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.

in 4 days he will run   12/14

step-by-step explanation:

b

step-by-step explanation:

a difference is the result of subtraction. while each of these expressions has subtraction in it, a difference will be the result when subtraction is the last operation applied in the order of operations for the expression. this means subtraction must be outside of the parenthesis. only b has this.