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

(New page: ==Problem== The graph of the line <math>y=mx+b</math> is shown. Which of the following is true? Image:2004 AMC 12A Problem 5.png <math>\mathrm {(A)} mb<-1 \qquad \mathrm {(B)} -1<mb<...)
 
(Solution)
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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> up. <math>\dfrac{4}{5}\cdot \dfrac{-1}{2}=\dfrac{-4}{10} \Rightarrow \mathrm {(B)}</math>
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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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<math>\dfrac{4}{5}\cdot \dfrac{-1}{2}=\dfrac{-4}{10} \Rightarrow \mathrm {(B)}</math>
  
 
==See also==
 
==See also==

Revision as of 07:44, 15 April 2008

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

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