Gabriella volenteers at the hospital every 4th day. she volenteers at the food pantry every 6th day. when will she volenteer at te hospital and the food pantry at the same day?

On every 12th day

Step-by-step explanation:

You must find the least common multiple of her schedules.

One way to find the LCM of 4 and 6 is to use the division method.

Write your numbers (4 and 6) in the top row of a table. Then keep dividing them by prime numbers until you can divide no more.

The LCM is the product of the numbers in the first column.

LCM = 2 ×3 × 3 = 12

Gabriella will be volunteering for both the hospital and the food pantry on Day 12 and every 12th day afterwards.

The number line below shows that the first time her volunteering coincides is Day 12.

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

a is the correct