Difference between revisions of "2022 AMC 8 Problems/Problem 2"
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<cmath> =((a^2-b^2) \, \bigstar \, c) </cmath> | <cmath> =((a^2-b^2) \, \bigstar \, c) </cmath> | ||
<cmath> =(a^2-b^2-c)^2 </cmath> | <cmath> =(a^2-b^2-c)^2 </cmath> | ||
− | <cmath> =a^4+b^4-(a^2)(b^2)-2(a^2)(c)-(b^2)(a^2)+2(b^2)(c)+ | + | <cmath> =a^4+b^4-(a^2)(b^2)-2(a^2)(c)-(b^2)(a^2)+2(b^2)(c)+c^2 </cmath> |
− | <cmath> =5^4+3^4-(5^2)(3^2)-2(5^2)(6)-(3^2)(5^2)+2(3^2)(6)+ | + | <cmath> =5^4+3^4-(5^2)(3^2)-2(5^2)(6)-(3^2)(5^2)+2(3^2)(6)+6^2 </cmath> |
<cmath> =625+81-225-300-225+108+36 </cmath> | <cmath> =625+81-225-300-225+108+36 </cmath> | ||
<cmath> =\boxed{\textbf{(D) } 100} </cmath> | <cmath> =\boxed{\textbf{(D) } 100} </cmath> | ||
− | ~ | + | ~megaboy6679 |
==Video Solution 1 by Math-X (First understand the problem!!!)== | ==Video Solution 1 by Math-X (First understand the problem!!!)== | ||
Line 61: | Line 61: | ||
~harungurcan | ~harungurcan | ||
+ | |||
+ | ==Video Solution 7 by Dr. David== | ||
+ | |||
+ | https://youtu.be/ItPIuHdgyNk | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=1|num-a=3}} | {{AMC8 box|year=2022|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:47, 19 November 2024
Contents
Problem
Consider these two operations: What is the output of
Solution 1
We can substitute , , and into the functions' definitions: ~pog ~MathFun1000 (Minor Edits)
Solution 2
We can find a general solution to any .
~megaboy6679
Video Solution 1 by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=JlQdlwkdbIYFNJXj&t=123
~Math-X
Video Solution 2 (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution 3
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=91 ~Interstigation
Video Solution 4
~savannahsolver
Video Solution 5
https://youtu.be/Q0R6dnIO95Y?t=53
~STEMbreezy
Video Solution 6
https://www.youtube.com/watch?v=c123kPqd11I
~harungurcan
Video Solution 7 by Dr. David
See Also
2022 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.