3 years ago

2 numbers that mulitply to be negative 486 and add to be 9

Mathematics

1 answer:

SVEN [57.7K]3 years ago

3 0

The answer to your question is:

-18 and 27

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What is the cube root of 27a^l2? -3a4 -3a 3a 3a*

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Answer:

$3a^4$

Step-by-step explanation:

What is the cube root of $27a^{12}$ ? This is the question.

We can write:

$\sqrt[3]{27a^{12}}$

We will use the below property to simplify:

$\sqrt[n]{a*b}=\sqrt[n]{a} \sqrt[n]{b}$

So, we have:

$\sqrt[3]{27a^{12}} =\sqrt[3]{27} \sqrt[3]{a^{12}}$

We will now use below property to further simplify:

$\sqrt[n]{x} =x^{\frac{1}{n}}$

Thus, we have:

$\sqrt[3]{27} \sqrt[3]{a^{12}} =3*(a^{12})^{\frac{1}{3}}$

We know power to the power rule: $(a^z)^b=a^{zb}$

Now, we have:

$3*(a^{12})^{\frac{1}{3}}\\=3*a^{\frac{12}{3}}\\=3a^4$

This is the correct answer: $3a^4$

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