Answer: (5.5, -4)
Step-by-step explanation:
You can use the midpoint formula to find the midpoint of point A and point B.

x₁ = 8
x₂ = 3
y₁ = -7
y₂ = -1





Midpoint = (5.5, -4)
Hope this helps!
Answer:

Step-by-step explanation:
Given

Required
Determine the solution
Since b is a perfect square, the equation can be expressed as:

Apply difference of two squares:

Split:

Remove brackets:

Make a the subject in both equations

The solution can be represented as:

Answer: -9.999e+23
I don't know what you're up to.
Answer:
B and D
Step-by-step explanation:


Answer:
2 kids plan to do cross country
Step-by-step explanation: