Mathematics
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OpenStudy (anonymous):
Square root of 8y^5 X the square root of 40y^2
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OpenStudy (anonymous):
to multiply variables with exponents, you add the exponents together
OpenStudy (anonymous):
oh there's a square root
OpenStudy (anonymous):
\[8y^5(\sqrt{40y^2})?\]
OpenStudy (anonymous):
you can pull a y out of the square root because \[\sqrt{y^2}=y\]
OpenStudy (anonymous):
you can also pull a 2 out of the square root because of \[\sqrt{2^2*10}\]
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OpenStudy (anonymous):
\[\sqrt{8y^{5}} * \sqrt{40y ^{2}}\]no its
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
you can put it all under one square root
OpenStudy (anonymous):
such as this \[\sqrt{(8y^5)(40y^2)}\]
OpenStudy (anonymous):
then i got \[\sqrt{320y ^{7}}\]
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OpenStudy (anonymous):
yep
OpenStudy (anonymous):
now you have to pull numbers and variables out
OpenStudy (anonymous):
as a hint, 64*5=320
OpenStudy (anonymous):
then change 320 to \[8\sqrt{5} \] then \[8\sqrt{5y ^{7}}\] but i dont know what to do next
OpenStudy (anonymous):
yes \[8\sqrt{y^7}\]
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OpenStudy (anonymous):
im sorry \[8\sqrt{5y^7}\]
OpenStudy (anonymous):
now you can factor y
OpenStudy (anonymous):
if there was a y^2 in there, you would just take out the y, correct?
OpenStudy (anonymous):
into 4 and 3
OpenStudy (anonymous):
yeah that too
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OpenStudy (anonymous):
so what if you factored \[y^7\] as \[y^6y\]
OpenStudy (anonymous):
i thought i have to factor it into a perfect square?
OpenStudy (anonymous):
\[y^6=(y^3)^2\], if you remember the rules of exponents
OpenStudy (anonymous):
so can you pull y^6 out now?
OpenStudy (anonymous):
\[8\sqrt{5y ^{6}}y\]
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OpenStudy (anonymous):
yes, but now \[y^6=(y^3)^2\], which is a perfect square
OpenStudy (anonymous):
\[8\sqrt{5y^6y}\]
OpenStudy (anonymous):
\[=8\sqrt{5(y^3)^2y}\]
OpenStudy (anonymous):
if i said y^3=x, then you would see an x^2 inside of the square root
OpenStudy (anonymous):
and you'd pull the x out of \[\sqrt{x^2}\]
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OpenStudy (anonymous):
see what i'm talking about?
OpenStudy (anonymous):
but then i have 8y^3 square root of 5y^3y
OpenStudy (anonymous):
nope, you'd pull the y^3 out of the root, leaving you with only the other y
OpenStudy (anonymous):
\[8\sqrt{5(y^3)^2y} = 8y^3\sqrt{5y} = \]
OpenStudy (anonymous):
forget that last = sign, that is it. it's done now
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OpenStudy (anonymous):
there's no way to make perfect squares from factors of 5 or y
OpenStudy (anonymous):
understand that when you raise a power to a power, you multiply the exponents
OpenStudy (anonymous):
i kind of get it i think..i just have to have more practice because it's still kind of confusing but ill get it
OpenStudy (anonymous):
yeah that's all it takes is practice