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45/3^4 * 4^3/5^2 * 27/8^2
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21 Jul 2025, 08:10
21 Jul 2025, 08:10
Expert Reply
00:00
Question Stats:
0% (00:00) correct
100% (01:46) wrong based on 2 sessions
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Quantity A |
Quantity B |
\(\dfrac{45}{3^4} \times \dfrac{4^3}{5^2} \times \dfrac{27}{8^2}\) |
\(\dfrac{3^2}{4^3} \times \dfrac{48}{5^2} \times \dfrac{75}{3^3}\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Re: 45/3^4 * 4^3/5^2 * 27/8^2
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21 Jul 2025, 08:24
21 Jul 2025, 08:24
Wouldn't the answer here be A since LHS is 3/5 and RHS 3/20?
45/3^4 * 4^3/5^2 * 27/8^2
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21 Jul 2025, 08:54
21 Jul 2025, 08:54
Expert Reply
The best way is to go through the process. Exponents can be tricky
A) \(\dfrac{5*3^2}{3^4} \times \dfrac{2^6}{5^2} \times \dfrac{3^3}{2^6} = \dfrac{3}{5}=0.6\)
B) \(\dfrac{3^2}{2^6} \times \dfrac{2*3*2^3}{5^2} \times \dfrac{3*5^2}{3^3} = \dfrac{3}{4}=0.75\)
\(B > A\)
A) \(\dfrac{5*3^2}{3^4} \times \dfrac{2^6}{5^2} \times \dfrac{3^3}{2^6} = \dfrac{3}{5}=0.6\)
B) \(\dfrac{3^2}{2^6} \times \dfrac{2*3*2^3}{5^2} \times \dfrac{3*5^2}{3^3} = \dfrac{3}{4}=0.75\)
\(B > A\)
