Mathematics
8 Online
OpenStudy (anonymous):
that is so not what i wanted
14 years ago
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OpenStudy (saifoo.khan):
o.O
14 years ago
OpenStudy (anonymous):
ok here goes
14 years ago
OpenStudy (anonymous):
\[\sqrt[8]{y^8} simplify\]
14 years ago
OpenStudy (saifoo.khan):
y^8/3
14 years ago
OpenStudy (anonymous):
huh?
14 years ago
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OpenStudy (saifoo.khan):
y raise to the power 8 over 3
14 years ago
OpenStudy (anonymous):
where did u get 3?
14 years ago
OpenStudy (saifoo.khan):
is it 3rd root?
or 8th root?
14 years ago
OpenStudy (anonymous):
i am so sorry. I dont know what you are talking about root.?
14 years ago
OpenStudy (saifoo.khan):
write your ques in words!
14 years ago
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OpenStudy (saifoo.khan):
whts on the LEFT TOP?
14 years ago
OpenStudy (anonymous):
8
14 years ago
OpenStudy (saifoo.khan):
oh,
14 years ago
OpenStudy (saifoo.khan):
then your answer is:
y
14 years ago
OpenStudy (anonymous):
so the square root of 8 is 1 = y?
14 years ago
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OpenStudy (saifoo.khan):
its 8th root of 8
14 years ago
OpenStudy (saifoo.khan):
not squared coot. :p
14 years ago
OpenStudy (anonymous):
umm....
14 years ago
OpenStudy (anonymous):
how do you get y? say the numbers are changed to 6 and 12. how do i do that?
14 years ago
OpenStudy (saifoo.khan):
which numbers?
14 years ago
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OpenStudy (anonymous):
\[\sqrt[6]{y ^ 12}\]
14 years ago
OpenStudy (anonymous):
y 12
14 years ago
OpenStudy (saifoo.khan):
can u join me @ twiddlA?
14 years ago
OpenStudy (anonymous):
yeah
14 years ago
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jimthompson5910 (jim_thompson5910):
In general, \[\sqrt[n]{x^n}=|x|\] where n is an even number and is any real number. If we force x to be nonnegative, then \[\sqrt[n]{x^n}=x\]
So if we assume that y is nonnegative, then \[\sqrt[8]{y^8}=y\]
If y is any real number, then \[\sqrt[8]{y^8}=|y|\]
14 years ago