Find the indicated values where g(t)=t^2-t and f(x)=1+x g(f(0))+f(g(0))

1

Step-by-step explanation:

First find f(0) and g(0). These are the values where x=0 in each function.

f(0) = 1+0 = 1

g(0) = 1^2 - 1 = 1-1 = 0

So f(0) = 1 and g(0) = 0.

Now substitute f(0) = 1 into g(t).

g(1) = 1^2 -1 = 1-1 = 0.

So g(f(0)) = 0.

Now substitute g(0) = 0 into f(t).

f(0) = 1 + 0 = 1.

So f(g(0)) = 1.

Add the values 0 and 1 to get 0+1 = 1.

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

answer: its c because when it has a magnet next to it it will move away from the mixture. hope this

step-by-step explanation: