How do you simplify this expression? 3 square root 81 ÷ 5 square root 729
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calendarEducator since 2011
write5,349 answers
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You need to simplify the following expression such that:
`(3sqrt(81))/(5sqrt729)`
You need to convert the square root into a power such that:
`sqrt 81 = 81^(1/2)`
You should find the prime factors of 81 such that:
`81 = 1*3^4`
Substituting `3^4` for 81 yields:
`sqrt 81 = (3^4)^(1/2)`
You need to use the property of exponential and multiply the powers such that:
`(a^b)^c = a^(b*c)`
Reasoning by analogy yields:
`(3^4)^(1/2) = 3^(4*(1/2)) = 3^(4/2) => sqrt81 = 3^2 = 9`
You need to convert `sqrt729` into a power such that:
`sqrt(729) = 729^(1/2)`
You should find the prime factors of 729 such that:
`729 = 3^6 => 729^(1/2) = (3^6)^(1/2) => 729^(1/2) = 3^(6/2) => 729^(1/2) = 3^3`
Hence, substituting 9 for `sqrt 81` and 27 for `sqrt 729` in the given expression yields:
`(3sqrt(81))/(5sqrt729) = (3*9)/(5*27) => (3sqrt(81))/(5sqrt729) = 27/(5*27) = 1/5`
Hence, performing the simplification yields `(3sqrt(81))/(5sqrt729) = 1/5.`
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calendarEducator since 2011
write7 answers
starTop subject is Math
(3*sqrt(81)) / (5 *sqrt(729))
square root 81 = square root (9 * 9) = 9
square 729 - 27 * 27
thus, we have the following:
(3 * 9) / (5 * 27)
27 = 3 * 9
so now we have
(3 * 9) / (5 * 3 * 9)
the 3 and 9 in the numerator cancel the 3 and 9 in the denominator.
and the simplified result is:
1/5 answer
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