Difference between revisions of "2006 AMC 8 Problems/Problem 10"
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The length of the rectangle will relate invertly to the width, specifically using the theorem <math> l=\frac{12}{w} </math>. The only graph that could represent a inverted relationship is <math> \boxed{\textbf{(A)}} </math>. (The rest are linear graphs that represent direct relationships, therefore they are incorrect.) | The length of the rectangle will relate invertly to the width, specifically using the theorem <math> l=\frac{12}{w} </math>. The only graph that could represent a inverted relationship is <math> \boxed{\textbf{(A)}} </math>. (The rest are linear graphs that represent direct relationships, therefore they are incorrect.) | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/MzKpqmX0xfI | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2006|num-b=9|num-a=11}} | {{AMC8 box|year=2006|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:23, 29 October 2024
Problem
Jorge's teacher asks him to plot all the ordered pairs 
of positive integers for which 
is the width and 
is the length of a rectangle with area 12. What should his graph look like?
![[asy] size(75); draw((0,-1)--(0,13)); draw((-1,0)--(13,0)); dot((1,12)); dot((2,6)); dot((3,4)); dot((4,3)); dot((6,2)); dot((12,1)); label("$l$", (0,6), W); label("$w$", (6,0), S);[/asy]](http://latex.artofproblemsolving.com/3/a/0/3a01c402526b29bd19717ac27a964c97c7c90256.png)
![[asy] size(75); draw((0,-1)--(0,13)); draw((-1,0)--(13,0)); dot((1,1)); dot((3,3)); dot((5,5)); dot((7,7)); dot((9,9)); dot((11,11)); label("$l$", (0,6), W); label("$w$", (6,0), S);[/asy]](http://latex.artofproblemsolving.com/5/b/a/5bac0fbc6b424579dd878f2c9b6fd25e5fdc57bc.png)
![[asy] size(75); draw((0,-1)--(0,13)); draw((-1,0)--(13,0)); dot((1,11)); dot((3,9)); dot((5,7)); dot((7,5)); dot((9,3)); dot((11,1)); label("$l$", (0,6), W); label("$w$", (6,0), S);[/asy]](http://latex.artofproblemsolving.com/d/1/6/d16c6eb0a4feae2b575a49deae60ef5f2dc24924.png)
![[asy] size(75); draw((0,-1)--(0,13)); draw((-1,0)--(13,0)); dot((1,6)); dot((3,6)); dot((5,6)); dot((7,6)); dot((9,6)); dot((11,6)); label("$l$", (0,6), W); label("$w$", (6,0), S);[/asy]](http://latex.artofproblemsolving.com/7/8/2/7822b4158ccde67dd21fb1750f181c9088a739f2.png)
![[asy] size(75); draw((0,-1)--(0,13)); draw((-1,0)--(13,0)); dot((6,1)); dot((6,3)); dot((6,5)); dot((6,7)); dot((6,9)); dot((6,11)); label("$l$", (0,6), W); label("$w$", (6,0), S);[/asy]](http://latex.artofproblemsolving.com/2/8/6/286b49157b5fb1c448956dfccefe8e5ad2ff0cce.png)
Solution
The length of the rectangle will relate invertly to the width, specifically using the theorem 
. The only graph that could represent a inverted relationship is 
. (The rest are linear graphs that represent direct relationships, therefore they are incorrect.)
Video Solution by WhyMath
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
