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9^2*7^3*2^5
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07 Sep 2021, 07:09
07 Sep 2021, 07:09
Expert Reply
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Question Stats:
96% (01:08) correct
3% (02:15) wrong based on 27 sessions
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Re: 9^2*7^3*2^5
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07 Sep 2021, 08:28
07 Sep 2021, 08:28
1
\(\frac{9^2*7^3*2^5}{84^{3}}\)
\(\frac{3^4*7^3*2^5}{(4*3*7)^3}\)
\(\frac{3^4*7^3*2^5}{2^6*3^3*7^3}\)
\(\frac{3}{2}\)
Answer D
\(\frac{3^4*7^3*2^5}{(4*3*7)^3}\)
\(\frac{3^4*7^3*2^5}{2^6*3^3*7^3}\)
\(\frac{3}{2}\)
Answer D
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9^2*7^3*2^5
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10 Sep 2021, 07:23
10 Sep 2021, 07:23
Carcass wrote:
\(\frac{9^2*7^3*2^5}{84^{3}}=\)
A. 1/3
B. 2/3
C. 1
D. 3/2
E. 3
A. 1/3
B. 2/3
C. 1
D. 3/2
E. 3
\(\frac{9^2*7^3*2^5}{84^{3}}=\frac{(3^2)^2*7^3*2^5}{(2^2*3*7)^{3}}\) [Found the prime factorization of all non-prime values:]
\(=\frac{3^4*7^3*2^5}{2^6*3^3*7^3}\) [Applied the Power of a Power law as well as the Power of a Product law]
\(=\frac{3^4*2^5}{2^6*3^3}\) [Canceled the \(7^3\) term in both the numerator and denominator]
\(=\frac{3}{2}\) [Simplified the remaining terms]
Answer: D
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