The midpoint of (x1,y1) and (x2,y2) is
just average them
(4,1) and (-2,4)
the midpoint is
(1,-1.5)
All we need to do is start with the volume formula. Then, input the volume and solve the equation for the radius. And double it to get the diameter.
V = (4/3)pi(r^3)
12114.7 = (4/3)pi(r^3) Divide both sides by (4/3) and pi
2893.64 = r^3 Take the cube root of both sides
14.25 = r
So the diameter is about 28.5 cm (14.25 x 2 = 28.5)
Answer:
Step-by-step explanation:
2005 AMC 8 Problems/Problem 20
Problem
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$
Solution
Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.
See Also
2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 Followed by
Problem 21
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All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
In order to get an 81 average,
the student has to get 81x5=405
71+96+90+73+?=405
330+?=405
405-330=75
he needs to get an 75
Unit rate is when a rate is simplified so it has a denominator of 1 (x = 1) and y≠ 0
unit rate = y/x
Answer: Point C <span>represents the unit rate</span>
