For the triangle the value for w would be 11. Im not 100% sure for the other ones though. Sorry
A statement which proves that the diagonals of square PQRS are perpendicular bisectors of each other is: option D.
<h3>How to calculate the slope of a line?</h3>
Mathematically, the slope of a line is given by the following formula;

For line RP, we have:

Slope RP = 7.
For line SQ, we have:

Slope SQ = negative one-sevenths.
For the midpoint, we have:
In order to determine the midpoint of a line segment with two (2) endpoints, we would add each point together and divide by two (2).
Midpoint on x-coordinate = (8 + 1)/2 = 9/2 = 4.5.
Midpoint on y-coordinate = (9 + 2)/2 = 11/2 = 5.5.
In conclusion, a statement which proves that the diagonals of square PQRS are perpendicular bisectors of each other is that the midpoint of both diagonals is (4.5, 5.5), the slope of RP is 7, and the slope of SQ is negative one-sevenths.
Read more on squares here: brainly.com/question/2882032
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Yes it is equal
75/100=15/20=3/4
Answer:
(a) m= 3
(b) b= -2
(c) y=3x-2 or y=3x+(-2)
Step-by-step explanation:
(a) to find the slop you do rise over run or change in y over change in x (from point to point it goes up 3 over 1 so the slope would be 3/1 = 3)
(b) the y intercept is wherever the line intercepts (touches/crosses) the y (vertical/up and down)axis
(c) the equation formula is y=mx+b where m = the slope and b = y intercept. All you have to do is plug in the information to get your answer. You can also either add a negative or just subtract which is why there are two viable optional answers
hmmm we know B (2 , 1), and let's say the y-coordinate is "y" for A, so A (-10 , y).
![~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad A(\stackrel{x_2}{-10}~,~\stackrel{y_2}{y})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BA=\sqrt{[-10 - 2]^2 + [y - 1]^2}\implies 13=\sqrt{(-12)^2+(y^2-2y+1)} \\\\\\ 13^2=(-12)^2+(y^2-2y+1)\implies 169=144+y^2-2y+1 \\\\\\ 169=y^2-2y+145\implies 0=y^2-2y-24 \\\\\\ 0=(y-6)(y+4)\implies y= \begin{cases} 6\\ -4 \end{cases}~\hfill A= \begin{cases} (-10~,~6)\\ (-10~,~-4) \end{cases}](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20B%28%5Cstackrel%7Bx_1%7D%7B2%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5Cqquad%20A%28%5Cstackrel%7Bx_2%7D%7B-10%7D~%2C~%5Cstackrel%7By_2%7D%7By%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20BA%3D%5Csqrt%7B%5B-10%20-%202%5D%5E2%20%2B%20%5By%20-%201%5D%5E2%7D%5Cimplies%2013%3D%5Csqrt%7B%28-12%29%5E2%2B%28y%5E2-2y%2B1%29%7D%20%5C%5C%5C%5C%5C%5C%2013%5E2%3D%28-12%29%5E2%2B%28y%5E2-2y%2B1%29%5Cimplies%20169%3D144%2By%5E2-2y%2B1%20%5C%5C%5C%5C%5C%5C%20169%3Dy%5E2-2y%2B145%5Cimplies%200%3Dy%5E2-2y-24%20%5C%5C%5C%5C%5C%5C%200%3D%28y-6%29%28y%2B4%29%5Cimplies%20y%3D%20%5Cbegin%7Bcases%7D%206%5C%5C%20-4%20%5Cend%7Bcases%7D~%5Chfill%20A%3D%20%5Cbegin%7Bcases%7D%20%28-10~%2C~6%29%5C%5C%20%28-10~%2C~-4%29%20%5Cend%7Bcases%7D)