The answer to this question is 25
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:
Answer:
163/20 or 8.15
Step-by-step explanation:
well we are going to do PEMDAS
parentheses
exponents
multiplication/ division
addition/ subtraction
2 + 2/5 x 23/4 (i made the fractions into decimals)
2 + 6.15
163/20 or 8.15
Answer:
19.5
Step-by-step explanation:
3r+(7+8)/(t-3)
3(4)+(15)/(5-3)
12+(15/2)
12+7.5
19.5