Difference between revisions of "2023 AMC 12B Problems/Problem 4"
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~Mintylemon66 | ~Mintylemon66 | ||
− | ==Solution 2(Standard Form)== | + | ==Solution 2 (Standard Form)== |
− | 6.5 millimeters can be represented as 65 \times 10^{-2} centimeters. 25 meters is 25 \times 10^{2} centimeters. Multiplying out these results in (65 \times 10^{-2}) \times (25 \times 10^{2}) which is 65 \times 25 | + | 6.5 millimeters can be represented as <math>65 \times 10^{-2}</math> centimeters. 25 meters is <math>25 \times 10^{2}</math> centimeters. Multiplying out these results in <math>(65 \times 10^{-2}) \times (25 \times 10^{2})</math>, which is <math>65 \times 25</math> making the answer <math>\boxed{\textbf{(C) 1,625}}</math>. |
~darrenn.cp | ~darrenn.cp |
Revision as of 05:42, 16 November 2023
- The following problem is from both the 2023 AMC 10B #4 and 2023 AMC 12B #4, so both problems redirect to this page.
Contents
Problem
Jackson's paintbrush makes a narrow strip with a width of 6.5 millimeters. Jackson has enough paint to make a strip 25 meters long. How many square centimeters of paper could Jackson cover with paint?
Solution 1
6.5 millimeters is equal to 0.65 centimeters. 25 meters is 2500 centimeters. The answer is , so the answer is .
~Failure.net ~Mintylemon66
Solution 2 (Standard Form)
6.5 millimeters can be represented as centimeters. 25 meters is centimeters. Multiplying out these results in , which is making the answer .
~darrenn.cp
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=SUnhwbA5_So
See also
2023 AMC 10B (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 10 Problems and Solutions |
2023 AMC 12B (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.