Difference between revisions of "2015 AMC 10B Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | The first rotation moves the base of the <math>F</math> to the negative <math>y</math>-axis, and the stem to the positive <math>x</math>-axis. The reflection then moves the stem to the negative <math>x</math>-axis, with the base unchanged. Then the half turn moves the stem to the positive <math>x</math> axis and the base to the positive <math>y</math>-axis, choice <math>\boxed{\mathbf{( | + | The first rotation moves the base of the <math>F</math> to the negative <math>y</math>-axis, and the stem to the positive <math>x</math>-axis. The reflection then moves the stem to the negative <math>x</math>-axis, with the base unchanged. Then the half turn moves the stem to the positive <math>x</math> axis and the base to the positive <math>y</math>-axis, choice <math>\boxed{\mathbf{(E)}}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2015|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:52, 25 September 2016
Problem
The letter F shown below is rotated 
clockwise around the origin, then reflected in the 
-axis, and then rotated a half turn around the origin. What is the final image?
Solution
The first rotation moves the base of the 
to the negative -axis, and the stem to the positive 
-axis. The reflection then moves the stem to the negative -axis, with the base unchanged. Then the half turn moves the stem to the positive axis and the base to the positive -axis, choice 
.
See Also
| 2015 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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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