The sum of three integers is 12. The sum of the first and second integers exceeds the third by 32. The third integer is 15 less than the first. Find the three integers.

Step-by-step explanation:

So we know the sum of three integers is 12

x+y+z=12

Next we know that the sum of the first and second integer is more than the third by 32

x+y=z+32

Finnaly, we can see that the third is 15 less than the first.

x-15=z

Now time to put it all together.

x+(x-15+32-x)+(x-15)=12

2x-15-15+32=12

2x-30+32=12

2x+2=12

2x=14

x=7

Now time to plug it all in

7 also known as x minus 15 = z

7-15=-8

z=-8

Next we can plug it in the first equation.

7+y-8=12

-1+y=12

y=11

Now time to check if I am correct(spoiler, if it's posted it means I am correct :D)

x=7,y=11,z=-8

7+11-8=12

Yay there is ur answer.

First Integer Is 7, 2nd is 11, third is -8

P.S. Sorry if it is correct, my first answer.

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

the answer will be