Answer: -8 and -4
This is something you do through trial and error. Making a list or a table like shown below might help.
Answer:
C) The angles in a square measure 90 degrees. The rhombus has obtuse and
acute angles.
Step-by-step explanation:
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:
Put into slope intercept form, y=mx+b where m=slope
p=-2q+6
slope is -2
subsitute values for q and get values for p
q=0, p=6
(q,p)
(0,6)
(1,4)
p=q+4
slope is 1
subsitutte
q=0,p=4
(q,p)
(0,4)
(1,5)
below are the graphs
Answer:
11- All 0's are on the x and y planes
12- All positive points/numbers are where the x and y axis are both positive (Quadrant 1, 2, and 4)
13- All negative points/numbers would be wherever the x and y axis are negative (Quadrants 2, 3, and 4)