Difference between revisions of "2004 AMC 12A Problems/Problem 5"
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==Solution== | ==Solution== | ||
− | It looks like it has a slope of <math>-\dfrac{1}{2}</math> and is shifted <math>\dfrac{4}{5}</math> | + | It looks like it has a slope of <math>-\dfrac{1}{2}<m<0</math> (looking at the intercepts) and is shifted <math>0<b<1</math> up. Try <math>m=-\dfrac{1}{3}</math> and <math>b=\dfrac{4}{5}</math>. |
− | <math>\dfrac{4}{5}\cdot \dfrac{-1}{ | + | <math>\dfrac{4}{5}\cdot \dfrac{-1}{3}=\dfrac{-4}{15} \Rightarrow \mathrm {(B)}</math> |
==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}} |
Revision as of 22:14, 3 February 2013
Problem
The graph of the line is shown. Which of the following is true?
Solution
It looks like it has a slope of (looking at the intercepts) and is shifted up. Try and .
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 |