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10 Sep 2017, 05:03
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
75% (03:06) correct
25%
(02:41)
wrong
based on 186
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Which of the following is the result when \((6x^3y^3)^4 + (4x^4)^3y^6\) is divided by \((2x^4y^6)^2\), if \(xy \neq 0\)?
A. \(2^23^4x^4 + 2^4x^4y^{(–6)}\)
B. \(3^2x^4 + 4^2x^4y^{(–6)}\)
C. \(2^{10}x^4y + 2^4x^4y^4\)
D. \(2^33^4x^4 + 4^3x^{12}y^{16}\)
E. \(3^2x^4 + 4^3x^{12}y^{16}\)
A. \(2^23^4x^4 + 2^4x^4y^{(–6)}\)
B. \(3^2x^4 + 4^2x^4y^{(–6)}\)
C. \(2^{10}x^4y + 2^4x^4y^4\)
D. \(2^33^4x^4 + 4^3x^{12}y^{16}\)
E. \(3^2x^4 + 4^3x^{12}y^{16}\)
10 Sep 2017, 10:00
Kudos
Bookmarks
Bunuel
Writing this solution out makes it look a lot harder than it is.
Also, when solving, I combined steps that, for clarity's sake, I broke out here.
The concept is simply \(\frac{a^{x}}{a^{y}} = a^{(x-y)}\)
\(\frac{(6x^3y^3)^4 + (4x^4)^3y^6}{(2x^4y^6)^2}\) = ?
Split the numerator:
\(\frac{(6x^3y^3)^4}{(2x^4y^6)^2} + \frac{(4x^4)^3y^6}{(2x^4y^6)^2}\)
Distribute the exponents
\(\frac{(6^4x^{12}y^{12})}{(2^2x^8y^{12})} + (\frac{(4^3x^{12}y^6)}{(2^2x^8y^{12})}\)
In the numerators, factor \(6^4\) and \(4^3\) (they have factors of 2, and will be divisible by the 2 in the denominator) and then substitute the results into respective numerators
(\(6^4\)) = \(2^43^4\)
(\(4^3\)) = \((2^2)^3 = 2^6\)
\(\frac{(2^{{43}^4})x^{12}y^{12}}{2^{{2x}^{8y}^{12}}} + \frac{(2^6)x^{12}y^6}{2^{{2x}^{8y}^{12}}\)
Simplify - divide similar terms in the numerator and denominator of both fractions by subtracting exponents of identical bases
\(2^{(4-2)}3^{4x^{(12-8)}}y^{(12-12)} + 2^{(6-2)})x^{(12-8)}y^{(6-12)}\) =
\(2^23^4x^4 + 2^4x^4y^{(–6)}\)
Answer
20 Jun 2018, 06:13
Kudos
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Let x=y=1, then when \((6x^3y^3)^4 + (4x^4)^3y^6\) is divided by \((2x^4y^6)^2\) = \((6^4+4^3)/2^2\)=\(2^2*3^4+2^4\)
Only A. \(2^23^4x^4 + 2^4x^4y^{(–6)}\) gives us this result.
Answer A.
Only A. \(2^23^4x^4 + 2^4x^4y^{(–6)}\) gives us this result.
Answer A.
