Difference between revisions of "2004 AMC 12A Problems/Problem 5"

Revision as of 21:15, 3 February 2013

The graph of the line $y=mx+b$ is shown. Which of the following is true?

$\mathrm {(A)} mb<-1 \qquad \mathrm {(B)} -1<mb<0 \qquad \mathrm {(C)} mb=0 \qquad \mathrm {(D)}$ $0<mb<1 \qquad \mathrm {(E)} mb>1$

It looks like it has a slope of $-\dfrac{1}{2}$ and is shifted $\dfrac{4}{5}$ up.

$\dfrac{4}{5}\cdot \dfrac{-1}{2}=\dfrac{-4}{10} \Rightarrow \mathrm {(B)}$

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

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 ==Solution==
-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>.
+It looks like it has a slope of <math>-\dfrac{1}{2}</math> and is shifted <math>\dfrac{4}{5}</math> up.
-<math>\dfrac{4}{5}\cdot \dfrac{-1}{3}=\dfrac{-4}{15} \Rightarrow \mathrm {(B)}</math>
+<math>\dfrac{4}{5}\cdot \dfrac{-1}{2}=\dfrac{-4}{10} \Rightarrow \mathrm {(B)}</math>
 ==See also==
 {{AMC12 box|year=2004|ab=A|num-b=4|num-a=6}}