If I’m assuming that symbol next to the C at the end of the question means C mirrored the the answer would be (3,1) but I’m not quite sure what it means
Answer:
- B. Quadrilateral FLAG is a parallelogram because opposite sides are parallel.
Step-by-step explanation:
<em>Refer to attached.</em>
Plot the point and connect in order FLAG.
<u>We see </u>
- FL ║ GA
- GF ║ AL
- FA and GL are not perpendicular
<u>All the above can be confirmed by calculating the slopes:</u>
- m(FL) = (3 - 5)/(6 - 1) = -2/5
- m(GA) = (1 + 1)/(-2-3) = -2/5
- These are same.
- m(GF) = (5 - 1)/(1 + 2) = 4/3
- m(AL) = (3 + 1)/(6 - 3) = 4/3
- These are same
- m(FA) = (-1 - 5)/(3 - 1) = -3
- m(GL) = (3 - 1)/(6 +2) = -1/4
- These are negative but not reciprocal
The quadrilateral is a parallelogram.
Correct choice is B
-2/3x+9=4/3x-3 original problem add 2/3x to both sides
9=6/3x-3 simplify 6/3 to 2 and add 3 to both sides
12=2x divide both sides by 2
x=6
hope that helps
Answer:
10.95 (if you need to round 11)
Step-by-step explanation:
:p
We know that the area of a parallelogram is the base * the height of it, so if we divide both sides by the height, then area/height=base. Therefore, we must divide the area by the height. To divide using polynomials, we first set it up similar to a regular long division problem:
______________________
2x+3 | 2x²+13x+15
Next, we take the first component of the numerator (2x² in this case) and divide it by the first component of the denominator (2x) to get x. That will form the start of our answer, and at the bottom, we will subtract our numerator by the denominator (2x+3) multiplied by the start of our answer (x). Therefore, we have
_x_____________________
2x+3 | 2x²+13x+15
-(2x²+3x)
_______________
10x+15
We then repeat the process until we finish, and whatever's left at the top is our answer. If there's something left, that's our remainder.
_x+5_____________________
2x+3 | 2x²+13x+15
-(2x²+3x)
_______________
10x+15
- (10x+15)
_________________
0
Therefore, our base has a length of x+5.
Feel free to ask further questions, and Happy Holidays!
