Find 3 consecutive integers where the sum of the first 2 equals 3 times the third.

-5,-4,-3

Step-by-step explanation:

Let n be an integer.

The next integer would have to be n+1.

The integer before n would have to be n-1.

So the numbers in order are n-1, n , n+1

The sum of the first 2 equals 3 times the third:

n-1     +     n                      =     3(n+1)

2n-1                                 =      3(n+1)

2n-1=3n+3

Subtract 2n on both sides

    -1=n+3

Subract 3 on both sides

    -4=n

If n=-4

then n-1=-5

and n+1=-3

So the three numbers are -5,-4,-3 .

+25 = x

step-by-step explanation:

here x is the hypotenuse of a right triangle whose legs are 7 and 24.   by the pythagorean theorem, 7² + 24² must equal x²:

x² = 49 + 576 = 625.   thus, x = +√625 = +25 = x

on a coordinate grid, point a is at (−4.0, −2.4) and point b is at (4.0, −2.4). point b is a reflection of point a across the –axis. input either a lowercase x or i think is the right awnser

step-by-step explanation: