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3 years ago

What are the missing parts that correctly complete the proof

Mathematics

1 answer:

bonufazy [111]3 years ago

5 0

Answer:

2. $\overline{AX}\cong \overline{BX}$

3. $PX \perp AB$ - definition of perpendicular

4. $\angle PXA \cong \angle PXB$ - all right angles are congruent

6. $\triangle AXP\cong \triangle BXP$

7. $\overline{PA} \cong \overline{PB}$

Step-by-step explanation:

Given: Point P is the perpendicular bisector of AB

Prove: P is equidistant from the endpoints AB

Proof.

1. Point P is on the perpendicular bisector of AB - given

2. $\overline{AX}\cong \overline{BX}$ - definition of bisector

3. $PX \perp AB$ - definition of perpendicular

4. $\angle PXA \cong \angle PXB$ - all right angles are congruent

5. $\overline{PX} \cong \overline{PX}$ - reflexive property of congruence

6. $\triangle AXP\cong \triangle BXP$ - SAS congruency postulate

7. $\overline{PA} \cong \overline{PB}$ - corresponding parts of congruent triangles are congruent

8. Point P is equidistant from the endpoints of AB - definition of equidistant

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Prove the following statement.

jolli1 [7]

<h2>Step-by-step explanation:</h2>

As per the question,

Let a be any positive integer and b = 4.

According to Euclid division lemma , a = 4q + r

where 0 ≤ r < b.

Thus,

r = 0, 1, 2, 3

Since, a is an odd integer, and

The only valid value of r = 1 and 3

So a = 4q + 1 or 4q + 3

Case 1 :- When a = 4q + 1

On squaring both sides, we get

a² = (4q + 1)²

= 16q² + 8q + 1

= 8(2q² + q) + 1

= 8m + 1 , where m = 2q² + q

Case 2 :- when a = 4q + 3

On squaring both sides, we get

a² = (4q + 3)²

= 16q² + 24q + 9

= 8 (2q² + 3q + 1) + 1

= 8m +1, where m = 2q² + 3q +1

Now,

We can see that at every odd values of r, square of a is in the form of 8m +1.

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3 years ago

Drag values to complete each equation.

Alex777 [14]

Answer:

1. A. 1

2. D. $17^9$

Step-by-step explanation:

Properties used:

Power of a Power Property
Product Property
Quotient Property
Zero Exponent Property

1. First, let's deal with the numerator.

$(17^3)^6$ can be turned into $17^1^8$ by using the Power of a Power Property.

And then use the Product Property, $(17^1^8)(17^-^1^0) = 17^8$

So now, our fraction is this: $\frac{17^8}{17^8}$

All number over itself in a fraction is equal to 1. But you can also do this the mathmatic way using the Quotient Property: $\frac{a^m}{a^n} = a^m^-^n$ or $\frac{1}{a^(^n^-^m^)}$ . Which then you plug the numbers in: $\frac{17^8}{17^8} = 17^8^-^8 = 17^0$ . And since we know that in Zero Exponent Property: $a^0 = 1$ , we can see that $17^0 = 1$ . So either way, we get 1.