Question

Medal&Fan!!!!!!!!!!

Answer 1

http://prntscr.com/45y85h

Answer 2

@ganeshie8 @Hero @KlOwNlOvE @radar @OOOPS

Answer 3

can exponents be negative @jagr2713 ?

Answer 4

yes

Answer 5

\[\sqrt[3]{1^{6}}\]

Answer 6

dont give me the answer. can you explain please

Answer 7

im reading while doing this been awhile since i did this that wasnt the answer thats what itll be when you change the exponent to positive

Answer 8

ohoh ik how to do it from here i think

Answer 9

You can use the following law to simplify the denominator:
\[\large \sqrt[b]{a}=x ^{\frac{a}{b}}\]

Answer 10

doesnt it change to x^6/3

Answer 11

wow u just said the krop

Answer 12

is that right

Answer 13

The denominator becomes:
\[\Large \sqrt[3]{x ^{-6}}=x ^{\frac{-6}{3}}\]

Answer 14

so thats the answer right

Answer 15

Not really. At this stage you have:
\[\Large\frac{1}{\sqrt[3]{x ^{-6}}}=\frac{1}{x ^{\frac{-6}{3}}}\]
You can simplify the denominator further. Can you do that before the final step?

Answer 16

i dont think so and the one get removed and becomes

Answer 17

What is:
\[\Large \frac{-6}{3}=?\]

Answer 18

ohohoho -2

Answer 19

Correct: So now we have:
\[\Large \frac{1}{\sqrt[3]{x ^{-6}}}=\frac{1}{x ^{-2}}\]
Now multiply numerator and denominator by x^ to get the solution:
\[\Large \frac{1}{x ^{-2}}=\frac{1\times x ^{2}}{x ^{-2} \times x ^{2}}=you\ can\ calculate\]

Answer 20

by x^2 *

Answer 21

i have to calculate it now

Answer 22

Remember that:\[\Large x ^{a} \times x ^{-a}=1\]

Answer 23

idk lol

Answer 24

\[\Large \frac{1}{x ^{-2}}=\frac{1\times x ^{2}}{x ^{-2} \times x ^{2}}=x ^{2}\]