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3 years ago

Which diagram could be used to prove △ABC ~ △DEC using similarity transformations?

Mathematics

2 answers:

NeX [460]3 years ago

6 0

The answer is A just took the test.

trapecia [35]3 years ago

4 0

Answer:

Step-by-step explanation:

(A) From the given figure, ∠BAC=∠CDE and ∠ACB=∠DCE=90°.

Thus, from ΔABC and ΔDEF, we have

∠BAC=∠CDE (Given)

∠ACB=∠DCE=90°

Thus, by AA similarity rule,

ΔABC is similar to ΔDEF.

Therefore, this option is correct.

(B) From the given figure, ∠ABC=∠DCE and ∠ACB=∠DEC

But, there is no angle and side is similar to the other angle or side of the triangle. Thus, this option is not correct.

But, there is no angle or side is similar to the other angle or side of the triangle. Thus, this option is not correct.

(D) From the given figure, ∠BAC=∠CDE

But, there is no another angle or side is similar to the other angle or side of the triangle. Thus, this option is not correct.

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3 years ago

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Answer:

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Step-by-step explanation:

Given:

The expression to simplify is given as:

$\frac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}$

In order to simplify this, we have to use the law of indices.

1. $(ab)^m=a^mb^m$

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Substitute this value in the above expression. This gives,

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Now, we use another law of indices.

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Substitute these values in the above expression. This gives,

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