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(1/p)^ 2 − (1/p)(1/q) =
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22 Nov 2021, 08:34
22 Nov 2021, 08:34
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\((\frac{1}{p})^ 2 − (\frac{1}{p})(\frac{1}{q}) =\)
(A) \(\frac{q−p}{p^2q}\)
(B) \(\frac{p−q}{p^2q}\)
(C) \(\frac{q−p}{q^2p}\)
(D) \(\frac{q−p}{p^2−pq}\)
(E) \(\frac{1}{p^2−pq}\)
(A) \(\frac{q−p}{p^2q}\)
(B) \(\frac{p−q}{p^2q}\)
(C) \(\frac{q−p}{q^2p}\)
(D) \(\frac{q−p}{p^2−pq}\)
(E) \(\frac{1}{p^2−pq}\)
Re: (1/p)^ 2 − (1/p)(1/q) =
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22 Nov 2021, 08:58
22 Nov 2021, 08:58
\((\frac{1}{p})^2 - (\frac{1}{p})(\frac{1}{q}) = \frac{1}{p}(\frac{1}{p} - \frac{1}{q}) = \frac{1}{p}(\frac{q-p}{pq}) = \frac{q-p}{p^2q}\)
Hence, Answer is A
Hence, Answer is A
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Re: (1/p)^ 2 − (1/p)(1/q) =
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22 Nov 2021, 09:36
22 Nov 2021, 09:36
1
Carcass wrote:
\((\frac{1}{p})^ 2 − (\frac{1}{p})(\frac{1}{q}) =\)
(A) \(\frac{q−p}{p^2q}\)
(B) \(\frac{p−q}{p^2q}\)
(C) \(\frac{q−p}{q^2p}\)
(D) \(\frac{q−p}{p^2−pq}\)
(E) \(\frac{1}{p^2−pq}\)
(A) \(\frac{q−p}{p^2q}\)
(B) \(\frac{p−q}{p^2q}\)
(C) \(\frac{q−p}{q^2p}\)
(D) \(\frac{q−p}{p^2−pq}\)
(E) \(\frac{1}{p^2−pq}\)
Given: \((\frac{1}{p})^ 2 − (\frac{1}{p})(\frac{1}{q}) \)
Rewrite as: \(\frac{1}{p^2} − \frac{1}{pq}\)
Rewrite fractions with common denominators: \(\frac{q}{p^2q} − \frac{p}{p^2q}\)
Combine numerators: \(\frac{q-p}{p^2q} \)
Answer: A
