Difference between revisions of "2005 AMC 10B Problems/Problem 4"
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<math>(5 \diamond 12) \diamond ((-12) \diamond (-5))</math>? | <math>(5 \diamond 12) \diamond ((-12) \diamond (-5))</math>? | ||
− | <math>\ | + | <math>\textbf{(A) } 0 \qquad \textbf{(B) } \frac{17}{2} \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13\sqrt{2} \qquad \textbf{(E) } 26</math> |
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== Solution == | == Solution == | ||
− | < | + | <cmath> |
+ | \begin{align*} | ||
+ | (5 \diamond 12) \diamond ((-12) \diamond (-5))&=(\sqrt{5^2+12^2}) \diamond (\sqrt{(-12)^2+(-5)^2})\ | ||
+ | &=(\sqrt{169})\diamond(\sqrt{169})\ | ||
+ | &=13\diamond13\ | ||
+ | &=\sqrt{13^2+13^2}\ | ||
+ | &=\sqrt{338}\ | ||
+ | &=\boxed{\mathrm{(D)\,13\sqrt{2}}}\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Note that the negative signs did not matter and any number squared times two is that number times the square root of 2. | ||
+ | |||
== See Also == | == See Also == | ||
{{AMC10 box|year=2005|ab=B|num-b=3|num-a=5}} | {{AMC10 box|year=2005|ab=B|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:58, 24 July 2022
Problem
For real numbers and , define . What is the value of
?
Solution
Note that the negative signs did not matter and any number squared times two is that number times the square root of 2.
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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