What is the minimum number of candy pieces that alan has if there are at least 47 pieces of candy among both noah and alan

33

Step-by-step explanation:

Let's just give things vaiables and see how it goes.  let's call the number of candy pieces Alan has A and the ones Noah has N.  now, it tells us A = 5 + 2N, five more than twice Noah's.  it also tells us that A + N is at least 47.  or in other words A + N ≥ 47

Well, we know what A is in terms of N so let's use that instead.

A + N ≥ 47

5 + 2N + N ≥ 47

Now let's solve with algebra.

5 + 2N + N ≥ 47

5 + 3N ≥ 47

3N ≥ 47 - 5

3N ≥ 42

N ≥ 14

Then going back to Alan's candy let's plug 14 in for N, because we know that Noah has to have at least 14

A + N ≥ 47

A + 14 ≥ 47

A ≥ 33

So that tells us Alan must have at least 33.  If there's something you didn't get don't hesitate to ask.  

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

12 cups   is the answer

step-by-step explanation: