Five consecutive multiples of 11 have a sum of 220. what is the greatest of these numbers
a. 33
b. 44
c. 55
d. 66

66

Step-by-step explanation:

11,22,33,44, and 55 are 5 consecutive multiples of 11.

11=11(1)

22=11(2)

33=11(3)

44=11(4)

55=11(5)

You can see  consecutive multiples of 11 where we don't know the actual multiples will look like:

11n,11(n+1),11(n+2),11(n+3),11(n+4).

Now we are given the sum of the numbers I just mentioned is 220.

This means,

11n+11(n+1)+11(n+2)+11(n+3)+11(n+4)=220

Each term 11n,11(n+1),11(n+2),11(n+3),11(n+4), and 220 all have a common factor of 11 so divide both sides by 11:

1n+1(n+1)+1(n+2)+1(n+3)+1(n+4)=20

1 times anything is still just that anything:

n+n+1+n+2+n+3+n+4=20

Combine the like terms:

n+n+n+n+n+1+2+3+4=20

Simplify the combining:

5n+10=20

Subtract 10 on both sides:

5n     =10

Divide both sides by 5:

 n      =10/5

Simplify right hand side:

 n      =2

So if n=2, then the multiples of 11 in question look like this:

11n=11(2)=22

11(n+1)=11(3)=33

11(n+2)=11(4)=44

11(n+3)=11(5)=55

11(n+4)=11(6)=66

Add up to see if sum is actually 220.

Putting into my calculator gives me a result of 220.

We are good.

Now you just have to determine what the greatest of the number 22,33,44,55, and 66 is...

The greatest listed here is 66.

its not b

step-by-step explanation:

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.