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KatRina [158]
2 years ago
11

What is the following sum?

Mathematics
1 answer:
Angelina_Jolie [31]2 years ago
4 0

\sqrt[3]{125x^{10}y^{13}  } +\sqrt[3]{27x^{10}y^{13}  }=5\sqrt[3]{x^{10} y^{13} } +3\sqrt[3]{x^{10}y^{13}  } = 8\sqrt[3]{x^{10}y^{13}  }

<h3>How to solve an expression?</h3>

\sqrt[3]{125x^{10}y^{13}  } +\sqrt[3]{27x^{10}y^{13}  }

Therefore,

125 = 5³

27 = 3³

Hence,

5\sqrt[3]{x^{10} y^{13} } +3\sqrt[3]{x^{10}y^{13}  }

Hence,

Both cube root are the same now,

Therefore, we have to add them together.

5\sqrt[3]{x^{10} y^{13} } +3\sqrt[3]{x^{10}y^{13}  } = 8\sqrt[3]{x^{10}y^{13}  }

learn more on expression here; brainly.com/question/19221566

#SPJ1

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Please select the word from the list that best fits the definition
Ann [662]

Answer:

-line graph

Step-by-step explanation:

We are going to take the definition of all the terms and then deduced our correct answer.

<u>Pictograph</u> : Pictograph also known as pictogram uses picture to represent data.

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6 0
3 years ago
PLEASE HELP ASAP AND PLEASE SHOW ME THE WORK YOU DID TOO I NEED THISSS
zmey [24]

Answer:

Jeff has to walk 13.9 m north.

Jeff has to walk 14.3 m futher than Gail.

Step-by-step explanation:

Gail takes a short cut, which is 14.9 m, so Gail moves by the hypotenuse.

Jeff goes the long way. He starts out walking east for 5.3 m, and then turns north, so Jeff mowes by two legs. Let x m be the distance Jeff goes north.

By the Pythagorean theorem,

14.9^2=5.3^2+x^2\\ \\x^2=14.9^2-5.3^2\\ \\x^2=193.92\\ \\x=13.9\ m

So, Jeff has to walk 13.9 m north. In total, the long way is

5.3\ m +13.9\ m=19.2\ m

Jeff has to walk

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futher than Gail.

6 0
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