Find five consecutive integers such that:
“the sum of the first and 3 times the fourth is equal to 39 less than twice the sum of the second, third, and fifth.”
make sure to list all five integers in your answer and show all of your

  17, 18, 19, 20, 21

Step-by-step explanation:

Let x represent the third (middle) integer. Then the first is (x-2) and the given relation is ...

  (x-2) + 3(x+1) = -39 +2((x-1)+x+(x+2))

  4x +1 = 6x -37 . . . . simplify

  38 = 2x . . . . . . . . . add 37-4x

  x = 19 . . . . . . . . . . divide by 2

The integers are ...

  17, 18, 19, 20, 21

Check

  17 + 3·20 = -39 +2(18 +19 +21)

  17 +60 = -39 +2(58)

  77 = -39 +116 . . . . . yes

the slope of the first function is -3. (3/1)

the slope of the second function is 1. (13-10/6-3).

since |-3| is greater than |1|, the function with the steepest slope is the first.

step-by-step explanation:

the multiplicative inverse of a fraction a/b is b/a. for the multiplicative inverse of a real number, divide 1 by the number. for example, the reciprocal of 5 is one fifth 1/5 or 0.2, and the reciprocal of 0.25 is 1 divided by 0.25, or 4.