Difference between revisions of "2006 AMC 8 Problems/Problem 10"
(→See Also) |
|||
Line 71: | Line 71: | ||
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 14:23, 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?
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.