2004 AMC 10A Problems/Problem 2: Difference between revisions
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== Problem == | == Problem == | ||
For any three real numbers <math>a</math>, <math>b</math>, and <math>c</math>, with <math>b\neq c</math>, the operation <math>\otimes</math> is defined by: | For any three real numbers <math>a</math>, <math>b</math>, and <math>c</math>, with <math>b\neq c</math>, the operation <math>\otimes</math> is defined by: | ||
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<math> \mathrm{(A) \ } -\frac{1}{2}\qquad \mathrm{(B) \ } -\frac{1}{4} \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac{1}{4} \qquad \mathrm{(E) \ } \frac{1}{2} </math> | <math> \mathrm{(A) \ } -\frac{1}{2}\qquad \mathrm{(B) \ } -\frac{1}{4} \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac{1}{4} \qquad \mathrm{(E) \ } \frac{1}{2} </math> | ||
== Solution == | == Solution == | ||
<math>\otimes \left(\frac{1}{2-3},\frac{2}{3-1},\frac{3}{1-2}\right)=\otimes(-1,1,-3)=\frac{-1}{1+3}=-\frac{1}{4}\Longrightarrow\boxed{\mathrm{(B)}\ -\frac{1}{4}}</math> | <math>\otimes \left(\frac{1}{2-3},\frac{2}{3-1},\frac{3}{1-2}\right)=\otimes(-1,1,-3)=\frac{-1}{1+3}=-\frac{1}{4}\Longrightarrow\boxed{\mathrm{(B)}\ -\frac{1}{4}}</math> | ||
== See | |||
==Video Solution == | |||
https://youtu.be/KfjB4--G-Lc | |||
Education, the Study of Everything | |||
== See Also == | |||
{{AMC10 box|year=2004|ab=A|num-b=1|num-a=3}} | {{AMC10 box|year=2004|ab=A|num-b=1|num-a=3}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:13, 21 April 2021
Problem
For any three real numbers 
, 
, and 
, with 
, the operation 
is defined by:
![\[\otimes(a,b,c)=\frac{a}{b-c}\]](http://latex.artofproblemsolving.com/8/c/0/8c07fc892fc41e343f30a1c44dea8ea14edce4c9.png)
What is 
?
Solution
Video Solution
Education, the Study of Everything
See Also
| 2004 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
