The sum of 3 consecutive even integers is 50 more than the smallest one. find the integers.

Let's imagine the integers are a, b, and c. Since we know the sum of these numbers is 50 more than the smallest (which is a), our equation would look like this:

a + b + c = a + 50

We need a common variable, though, which we don't have right now. Since we know that these are 3 consecutive even integers, we know that:

b = a +2, which means that the second number is 2 more than the first number.
Using that same logic, we know that c = b +2 and since we know what b equals, we can sub that into this:

c = b + 2
c = (a + 2) + 2
c = a +4

So now instead of having this:

a + b + c = a + 50 

We have this:

a + (a + 2) + (a + 4) = a + 50
3a + 6 = a + 50

Subtract a + 6 from both sides:

3a + 6 - (a + 6) = a + 50 - (a + 6)
2a = 44

Divide both sides by 2

2a/2 = 44/2
a = 22

The smallest integer is 22 which means the set would be:
22, 24, and 26

C. i believe it is (5×(-20))+(100×(0.7))