Difference between revisions of "2007 AMC 12A Problems/Problem 13"
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| + | )<math> to </math>y=-5x+18<math>. Then the [[slope]] of the line that passes through the cheese and </math>(a,b)<math> is the negative reciprocal of the slope of the line, or </math>\frac 15<math>. Therefore, the line is </math>y=\frac{1}{5}x+\frac{38}{5}<math>. The point where </math>y=-5x+18<math> and </math>y=\frac 15x+\frac{38}5<math> intersect is </math>(2,8)<math>, and </math>2+8=10\ (B)$. | ||
==See also== | ==See also== | ||
Revision as of 06:06, 18 August 2020
Problem
A piece of cheese is located at 
in a coordinate plane. A mouse is at 
and is running up the line 
. At the point 
the mouse starts getting farther from the cheese rather than closer to it. What is 
?
Solution
We are trying to find the
)
y=-5x+18![$. Then the [[slope]] of the line that passes through the cheese and$](http://latex.artofproblemsolving.com/e/5/2/e52e2e0b6edca344cb100cac076f80e9aec25e67.png)
(a,b)
\frac 15
y=\frac{1}{5}x+\frac{38}{5}
y=-5x+18
y=\frac 15x+\frac{38}5
(2,8)
2+8=10\ (B)$.
See also
| 2007 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
