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Gwar [14]
3 years ago
12

How to factor 169m^2-144?

Mathematics
1 answer:
ikadub [295]3 years ago
6 0
I hope this helps you

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What is the dashed green part of the figure below?
Shtirlitz [24]
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6 0
3 years ago
Since it is given that AB ≅ AC, it must also be true that AB = AC. Assume ∠B and ∠C are not congruent. Then the measure of one a
BartSMP [9]

Answer: The required conclusion is

"if in triangle ABC, AB ≅ AC, then ∠B and ∠C must be congruent".

Step-by-step explanation:  Given that in triangle ABC,  AB ≅ AC, implies that AB = AC must be true. We are given to assume ∠B and ∠C are not congruent. Then the measure of one angle is greater than the other.

If m∠B > m∠C, then AC > AB because of the triangle parts relationship theorem.

For the same reason, if m∠B < m∠C, then AC < AB.

This is a contradiction to the given information.

We are to state the conclusion.

Since in the beginning, it is given that AB ≅ AC and we have assumed that ∠B and ∠C are not congruent, so

our assumption must be wrong.

That is, ∠B and ∠C must be congruent.

Thus, the required conclusion is

if in triangle ABC, AB ≅ AC, then ∠B and ∠C must be congruent.

5 0
3 years ago
Read 2 more answers
What is the approximate value of Log Subscript 6 Baseline 25? 0.233 0.557 0.620 1.796
Sedaia [141]

Answer:

D. on edge

Step-by-step explanation:

8 0
3 years ago
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SOMEONE HELP ME IM FREAKING OUT I LITERALLY CANT WITH THIS QUESTION IM PRAYING PLEASE HELP ME IM SO SERIOUS IM GONNA END IT PLS
antiseptic1488 [7]

Answer:

\sf -11+7\sqrt{2}

Step-by-step explanation:

Given expression:

\sf \dfrac{3-\sqrt{32}}{1+\sqrt{2} }

Rewrite 32 as 16 · 2:

\sf \implies \dfrac{3-\sqrt{16 \cdot 2}}{1+\sqrt{2} }

Apply radical rule \sf \sqrt{a \cdot b}=\sqrt{a}\sqrt{b}

\sf \implies \dfrac{3-\sqrt{16}\sqrt{2}}{1+\sqrt{2} }

As \sf \sqrt{16}=4:

\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} }

Multiply by the conjugate:

\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} } \times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }

\sf \implies \dfrac{(3-4\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}

\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4\sqrt{2}\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}

As \sf \sqrt{2}\sqrt{2}=\sqrt{4}=2:

\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4 \cdot 2}{1-\sqrt{2}+\sqrt{2}-2}

\sf \implies \dfrac{3-7\sqrt{2}+8}{1-2}

\sf \implies \dfrac{11-7\sqrt{2}}{-1}

\sf \implies -11+7\sqrt{2}

7 0
2 years ago
Simplify this expression <br> 7z+5+6z-9
Simora [160]

Answer:

7z + 5 + 6z - 9  (Add 7z and 6z, subtract 9 from 5)

<em>13z - 4  </em>(Can't be simplified further)

8 0
2 years ago
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