Difference between revisions of "2004 AMC 10A Problems/Problem 2"
m (minor) |
Hashtagmath (talk | contribs) |
||
(7 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | For any three | + | 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: |
− | < | + | <cmath>\otimes(a,b,c)=\frac{a}{b-c}</cmath> |
− | \otimes(a,b,c)=\frac{a}{b-c} | + | What is <math>\otimes(\otimes(1,2,3),\otimes(2,3,1),\otimes(3,1,2))</math>? |
− | </ | ||
− | What is <math>\otimes | ||
<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 | + | <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> |
+ | |||
+ | |||
+ | ==Video Solution == | ||
+ | https://youtu.be/KfjB4--G-Lc | ||
+ | |||
+ | Education, the Study of Everything | ||
+ | |||
− | == See | + | == 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}} |
Latest revision as of 13:13, 21 April 2021
Contents
Problem
For any three real numbers , , and , with , the operation is defined by: 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.