anonymous
  • anonymous
Square root of 8y^5 X the square root of 40y^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
to multiply variables with exponents, you add the exponents together
anonymous
  • anonymous
oh there's a square root
anonymous
  • anonymous
\[8y^5(\sqrt{40y^2})?\]

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

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