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anonymous
 4 years ago
Square root of 8y^5 X the square root of 40y^2
anonymous
 4 years ago
Square root of 8y^5 X the square root of 40y^2

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to multiply variables with exponents, you add the exponents together

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh there's a square root

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[8y^5(\sqrt{40y^2})?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can pull a y out of the square root because \[\sqrt{y^2}=y\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can also pull a 2 out of the square root because of \[\sqrt{2^2*10}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{8y^{5}} * \sqrt{40y ^{2}}\]no its

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can put it all under one square root

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0such as this \[\sqrt{(8y^5)(40y^2)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then i got \[\sqrt{320y ^{7}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now you have to pull numbers and variables out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then change 320 to \[8\sqrt{5} \] then \[8\sqrt{5y ^{7}}\] but i dont know what to do next

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im sorry \[8\sqrt{5y^7}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if there was a y^2 in there, you would just take out the y, correct?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so what if you factored \[y^7\] as \[y^6y\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i thought i have to factor it into a perfect square?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y^6=(y^3)^2\], if you remember the rules of exponents

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so can you pull y^6 out now?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, but now \[y^6=(y^3)^2\], which is a perfect square

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[=8\sqrt{5(y^3)^2y}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if i said y^3=x, then you would see an x^2 inside of the square root

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and you'd pull the x out of \[\sqrt{x^2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0see what i'm talking about?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but then i have 8y^3 square root of 5y^3y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nope, you'd pull the y^3 out of the root, leaving you with only the other y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[8\sqrt{5(y^3)^2y} = 8y^3\sqrt{5y} = \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0forget that last = sign, that is it. it's done now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there's no way to make perfect squares from factors of 5 or y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0understand that when you raise a power to a power, you multiply the exponents

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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
 4 years ago
Best ResponseYou've already chosen the best response.0yeah that's all it takes is practice
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