Jen butler has been pricing speed-pass train fares for a group trip to new york. three adults and four children must pay $ 99. two adults and three children must pay $ 70. find the price of the adult's ticket and the price of a child's ticket.

An adult ticket is 17

A childs ticket costs 12 dollars

Step-by-step explanation:

Let a = adult price

c = child price

3a + 4c = 99

2a + 3c = 70

Multiply the first equation by 2

2(3a + 4c) = 99*2

6a + 8c = 198

Multiply the second equation by -3

-3 (2a + 3c) = 70*-3

-6a -9c = -210

Add these new equations together

6a + 8c = 198

-6a -9c = -210

     -c = -12

Multiply by -1

        c = 12

A childs ticket costs 12 dollars

Now we need to find the price of an adult ticket

2a + 3c = 70

2a + 3(12) = 70

2a +36 = 70

Subtract 36 from each side

2a +36-36 =70-36

2a = 34

Divide by 2

2a/2 = 34/2

a = 17

An adult ticket is 17

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

kk.bj.

step-by-step explanation:

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