Difference between revisions of "G285 2021 MC10B"

Revision as of 15:52, 12 May 2021

Problem 1

Find $\left \lceil {\frac{3!+4!+5!+6!}{2+3+4+5+6}} \right \rceil$

$\textbf{(A)}\ 42\qquad\textbf{(B)}\ 43\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 45\qquad\textbf{(E)}\ 46$

Solution

Problem 2

If $deg(Q(x))=3$ , and $deg(K(x))=2$ , and $Q(x)=(x-2)K(x)$ , what is $deg(Q(x)-2K(x))$ ?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

Problem 3

A convex hexagon of length $s$ is inscribed in a circle of radius $r$ , where $r \neq s$ . If $\frac{s}{2r}=\frac{21}{29}$ , and $rs=58$ , find the area of the hexagon.

$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 120\qquad\textbf{(D)}\ 240\qquad\textbf{(E)}\ 480$

Solution

@@ Line 2: / Line 2: @@
 Find <math>\left \lceil {\frac{3!+4!+5!+6!}{2+3+4+5+6}} \right \rceil</math>
-<math>\textbf{(A)}\ \frac{290}{7}\qquad\textbf{(B)}\ \frac{890}{21}\qquad\textbf{(C)}\ \frac{89}{2}\qquad\textbf{(D)}\ \frac{87}{2}\qquad\textbf{(E)}\ \frac{223}{5}</math>
+<math>\textbf{(A)}\ 42\qquad\textbf{(B)}\ 43\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 45\qquad\textbf{(E)}\ 46</math>
 [[G285 MC10B Problems/Problem 1|Solution]]

Page

Toolbox

Search

Difference between revisions of "G285 2021 MC10B"

Revision as of 15:52, 12 May 2021

Problem 1

Problem 2

Problem 3