Question

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Answer 1

Convert them to exponents and then divide.
\[\sqrt[n]{a} = a^{1/n}\]

Answer 2

So for example
\[\sqrt[4]{x^3}\text{ divided by }\sqrt[3]{x^2} \]
would be
\[\frac{x^{\frac{3}{4}}}{x^{\frac{2}{3}}}\]
or \[x^{\frac{3}{4}-\frac{2}{3}}\]

Answer 3

Which would then be
\[x^{\frac{1}{12}}=\sqrt[12]{x}\]

Answer 4

Sometimes the test wants you to give you answer in terms of a radical, then you might have to convert it back. If they will accept the exponent version, then that is the way to go.

Answer 5

yes, what is \[\frac{x^{5/12}}{x^{1/6}}\]
Just because they want it in radical form as the answer doesn't mean that you can't do all the work in exponent form and then reconvert.
\[\frac{x^a}{x^b}=x^{a-b}\] This seems easier to me than trying to figure out the radical forms.

Answer 6

Um. No. What is \[\frac{5}{12}-\frac{1}{6}\]?

Answer 7

You are forgetting to put them into common denoms.
5\12 - 1\6 = 5\12 -2\12 = 3\12 = 1\4

Answer 8

When you divide same base different exponents you subtract the exponents from one another.

\[\frac{aaaaa}{aa}=\frac{a^5}{a^2}\]
\[\frac{aaa}{1}\frac{aa}{aa}=aaa = a^3= a^{5-2}=a^3\]

Answer 9

No, x^2/3 is wrong.

Sorry. If you want to do it in radical form you can.
\[\frac{\sqrt[12]{x^5}}{\sqrt[6]{x}}\]
can also be done realizing that
\[\sqrt[5]{a} = \sqrt[10]{a^2}\] so you can do it that way. You need to convert the index 6 to a twelve

Answer 10

Sure.

Answer 11

index 12√x^5 x index 6√x
doesn't it mean,
\(\huge \sqrt[12]{x^5} \times\sqrt[6]x\)

Answer 12

It does say to divide the expressions though.

\[\frac{\sqrt[12]{x^5}}{\sqrt[6]{x}}\to\frac{\sqrt[12]{x^5}}{\sqrt[12]{x^2}}\to \]
\[\sqrt[12]{\frac{x^5}{x^2}}\to\sqrt[12]{x^{5-2}}\to\sqrt[12]{x^3}\to \sqrt[4]{\sqrt[3]{x^3}}\to\sqrt[4]{x}\]

Answer 13

The main methods you need to know is that
1) \[\sqrt[5]{x} = \sqrt[10]{x^2}\]
2) \[\frac{\sqrt[12]{a}}{\sqrt[12]{b}}=\sqrt[12]{\frac{a}{b}}\]
Yes the fourth root of x is the answer in radical form.

Answer 14

The reason why I suggested immediately to use your exponent conversion is that, if this is algII, you probably learned the exponent rules prior to this and can then use them rather than memorize another set of rules for radicals.

Answer 15

for questions like these, you need to be able to write down the 4 or so rules of exponents and the radical methods as well. Good luck and don't give up. exponents give a lot of people confusion.

Answer 16

i hope to question was avtually to divide... its bit unclear....

Answer 17

*actually

Answer 18

yes it was. I am working on the fact that it said to divide the two expressions.

Answer 19

the 'x' in middle is throwing me off...

Answer 20

can u confirm the question abbie ?