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Re: A=(0.05)^2-(0.03)^2 [#permalink]
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From B and A, B > A as both are in squares so positive terms

C = 1/B and D = 1/A

if B > A then D > A because in fractions with same numerator, smaller denominator will be large value

Option B
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Re: A=(0.05)^2-(0.03)^2 [#permalink]
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Carcass wrote:
\(A=(0.05)^2-(0.03)^2\)

\(B= (0.05)^2+(0.03)^2\)

\(C= \dfrac{1}{(0.05)^2+(0.03)^2}\)

\(D= \dfrac{1}{(0.05)^2-(0.03)^2}\)

Which of the following is true?

B > D > C > A
D > C > B > A
C > D > B > A
C > B > D > A
None of the above


Options A and B are each equal to a value between 0 and 1.
Whereas option B adds the two terms. option A subtracts them.
Implication:
B > A

Options C and D are the reciprocals of B and A, respectively.
To make the four options easier to compare, test easy values for B and A such that both values are between 0 and 1 and B > A.
If B=1/2 and A=1/4, then C=2 and D=4.
Thus:
D > C > B > A

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Re: A=(0.05)^2-(0.03)^2 [#permalink]
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