Which type of transformation maps triangle GFH onto triangle G'F'H'?

step-by-step explanation:

so first find out how much each is worth.

quarter= 25 cents

dime= 10 cents

pennies= 1 cent

then multiply each by how much each is worth.

200                     150                   300

x                         x                     x   1

  25                       10              

                                  300

1000                     1500

+                          

4000                                      

5000                                  

then count 2 from the back and add a period.

$50.00

$15.00

$3.00

and then add them

  $50.00

+

  $15.00

+

    $3.00

$68.00

so sunghee has $68.00

read more on -

all but the second one

step-by-step explanation:

cuz that one goes over 180

mark brainliest

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