I'm assuming you meant to say a^3. If so, then
a^3 = a*a*a
a^3 = 5*5*5
a^3 = 125
The known endpoint is P = (-16,0)
Let Q = (x,y) be the other endpoint. It is unknown for now.
Looking at the x coordinates of P and Q, we see that they are -16 and x respectively. Adding these values up gives -16+x. Dividing that result by 2 gives (-16+x)/2. This result is exactly equal to the midpoint x coordinate, which is the x coordinate of M (0).
So we have this equation (-16+x)/2 = 0. Let's solve for x
(-16+x)/2 = 0
2*(-16+x)/2 = 2*0
-16+x = 0
x-16 = 0
x-16+16 = 0+16
x = 16
Therefore the x coordinate of point Q is 16.
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Let's do something similar for the y coordinates.
The y coordinates of P and Q are 0 and y respectively. Add them up and divided by 2, then set the result equal to -16 (y coordinate of midpoint M) getting this equation (0+y)/2 = -16
Solve for y
(0+y)/2 = -16
y/2 = -16
2*y/2 = 2*(-16)
y = -32
The y coordinate of point Q is -32
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The point Q goes from (x,y) to (16, -32)
Final Answer: (16, -32)
Answer:
you will get back 167$
Step-by-step explanation:
Answer:
The absolute value of 4 is just 4
Step-by-step explanation:
