Prove: every point on the perpendicular bisector of a segment is equidistant from the ends of the segment.

ap = bp

(fill in the blanks of the equation in the second picture with the correct number/letter/sign based off the first picture.)

Answer with explanation:

Here ,let line, x=0 intersect ,segment AB at point M.

So, PM ⊥ AB.

And, AM =BM →→→∵ y axis or segment PM , is Perpendicular bisector of segment AB.

In Δ AMP and ΔB MP

∠AMP = ∠ B MP=90°[→→ Each being 90°]

AM = BM →Line PM , is perpendicular Bisector.

Side MP , is Common.

Δ AMP ≅ ΔB MP→→→[S AS]

So, AP= BP →→[C P C T]

2. We will use distance formula to find AP and BP.

Distance between two points is given by

Hence, AP = BP

Where ,Point P, can be located anywhere on y axis.

based on context clues you gave i'm assuming the answer would be c.

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)