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blagie [28]
3 years ago
12

Would you please help me solve for x it should be a degrees but I’m not sure what it is

Mathematics
1 answer:
Leno4ka [110]3 years ago
6 0

Answer:

x=15º

Step-by-step explanation:

A triangle's angles add up to 180º.

∠A=30º

∠C=180º-45º

∠B=xº

To find ∠B we have to first find ∠C.

Since ∠C does not have an angle, knowing that a straight line is 180º we can subtract it from 45º to find ∠C.

180-45=135

∠C=135º

Now add ∠A and ∠C.

30+135=165

Subtract the sum from 180º

180-165=x

x=15º

Hope this helps :)

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After solving \sqrt[5]{128x^8y^2} we get 2x\sqrt[5]{4}\sqrt[5]{x^3}\sqrt[5]{y^2}

Step-by-step explanation:

We need to solve \sqrt[5]{128x^8y^2}

Applying the rule: \sqrt[a]{xy}=\sqrt[a]{x}.\sqrt[a]{y}

\sqrt[5]{128x^8y^2}\\=\sqrt[5]{128}\sqrt[5]{x^8}\sqrt[5]{y^2}\\We\,\,know\,\,that\,\,128=2\times2\times2\times2\times2\times2\times2=2^7\,\,or\,\,2^5.2^2\\=\sqrt[5]{2^5.2^2}\sqrt[5]{x^5.x^3}\sqrt[5]{y^2}\\=(2^5)^{\frac{1}{5}}\sqrt[5]{2^2}\sqrt[5]{x^5}\sqrt[5]{x^3}\sqrt[5]{y^2}\\=2x\sqrt[5]{4}\sqrt[5]{x^3}\sqrt[5]{y^2}

So, After solving \sqrt[5]{128x^8y^2} we get 2x\sqrt[5]{4}\sqrt[5]{x^3}\sqrt[5]{y^2}

Keywords: Solving with Exponents

Learn more about Solving with Exponents at:

  • brainly.com/question/4934417
  • brainly.com/question/13174254
  • brainly.com/question/13174255

#learnwithBrainly

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