Difference between revisions of "2004 AMC 12A Problems/Problem 5"
(→Problem) |
m (wording) |
||
(3 intermediate revisions by 2 users not shown) | |||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | + | The line appears to have a slope of <math>-\dfrac{1}{2}</math> and y-intercept of <math>\dfrac{4}{5}</math> up. | |
<math>\dfrac{4}{5}\cdot \dfrac{-1}{2}=\dfrac{-4}{10} \Rightarrow \mathrm {(B)}</math> | <math>\dfrac{4}{5}\cdot \dfrac{-1}{2}=\dfrac{-4}{10} \Rightarrow \mathrm {(B)}</math> | ||
Line 13: | Line 13: | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2004|ab=A|num-b=4|num-a=6}} | {{AMC12 box|year=2004|ab=A|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:48, 12 July 2017
Problem
The graph of the line is shown. Which of the following is true?
Solution
The line appears to have a slope of and y-intercept of up.
See also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
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.