# School of Mathematics Help Desk

Welcome to the School of Mathematics Help Desk!

This is the place to ask any questions you have about mathematics, mathematics homework, or anything else math related.

Note: You can use TeX mathematics markup to make formulae easier to read

## Equal Temperament Question

(Moved to bottom of this page.) StuRat 16:07, 28 May 2007 (UTC)

## Homework Questions

How can N dots be placed on the surface of a sphere, equally distanced or evenly spaced ? What kind of math is needed for this kind of problems ?

What does evenly spaced mean? The simplest example - put them all on the equator. Are you talking about the w:Platonic solids?
Find area of the sphere and divide it by N. If an area element can be devised that has this area with a dot in the middle then you can divide and place dots without pesky off by one errors. Square root of the area element might give an approximatation of the size of the area element. This type of approach is related to limits and calculus. Mirwin 06:23, 20 September 2007 (UTC)
This Q was discussed in depth here: w:Reference_desk/Archives/Mathematics/2007_August_29#Distributing_points_on_a_sphere. StuRat 14:32, 27 November 2009 (UTC)

What is a complex plane?

Expanding a bit. A complex number has two components which are different types of numbers. A common representation is a real number plus an imaginary number (a real number multiplied by the square root of minus one or i). The set of complex numbers x + y(i) represented as a plane can be plotted two dimensionally by treating the x as an x coordinate and the y(i) as a y-coordinate. Mirwin 11:41, 25 January 2007 (UTC)
"The" complex plane, rather than "a" complex plane, I'd say better. --Jorge 02:22, 6 February 2007 (UTC)
Check out complex numbers I will explain all there Graemeb1967 11:12, 27 November 2009 (UTC)

How do you find the capacity of a cylinder in litres?

The formula for the volume of a right cylinder is V =Ah. That is, the volume equals the area of one of the bases times the cylinder's height. In the case of a circular cylinder, the area of a base is A = πr2. So, the volume of a right circular cylinder is V = πr2h. Now, if you start with radius and height in centimeters, this will give you cubic centimeters. If you start with inches, this will give you cubic inches. The final step is to convert these volume units into liters. This site lists the conversion factors needed: [1]. StuRat 21:13, 17 April 2007 (UTC)

If there are twenty-five people in a room and everyone shakes hands with everyone else once, how many handshakes are there? (I got three hundred but my textbook says three hundred and twenty-five which I think is wrong. If you also got any of these two answers please tell me and explain how.) (The preceding unsigned comment was added by Ricky Di Don (talkcontribs) 16 October 2016‎)

See http://www.mathcircles.org/node/835 for the solution to this problem. -- Dave Braunschweig (discusscontribs) 03:34, 17 October 2016 (UTC)
Yes, seems to me that Ricky Di Don is right and the textbook is incorrect. It would be 25 choose 2 and not 26 choose 2. Jjjjjjjjjj (discusscontribs) 07:34, 28 April 2017 (UTC)

## Petroleum Tank Measuring

How do you know how many gallons are in a petroleum tank that is horizontal ? What is the math equation. Note, I am not a math wizard so please use examples that a lay person can understand.

What shape is the tank? Also, do you have any more details about the tank (for example, do you know the height of the petroleum as measured from the bottom of the tank)? Deltinu 05:04, 19 January 2007 (UTC)

## Interested in Brushing Up, Where Do I Start

I'm interested in brushing up on my math and have "Calculus by Discovery" on loan from a friend. I'm looking for something here that would complement that. I found the Intro to Calculus page very confusing. My question is where do I start, and is there someone to whom I can pose questions or talk to to make sure I'm understanding everything correctly.

Also, on an only tangentially related note, has there been any discussion of a mathematics-through-application course? One that would pose (word) problems and then explain the math in them. I ask because I'm 27 and it took until now for numbers to mean more to me than simply a grade on a report card.

Feel free to pose any question at my Talk Page or just right here. I'll try to answer them to my best. About the course you mention, I don't know if anything has been talked. --Jorge 07:36, 9 February 2007 (UTC)

A section on "Word problems" would be great! Feel to start it or continue asking word problems here. Eventually someone will probably edit the best of this archive into a page such as you suggest. Great idea! Thanks for contributing here. Mirwin 06:27, 20 September 2007 (UTC)

I think that I'm in much the same boat as the one who started this entry. I, too, would like to brush up on my math skills, but I'm pretty sure that I'm further behind than he is (calculus comes after trig, right? ) The trouble is that I'm not just uncertain about how and where to begin; I'm also uncertain of where, exactly, I'm at. I was hoping someone could direct me to some kind of online self-assesment site, test, or application to figure out where I fell behind. I've been out of school sometime now, and it seems like everything after 8th grade is just a blur. I'm primarily interested in algebra, trigonometry and geometry, but I think it makes sense to assess where I'm at before I start looking for learning materials. Any advice?

Khanacademy.org has lots of content including, most notably, a collection of video lectures covering different topics. It might be helpful. 173.154.158.197 06:51, 22 March 2011 (UTC)

## Equal Temperament Question

Hi, can you outline the steps to get the answer 261.626 by the below equation on a scientific calculator?

${\displaystyle P_{40}=440_{Hz}\times 2^{\frac {40-49}{12}}\approx 261.626_{Hz}}$

If possible, using this online calculator: http://www.calculator.com/calcs/calc_sci.html. Thank you.

You can do this, as is, using parens, but it's far easier if you do part of it in your head and/or use the calculator memory (or write down) parts of it. Here are some ways to do it:

Here's the long way:

440 × 2 yx ( ( 40 - 49 ) ÷ 12 ) =


440 × 2 yx ( 9 ± ÷ 12 ) =


Here's an even shorter way (doing the division in your head, too), which eliminates the need for parens:

440 × 2 yx .75 ± =


Note that the ± key is labeled +/- on the calculator at that site, and is located right under the + sign. It toggles the current displayed value between positive and negative. That calculator doesn't appear to do rounding, so you will need to round off to 3 decimal places yourself. StuRat 16:22, 28 May 2007 (UTC)

## Different Fonts

Ergo, ${\displaystyle g=Ar=R\omega ^{2}=L(2\pi /T)^{2}}$

• and ${\displaystyle T=2\pi {\sqrt {(}}L/g)}$

Why are the above equations in two different fonts? How can I change the top equation to the font size of the bottom eqation? Calgea 18:17, 28 May 2007 (UTC)

If the equation is simple enough, it will be rendered as HTML. There's an option in your preferences page to change this. You can also force PNG rendering by inserting and removing a space (\,\;) -- take a look: ${\displaystyle g=Ar=R\omega ^{2}=L(2\pi /T)^{2}\,\;}$. See also m:Help:Displaying a formula. HTH. --HappyCamper 18:38, 28 May 2007 (UTC)
I agree with the above except you only need the "\;" part. So, this formula:

$g = Ar = R\omega^2 = L(2\pi/T)^2 \;$

gives you this:

${\displaystyle g=Ar=R\omega ^{2}=L(2\pi /T)^{2}\;}$

StuRat 02:16, 29 May 2007 (UTC)

## Wikiversity Talk: Naming Conventions # Removing Course Numbering Scheme

At "Removing Course Numbering Scheme" Talk #1 & "Removing Course Numbering Scheme" Talk #2 are Discussions in which the Undersigned would wish to have the benefit of the thinking of our Mathematicians. Could some Mathematicians please weigh in on these Discussions. Thank you in advance for your assistance.

Best regards,

(s) Dionysios (talk), a Participant in the Wikiversity School of Advanced General Studies, Date: 2007-07-23 (July 23, 2007) Time: 1859 UTC

## constant growth

how does this happen:

• ${\displaystyle V_{0}={\frac {D_{0}(1+g)}{(1+k)}}+{\frac {D_{0}(1+g)^{2}}{(1+k)^{2}}}+...+{\frac {D_{0}(1+g)^{n}}{(1+k)^{n}}}}$

simplifies into this:

• ${\displaystyle V_{0}={\frac {D_{0}(1+g)}{k-g}}}$

what's the rule or whatever? thanks.

-kv2 00:00, 27 August 2007 (UTC)

### Power series

See w:Power series. Is this a homework problem?

${\displaystyle \sum _{n=0}^{\infty }x^{n}={\cfrac {1}{1-x}}\quad {\text{for}}~|x|<1}$

If you case, assuming that ${\displaystyle n\rightarrow \infty }$ and ${\displaystyle g you get a series

${\displaystyle {\cfrac {D_{0}(1+g)}{1+k}}\sum _{n=0}^{\infty }x^{n}}$

where

${\displaystyle x:={\cfrac {1+g}{1+k}}}$

You result follows after a another step of algebra. -- Banerjee 02:13, 27 August 2007 (UTC)

## These 2 phrases are equivalent?

Given that f(x) is a boolean function, if

${\displaystyle g(x)\in \mathbb {R} \implies f(g(x))}$

Can we imply the following phrase?

${\displaystyle x\in \mathbb {R} \implies f(x)}$

Are they equivalent? --Ans 11:19, 23 February 2009 (UTC)

Yes if g is sujective.

With, ${\displaystyle g(x)\in \mathbb {R} \implies f(g(x))}$ (1) and ${\displaystyle y\in \mathbb {R} \implies f(y)}$ (2)

Indeed, to prove (2) from (1) : if ${\displaystyle g}$ is surjective, for any ${\displaystyle y}$ you can find an ${\displaystyle x}$ such that ${\displaystyle y=g(x)}$ then apply (1).

## Rectangle packing?

(posted here as I'd prefer an algorithmic explanation than program code)

For a set of small rectangles whose dimensions are random, place these rectangles within the minimum number of fixed size larger rectangles? Any thoughts? If it helps the context is image printing... Sfan00 IMG 23:47, 21 March 2011 (UTC)

Just to try to understand better, I have some questions:
1) Are each of the small rectangles the same size as each other ?
2) Are each of the large rectangles the same size as each other ?
For now, I'm assuming that the small rectangles vary in size, while the large rectangles do not. Some general thoughts:
A) This is going to require lots of trial-and-error, so a computer program probably is the best way to go, if you have lots of rectangles.
B) I think the edges of all the rectangles should always be parallel, for ideal packing. Thus, the only possible rotation is 90 degree (180 or 270 just give you the same thing as 0 and 90). Only rotate the small rectangles.
C) Start by placing the largest of the small rectangles (if they are different sizes). Then place the next largest small rectangle, etc., until all have been placed, trying the 90 degree rotation at each step.
D) Calculate the "packing factor" (percent of space used) at each step, then try the next arrangement, until you find the best arrangement. StuRat 00:13, 26 March 2011 (UTC)
For context,
• The smaller rectangles are not all the same size, and the sizes are random. (representing bitmap graphics areas). However it is a finite set of rectangles.
• the larger rectangles are a fixed size ( representing a sheet of printer paper , for example A4)

I don't need a 100% optimal soloution ( which would be rather hard to compute for any given set of random rectangles I'm told) You are correct about the simplification you note in (B), that the only possible orientations are landscape or portrait..

Starting with the largest rectangle is where I would have started as well.

In fact, I would probably go further and arrange the incoming rectangles so that all have the same orientation, either portrait or landscape , with the longest side of the smaller rectangles parllel to the longest side of the larger one.

In reading around I've come across some demonstrations of shelf-packing algorithms here - http://users.cs.cf.ac.uk/C.L.Mumford/

Sfan00 IMG 11:53, 26 March 2011 (UTC)

I don't think there's any reason why arranging all the rectangles in the same orientation is likely to produce optimal packing. Here's a counter-example:
+------------------+
| +--------------+ |
| |              | |
| |              | |
| +--------------+ |
| +------+ +-----+ |
| |      | |     | |
| |      | |     | |
| |      | |     | |
| |      | |     | |
| |      | |     | |
| |      | |     | |
| +------+ +-----+ |
+------------------+

StuRat 20:52, 26 March 2011 (UTC)