1. given
2. addition property of equality
3. division property of equality
The standard form of a quadratic equation is

, while the vertex form is:

, where (h, k) is the vertex of the parabola.
What we want is to write

as

First, we note that all the three terms have a factor of 3, so we factorize it and write:

.
Second, we notice that

are the terms produced by

, without the 9. So we can write:

, and substituting in

we have:
![\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11]](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20y%3D3%28x%5E2-6x-2%29%3D3%5B%28x-3%29%5E2-9-2%5D%3D3%5B%28x-3%29%5E2-11%5D)
.
Finally, distributing 3 over the two terms in the brackets we have:
![y=3[x-3]^2-33](https://tex.z-dn.net/?f=y%3D3%5Bx-3%5D%5E2-33)
.
Answer:
So the formula is
7*(1/3+1/5) = x
First find the common denominator of the fractions. In this case the common denominator of 3 and 5 is 15 so we must convert them to 15ths. To do that we divide 15 by the denominator and take that answer and multiply the numerator by it.
So for 1/3 we take 15/3(denominator) = 5 and multiply the numerator (1) by it
5 * 1 = 5 so 1/3 converts to 5/15
Do the same for 1/5 we take 15/5 = 3 and multiply the numerator (4) by it
3*4 = 12 so 4/5 converts to 12/15
Now we add 5/15 and 12/15 = 17/15 so we have our formula now to
7*17/15
We make seven a fraction 7/1 and multiply across 7*17 = 119 and 15 * 1 = 15
We have 119/15 which when we divide and get 7 14/15 as your answer
You can use liters, cups, etc i would suggest kiters