Answer:
The area of this triangle would be 24.
Step-by-step explanation:
In order to find this, we need to first determine what to use as the base of the triangle. Since when we draw it, the line between G and H is a flat line, it is the easiest to use as a base. Finding the length of the base is easy because we are simply looking for the difference in the x values.
6 - -2 = 8 = base
Now that we have the base, we need to find the height. The height is always the perpendicular line from the 3rd point to the base line. In this case, this would just be the change in y from the other two.
5 - -1 = 6 = height
Now that we have those two distances, we can just use the triangle formula.
A = 1/2bh
A = 1/2(8)(6)
A = 24
The answers are
1: False
2: True
3: True
4: False
Answer:
16 Days
Step-by-step explanation:
$8 x 1 = 8
$8 x 16 = 128
16 Days!!!!
If Alicyn makes $8 everyday for babysitting her little sister, It'll take her 16 days to make $128.
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
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.
