Alan

-1 = 5 + x/6

Subtract 5 from both sides

-6 = x/6

Multiply both sides by 6

-36 = x  (or x = -36)

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Another approach:

$$\\
\text{Let }\theta=\cos^{-1}x\\\\
x=\cos\theta\quad\text{so }dx=-\sin\theta.d\theta\\\\
\sqrt{1-x^2}=\sqrt{1-\cos^2\theta}=\sin\theta\\\\
\text{So }\\\\
\int\frac{(\cos^{-1}x)^2}{\sqrt{1-x^2}}dx=-\int\frac{\theta^2.\sin\theta}{\sin\theta}d\theta=-\int\theta^2d\theta=-\frac{\theta^3}{3}=-\frac{(\cos^{-1}x)^3}{3}$$

There is a constant to be added of course.

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Let b =initial number of blue tiles and w = initial number of white tiles.

Initially

b/w = 5/9  or b = 5w/9   (1)

After

(b+9)/(w-9) = 13/15  or 15(b+9) = 13(w-9)    (2)

Using (1) in (2)

15(5w/9 + 9) = 13(w - 9)

15*5*w/9 + 15*9 = 13*w - 13*9

25w/3 + 135 = 13w - 117

Multiply through by 3

25w + 405 = 39w - 351

39w - 25w = 405 + 351

14w = 756

w = 756/14 = 54

Put this back into (1)

b = 5*54/9 = 30

Hence total number of tiles: b + w = 30 + 54 = 84

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The state of the Markov chain after a time t depends on the initial state.  The occupancy of state 0 after time t is its initial occupancy (1/3) multiplied by the sum of the probabilities of the transitions to and from that state from the other states (the first column of P(t)).

Notice that the sum of the terms in the first column of P(t) given in the answer to part 4 is divided by 3 (i.e. multiplied by 1/3).

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Does this help?

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Are you sure the answer given for part 1 is correct?  Looks to me like the 3 on the bottom line of Q should be -3.

If you then calculate P(0.1) you get the numbers in the first column in part 4 the same as given in the answer.

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A loonie is a Canadian one-dollar coin.

A toonie is a Canadian two-dollar coin.

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Let the cursor hover over the log button.  You will see a mini menu appear on which is ln (log to the base e).

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k² + 8k + 15 → (k + 3)(k + 5)

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Suppose the rational number is m/n, where m and n are integers.

Let the irrational number be t

Assume the product t*m/n is rational; then it could be represented by a/b, where a and b are integers.

So we would have a/b = t*m/n

Multiply both sides by n/m to get:  t = a*n/(b*m)

a and n are integers so a*n = p, another integer.

Similarly, b*m = q, an integer.

But this means that t = p/q, the ratio of two integers.  In other words, t, our irrational number turns out to be rational!  

This is a contradiction of course, which means that our initial assumption, that an irrational number multiplied by a rational number is rational, is false.

In other words, the product of an irrational number and a rational number is irrational.

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