Ask
your own question, for FREE!
Mathematics
9 Online
OpenStudy (anonymous):
√((125²)^-⅓)
solve without using calculator
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (mrnood):
note that 125 = 5^3
also note that sqrt(x) is the same as x^1/2
now use properties of exponents to simplify
OpenStudy (unklerhaukus):
\[\sqrt{(125^2)^{-1/3}}=((125^2)^{-1/3})^{1/2}=125^{(2\times-\tfrac13\times\tfrac12)}\]
OpenStudy (anonymous):
To be polite both solutions are wrong
OpenStudy (anonymous):
i promise to follow the person that can get this answer right
OpenStudy (unklerhaukus):
\[\sqrt{(125^2)^{-1/3}}=((125^2)^{-1/3})^{1/2}=125^{(2\times-\tfrac13\times\tfrac12)}=125^{-1/3}=(5^3)^{-1/3}=...\]
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
what is the answer
OpenStudy (unklerhaukus):
\[(5^3)^{-1/3}=5^{3\times-1/3}=...\]
OpenStudy (anonymous):
its wrong
OpenStudy (anonymous):
who needs a clue
OpenStudy (mrnood):
@daniel56K
I don't need a clue - I have worked out ht eanswer correctly
I have given oyu the correct way to do it
hanve YOU worked it out?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (mrnood):
125^2 = 5^3^2 = 5^6
5^6 ^-1/3 = 5 ^-2
5^-2 ^ 1/2 = 5^-1
= 1/5 = 0.2
OpenStudy (anonymous):
1/5
OpenStudy (anonymous):
That is correct
OpenStudy (mrnood):
@Daniel56k
if oyu look at @UnkleRhaukus answer you will see that he already did that
his answer is the SAME as 1/5 (as is mine)
OpenStudy (unklerhaukus):
yea
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
oh yes
OpenStudy (anonymous):
But i said i need answers not solution
OpenStudy (unklerhaukus):
is -0.2 also valid?
OpenStudy (anonymous):
no it is a positive
OpenStudy (anonymous):
Am closing this now u all should be waiting for another question
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (unklerhaukus):
thanks for playing
OpenStudy (anonymous):
No. thank u for participating
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!