water that is treated as an in compressible fluid at room temperature enters a nozzle who's inlet area is 0.5 meters and outlet area is 0.1 meters at a constant flowrate of 15 cubic meters/second what is the exit velocity of the nozzle in meters/second ?
first we define flow rate Q=A*V=Ai*Vi=Ao*Vo since the flow rate is given as Q=15 and the Ai= 0.5 Vi=Q/Ai=15/0.5= 30 meters/ second so Vo=(Ai/Ao)*Vi= (0.5/0.1)*30= 150 meters/second.
Take the derivative and integral of the following function y=3(x^5)+2(x^4)+(x^3)+5)x^2)+9x+7
the derivative dy/dx=(3*5)(x^(5-1)))+((2*4)(x^(4-1)))+3x^(3-1)+((5*2)(x^(2-1)))+9(x^(1-1))+0=15x^4 +8x^3 + 3x^2 + 10x +9. the integral is Y= (3/(5+1))x^(5+1) +(2/(4+1))x^(4+1) +(1/(3+1))x^(3+1) +(5/(2+1))x^(2+1) + (9/(1+1))x^(1+1) +(7/(0+1))x^(0+1)= 0.5x^6 + 0.4x^5 + 0.25x^4 + 1.67x^3 + 4.5x^2 + 7x
A line passes through the following points (4,7) and (8,2) what are the equation of the line in both slope intercept and point slope form and the x and y intercepts.
first you have to calculate the slope(m) of the function by taking the change in the y(dy) and divide it by the change in the x(dx) dy=2-7=-5, dx=8-4=4, m=dy/dx=-5/4=-1.5 point slope equation is y-7=m(x-4) or y-7=-1.25(x-4) slope intercept for solve for y so y= -1.25x+5+7=-1.25x+12 so y intercept is 12 and the x intercept is x= -12/-1.25= 12*(4/5)=48/5=9.6