Alan

Here is a brief version of how I approached it - pretty much as Tiggsy explained, though I did a numerical solution first to get a feel for the shape of the velocity as a function of time:

Note: Tiggsy approach from scratch is the better way to do this.

I think the solutions can be written as:

\(v=\sqrt{\frac{k_1}{k_2}}\tanh{(\sqrt{k_1k_2}t)}\text{ and }x=\ln{(\cosh{(\sqrt{k_1k_2}t)})}/k_2\)

which I obtained by guessing a few forms for the velocity and manipulating them until they matched numerical solutions for all the values of k1 and k2 that I tried!!   I've no doubt there is a more straightforward way of doing it!!

I interpret this question to mean the probability of the experiment ending exactly on the fifth throw. Melody has calculated the probability that it could end on any throw from the second to the fifth (I think!).

If my interpretation is correct, then the fourth and fifth throws must be 4s, and the third throw must be not4.   This means the possible combinations for the first two throws are (1) 4 followed by not4, (2) not4 followed by 4, (3) not4 followed by not4.

Putting all this in probability terms we have p = (1/6²)(5/6)[ (1/6)(5/6) + (5/6)(1/6) + (5/6)(5/6) ]

or p = 175/7776 ≈ 0.0225

This is borne out by a MonteCarlo simulation (which also confirms Melody’s calculation using her interpretation of the question).

The curve passes through (0, -13) so -13 = a*0²+b*0+c or c = -13

Hence y = ax² + bx -13

Maximum occurs when  y = 5 and x = 3, so 5 = 9a + 3b -13

or 9a + 3b = 18, or 3a + b = 6 ...(1)

The slope is given by slope = 2ax + b, and this must be zero at a maximum,

so 2a*3 + b = 0 or 6a + b = 0 ...(2)

You can use equations (1) and (2) to find a and b, then use

m = a*4² + 4b - 13 to calculate m.

You don’t show the original triangle, so I can only guess that you’ve got the lengths p and q interchanged.

Because sin(65)*22/18 is bigger than 1, and there is no angle whose sine is bigger than 1.

The equation of the red dotted line is of the form y = mx + c.

It passes through points (3,1) and (1, -2), so the slope, m, is m = (1-(-2))/(3-1) or m = 3/2.

Since it passes through (3,1) we have 1 = (3/2)*3 + c, so c = -7/2

Hence The inequality describing the yellow region is y >= (3/2)x - 7/2

.

Like so:

Two solutions:

Not sure which you meant by the third (including or excluding the one I did!), so: