If two secants intersect from a point outside of the circle, then the product of the lengths of the secant and its external segment equals the product of the other secant and its external segment.
#1
5(x+5) = 6(4+6)
5x + 25 = 6 * 10
5x = 60 - 25
5x = 35
x = 7
#2
4(x+4) = 3(5+3)
4x + 16 = 3 * 8
4x = 24 - 16
4x = 8
x = 8/4
x = 2
Answer:
294 cars.
Step-by-step explanation:
Let x be the number of cars and y be the number of trucks.
We have been given that the first dealership sells a total of 164 cars and trucks. We can represent this information as:
The second dealership sells twice as many cars and half as many trucks as the first dealership. So the number of cars sold by 2nd dealership will be 2x and number of trucks sold by 2nd dealership will be y/2.
Further, the 2nd dealership sold a total of 229 cars and trucks. We can represent this information as:
We can see that total number of cars sold on two dealerships will be 
.
We will use substitution method to solve for x. From equation (1) we will get,
Substituting this value in equation (2) we will get,
Now let us have a common denominator.
Upon multiplying both sides of our equation by 2 we will get,
Therefore, the total number of cars sold by two dealerships is 294.
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.
Answer:
The answer is B..... The corresponding is the identical number between all
The is 7 in each...
