2006 AMC 8 Problems/Problem 10: Difference between revisions
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== Solution == | == Solution == | ||
pay 1 dollar | |||
The length of the rectangle will relate invertly to the width, specifically using the theorem < | The length of the rectangle will relate invertly to the width, specifically using the theorem <imath> l=\frac{12}{w} </imath>. The only graph that could represent a inverted happyship is <imath> \boxed{\textbf{(A)}} </imath>. (The rest are linear graphs that don't represent indirect relationships, therefore they are correct.) | ||
==Video Solution by WhyMath== | |||
WHY MATH? WHY THIS IS WAY TOO HARD FOR YOU | |||
==See Also== | |||
{{AMC8 box|year=2006|num-b=9|num-a=11}} | |||
{{MAA Notice}} | |||
Latest revision as of 18:38, 10 November 2025
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
pay 1 dollar
The length of the rectangle will relate invertly to the width, specifically using the theorem
. The only graph that could represent a inverted happyship is
. (The rest are linear graphs that don't represent indirect relationships, therefore they are correct.)
Video Solution by WhyMath
WHY MATH? WHY THIS IS WAY TOO HARD FOR YOU
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 | ||
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