I take it that you mean "sqrt (32 sqrt 2)".
This expression can be rewritten as sqrt (2 ^ 5 * 2 ^(1/2)),
which = (2 ^ (11/2)) ^ (1/2),
= 2 ^ (11/4),
which a calculator, or Excel, e.g., can evaluate to 6.727171322 Or you could start from the original expression at the top -- but include an asterisk, as in the second line!
2007-05-29 16:08:20
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answer #1
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answered by Keith A 6
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I am assuming your question is what is the sqrt of (32 times sqrt of 2). if so, Take the sqrt of 32 which is 4 time sqrt 2. You now will move that two under the inside sqrt by squaring it which leaves you with 4 times the sqrt of the srqt of 4*2 or 8. The sqrt of a sqrt is the fourth root, so the answer is 4 times the fourth root of 8
2007-05-29 23:02:50
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answer #2
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answered by Stealth555 2
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using exponents to help simplify: sqrt 2 = 2 ^1/2 and 32 = 2 ^5, so 32 sqrt 2 = 2^5 * 2^(1/2) = 2 ^ (11/2). sqrt 2 ^ (11/2) = (2 ^ (11/2))^(1/2) = 2 ^ (11/4) or 4th root of 2 ^ 11 which is 4 * 4th root of 8.
2007-05-29 23:04:22
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answer #3
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answered by amac577 2
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it equals the square root of 32 * 2 ^ 1/2
the final answer is (32^1/2)(2^ 1/4)
or writing it in a different way, 4sqrt 2 * 4th root of 2
2007-05-29 22:56:14
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answer #4
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answered by r 3
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sqrt 32 sqrt 2 = sqrt (32 *sqrt 2 ) = 4sqrt(2*sqrt(2))
2007-05-29 22:58:02
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answer #5
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answered by Olu 1
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it appears it should amount to 4sqrt(2sqrt(2)) assuming the direction is for simplification. Starting from sqrt(32sqrt(2)), we extract sqrt(16) and bring it outside the sqrt as 4 and thus we retain inside the following: sqrt(2sqrt(2)).
2007-05-29 22:58:49
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answer #6
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answered by rales_lds 1
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sqrt(32 * 2)
sqrt(64) = 8
2007-05-29 22:54:40
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answer #7
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answered by Anonymous
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clarify please are we solving for the sqrt of 32 what are we solving for?
2007-05-29 22:56:08
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answer #8
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answered by metzilla24 1
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is that a problem or an answer? try a calculator
2007-05-29 22:54:22
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answer #9
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answered by *Forget me not* 5
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can u please explain this better, ill try to help u if u do, but i dont understand what u r asking
2007-05-29 22:54:16
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answer #10
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answered by Anonymous
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