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==
It looks like it has a slope of <math>-\dfrac{1}{2}</math> and is shifted <math>\dfrac{4}{5}</math> up.
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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>
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==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}}
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{{MAA Notice}}

Latest revision as of 15:48, 12 July 2017

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

The line appears to have a slope of $-\dfrac{1}{2}$ and y-intercept of $\dfrac{4}{5}$ 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
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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