The area of a square is increasing at the constant rate of 16sq. ft./min. When the perimeter of the square is 36ft, how fast is the perimeter of the square increasing?

The perimeter of square is increasing by 3.76ft/min and then by 3.4 ft/min.

Step-by-step explanation:

Given that area of square is increasing at a rate of 16 sq ft/min.

Given that final perimeter is 36ft

Perimeter of a square = 4 side = 36

So, side, a' = 9 ft

We know that area of a square is given by the formula:

(If we let side = a units)

Change in area =

So, side got changed from 8.06ft to 9 ft.

So, perimeter when side was 8.06 ft:

Hence, increase in the perimeter when perimeter is 36 ft is = 36 - 32.24 = 3.76 ft

For finding Next increase:

area gets changed from 81 sq ft to 81+16 = 97 sq ft

So, new side = ft = 9.85 ft

Next increase in perimeter = 4 (New side - Old side)

= 4 (9.85 - 9)

= 3.4 ft/min

she will most likely have 33  

step-by-step explanation:

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2