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gregori [183]
2 years ago
6

PLEASE SOLVE I WILL MARK BRAINLIEST HURRY PLEASE

Mathematics
1 answer:
suter [353]2 years ago
3 0

Applying the Pythagorean theorem, the missing lengths in the right triangles are:

1. √19 ≈ 4.4

2. √96 ≈ 9.8

3. √60 ≈ 7.7

4. √231 ≈ 15.2

5. √21 ≈ 4.6

6. √24 ≈ 4.9

<h3>What is the Pythagorean Theorem?</h3>

The Pythagorean theorem is a formula that can be used to find the leg of a  right triangle, and is given as: c² = a² + b², where a and b are the smaller legs and c is the longest leg/hypotenuse.

1. missing leg = √(10² - 9²) = √19 ≈ 4.4

2. missing leg = √(11² - 5²) = √96 ≈ 9.8

3. missing leg = √(8² - 2²) = √60 ≈ 7.7

4. missing leg = √(16² - 5²) = √231 ≈ 15.2

5. missing leg = √(5² - 2²) = √21 ≈ 4.6

6. missing leg = √(7² - 5²) = √24 ≈ 4.9

Learn more about the Pythagorean theorem on:

brainly.com/question/343682

#SPJ1

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nata0808 [166]

To simplify

\sqrt[4]{\dfrac{24x^6y}{128x^4y^5}}

we need to use the fact that

\sqrt[4]{x^4}=|x|

Why the absolute value? It's because (-x)^4=(-1)^4x^4=x^4.

We start by rewriting as

\sqrt[4]{\dfrac{2^23x^6y}{2^6x^4y^5}}

\sqrt[4]{\dfrac{2^23x^4x^2y}{2^42^2x^4y^4y}}

Since x\neq0, we have \dfrac xx=1, and the above reduces to

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

Then we pull out any 4th powers under the radical, and simplify everything we can:

\dfrac1{\sqrt[4]{2^4y^4}}\sqrt[4]{\dfrac{3x^2y}{y}}

\dfrac1{|2y|}\sqrt[4]{3x^2}

where y>0 allows us to write \dfrac yy=1, and this also means that |y|=y. So we end up with

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

making the last option the correct answer.

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3 years ago
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Step-by-step explanation:

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Part B:
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