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.

2.

according to newton’s cannonball model, what happens if the cannonball's velocity is enough to overpower the pull of earth’s gravitystep-by-step explanation:

207/8 is the right