Difference between revisions of "1965 AHSME Problems/Problem 18"
(Created page with "<math> \frac{1}{1 + y} = \frac{1 - y}{1 - y^2}</math> <math> \frac{1 - y}{\frac{1 - y}{1 - y^2}} = 1 - y^2</math> <math> \boxed{(B) y^2}</math>") |
Tecilis459 (talk | contribs) (Add problem statement) |
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+ | == Problem == | ||
+ | |||
+ | If <math>1 - y</math> is used as an approximation to the value of <math>\frac {1}{1 + y}, |y| < 1</math>, the ratio of the error made to the correct value is: | ||
+ | |||
+ | <math>\textbf{(A)}\ y \qquad | ||
+ | \textbf{(B) }\ y^2 \qquad | ||
+ | \textbf{(C) }\ \frac {1}{1 + y} \qquad | ||
+ | \textbf{(D) }\ \frac{y}{1+y}\qquad | ||
+ | \textbf{(E) }\ \frac{y^2}{1+y}\qquad </math> | ||
+ | |||
+ | == Solution == | ||
+ | |||
<math> \frac{1}{1 + y} = \frac{1 - y}{1 - y^2}</math> | <math> \frac{1}{1 + y} = \frac{1 - y}{1 - y^2}</math> | ||
Revision as of 13:51, 16 July 2024
Problem
If is used as an approximation to the value of
, the ratio of the error made to the correct value is:
Solution