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April 4

Foot and Penis Size

Hi, I heard that if you take 3 quaters of the measurement of your foot then times it by 3, it shows your erect size. Is this true? Thank you —The preceding unsigned comment was added by 202.161.2.238 (talk) 06:54, 4 April 2007 (UTC).[reply]

This reference desk is for questions about Mathematics, Calculus, and Accounting. --Lambiam ^Talk 08:05, 4 April 2007 (UTC)[reply]

No, it's not true, see this mention on Snopes.com for further information. - Mgm|^(talk) 09:30, 4 April 2007 (UTC)[reply]

If you take 3/4 of the length of a typical adult male foot, say 10 inches, then multiply by 3, you get 9/4 of 10 or 22.5 inches. Does that sound like the proper length to you ? StuRat 20:51, 4 April 2007 (UTC)[reply]

BTW, doesn't this question sound like it would also work under the title of the previous question ? :-) StuRat 20:53, 4 April 2007 (UTC)[reply]

Sounds proper to me! :D Splintercellguy 22:50, 4 April 2007 (UTC)[reply]

HAHA. --JianLi 01:24, 8 April 2007 (UTC)[reply]

Differentation

${\cfrac {(x+\delta x)^{2}-x^{2}}{\delta x}}$

Then...

${\cfrac {x^{2}+2x\delta x+\delta x^{2}-x^{2}}{\delta x}}$

Then...

${\cfrac {2x\delta x+\delta x^{2}}{\delta x}}$

According to my book, by simplification, we get:

$2x+\delta x\,$

But I notice that two $\delta x$ have been removed from the numerator when there is only one in the denominator. What have I missed here? —Preceding unsigned comment added by 164.11.204.51 (talk • contribs)

A single factor of

\delta x

has been removed from each of the terms in the numerator. Filling in the missing step, we have:

{\cfrac {2x\delta x+\delta x^{2}}{\delta x}}={\cfrac {\delta x(2x+\delta x)}{\delta x}}=2x+\delta x

... Gandalf61 10:16, 4 April 2007 (UTC)[reply]

largest prime number

i read in wikipedia that there are prizes to find the largest prime number. can we put (10^n)+1,n,is integer.then we start to pick up n`s as large as possible to find the lrgest prime number???is`nt 100001,10000000000001 are prime numbers??? 80.255.40.168 12:39, 4 April 2007 (UTC)fwfabii[reply]

Prime number records are typically created in this fashion. Take a look at Mersenne prime for example. (note that not all numbers of the form 10^n+1 are prime, for example 1001 is not. Sander123 12:45, 4 April 2007 (UTC)[reply]

To be precise, there is no such thing as the largest prime number. Euclid already proved that there are infinitely many prime numbers. The prizes are for very large prime numbers. The example numbers you give are not primes: 100001 = 11 × 9091 and 10000000000001 = 11 × 859 × 1058313049. They are also not very large; the largest of these two has 14 digits. To win the least of the prizes, the prime needs to have at least 10000000 digits; printed in a book that would take up something like one thousand pages. Although we have "fast" primality tests, they are not nearly fast enough to cope with primes that large. For that we need some mathematical breakthrough. --Lambiam ^Talk 13:05, 4 April 2007 (UTC)[reply]

Actually, there are known primality tests that can test numbers of special forms above 10,000,000 digits, for example the Lucas–Lehmer primality test. A test takes a long time and a lot of tests are probably needed. Great Internet Mersenne Prime Search (GIMPS) has tested many candidates with it and found the largest known prime which currently has 9,808,358 digits. Note that 10ⁿ+1 for a positive integer n is known to be composite when n is not a power of 2. And

10^{2^{m}}+1

grows so fast that it appears likely that 11 and 101 (for m = 1 and 2) are the only primes of that form. PrimeHunter 02:03, 5 April 2007 (UTC)[reply]

Angle problem

I was e-mailed this angle problem and I'm stuck trying to figure it out. It consists of a regular hexagon inside a circle. Below the image is what I've been able to figure out. Any help with this brainteaser would be greatly appreciated. Cheers. --MZMcBride 22:19, 4 April 2007 (UTC)[reply]

$\angle {1}$ = 30˚
$\angle {2}$ = 70˚
$\angle {3}$ = 50˚
$\angle {4}$ = 30˚
$\angle {5}$ = 45˚
$\angle {6}$ = 45˚

$\angle {7}$ = 120˚
$\angle {8}$ = 60˚
$\angle {9}$ = 90˚
$\angle {10}$ = 100˚
$\angle {11}$ = 50˚
$\angle {12}$ = 30˚

$\angle {13}$ = 60˚
$\angle {14}$ = 120˚
$\angle {15}$ = 45˚
$\angle {16}$ = 45˚
$\angle {17}$ = 60˚
$\angle {18}$ = 120˚

$\angle {19}$ = 90˚
$\angle {20}$ = 105˚
$\angle {21}$ = 75˚
$\angle {22}$ = 55˚
$\angle {23}$ = 125˚
$\angle {24}$ = 110˚

$\angle {25}$ = 70˚
$\angle {26}$ = 100˚
$\angle {27}$ = 80˚

Once you have found the answer for any of the blank slots, the remaining ones should be easy. Let us give a name to the point on the side AD of the squares serving as the vertex of angles 10, 11 and 12, and call it J. Also, for simplicity, set the length of the sides of the square to 1, and think of the diagram as embedded in the Euclidean plane with Cartesian coordinates, with B = (0, 0), A = (1, 0) and C = (0, 1). First, work out the length of the sides of the hexagon. That gives you the coordinates of H, and therefore the length of AH. Then find the length of AJ, and thereby the coordinates of J. Also find the coordinates of E. You then have the slope of the line EJ, and thereby the angle it makes with the horizontal. I performed these calculations, and unless I made a mistake, the answers are not nice round numbers. Once you've solved it, you can return the favour by posing essentially the same question with a pentagon. --Lambiam ^Talk 23:01, 4 April 2007 (UTC)[reply]

Just glancing over the data filled in, clearly angle 20 is wrong, and angle 13 is obvious (both for the same reason). When two lines cross, we have two pairs of equal angles, which will help. My suggestion is to use the theorem on external angles of a polygon. --KSmrq^T 00:29, 5 April 2007 (UTC)[reply]

Great catch on angle 20! I put the answers that I got up. The slope method seemed to work really well, the answer I got for the slope was 10˚. I wasn't really sure what you meant about using a pentagon instead, but I made one below anyway. Thanks for all the help. --MZMcBride 01:00, 5 April 2007 (UTC)[reply]

I meant sending it as a problem to whoever sent the first problem to you. Another variant would be one in which the upper vertex of the pentagon lies on the upper side of the square. --Lambiam ^Talk 11:00, 5 April 2007 (UTC)[reply]