Alan

cosx - cosxsin²x → cosx(1 - sin²x) → cosx*cos²x → cos³x

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There are 20 cards that are black and numbered, so the probability of drawing one of these is 20/52 or   5/13.

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Let T be total number of votes.

685 = 2T/3 + 125

560 = 2T/3

3*560/2 = T

T = 840

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Let x = 0.1919191919.....

100x = 19.19191919....

Subtract x from 100x to get 99x = 19

Hence x = 19/99

If x = 0.199999999.....

10x = 1.999999999...

100x = 19.9999999...

Subtract 10x from 100x so 90x = 18 

x = 18/90 → 1/5

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Hmm.  Guest #1 seems to have forgotten the initial 3.

3 - x/(x-3) = 2x/(x+6)

Multiply by (x-3)(x+6):

3(x-3)(x+6) -x(x+6) = 2x(x-3)

3(x^2 + 3x - 18) - x^2 - 6x  = 2x^2 - 6x

2x^2 + 3x - 54 = 2x^2 - 6x

9x = 54

x = 6

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Because the function is a straight line the range will be given by the value of the function at the lowest value of the domain, namely f(-2), and the value of the function at the highest value of the domain, namely f(7).

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Imagine a right angled triangle with hypotenuse of length 6 and opposite side to angle t of length sqrt(7).  Then the adjacent length is sqrt(6^2 - sqrt(7)^2) or sqrt(36 - 7) or sqrt(29).  Hence cos(t) = sqrt(29)/6

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N = N₀e^(ln(2)*t/T) where t is time and T is half life

250 = 600e^(-ln(2)*20/T)

Divide by 600 and take logs

ln(25/60) = -ln((2)*20/T

Rearrange

T = -ln(2)*20/ln(25/60) years. ≈ 15.8 years

10/(2/3) → 10*3/2 → 5*3 → 15

You will need to become pretty familiar with Bessel functions to answer Guest #3's new question Melody!  I suggest you just enter it into WolframAlpha.