Alan

Let x = 0.572727272.....     (1)

Multiply (1) by 1000:

1000x = 572.727272.....     (2)

Multiply (1) by 10:

10x = 5.72727272.....         (3)

Subtract (3) from (2)

990x = 567          (4)

Divide both sides of (4) by 990;

x = 567/990

or x = 63/110

.

"It wasn't me Alan, I didn't put anything but I also noticed that error, I suppose I should have said so :))

Could also have one die with faces numbered 1 to 6 as normal, and the other die with half the faces numbered 6 and half numbered zero.

Or one die with faces numbered 7 to 12, and the other die with half numbered -6 and half zero.

Or one die with faces numbered 13 to 18 and the other with half numbered -12 and half numbered -6.

Or one die with ...!!!

.

This graph doesn't constitute a proof, but simply suggests that it's likely to be true (though you should have put 0.07 rather than 7 Melody!):

.

It's not clear to me what you are asking. 

y is used as part of the calculation of r using Pythagoras's theorem.

dy is used to indicate a small (infinitesimal) region of the linear charge distribution.

.

1. Yes, you are correct.

2. When log₅x = 0 then log₂₅y = 2

This means that when x = 5⁰ then y = 25²  or when x = 1 then y =625

If, in general, y = k*xⁿ then we must have 625 = k*1ⁿ or just k = 625

.

The geometric mean only applies to positive numbers.

It has a geometric interpretation.  Image a rectangle with sides 10 and 30.  It has an area of 300.  A square of side √(10*30), i.e. one with side lengths equal to the geometric mean of the two rectangle lengths, has the same area.  

Similarly, for higher order geometric means. e.g. a cuboid with side lengths u, v and w has the same volume as a cube with side lengths ³√(u*v*w), etc.

.

For interest's sake then, let's suppose the two cars collide elastically, conserving both energy and momentum.

Momentum before:  1300*30 - 900*15 kg.m/s = 

Momentum after: 1300v1 + 900v2   

so 1300v1 + 900v2 = 25500   ...(1)

Kinetic energy before:  1300*30²/2 + 900*15²/2 = 686250 J

Kinetic energy after:  1300*v1²/2 + 900*v2²/2

so 1300*v1²/2 + 900*v2²/2 = 686250   ...(2)

Equations (1) and (2) can now be solved for v1 and v2. (v1 = -75/11 m/s ≈ -6.8 m/s;  v2 = 420/11 m/s ≈ 38.2 m/s).

.

Except that P(A∩B) is given as 3!!!  Perhaps this was meant to be 0.3.

.

Here's an approximate geogebra confirmation of Melody's result.

Scale everything so that h = 1.  Look down from above to see the following:

The distances DE and AE are just 3/4 of distance AD (KD is √3).

The radius of the circle centred on K and going through E is approximately 2.04, so the angle of elevation from E to the top of the post is tan^-1(1/2.04)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{2.04}}}}\right)} = {\mathtt{26.113\: \!912\: \!630\: \!291^{\circ}}}$$

or approximately 26°

.