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creativ13 [48]
3 years ago
7

If a is a nonnegative real number and n is a positive integer, then . a.true b.false

Mathematics
2 answers:
pishuonlain [190]3 years ago
8 0
I added a screenshot of the complete question.

<u><em>Answer:</em></u>
True

<u><em>Explanation:</em></u>
<u>Rules of the root are as follows:</u>
\sqrt[n]{x} =  x^{ \frac{1}{n}}  \\  \sqrt[n]{x} * \sqrt[n]{y} = \sqrt[n]{xy}  \\  \sqrt[n]{ \frac{x}{y}} =   \frac{ \sqrt[n]{x} }{ \sqrt[n]{y} }

<u>Now the given is:</u>
\sqrt[n]{a} =  a^{ \frac{1}{n}}
This rule is the same as the first rule mentioned above, therefore, it is correct.
We use this rule to convert roots into powers which facilitates solving the problems given

Hope this helps :0

HACTEHA [7]3 years ago
5 0

Answer:

Just like the other guy said, the answer is true

Step-by-step explanation:

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3 0
3 years ago
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Please help I have no idea what I’m doing
Brums [2.3K]

Answer:

=12b^6+6b³-18b²

Step-by-step explanation:

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5 0
3 years ago
One year there was a total of 59 commercial and noncommercial orbital launches worldwide. In addition, the number of noncommerci
Lina20 [59]

Answer:

The number of commercial orbital launches is equal to 19 and the number of noncommercial orbital launches is equal to 40.

Step-by-step explanation:

IMPORTANT INFORMATION: The number of noncommercial orbital launches was two more than twice the number of commercial orbital launches.

Let's say that the number of commercial orbital launches is equal to x. Now, the number of noncommercial orbital launches is equal to 2x+2.

We are told that if we add the number of commercial and noncommercial orbital launches together we will get 59.

Therefore, if we add 2x+2 with x we should get 59:

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x=19

This is how we can come to the answer that the number of commercial orbital launches is equal to 19 and the number of noncommercial orbital launches is equal to 40 (2x+2=2*19+2=38+2=40).

Hope this helps!

4 0
3 years ago
4 2/3 ÷ 2 1/3 ? what is the quotient?​
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Answer: 42/21 or 2

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7 0
3 years ago
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