Find three consecutive even interfere such that their sum is 50 more than the largest interfere

Hm. This is an interesting problem.

Let's see if we can express each of these numbers in terms of x and make an algebra equation to help us solve.

Consecutive even numbers increase by 2.

Let's allow x to equal our first number.

Let's allow x + 2 to equal our 2nd number.

Let's allow x + 4 to equal our 3rd number.

When we add those together, their sum need to equal 50 more than the largest integer (our x + 4)

Let's set it up!

x + (x +2) + (x + 4) = (x + 4) + 50

Simplify!

3x + 6 = x + 54

Let's reorganize the left and right sides.

3x - x = 54 - 6

2x = 48

x = 24 (our 1st even number)

Now, let's check our answer!

24 + 26 + 28 = 78

78 - 50 = 28 Our highest integer.

The sum of consecutive integers 24, 26, 28 is 50 more than the highest integer.

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.