Math Is Fun Forum

I like this classic one:

A neutron walks into a bar and asks for a drink. “How much is it?” he asks. The bartender replies: “For you, no charge.”

And this one:

1st hydrogen atom: Help! I’ve lost my electron!
2nd hydrogen atom: No, you can’t be serious!
1st hydrogen atom: Oh yes! I’m positive!

Well then, here’s one I just made up. It should be quite easy, I hope.

Don’t worry, it’s just a small mistake.

BTW, I’ll also appreciate it if people correct any mistakes they find in my posts, even my non-mathematical ones. I’ll be very grateful.

I already gave you a hint for #1 above. See earlier in the thread. To get each of the distances AC and BC, multiply the speed by the time and divide by 2 (which is what you’ve already done).

For #2, my answer is also 12:7. The length of each side of ∆LMN should be √21.

For #3, find the legnth PR, which is the sum of PT and RT. Since ∆PST, ∆SRT and ∆PRS are similar (they all have the same interior angles), this should be straightforward to work out. Next, calculate QR. Using the hint that ∆PRQ and ∆QRT are similar, you should get QR:PR = RT:QR. The answer for QR should be √3. Now, if U is the point on PQ such that PSRU is a rectangle, you know that PU = SR and UR = PS. Calculate UR and you’re done.

I have no idea how to solve it but I found the general solution on the Internet. Your equation is a so-called special Riccati equation, and – my word! The general solution involves Bessel functions of both the first and second kinds! Well, maybe the particular solution you’re looking for is simpler – but, even so …

Here it is, if you want to make sense of it:
http://eqworld.ipmnet.ru/en/solutions/ode/ode0106.pdf (PDF document)

It seems that the general Riccati equation is easier to solve than the special Riccati equation.

No, +2 (mod 19) and −2 (mod 19) are different. In fact, −2 ≡ 17 (mod 19), so 13[sup]2[/sup] is 17 in

For #1:

Let the point C be

You can even specify exactly how much spacing you want. For example, if you want exactly 30 mm of white space,

\hspace{30mm}

Thus:

Umm, 18[sup]2[/sup] is 1 in

Note that if x=n is a solution to x[sup]2[/sup] = 2 in

Similarly for

EDIT: Wait. I don’t think there’s a solution at all in

Technically, the big X is called a random variable. In this case, it represents the number of sixes from the throwing of m dice.

I would think of a straight line as having infinite (rather than zero) radius of curvature. On the other hand, I would regard a point of inflexion (where the curvature is also 0) as having zero radius of curvature. This is just based on my intuitive geometric visualization, of course.