Caleb and Eva are running around a track. They both start at the same time. It takes Caleb 18 minutes to complete 1 lab. It takes Eva 14 minutes to complete 1 lab. How many labs will each of them have run before they meet at the start again?

For x=7 and y=9 Caleb and Eva will meet again. That is, Caleb will have made 7 labs and Eva 9 labs.

Step-by-step explanation:

In order to find the number of labs that Caleb and Eva must complete, to meet again at the start, you can use the following equation:

     (1)

where x is the number of labs made by Caleb and y the number of labs made by Eva. The equation (1) means that for the specific values of x and y, Caleb and Eva will spend the same amount of minutes on the labs.

Then, you need to find the least integers values of x and y.

First, you solve the equation (1) for y:

     (2)

By inspection you can find that the least value of x that makes possible that y is an integer, is 7

In fact, you have:

Hence, for x=7 and y=9 Caleb and Eva will meet again. That is, Caleb will have made 7 labs and Eva 9 labs.

(-1,-0.5)

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