Fthe height remains fixed and the side of the base is decreasing by 0.002 meter/yr, what rate is the volume decreasing when the height is 180 meters and the width is 200 meters?

The original volume equation looks like this: V = 1/3 * h * (x^2) 
After the side is reduced by 0.002, the new volume would look like V1 = 1/3 * h * (x-0.002) ^ 2 

Then we have: 
V-V1 = 1/3*h*(x^2) - 1/3*h*(x – 0.002) ^2 

= 1/3 * h *(x^2 - (x – 0.002) ^2) 

= 1/3 * h * (0.004x - 0.00004) 

The rate of decreasing is computed by:

(V-V1)/V * 100% = [1/3 * h *(0.004x - 0.00004)] / [1/3 * h * (x ^ 2)] * 100% this would be equal to (0.004x - 0.00004) / (x^2) * 100% 

So replace x by 200, you’ll get:

(0.004(200) - 0.00004) / (200^2) * 100% 

= 0.001999% is the rate of decreasing.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: