Let f(x)=25x2−2 . the function g(x) is a vertical stretch of f(x) by a factor of 2. what is the equation of g(x)? enter your answer in the box. g(x) =

We are given function f(x) = 25x^2 -2.

g(x) is a vertical stretch of f(x) by a factor of 2.

We know, the transformation rule

y= k f(x), vertical stretch by a factor k. For k>0.

Because function f(x) is being vertical stretched by a factor of 2. So, we need to multiply f(x) function by 2.

Therefore, g(x) = 2 f(x) = 2*(25x^2-2).

Therefore, g(x) = 2*(25x^2-2) or g(x) = 50x^2-4.

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

(a) the volume of the cone is 20/3 in3

hence, option a is correct.