Difference between revisions of "1953 AHSME Problems/Problem 38"

Latest revision as of 00:42, 4 February 2020

If $f(a)=a-2$ and $F(a,b)=b^2+a$ , then $F(3,f(4))$ is:

$\textbf{(A)}\ a^2-4a+7 \qquad \textbf{(B)}\ 28 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 11$

We find $f(4)=(4)-2=2$ , so $F(3,f(4))=F(3,2)=(2)^2+3=\boxed{\textbf{(C) }7}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 37	Followed by Problem 39
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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 We find <math>f(4)=(4)-2=2</math>, so <math>F(3,f(4))=F(3,2)=(2)^2+3=\boxed{\textbf{(C) }7}</math>.
+==See Also==
+{{AHSME 50p box|year=1953|num-b=37|num-a=39}}
+{{MAA Notice}}