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Here what we can do is we can make the denominators constant and compare the fractions w.r.t their numerators.
The fraction with highest numerator will the highest one.
Lets make the dominator common

Let 3^2*5^2 be p

A) 1/p
B)2/p
C)(7/3 )/p
D)3/p
E)[75/(9*125)]/p

Numerator of D is the highest hence D is the Greatest fraction.

Hence D
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Multiply all by 3^4 * 5^5 (denominator of answer E)

A) 3^2 * 5^3 = 9 * 5^3
B) 2 * 3^2 * 5^3 = 18 * 5^3
C) 7 * 3 * 5^3 = 21 * 5^3
D) 45 * 3 * 5^2 = 27 * 5^3
E) 3 * 5^2

D is the largest value.
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Which of the following fractions has the greatest value?

(A) \(\frac{1}{(3^2)(5^2)}\)
(B) \(\frac{2}{(3^2)(5^2)}\)
(C) \(\frac{7}{(3^3)(5^2)}\)
(D)\(\frac{45}{(3^3)(5^3)}\)
(E) \(\frac{75}{(3^4)(5^5)}\)
Solution:

First, we can eliminate choice A since it’s less than choice B. We can also reduce the two fractions in the last two choices:

D) 45 / [(3^3)(5^3)] = (3^2)(5) / [(3^3)(5^3)] = 1 / [(3)(5^2)]

E) 75 / [(3^4)(5^5)] = (3)(5^2) / [(3^4)(5^5)] = 1 / [(3^2)(5^3)]

We can eliminate choice E since it’s less than choice D.

Now, we can compare the ones in choices B, C and D by expressing them in terms of their least common denominator, which is (3^3)(5^2):

B) 2 / [(3^2)(5^2)] = 6 / [(3^3)(5^2)]

C) 7 / [(3^3)(5^2)] = 7 / [(3^3)(5^2)]

D) 1 / [(3)(5^2)] = 9 / [(3^3)(5^2)]

We see that choice D has the greatest value.

Answer: D
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\(a)\) \(\frac{1}{(3^2)(5^2)}\)

\(b)\) \(\frac{2}{(3^2)(5^2)}\)

\(c)\) \(\frac{7}{(3^3)(5^2)}\)

\(d)\) \(\frac{45}{(3^3)(5^3)}\) = \(\frac{45}{(3)(5^2)(45)}\) = \(\frac{1}{(3)(5^2)}\)

\(e)\) \(\frac{75}{(3^4)(5^5)}\) = \(\frac{3*25}{(3^3)(3)(5^3)(5^2)}\) = \(\frac{1}{(3^3)(5^3)}\)

Let, \(\frac{1}{(3)(5^2)}\) = k

\(a)\) \(\frac{1}{(3^2)(5^2)}\) = \(\frac{1k}{3}\)

\(b)\) \(\frac{2}{(3^2)(5^2)}\) = \(\frac{2k}{3}\)

\(c)\) \(\frac{7}{(3^3)(5^2)}\) = \(\frac{7k}{9}\)

\(d)\) \(\frac{1}{(3)(5^2)}\) = \(k\) \((D)\)

\(e)\) \(\frac{1}{(3^3)(5^3)}\) = \(\frac{1k}{45}\)
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­Looking for the smallest denominator comparatively:

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