2004 AMC 12A Problems/Problem 5: Difference between revisions

Revision as of 17:37, 25 December 2011

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
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All AMC 12 Problems and Solutions

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 [[Image:2004 AMC 12A Problem 5.png]]
-<math>\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</math>
+<math>\mathrm {(A)} mb<-1 \qquad \mathrm {(B)} -1<mb<0 \qquad \mathrm {(C)} mb=0 \qquad \mathrm {(D)}</math> <math> 0<mb<1 \qquad \mathrm {(E)} mb>1</math>
 ==Solution==