1987 AJHSME Problems/Problem 5: Difference between revisions

Revision as of 13:41, 29 January 2009

Problem

The area of the rectangular region is

$[asy] draw((0,0)--(4,0)--(4,2.2)--(0,2.2)--cycle,linewidth(.5 mm)); label(".22 m",(4,1.1),E); label(".4 m",(2,0),S); [/asy]$

$\text{(A)}\ \text{.088 m}^2 \qquad \text{(B)}\ \text{.62 m}^2 \qquad \text{(C)}\ \text{.88 m}^2 \qquad \text{(D)}\ \text{1.24 m}^2 \qquad \text{(E)}\ \text{4.22 m}^2$

Solution

$(.4) \cdot (.22) = \frac{4}{10} \cdot \frac{22}{100} = \frac{4\cdot 22}{10\cdot 100} = \frac{88}{1000} = \boxed{0.088}.$

@@ Line 13: / Line 13: @@
 ==Solution==
-{{Solution}}
+<math>(.4) \cdot (.22) = \frac{4}{10} \cdot \frac{22}{100} = \frac{4\cdot 22}{10\cdot 100} = \frac{88}{1000} = \boxed{0.088}.</math>
 ==See Also==
 [[1987 AJHSME Problems]]

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1987 AJHSME Problems/Problem 5: Difference between revisions

Revision as of 13:41, 29 January 2009

Problem

Solution

See Also