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Katen [24]
3 years ago
5

What else needs to be given as congruent in order to

Mathematics
2 answers:
insens350 [35]3 years ago
8 0
D it’s the answer because it sounds right and to be honest I need points
ale4655 [162]3 years ago
6 0

<u>Included sides AE and BE</u>  need to be given as congruent to prove that triangle AEC is congruent to triangle BED by the Angle-Side-Angle (ASA) Congruence Theorem.

According to the Angle-Side-Angle (ASA) Congruence Theorem, if two angles and an included side of a triangle is congruent to corresponding two angles and an included side of another triangle, both triangles can be proven to be equal or congruent to each other.

We are know the following from the given image:

<AEC = <BED (vertical angles are congruent)

<EAC = <EBD (congruent angle)

This implies that two angles (<AEC and <EAC) in triangle AEC are congruent to two corresponding angles (<BED and <EBD) in triangle BED.

Therefore, to prove that both triangles are congruent by ASA, we need to be given that the included sides AE and BE are congruent.

Learn more about Angle-Side-Angle (ASA) Congruence Theorem here:

brainly.com/question/23968808

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Answer:

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Step-by-step explanation:

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12 yellow balls and 9 red balls are placed in an urn. Two balls are then drawn in succession without replacement. What is the pr
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Answer:

\frac{9}{35}

Step-by-step explanation:

When the first draw is done there are 9 red balls in a sample size of 21. So there probability of drawing a red ball will be \frac{9}{21}

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The probability of them happening in this order is the product of both probabilities:

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3 years ago
Simplify:<br>1/1-x^(b-a)+1/1-x^(a-b)​
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Answer:

<h3>1</h3>

Step-by-step explanation:

simplify:

1/1-x^(b-a)+1/1-x^(a-b)​

\frac{1}{1-x^{b-a}} + \frac{1}{1-x^{a-b}}\\= \frac{1-x^{a-b} + 1-x^{b-a}}{(1-x^{b-a})(1-x^{a-b})}\\=  \frac{1-x^{a-b} + 1-x^{b-a}}{(1-x^{b-a}-x^{a-b}+x^{b-a+a-b})}\\= \frac{1-x^{a-b} + 1-x^{b-a}}{(1-x^{b-a}-x^{a-b}+1)}\\= \frac{2-x^{a-b}-x^{b-a}}{2-x^{b-a}-x^{a-b}}\\= 1\\ \\

Hence the required answer is 1

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3 years ago
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