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February 23

Math Problem

5≤ -s/12 -3 — Preceding unsigned comment added by 68.80.157.180 (talk) 02:17, 23 February 2012 (UTC)[reply]

Did you have a question? RudolfRed (talk) 02:48, 23 February 2012 (UTC)[reply]

I added a title. I think we can assume you want us to help solve the inequality for s. It works pretty much like solving an equality, with one big exception: "When you multiply or divide both sides of an inequality by a negative number, you also flip the sign of the inequality". Thus, you will flip ≤ to ≥. (There are other differences in solving inequalities, such as dealing with square roots, but those don't come into play here.) If you show us your work, we would be glad to check it for you. StuRat (talk) 03:09, 23 February 2012 (UTC)[reply]

Try www.wolframalpha.com : [1] Bo Jacoby (talk) 08:36, 23 February 2012 (UTC).[reply]

Trig function help

Hello, how would I accomplish this trig problem?

cos (arcsin (5/13) - arctan (3/4))

thanks.--Prowress (talk) 03:43, 23 February 2012 (UTC)[reply]

Hint: determine the missing lengths of the triangles involved, then use the identity $\cos {(A-B)}=\cos {A}\cos {B}+\sin {A}\sin {B}$ . --Kinu ^t/_c 04:19, 23 February 2012 (UTC)[reply]

Difference between a series and a progression

Could you please explain the difference between a Series and a progression with examples? Kasiraoj (talk) 08:09, 23 February 2012 (UTC)[reply]

You could start by working out the difference between a sequence and a series, and then seeing how the listed examples of a progression fit into the picture. — Quondum ^☏✎ 09:04, 23 February 2012 (UTC)[reply]

Uniqueness of injective/flat resolutions

The article resolution (algebra) article mentions projective resolutions are unique up to chain homotopy. I would guess the same can be said for injective and flat resolutions, but I'm not familiar enough with the subject matter to verify. Is this the case? Thanks! Rschwieb (talk) 18:27, 23 February 2012 (UTC)[reply]

Yes injective resolutions are unique up to homotopy, however I don't know about flat ones, although I think they're unique up to quasi-isomorphism. Money is tight (talk) 22:48, 25 February 2012 (UTC)[reply]

If you have the time to point me to a reference that I can look at, I can take care of the footwork to put in the info :) Thanks! Rschwieb (talk) 14:16, 27 February 2012 (UTC)[reply]

Heart-shaped function

I found it quite interesting to find out that the function:

$(\cos(100x){\sqrt {\cos(x)}}+{\sqrt {|x|}}-0.7)(4-x^{2})^{0.01}$

gives a good approximation of a heart shape when viewed graphically (e.g. here).

How would this fairly complicated equation have been derived? Trial and error, or chance? Is there software for generating a function that will deliver a certain shape? --Iae (talk) 20:01, 23 February 2012 (UTC)[reply]

I don't know, but a Limaçon can be heart-shaped. Bubba73 ^{You talkin' to me?} 20:20, 23 February 2012 (UTC)[reply]

You might also be interested in [2]. There was probably some trial and error involved in finding the function but there is some cleverness involved as well. For example the cos 100x produces the up and down squiggles that fill in the shape. The sqrt(cos x) gives the rounded bits on either side and the sqrt(|x|) produces the V shape in the middle. You might try playing with the formula to see what effects small changes have, for example to find out what happens if you change the .7 to a .8 or a .6.--RDBury (talk) 22:45, 23 February 2012 (UTC)[reply]

MATH GESSUING

I am a two digit number over 50.When you put me in groups of 7,2 are left over.The sum of my digits are 11. What number am I ? — Preceding unsigned comment added by 24.163.1.170 (talk) 22:01, 23 February 2012 (UTC)[reply]

You are not a number! You are a free man. --Trovatore (talk) 22:17, 23 February 2012 (UTC)[reply]

The answer is 65. The title of this question is misleading, though. --COVIZAPIBETEFOKY (talk) 22:50, 23 February 2012 (UTC)[reply]

I'm still trying to figure out why the "over 50" caveat is given. Number is of the form 7n+2 from the one criteria and a subset of those of the form 9n+2 from the other so the answers are a subset of the numbers of the form 63n+2, for n=0 the answer is 2 (which doesn't fit the second criteria exactly, n=1 gives 65 and n=2 gives 128 which is too big. Without the requirement that it be a two digit number, the numbers start with (marked * sum to another 9n+2 other than 11) are 2*,65, 128, 191, 254, 317, 380, 443, 506, 569*, 632, 695*, 758*, 821, 884*, 947*, 1010*, 1073, etc. I don't know if there is a clean way to find out what the last number of the form 63n+2 whose digits sum to 11, but we know it can't be more than 11111111111.Naraht (talk) 03:52, 24 February 2012 (UTC)[reply]

What about 10000001111111111, 10000000000001111111111, 10000000000000000001111111111, ... ?--RDBury (talk) 12:01, 24 February 2012 (UTC)[reply]

There isn't a largest. 6 + 3 + 2 = 11, so setting n to any (positive integer) power of 10 gives you a number that fits the form. 632, 6302, 63002, 630002... Smurrayinchester 11:29, 27 February 2012 (UTC)[reply]

86.174.199.35 (talk) 16:40, 24 February 2012 (UTC)[reply]

Area enclosed

Hello. Suppose I have an arbitrary function defined by a parametric which on some interval of the parameter is closed and simple. How would I find the area enclosed by the curve? (without doing something ugly like splitting it into two regular functions and integrating each, then subtracting of course... *shudder*...) Thanks. 24.92.85.35 (talk) 23:50, 23 February 2012 (UTC)[reply]

See Green's theorem#Area_Calculation – b_jonas 10:08, 24 February 2012 (UTC) (Update: deleted the formula. It was wrong. – b_jonas 10:09, 24 February 2012 (UTC))[reply]

This means that if you have the point of the curve parametrized as (x(t), y(t)) where ~~(x runs)~~ t runs from a to b, and this is a closed curve so x(a) = x(b) and y(a) = y(b), then you can get the area as the absolute value of

\int _{a}^{b}y{\frac {\partial x(t)}{\partial t}}\,dt

– b_jonas 10:13, 24 February 2012 (UTC)[reply]

t runs from a to b.86.174.199.35 (talk) 16:40, 24 February 2012 (UTC)[reply]

Right. Fixed it. – b_jonas 17:08, 25 February 2012 (UTC)[reply]