There are two numbers that are twice as far away from 13 than 22. The sum of those numbers are?

Sum of those numbers are 50

Step-by-step explanation:

Let x represent a number that is twice as far away from 13 than 22.

The number can't be less than 13 but will be between 13 and 22 or slightly more than 22 otherwise it will be far away from both.

Thus;

|x - 13| = 2|x - 22|

It's in absolute form, so we will square both sides to get rid of the absolute brackets.

Thus;

(x - 13)² = 4(x - 22)²

x² - 26x + 169 = 4(x² - 44x + 484)

x² - 26x + 169 = 4x² - 176x + 1936

4x² - x² - 176x + 26x + 1936 - 169 = 0

3x² - 150 + 1767 = 0

Using quadratic formula, we have;

x = 19 or x = 31

Thus,the sum of the 2 numbers = 31 + 19 = 50

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240