in and b = 4in options are A. c = 2 in B. c = 3 in C. c = 4 in D. c = 6 in
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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• This slope-intercept game has ten multiple choice questions about the slope-intercept form of a linear equation. To learn more about the concepts involved in this lesson, watch this free online video. Return from the Slope-Intercept Game page to the Algebra Lessons page or to the Math Online Games.
• Improve your math knowledge with free questions in "Slope-intercept form: graph an equation" and thousands of other math skills.
• Content includes: - understanding slope and y-intercept - formula for an equation in slope-intercept form - steps for graphing a line - working with graph form, equation form, tables, and ordered pairs - special cases - what to do with equations that do not yet have y isolated - practice and examples Check out the preview for more detail about ...
• Y Intercept Form. Displaying all worksheets related to - Y Intercept Form. Worksheets are Graphing lines in slope intercept, Practice for slope y intertcept and writing equations, Slope intercept form word problems, Model practice challenge problems vi, Infinite algebra 1, Graphing lines in slope intercept form, Algebra i name block date y mx b practice a for use, Graphing lines.
• where m is the slope of the line and b is the y-intercept of the line, or the y-coordinate of the point at which the line crosses the y-axis.. To write an equation in slope-intercept form, given a graph of that equation, pick two points on the line and use them to find the slope.
• Dec 20, 2018 · Calculus Definitions >. The y-intercept is the point where a graph crosses the y-axis. On the graph below, the plotted line crosses the y-axis at y = 3: More Examples. To find the y intercept on a graph, just look for the place where the line crosses the y-axis (the vertical line).
• Write the slope-intercept form of the equation of each line given the slope and y-intercept. ... Find the slope and y-intercept of each line. Write the equation of ...
• Writing Linear Equations Date_____ Period____ Write the slope-intercept form of the equation of each line. ... Write the standard form of the equation of the line ...
• Sep 27, 2012 · Alternatively, we can determine the x-intercept and the y-intercept of the standard form linear equation by subtituting y = 0, then solve for x and substituting x = 0, then solve for y respectively.
• The slope-intercept form is , where is the slope and is the y-intercept. Find the values of and using the form . The slope of the line is the value of , and the y-intercept is the value of .
• where m is the slope of the line and b is the y-intercept of the line, or the y-coordinate of the point at which the line crosses the y-axis.. To write an equation in slope-intercept form, given a graph of that equation, pick two points on the line and use them to find the slope.
• In the slope-intercept form of a straight line, I have y, m, x, and b. They've given me the value for m, along with values for an x and a y. So the only thing I don't have so far is a value for is b (which gives me the y -intercept).
• When an equation is in slope-intercept form, the slope is the number in front of the x. This means the slope is 1/4. A line with a slope of 1/4 will go up 1 and over to the right 4. The y-intercept is the constant at the end. Pay close attention to the sign of the y-intercept. This equation has a minus sign, so the y-intercept is -1.
• Let's plot the y intercept and use the slope to form the line. we can see that b = 3, so the y intercept will be (0,3). The slope is m = -2 , so we can go down 2 and right 1 (-2/1) or up 2 and left 1 (2/-1) to find the next point on the line.
• Y intercept is found by setting x to 0: the equation becomes by=c, and therefore y = c/b. Y intercept is -2/2 = -1. Slope is -5/2 = -2.5. Equation in slope-intercept form: y=-2.5*x+-1.
• Plot the point corresponding to the y-intercept, (0,1) The m-value, the slope, tells us that for each step to the right on the x-axis we move 2 steps upwards on the y-axis (since m = 2) And once you have your second point you can just draw a line through the two points and extend it in both directions.
• When an equation is in slope-intercept form, the slope is the number in front of the x. This means the slope is 1/4. A line with a slope of 1/4 will go up 1 and over to the right 4. The y-intercept is the constant at the end. Pay close attention to the sign of the y-intercept. This equation has a minus sign, so the y-intercept is -1.
• where m is the slope of the line and b is the y-intercept of the line, or the y-coordinate of the point at which the line crosses the y-axis.. To write an equation in slope-intercept form, given a graph of that equation, pick two points on the line and use them to find the slope.
• Improve your math knowledge with free questions in "Slope-intercept form: graph an equation" and thousands of other math skills.
• Graphing in Slope Intercept Form is the easiest way to graph linear equations.Graphing in Slope Intercept Form is convenient because Slope Intercept Form already has the slope and y-intercept built into the form which means you just have to know how to use them.
• The equation will be in the form "y = m*x +b" where m is a number corresponding to the slope and b is a number corresponding to the y-intercept. Tip Remember, you type an equals sign as the first character in a cell where you want to enter a formula.
• The slope intercept form for this line is y = .5x + .5. This line crosses the y-axis at .5 and has a slope of .5, so this line rises one unit along the y-axis for every 2 units it moves along the x-axis. So, where would you ever use this? Here's an article on ways to use the Slope Intercept Form in Real Life.
• A parabola is a visual representation of a quadratic function. Each parabola contains a y-intercept, the point at which the function crosses the y-axis. Learn the tools you need to find the y-intercept using the graph of a quadratic function and the equation of a quadratic function.
• Score : Printable Math Worksheets @ www.mathworksheets4kids.com Name : Represent each equation in slope-intercept form and graph them. Graphing a Line
• There are a few different ways to write the equation of a line. One of the most common ways is called "slope-intercept" form. It's called this because it clearly identifies the slope and the y-intercept in the equation. The slope is the number written before the x. The y-intercept is the constant written at the end.
• This means that we have an undefined slope. If you were to graph the line, it would be a vertical line, as shown above. If your linear equation is written in this form, m represents the slope and b represents the y-intercept.
• The y-intercept is the intersection between the graph and the y-axis. And the y -axis is x = 0. So, to find the y -intercept, put x = 0 into the given linear equation: 3⋅0 + 4 y = 12.
• Y Intercept Form. Displaying all worksheets related to - Y Intercept Form. Worksheets are Graphing lines in slope intercept, Practice for slope y intertcept and writing equations, Slope intercept form word problems, Model practice challenge problems vi, Infinite algebra 1, Graphing lines in slope intercept form, Algebra i name block date y mx b practice a for use, Graphing lines.
• Content includes: - understanding slope and y-intercept - formula for an equation in slope-intercept form - steps for graphing a line - working with graph form, equation form, tables, and ordered pairs - special cases - what to do with equations that do not yet have y isolated - practice and examples Check out the preview for more detail about ...
• In analytic geometry, using the common convention that the horizontal axis represents a variable x and the vertical axis represents a variable y, a y-intercept or vertical intercept is a point where the graph of a function or relation intersects the y-axis of the coordinate system. As such, these points satisfy x = 0.
• The Slope Intercept Form Calculator is used to help you find the slope intercept form for the equation of the straight line that passes through two points. It also calculates the slope and the y-intercept of the line.
• Content includes: - understanding slope and y-intercept - formula for an equation in slope-intercept form - steps for graphing a line - working with graph form, equation form, tables, and ordered pairs - special cases - what to do with equations that do not yet have y isolated - practice and examples Check out the preview for more detail about ...
• 4. Find the equation of a line given its slope and y‐intercept. 5. Work with linear models in slope‐intercept form. 1. Use the slopeintercept form to identify the slope and yintercept of a line Slopeintercept Form‐is an equation of the form y =mx +b where _____ is the slope and
• to the standard form, the first thing to do is to multiply each term by 8. This gives you 8y = –3x + 56. To put it in standard form, you add 3x to each side; the standard form is 3x + 8y = 56. The slope of –3/8 and y-intercept of 7 were more obvious in the original form, but you can pick up the x-intercept by using C/A; the x-intercept is ...
• Standard Form to Slope-Intercept Form Write the slope-intercept form of the equation of each line. 1) x y 2) x y 3) x y 4) x y 5) x y 6) x y
• Slope-Intercept Form If you know the slope m , and y -intercept ( 0 , b ) of a line (the point where the line crosses the y -axis), you can write the equation of the line in slope-intercept form . y = m x + b
• Y intercept is found by setting x to 0: the equation becomes by=c, and therefore y = c/b. Y intercept is -2/2 = -1. Slope is -5/2 = -2.5. Equation in slope-intercept form: y=-2.5*x+-1.
• Slope-intercept form, y=mx+b, of linear equations, emphasizes the slope and the y-intercept of the line. Watch this video to learn more about it and see some examples.
• Welcome to The Converting from Standard to Slope-Intercept Form (A) Math Worksheet from the Algebra Worksheets Page at Math-Drills.com. This Algebra Worksheet may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math.
• The calculator will find the x- and y-intercepts of the given function, expression or equation.
• Feb 11, 2020 · The y-intercept is the point where the line intersects with the y-axis. Since the y-axis is located at x = 0, the x coordinate of the y-intercept is always 0. Example 1 (cont.): The y-intercept is at y = -2, so the coordinate point is (0, -2).
• The equation of a horizontal line is y = b where b is the y-intercept. Since the slope of a horizontal line is 0, the general formula for the standard form equation , y = mx + b becomes y = 0x +b y = b. Also,since the line is horizontal, every point on that line has the exact same y value.
• The slope intercept form is one method of expressing a linear equation. A linear equation is an equation of a straight line. In the slope intercept form you use the slope and the y intercept to express the line.
• There are a few different ways to write the equation of a line. One of the most common ways is called "slope-intercept" form. It's called this because it clearly identifies the slope and the y-intercept in the equation. The slope is the number written before the x. The y-intercept is the constant written at the end.
• The Y-Intercept of a line is the point where a line's graph intersects (crosses) the Y-axis. A y-intercept of 3 means that a line's graph intersects the Y-axis at the point (0,3). A y-intercept of -4 means that the graph of a line crosses the Y-axis at the point (0,-4).
• In mathematics, a linear equation is an equation that may be put in the form where are the variables (or unknowns or indeterminates ), and are the coefficients, which are often real numbers. The coefficients may be considered as parameters of the equation, and may be arbitrary expressions,...
• The slope-intercept form is the most "popular" form of a straight line. Many students find this useful because of its simplicity. One can easily describe the characteristics of the straight line even without seeing its graph because the slope and y-intercept can easily be identified.
• You don't say how you are intending to fit the probit model, but if it uses R's formula notation to describe the model then you can supply either + 0 or - 1 as part of the formula to suppress the intercept:
• Aug 18, 2009 · Slope Intercept Form y=mx+b, Point Slope & Standard Form, Equation of Line, Parallel & Perpendicular - Duration: 48:59. The Organic Chemistry Tutor 366,461 views 48:59
# Y intercept form
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The equation of any straight line, called a linear equation, can be written as: y = m x + b, where m is the slope of the line and b is the y-intercept. Since the y -intercept is where x =0, substituting this in shows us that the a and b terms drop out, leaving us with only the c value. Therefore, the c value is always the y -intercept. This is kind of cool, but substituting x =0 into the other forms to find the y -intercept is pretty easy too.
Free slope intercept form calculator - find the slope intercept form of a line given two points, a function or the intercept step-by-step This website uses cookies to ensure you get the best experience. Jul 20, 2018 · Slope Intercept Form Formula. For the equation, “y = mx + b”, m is the slope of the line that is multiplied by x and b is the y-intercept or we can say the point where the line will cross the vertical y-axis. This is a sensible equation that can also be named as the slope-intercept form.
Feb 06, 2018 · Equation of straight line in slope - intercept form can be written as: y=mx+c here, m= slope of the line and . c= y-intercept Now to your question: 2x+3y=5 The terms can be rearranged to write it as y=(-2/3)x+(5/3) Comparing with the standard equa... Y-intercept definition is - the y-coordinate of a point where a line, curve, or surface intersects the y-axis. the y-coordinate of a point where a line, curve, or surface intersects the y-axis… See the full definition
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Activity Overview: In this activity, students will look for patterns between the change in y values and the slope of a line and the y value when x equals zero and the y-intercept of a line.They will use the CT concepts of pattern recognition, data representation, and decomposition to move between representing a line in chart form, equation form ... The outcomes could signal whether the energy behind Bernie Sanders movement is contagious, or contained by the omnipresence of the presidential campaign.
### Algebra 2 helper
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Solve for y. Slope-Intercept Form of a Linear Equation The slope-intercept form of the equation of a line is y = mx + b, where m is the slope and b is the y-intercept Let (x, y) and (x1, y1) be points on line AB. .
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Calculator to plot lines in Slope y-intercept form and Standard form. Step by step explanations are provided. Gas cigarette lighter repair
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+0
# help
0
95
1
Allen has four times as many cards as Bob, and Jane has twice as many cards as Allen. Together, Allen and Jane have 468 cards. How many cards do Allen, Jane, and Bob EACH have?
Apr 11, 2020
#1
+20957
0
Bob has the fewest cards; call the amount that Bob has: x
Allen has four times as many cards as Bob; Allen has 4x
Jane has twice as many cards as Allen; Jane has 8x
Together, they have 468 cards: x + 4x + 8x = 468
13x = 468
x = 36
Bob: x = 36
Allen: 4x = ...
Jane: 8x = ...
Apr 11, 2020
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# (Solved) Math225 Week 2 Assignment: Frequency Tables
Question
A group of students were surveyed about the number of times they went to a movie theater last year. Their responses are summarized in the relative frequency table below.
What is the cumulative relative frequency of students that went to 7 or fewer movies?
Number of Movies Relative Frequency 0−1 0.08 2−3 0.20 4−5 0.30 6−7 0.24 8−9 0.10 10 or more 0.08
Question
Given the relative frequency table below, which of the following is the corresponding cumulative relative frequency table?
Value Relative Frequency 4 0.35 5 0.2 6 0.05 7 0.4
Question
A data set is summarized in the frequency table below. Using the table, determine the number of values less than or equal to 7 in the data set. Give your answer as a single number. For example if you found the number of values was 16, you would enter 16.
Value Frequency 3 8 4 4 5 3 6 2 7 2 8 7 9 3 10 6 11 3 12 7
Question
As the manager of a store, you wish to determine the amount of money that people who visit this store are willing to spend on impulse buys on products placed near the checkout register. You sample twenty individuals and records their responses. Construct a frequency table for grouped data using five classes.
8,18,15,10,29,4,15,2,4,9,16,14,13,8,25,25,27,1,15,24
Question
A group of students were surveyed about the number of siblings they have. The frequencies and relative frequencies of their responses are shown in the below.
Complete the cumulative relative frequency table.
Number of Siblings Relative Frequency 0 0.18 1 0.33 2 0.16 3 0.14 4 or more 0.19
Question
Given the relative frequency table below, which of the following is the corresponding cumulative relative frequency table?
Value Relative Frequency 4 0.28 5 0.24 6 0.04 7 0.2 8 0.24
Question
A small startup company wishes to know how many hours, per week, that its employees spend commuting to and from work. The number of hours for each employee are shown below. Construct a frequency table for grouped data using four classes.
9, 18, 18, 14, 13, 4, 12, 9, 13, 10, 20, 12, 19, 20, 13, 3, 5, 20, 17, 1
Question
William wishes to view a frequency table for grouped data using his monthly credit card statements for the last 20 months, shown below. Construct the table for William using six classes.
1312, 1303, 809, 1477, 1263, 1444, 894, 1051, 1485, 1433, 1132, 1221, 1179, 945, 995, 1179, 1172, 1373, 906, 955
# Solution:
A group of students were surveyed about the number of times they went to a movie theater last year. Their responses are summarized in the relative frequency table below.
What is the cumulative relative frequency of students that went to 7 or fewer movies?
Number of Movies Relative Frequency 0−1 0.08 2−3 0.20 4−5 0.30 6−7 0.24 8−9 0.10 10 or more 0.08
Given the relative frequency table below, which of the following is the corresponding cumulative relative frequency table?
Value Relative Frequency 4 0.35 5 0.2 6 0.05 7 0.4
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https://totaltrainerfactoryoutlet.com/qa/what-is-value-of-expression.html
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# What Is Value Of Expression?
## What is the value of expression 100 25 in python?
the value of given expression 100/25 = 4..
## How will Python evaluate the following expression?
In general, Python executes statements from top to bottom. Python executes a statement by evaluating its expressions to values one by one, then performing some operation on those values. Python evaluates an expression by first evaluating its sub-expressions, then performing an operation on the values.
## What happens to the value of the expression N 15 as N decreases?
As the variable n decreases in size, the value of the given expression will decrease in size as well.
## What is 5i equal to?
The imaginary number i is equal to the square root of -1. In other words, i2 equals -1. The square root of a negative number is not a real number and it is not a variable. For example, the square root of -25 is written as 5i because 5i times 5i equals 25 times -1 or -25.
## What does 4 Evaluate to in python?
What does ~4 evaluate to? Explanation: ~x is equivalent to -(x+1).
## What happens to the value of the expression 20 A as A increases?
Answer. With the increase of the value of a ,the value of the expression 20+a will also increase.
## What happens to the value of the expression 150 J as J decreases?
Answer. Step-by-step explanation: In the expression, 150-j, if we decrease the value of ‘j’, the value of the expression will increase as less is being subtracted from the answer.
## How do you find the value of an expression?
To evaluate an algebraic expression means to find the value of the expression when the variable is replaced by a given number. To evaluate an expression, we substitute the given number for the variable in the expression and then simplify the expression using the order of operations.
## How do I simplify an expression?
Here are the basic steps to follow to simplify an algebraic expression:remove parentheses by multiplying factors.use exponent rules to remove parentheses in terms with exponents.combine like terms by adding coefficients.combine the constants.
## What is value and example?
Familiar examples of values are wealth, loyalty, independence, equality, justice, fraternity and friendliness. Familiar examples of values are wealth, loyalty, independence, equality, justice, fraternity and friendliness.
## What is the value of percentage?
The percentage value or new value is calculated by multiplying the original value by the percent rate and dividing by 100%. The original value is calculated by dividing the amount already paid by the percentage rate and multiplying the result by 100.
## What does represent in Python?
The % symbol in Python is called the Modulo Operator. It returns the remainder of dividing the left hand operand by right hand operand. It’s used to get the remainder of a division problem. The modulo operator is considered an arithmetic operation, along with + , – , / , * , ** , // .
## What is the value of Expression 1000?
SOLUTION: investing: the expression 1000 (1.1)t represents the value of a \$1000 investment that earns 10% intrest per year. compounded annually for t years.
## What is the value of in math?
In math, value is a number signifying the result of a calculation or function. So, in the example above, you could tell your teacher that the value of 5 x 6 is 30 or the value of x + y if x = 6 and y = 3 is 9. Value can also refer to a variable or constant. … A constant is a fixed or well-defined number.
## What expression is equivalent to 100000?
Answer: The expression that is equivalent to 100,000 is C. .
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