Difference between revisions of "2010 AMC 10A Problems/Problem 11"
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== See also == | == See also == | ||
{{AMC10 box|year=2010|ab=A|num-b=10|num-a=12}} | {{AMC10 box|year=2010|ab=A|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:01, 12 March 2017
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
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 |
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