catmg

Responded in first post. 

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Adding on to what Melody said. :))

Let theta = arccos(-1/9)

cos(theta) = -1/9

Note that the range of arccos is [-pi/2, pi2].

Can you figure out what sin(theta) is?

Hint: look at Melody's hint with the triangle and see if you can construct another one. 

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Thank you, it's always nice to see the problem in a graph. :))

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Nice solution, however I also tried doing this problem and got a different answer. 

I viewed all the cards as different, so the number of ways to distribute the cards was 9!/3!/3!/3! = 1680. 

I think 9C3 is the number of ways to give out 3 cards to one person, not all 3.

SImilar to what you did, I then assumed everyone recieved a red card, 6 ways of doing this. 

The number of ways to distribute the remaining 6 cards was 6!/2!/2!/2! = 90.

90*6/1680 = 32.14%

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(10)^2005 * 5^2 * 3

Does writing it like this help?

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Let r be the radius of the circle. 

The circle must have a center of (3 - r, 0).

So our formula is (x - 3 + r)^2 + y^2 = r^2

9(x - 3 + r)^2 + 9y^2 = 9r^2

x^2 /9 + y^2 /4 = 1

4x^2 + 9y^2 = 36

Subtracting the two formulas. 

9(x - 3 + r)^2 - 4x^2 = 9r^2 - 36

5x^2 + (18r - 54)x + (117 - 54r) = 0

There is one solution when the discriminant is 0. 

(18 - 54r)^2 - 4(5)(117 - 54r) = 0

r = 2(2 + sqrt(130))/27 (note that r can't be negative so it's not the other option)

Hope this was helpful. 

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Nice work. :))

One small correction, log_2(3x) = 3, turns into 3x = 8, (2^3 = 8)

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Try expanding (x^2 - 1)(x - 1). 

See what happens. :))

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I think that perhaps using AM-GM gets the right answer.

(a + 5b + 15c)/3 >= cbrt(75*abc)

abc = a+5b+15c = x

x/3 >= cbrt(75*x)

And from there you can use algebra to simplify and solve. :))

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I don't think the question is referring abc as 100a + 10b + c since a, b, and c are positive real numbers, not integers from 0-9. 

I think abc is a*b*c. 

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