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svlad2 [7]
3 years ago
8

Which of the following is equivalent to 5 square root 13^3

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
7 0

For this case we must find an expression equivalent to the following:

5 \sqrt {13 ^ 3}

By definition of properties of powers and roots we have:

\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}

In addition, we know that:

13 ^ 3 = 13 ^ 2 * 13

Rewriting the given expression we have:

5 \sqrt {13 ^ 3} = 5 \sqrt {13 ^ 2 * 13} = 13 * 5 \sqrt {13} = 65 \sqrt {13}

Answer:

65 \sqrt {13}

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alexira [117]

Answer:

B. 3

Step-by-step explanation:

The degree of this polynomial is based on the highest power on the exponents (if there is more than one variable, it is based on the sum)

The highest power is 3, so the degree is 3

6 0
3 years ago
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egoroff_w [7]

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look at explanation

Step-by-step explanation:

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5 0
3 years ago
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For this case we have the following number:

600,000

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4 0
3 years ago
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1. Find the midpoint of AB<br> given A(-3, -4) and B(1,6)
emmainna [20.7K]

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7 0
3 years ago
Select True or False for each statement.
SSSSS [86.1K]

\left( \dfrac 1 {64} \right)^{- 5/6} =64^{5/6} = (\sqrt[6]{64})^5 = 2^5 =32

TRUE

\sqrt[5]{36^4}=36^{4/5}

which surely isn't 36.  FALSE

\sqrt{12} - \dfrac 2 5 \sqrt{75} = 2 \sqrt{3} -\dfrac 2 5 (5) \sqrt{3} = 0

FALSE

For the fourth one we have a

\sqrt{98b} + \sqrt{2b}

which isnt

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\dfrac{1}{(\sqrt 5 - \sqrt 6)^2}

= \dfrac{1}{(\sqrt 5 - \sqrt 6)^2} \cdot \dfrac{(\sqrt 5 + \sqrt 6)^2}{(\sqrt 5 + \sqrt 6)^2}

= \dfrac{(\sqrt 5 + \sqrt 6)^2}{ ( (\sqrt 5 - \sqrt 6)(\sqrt 5 + \sqrt 6))^2}

= \dfrac{(\sqrt 5 + \sqrt 6)^2}{( 5-6)^2}

=(\sqrt 5 + \sqrt 6)^2

No fractions in that one so FALSE.

3 0
3 years ago
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