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gtnhenbr [62]
2 years ago
5

What is the following quotient? StartFraction 6 minus 3 (RootIndex 3 StartRoot 6 EndRoot) Over RootIndex 3 StartRoot 9 EndRoot E

ndFraction.
Mathematics
1 answer:
Bezzdna [24]2 years ago
4 0

Exponent properties help us to simplify the powers of expressions.  The quotient of the given expression \dfrac{6 - 3(\sqrt[3]{6})}{\sqrt[3]{9}} is (2∛3 - ∛18).

<h3>What are the basic exponent properties?</h3>

{a^m} \cdot {a^n} = a^{(m+n)}\\\\\dfrac{a^m}{a^n} = a^{(m-n)}\\\\\sqrt[m]{a^n} = a^{\frac{n}{m}}\\\\(a^m)^n = a^{m\times n}\\\\(m\times n)^a = m^a\times n^a\\\\

Given to us

\dfrac{6 - 3(\sqrt[3]{6})}{\sqrt[3]{9}}

We will solve the problem using the basic exponential properties,

\dfrac{6 - 3(\sqrt[3]{6})}{\sqrt[3]{9}}\\\\ = \dfrac{6}{\sqrt[3]{9}} - \dfrac{3(\sqrt[3]{6})}{\sqrt[3]{9}}\\\\ = (6\cdot 3^{-\frac{2}{3}}) - [3 \cdot (2 \cdot 3)^{-\frac{2}{3}}3^{-\frac{2}{3}}]\\\\= (2 \cdot 3 \cdot 3^{-\frac{2}{3}}) - [3 \cdot 2^{-\frac{2}{3}} \cdot 3^{-\frac{2}{3}}3^{-\frac{2}{3}}]\\\\

= [2 \cdot 3^{(1-\frac{2}{3})}] - [2^{\frac{1}{3}}\cdot 3^{(1+\frac{1}{3} - \frac{2}{3})}]\\\\=  [2 \cdot 3^{(\frac{1}{3})}] - [2^{\frac{1}{3}}\cdot 3^{(\frac{2}{3})}]\\\\= 2\sqrt[3]{3} - \sqrt[3]{2}\sqrt[3]{9}\\\\=2\sqrt[3]{3} - \sqrt[3]{18}

Hence, the quotient of the given expression \dfrac{6 - 3(\sqrt[3]{6})}{\sqrt[3]{9}} is (2∛3 - ∛18).

Learn more about Exponents:

brainly.com/question/5497425

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x^2-64

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100cm

Step-by-step explanation:

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2 years ago
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ExtremeBDS [4]

1) Seven less than twice a number, n, is 32.

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Find the solutions of x^2+30 = 0<br><br> please give detailed steps!
murzikaleks [220]

Answer:

x= <em>i</em>√30

Step-by-step explanation:

<u>I'm going to go into this under the assumption that you've covered imaginary numbers based on the </u><u>question.</u><u> If I'm wrong then sorry about that.</u>

Okay, so first you want to subtract 30 from both sides

x^2=-30

Then you take the square root of each side.

√(x^2)=√-30

x=√-30

Since it's impossible to square a number to get a negative number, you'll end up with an imaginary number. You have to rewrite x=√-30 to get rid of the negative sign under the radical. Rewriting this will also indicate that it's an imaginary number.

<u>Final</u><u> </u><u>answer</u><u>:</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>i√30</u>

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3 years ago
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