The volume Vequalsfour thirds pi r cubed of a spherical balloon changes with the radius. a. At what rate (ftcubed/ft) does the volume change with respect to the radius when r equals 2 ft question mark b. Using the rate from part a, by approximately how much does the volume increase when the radius changes from 2 to 2.4 ft question mark

a)

b) Approximate error of the volume increase when the radius changes from 2 to 2.4 ft = 0.6 that is

Step-by-step explanation:

Explanation:-

a) Given Volume of spherical balloon

        …(i)

Differentiating with respective to the radius 'r'

At r=2

b) Step(1):-

Given Volume of spherical balloon

we property of logarithmic log(ab) = log a +log b

taking logarithmic on both sides , we get

…(ii)

Step2:-

we will use  

δ v = 0 + δ r

Given data  radius changes from 2 to 2.4 ft

r+ δ r = 2.4 =2+0.4

here r =2 and δ r =0.4

Now δ v =  

δ v = 0.6

Approximate error in Volume = 0.6 cubic/ft

Approximate error of the volume increase when the radius changes from 2 to 2.4 ft = 0.6 that is

y-1700=200(x+4)

step-by-step explanation:

200(x+4) is 200x + 800 she is saving 200 per week x is the weeks and the -1700 is. us that’s howmuch she already had

step-by-step explanation:

your answer is a