Question

evaluate (11^1/9)-^12

Answer 1

flip it; multiply exps; simplify to 4/3 and cbrt the denominator of 11^4

Answer 2

\[\frac{1}{\sqrt[3]{11^4}}\]

Answer 3

can 11 be cubed?

Answer 4

not that I can tell; nothing perfect cubes to 11

Answer 5

at least no integers do

Answer 6

so if it cant be cubed I just work out 11^4?

Answer 7

well, notice that 11^4 = 11^3 * 11

\[\frac1{11\sqrt[3]{11}}\]
\[\frac1{11\sqrt[3]{11}}*\frac{\sqrt[3]{11^2}}{\sqrt[3]{11^2}}\]
\[\frac{\sqrt[3]{11^2}}{121}\]
maybe

Answer 8

my book gives the answer 3/10

Answer 9

that seems a bit contrived; you sure your matching up the right problem/answer ?

Answer 10

yes doubled checked before posting

Answer 11

http://www.wolframalpha.com/input/?i=11^%28-4%2F3%29

Answer 12

sorry, i question is incorrect

Answer 13

http://www.wolframalpha.com/input/?i=%2811^%281%2F9%29%29^%28-12%29

lol

Answer 14

evaluate (11^1/9)-^1/2

Answer 15

still not 3/10

http://www.wolframalpha.com/input/?i=%2811^%281%2F9%29%29^%28-1%2F2%29

Answer 16

could my textbook be wrong?

Answer 17

its possible, but there tends to be a myriad of reasons as to what might be going awry

Answer 18

B^a^b = B^ab is the basic rule here; then simplify the results

Answer 19

\[(11^{1/9})^{-1/2}=11^{-1/18}=1/\sqrt[18]{11}\]

Answer 20

@ radar my textbook gives the answer 3/10

Answer 21

@ Hollywood_chrissy
Would it be possible to post a digital picture of the expression shown in your textbook?

Answer 22

i can try

Answer 23

Question 41.

Answer 24

oooooooooh is this
\[(11\tfrac{1}{9})^{-\frac{1}{2}}\]

Answer 25

then get rid of that stupid improper fraction and write
\[(\frac{100}{9})^{-\frac{1}{2}}=(\frac{9}{100})^{\frac{1}{2}}=\sqrt{\frac{100}{9}}=\frac{10}{3}\]

Answer 26

lol, brilliant, thank you

Answer 27

yw

Answer 28

see, that was one myraidecimal circumstance lol

Answer 29

sorry??

Answer 30

Thanks for clearing that up!