Difference between revisions of "1957 AHSME Problems/Problem 6"
(Created page with "The resulting metal piece looks something like this where the white parts are squares of length <math>x</math>: <asy> fill((0,4)--(4,4)--(4,0)--(6,0)--(6,4)--(10,4)--(10,10)-...") |
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From here, try to visualize the rectangular prism coming together and realize the height is <math>x</math>, the length is <math>14-2x</math>, and the width is <math>10-2x</math>. Therefore, the volume is <math>x(14-2x)(10-2x)=x(4x^2-48x+40)= | From here, try to visualize the rectangular prism coming together and realize the height is <math>x</math>, the length is <math>14-2x</math>, and the width is <math>10-2x</math>. Therefore, the volume is <math>x(14-2x)(10-2x)=x(4x^2-48x+40)= | ||
\boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}</math>. | \boxed{\textbf{(A) } 140x - 48x^2 + 4x^3}</math>. | ||
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+ | {{AHSME box|year=1957|ab=B|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Revision as of 01:05, 19 June 2019
The resulting metal piece looks something like this where the white parts are squares of length :
From here, try to visualize the rectangular prism coming together and realize the height is , the length is , and the width is . Therefore, the volume is .
1957 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.