Ludovica creates flashcards for her vocabulary words. she creates 3 more cards per verb than per noun. ludovica creates 2 fewer than twice as many flashcards per verb as she creates per noun.
how many flashcards does ludovica create per noun and how many per verb?

5 per noun, 8 per verb

Step-by-step explanation:

start by defining what each variable is

v = n+3

v = 2n-2

substitute v for one of the equations above

n+3 = 2n-2

subtract n from both sides

3 = n-2

add 2 to each side

5 = n

so there are 5 flashcards per noun

substitute 5 in for n in both equations

v = 5+3

and

v = 2(5)-2

solve both equations

v = 8 and v = 8

so there are 8 flashcards per verb

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

3 is greater then both of those numbers