Polygon ABCD is dilated by a scale factor of 2 with the center of dilation at the origin to create polygon A′B′C′D′. If the endpoints of line AB are located at (0, -7) and (8, 8), what is the length of line A'B'? Use the distance formula to help you decide (picture added)

A'B' = 34 units

Step-by-step explanation:

Since the dilatation is centred at the origin , then multiply the coordinates by 2

A' = (2(0), 2(- 7) = (0, - 14 )

B' = (2(8), 2(8) = (16, 16)

Calculate the length using the distance formula

d =

with (x₁, y₁ ) = A'(0, - 14) and (x₂, y₂ ) = B'(16, 16)

d =

   =

   =

   =

   = 34

78.59 is the mean

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