Each of the digits 7, 5, 8, 9, and 4 is used only once to form a three-digit integer and a two-digit integer. if the sum of the integers is 555, how many such pairs of integers can be formed?

4 such pairs of integers can be formed.

Step-by-step explanation:

Consider the provided information.

Let the first number is "abc" and the second number is "de".

The sum should be 555.

Let us first find the unit digit of "abc" and "de".

The sum of 8 and 7 end up with 5.

So the number of possible ways are:

ab8+d7 Or ab7+d8

Now we are left with only 3 digits 4, 5 and 9.

The first digit must be 4.

Now the number will look like "4b8+d7" Or "4b7+d8"

b and d can be 5 and 9 or vise-versa.

Now the possible number are: (498, 57), (458, 97), (497, 58) and (457, 98)

Therefore, 4 such pairs of integers can be formed.

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 think u have 12 apples sorry again i don't really know if it correct i'm just tryna hrlp people out.