Answer:
- <em><u>5.6875 in</u></em>
Explanation:
At the point of tangency, the <em>tangent </em>to a circle and the <em>radius</em> form a right triangle (the radius is perpendicular to the tangent).
Here you are given the length of the tangent (6in), and the distance from the bisected vertex to the circle (2.75 in)
I tried to upload the drawing but the tool is not allowing it now.
In the figure:
- The length of the tangent (6 in) is one leg of the triangle
- The distance from vertex and the circle (2.75in) along with the radius forms the hypotenuse of the right triangle: 2.75 + r.
- The other leg is the radius, r.
Then, you can use Pythagorean theorem:
Solve:
- r² + 36 = r² + 5.5r + 7.5625
The solution is in inches: r = 5.6875 inches ← answer
<span>1/r + 2/1-r = 4/r^2
1-r+2r/r(1-r)=4/r^2
(1+r)/r(1-r)=4/r^2 cancle r both side
1+r/1-r=4/r
cross multiply
r+r^2=4-4r
r^2+4r+r-4=0
r^2+5r-4=0
r^2+4r+r-4=0
solve it for r factor it...
</span>
So assuming that youmeans there were 35 MORE rose plants than sunflower seeds, then
Rose:Sunflower =8:3
rose=sunflower+35
therefor
8 units=rose
3 units=sunflowers
rose-35=sunflower
8unit-35=3unit
5unit-35=0
5unit=35
therefor
35=5unit
divide by 5
7=1unit
so 8 units of roses
7 times 8=56
3 units of sunflowers
3 times 7=21
rose=56
sunflower=21
Answer:
5
Step-by-step explanation:
k(x)=5 says I'm 5 no matter the value of x...
So therefore
k(-4)=5
k(56)=5
k(66378)=5
k(whatever)=5
k(x) is constantly 5 for whatever input x.
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.
