Ok, this is what you do,
7(^^^^^
84
7 goes into 8 1 time so it would be this
1
7(^^^^
84
- 7
and it leaves you with 14
1
7(^^^^
14
then how many time does 7 go into 14? 2 times
so it would be
12
7(^^^^
14
-14
AND YOUR ANSWER IS 12.
Answer:
13 miles
Step-by-step explanation:
The shortest distance would be a straight line between the two points. This would be a hypotenuse of a right triangle
We can use the Pythagorean theorem to determine the length
a^2+b^2=c^2
5^2 +12^2 = c^2
25+144= c^2
169 = c^2
Taking the square root of each side
sqrt(169) = sqrt(c^2)
13 = c
Answer:

Step-by-step explanation:
In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).
The formula we will use for this problem is the following:

where:


a=0

so the volume becomes:

This can be simplified to:

and the integral can be rewritten like this:

which is a standard integral so we solve it to:
![V=9\pi[tan y]\limits^\frac{\pi}{3}_0](https://tex.z-dn.net/?f=V%3D9%5Cpi%5Btan%20y%5D%5Climits%5E%5Cfrac%7B%5Cpi%7D%7B3%7D_0)
so we get:
![V=9\pi[tan \frac{\pi}{3} - tan 0]](https://tex.z-dn.net/?f=V%3D9%5Cpi%5Btan%20%5Cfrac%7B%5Cpi%7D%7B3%7D%20-%20tan%200%5D)
which yields:
]
Answer:
9+11s
Step-by-step explanation:
9+18s-7s
=9+11s
You can't add 9 and 11s because they are different variables
We keep the 2x^2 because we can only subtract something from that if it also is squared. From there, we subtract the 7x from 2x^2-11. Since there is no previous x's in the 2x^2-11, we just make it -7x. So, without even continuing the problem, we see that A) is the correct answer because it is the only one with -7x.