Answer:
(D)The midpoint of both diagonals is (4 and one-half, 5 and one-half), the slope of RP is 7, and the slope of SQ is Negative one-sevenths.
Step-by-step explanation:
- Point P is at (4, 2),
- Point Q is at (8, 5),
- Point R is at (5, 9), and
- Point S is at (1, 6)
Midpoint of SQ 
Midpoint of PR 
Now, we have established that the midpoints (point of bisection) are at the same point.
Two lines are perpendicular if the slope of one is the negative reciprocal of the other.
In option D
- Slope of SQ

Therefore, lines RP and SQ are perpendicular.
Option D is the correct option.
<span>Garreth and Elisa ran together, Luigi and Jasmine went the same distance, and Logan and Jasmine have the same displacement in the opposite direction. This is the answer which is C</span>
Answer:
$12.50
Step-by-step explanation:
I might be wrong but you can try this answer
Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.
We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)
N = -(sin7A - sinA) + sin5A - sin3A
= -2cos4A*sin3A + 2cos4A*sinA
= 2cos4A(sinA - sin3A)
= 2cos4A*2cos(2A)sin(-A)
= -4cos4A*cos2A*sinA
D = cos7A + cosA - (cos5A + cos3A)
= 2cos4A*cos3A - 2cos4A*cosA
= 2cos4A(cos3A - cosA)
= 2cos4A*(-2)sin2A*sinA
= -4cos4A*sin2A*sinA
Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
= cos2A/sin2A
= cot2A
This verifies the identity.
A) Let x stand for time, y stand for velocity.
We are given the points (2,50), (6, 54). We can make a line using the slope intercept form
y = mx + b.
slope is (54 - 50)/(6-2) = 4/4 = 1
y = 1x + b
plug in point (2,50) to find b
50 = 1(2) + b
50-2 = b
48 = b
the equation is y = 1x + 48
Make standard form.
<span>x - y = -48</span>