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Anika [276]
3 years ago
15

What is the area of a rectangle that is 4 x 4 1/2 ?

Mathematics
2 answers:
Yakvenalex [24]3 years ago
6 0
The area of the rectangle is 18
Klio2033 [76]3 years ago
5 0
4 x 4 1/2 = 8 4/8 = 8 2/4 = 8 1/2
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f(x) + g(x) = 5x - 3 - 2x + 7

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Hope this helps! If you have any questions, feel free to ask.

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Read 2 more answers
Can I please get some help with this
Marizza181 [45]

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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3 years ago
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