Difference between revisions of "2010 AMC 10A Problems/Problem 11"
m |
Whitelisted (talk | contribs) m (→Video Solution) |
||
Line 48: | Line 48: | ||
==Video Solution== | ==Video Solution== | ||
− | https:// | + | https://www.youtube.com/watch?v=dQw4w9WgXcQ |
~IceMatrix | ~IceMatrix |
Revision as of 09:38, 20 May 2021
Contents
Problem 11
The length of the interval of solutions of the inequality is . What is ?
Solution
Since we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and .
Subtract from all of the quantities:
Divide all of the quantities by .
Since we have the range of the solutions, we can make them equal to .
Multiply both sides by 2.
Re-write without using parentheses.
Simplify.
We need to find for the problem, so the answer is
Video Solution
https://www.youtube.com/watch?v=dQw4w9WgXcQ
~IceMatrix
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 10 Problems and Solutions |
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.