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

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[[Image:2004 AMC 12A Problem 5.png]]
 
[[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>
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<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==
 
==Solution==

Revision as of 18:37, 25 December 2011

Problem

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

2004 AMC 12A Problem 5.png

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

Solution

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

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

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AMC 12 Problems and Solutions
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