Answer:
Yes it does answer A_____
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:
Step-by-step explanation:
given
a = - 3
3a² = 3 * ( -3) ²
= 3 * 9
= 27
The value of 3a² is 27.
Answer:
second option, IQR = 13-6 = 7 and Range = 17-6 = 11
Step-by-step explanation:
The area of the trapezoid is given by:
A = (1/2) * (b1 + b2) * (h)
Where,
b1, b2: bases of the trapezoid
h: height
Substituting values we have:
91 = (1/2) * ((2 * 7) + b2) * (7)
Rewriting we have:
91 = (1/2) * (14 + b2) * (7)
(2/7) * 91 = 14 + b2
b2 = (2/7) * 91 - 14
b2 = 12 m
Answer:
The measure of the other base of the trapezoid is:
b2 = 12 m