Answer is B 1, -1
y = -2x + 5
when x = 2; y = -2(2) + 5 = 1
when x = 3; y = -2(3) + 5 = -1
Hi Vanessa
3x -1/9 (27) =18
3x - 27/9 =18
3x- 3 =18
Add 3 to both sides
3x-3+3=18+3
3x=21
Divide both sides by 3
3x/3= 21/3
x= 7
The value of x is 7
Now let's check if my answer is correct
To check it we gonna replace x by 7 and 27 for y
(3)(7) -1/9 (27) = 18
21 -1/9 (27)=18
21- 27/9 = 18
21- 3 = 18
18 = 18
The answer is good and I hope its help:0
8/12: 2/3 and 16/24
6/8: 3/4 and 12/16
9/15: 3/5 and 18/30
2/16: 1/8 and 4/32
Answer:
No, it is not a square
Step-by-step explanation:
If one wall is 19", that would mean the wall perpendicular to this wall is also 19" (in fact all of the walls would be 19"!) If this was a square, then the diagonal we draw at 20.62" would serve as the hypotenuse of a right triangle. One wall would serve as a leg, and another wall as another leg. If this is a square, then the Pythagorean's Theorem would be satisfied when we plug in the 2 wall measures for a and b, and the diagonal for c:

We need to see if this is a true statement. If the left side equals the right side, then the 2 legs of the right triangle are the same length, and the room, then is a square.
361 + 361 = 425.1844
Is this true? Does 722 = 425.1844? Definitely not. That means that the room is not a square.
I don't know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse.
(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently.)

Where

is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is:

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis.
The major axis in this case is that of the y-axis. In other words,
So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are:

&

I really hope this helped you! (Partially because I spent a lot of time on this lol)
Sincerely,
~Cam943, Junior Moderator