Please answer please and what do you do to solve this? 9+4(a+5)
a) 18+4a B)29+4a
C) 33+4a D) 23+4a

B) 29 +4a

Step-by-step explanation:

for this question we need to expand the brackets. to do this you would times everything in the brackets separately by 4, 4 times a which is 4a and 4 times 5 which is 20.

this would creat a sum of 9+4a+20

now we can add together the like terms which in this case are 20 and 9 making the whole sum 29 +4a

this is how you expand the brackets.

deserving of a brainliest?

B) 29+4a

Step-by-step explanation:

Distribute the parentheses first as multiplication comes before addition:

9+4(a+5)

9+4a+20

Add 9 and 20:

29+4a

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

i wish i could u

step-by-step explanation:

i wish i could u