Let A(t)A(t) be the area of a circle with radius r(t)r(t), at time tt in min. Suppose the radius is changing at the rate of drdt=3drdt=3 ft/min. Find the rate of change of the area at the moment in time when r=6r=6 ft.

At a point when r = 6 ft

The rate of change of Area with time is given as

(dA/dt) = 36π = 113.1 ft²/min

Step-by-step explanation:

Area of a circle, A = πr²

A = πr²

(dA/dt) = (dA/dr) × (dr/dt)

(dA/dr) = 2πr, (dr/dt) = 3

(dA/dt) = 2πr × 3 = 6πr

At a point when r = 6 ft

(dA/dt) = 6πr = 6π × 6 = 36π = 113.1 ft²/min

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!