Find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18?

Let, numbers = x,(x+2),(x+4),(x+6)

A.T.Q,

x+x+2+x+4 = x+6+18

3x+6 = x+24

3x-x= 24-6

2x = 18

x = 9

Numbers would be 9, 11, 13, 15

9,11,13,15

Step-by-step explanation:

Let our 4 numbers be

-n

-n+2

-n+4

-n+6

first, we have to write an expression and we gradually simplify.

n+n+n+2+4=n+6+18

combine like terms:

3n+6==n+24

subtract from both parts of the equation.

3n-n=24-6

now are finally finishing our problem.

2n=18

n=9

Now that we know our first integer, we need to consecute.

9,11,13,15

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!