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3 years ago

1.) Prove that triangles ABC and EDC are congruent.

2 answers:

6 0

I wrote my STATEMENTS and REASONS as follows:
STATEMENT: 1. Line segments AE & DB intersect at C.
REASON: 1. Given.
STATEMENT: 2. Line segments AC & EC are congruent.
REASON: 2. Given.
STATEMENT: 3. Line segments BC & DC are congruent.
REASON: 3. Given.
STATEMENT: 4. Angles ACB & ECD are congruent.
REASON: 4. Vertical angles are congruent (Theorem) VERTICALLY OPPOSITE ANGLES ARE EQUAL..THAT IS THE CORRECT STATEMENT.
STATEMENT: 5. Triangles ABC & EDC are congruent.
REASON: 5. (SSS are congruent to SSS). If three sides of one triangle are congruent, respectively, to three sides of a second triangle, then the triangles are congruent. (Postualte) ...NO THEOREM IS SAS..2 SIDES AND INCLUDED REPEAT INCLUDED ANGLES ARE EQUAL RESPECTIVELY THE CORRESPONDING SIDES AND INCLUDED ANGLE , THEN THE 2 TRIANGLES ARE CONGRUENT.
I'm not sure if Step 4 is correct or if I can even write that statement.IT IS OK BUT FOR THE SLIGHT MODIFICATION SUGGESTED.
REASON 5 IS TO BE CHANGED AS GIVEN .

Simora [160]3 years ago

3 0

C is the answer of number one

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Step-by-step explanation:

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2 years ago

What is the simplest form of ^4 sqrt 81x^8y^5

lara31 [8.8K]

ANSWER

$3 {x}^{2} y \sqrt[4]{y}$

EXPLANATION

We want to simplify:

$\sqrt[4]{81 {x}^{8} {y}^{5} }$

We can split the radical sign to obtain:

$\sqrt[4]{81} \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{5} }$

$\sqrt[4]{81} \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{4} \times y}$

$\sqrt[4]{81} \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{4}} \times \sqrt[4]{y}$

$\sqrt[4]{ {3}^{4} } \times \sqrt[4]{ {x}^{8} } \times \sqrt[4]{ {y}^{4}} \times \sqrt[4]{y}$

Recall that:

$\sqrt[n]{ {a}^{m} } = {a}^{ \frac{m}{n} }$

${3}^{4 \times \frac{1}{4} } \times {x}^{8 \times \frac{1}{4} } \times {y}^{4 \times \frac{1}{4} }\times \sqrt[4]{y}$

$3 {x}^{2} y \sqrt[4]{y}$

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Read 2 more answers

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Answer:

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Step-by-step explanation:

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