Hey there,
So the formula of a area of a square would be 1 side x's the other side so it would be L*W*H.
~Jurgen
Answer:
486
Step-by-step explanation:
just multiply and divide
8 x 27 / 4 x 9
Answer:
velocity v(t) = -2cos(t)+t+2
position s(t) = ![-2sin(t)+\frac{t^2}{2} +2t + 3](https://tex.z-dn.net/?f=-2sin%28t%29%2B%5Cfrac%7Bt%5E2%7D%7B2%7D%20%2B2t%20%2B%203)
average of position from 2 to 5 = 17.666
Step-by-step explanation:
velocity is anti-derivative/integral of acceleration
= -2cos(t)+t+C
v(0) = 0
0 = -2cos(0)+0+C = -2 +C; C = 2
v(t) = -2cos(t)+t+2
position is anti-derivative of velocity
= ![-2sin(t)+\frac{t^2}{2} +2t + C](https://tex.z-dn.net/?f=-2sin%28t%29%2B%5Cfrac%7Bt%5E2%7D%7B2%7D%20%2B2t%20%2B%20C)
s(0) = 3
3 = -2sin(0) + 0 + 0+ C
3 = C
s(t) = ![-2sin(t)+\frac{t^2}{2} +2t + 3](https://tex.z-dn.net/?f=-2sin%28t%29%2B%5Cfrac%7Bt%5E2%7D%7B2%7D%20%2B2t%20%2B%203)
average value of a function of a domain [a, b] is given by the equation
![\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bb-a%7D%20%5Cint%5Climits%5Eb_a%20%7Bf%28x%29%7D%20%5C%2C%20dx)
average of position from t = 2 to t = 5
![\frac{1}{5-2} \int\limits^5_2 {(-2sin(t)+\frac{t^2}{2} +2t + 3)} \, dt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5-2%7D%20%5Cint%5Climits%5E5_2%20%7B%28-2sin%28t%29%2B%5Cfrac%7Bt%5E2%7D%7B2%7D%20%2B2t%20%2B%203%29%7D%20%5C%2C%20dt)
= 2cos(5)+125/6+25+15 - 2cos(2) - 8/6 -4 + 6
=52.999 (use a calculator)
52.999/3 = 17.666
H = height of triangle
b = length of triangle base
A = area of triangle
Recall the formula for area of a triangle :
A = 1/2 bh
Plug in what was given in the problem :
h = b - 5
A = 42
A = 1/2 bh
42 = 1/2 b (b-5)