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cone surface area question by Molly Blackman - Thu, 31 Jul 2014 06:31:54 EST ID:Yu7owZ2i No.14254 Ignore Report Reply Quick Reply
File: 1406802714233.gif -(961964B / 939.42KB, 180x155) Thumbnail displayed, click image for full size. 961964
a water tank has the shape of a cone. the tank is 10 meters high and has a radius of 3 meters at the top. if the water is 5 meters deep ( in the middle), what is the area of the top of the water?

thanks guise

pic unrelated
Archie Blickleshit - Thu, 31 Jul 2014 18:56:53 EST ID:atHzm3qN No.14255 Ignore Report Quick Reply
Oh, c'mon. You're not even trying.

A = π3 (3+10*2+3*2)

If you only care about the surface area of the top ( a circle) you would just take the square of r (= 3) and multiply it by pi.
Angus Ferrywell - Fri, 01 Aug 2014 03:08:22 EST ID:Yu7owZ2i No.14257 Ignore Report Quick Reply
but the radius given is the maximum radius at the top, how do you calculate the radius of the top circle formed by the water when its 5 meters deep?
Ian Sagglewick - Fri, 01 Aug 2014 11:37:04 EST ID:5quJ2CR5 No.14260 Ignore Report Quick Reply
Its a cone, you know the radius at the top and bottom and the height. Its just lines bro.
Jack Sizzlewodge - Sun, 03 Aug 2014 13:36:02 EST ID:8NZnQ0yA No.14262 Ignore Report Quick Reply
1407087362061.jpg -(24067B / 23.50KB, 292x327) Thumbnail displayed, click image for full size.
Solve for the radius of the circle of the top of the water using similar triangles:

Solve for the area of the top of the water using the radius:
Fucking Bardridge - Mon, 04 Aug 2014 21:03:34 EST ID:uspvjvJI No.14263 Ignore Report Quick Reply
Just think about the geometry of it in your head. The radius at the top of the cone is 3, and the radius at the bottom is 0 because it's just a point. The radius increases from 0 to 3 proportionally with the height increasing from 0 to 10, so if the water height is half of the maximum, then the radius is also half of the maximum.

I need to know the background knowledge of these subjects, for science! by Albert Pammlelock - Fri, 01 Aug 2014 00:08:57 EST ID:FOjl8NEa No.14256 Ignore Report Reply Quick Reply
File: 1406866137068.png -(624362B / 609.73KB, 1600x900) Thumbnail displayed, click image for full size. 624362
Hi, I am a guy who needs a refresher on certain topics on mathematics so I can work on these:

Sadly, I am not so sure if these would translate into their english counter parts, so I left them like this.
I am working on pre-calculus on coursera, though and it works well enough

Linear Algebra I:
Gruppen, Körper, Vektorräume, lineare Unabhängigkeit, Basis, lineare Abbildungen, Dualraum, Matrizen (elementare Zeilentransformationen, Rang, Invertierbarkeit, Inverse, ...), lineare Gleichungssysteme, Determinante

Linear Algebra:
Groups, Bodies, Vectorrooms, linear independency, Base, linear mapping, Dualroom, Matrices, linear equation system, determinants.

Analysis I:
ganze Zahlen, vollständige Induktion, reelle und komplexe Zahlen, Folgen, Grenzwert, Reihen, Stetigkeit, Differentialrechnung, Taylorreihe, Integralrechnung, elementare Differentialgleichungen
Comment too long. Click here to view the full text.
Albert Pammlelock - Fri, 01 Aug 2014 03:52:38 EST ID:FOjl8NEa No.14258 Ignore Report Quick Reply
Oh shit, I worded it wrong.
I need the pre-requisites for it, just the pre-requisites
Basil Crummerbure - Fri, 01 Aug 2014 04:26:30 EST ID:VLYpS252 No.14259 Ignore Report Quick Reply

viel glück
Esther Chinningwater - Sat, 02 Aug 2014 11:52:45 EST ID:Gw2IN3ba No.14261 Ignore Report Quick Reply
>Am I on the right road?

So you're trying to teach this stuff to yourself? Why do you want to learn this? For linear algebra, you need calculus I and calculus II. For analysis, you need abstract vector spaces.

Quick question by Consumer Math - Tue, 29 Jul 2014 04:29:27 EST ID:BfGCwHN9 No.14252 Ignore Report Reply Quick Reply
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If I were to finance a bike at $520 for 12 months, what would my annual percentage rate be if the finance charge is at $35.67?
Lydia Clibbledale - Thu, 31 Jul 2014 04:03:52 EST ID:4UT2KZmc No.14253 Ignore Report Quick Reply
520/35.67=14.4780786815 a month

Greetings /π/ by Charles Murdworth - Sun, 20 Jul 2014 05:35:27 EST ID:OWPvoCXf No.14222 Ignore Report Reply Quick Reply
File: 1405848927891.jpg -(79370B / 77.51KB, 911x838) Thumbnail displayed, click image for full size. 79370
I'm here with a burning question, and I'm a dipshit and this shit flies right above my head.
Let's say what if this guy actually figured out a valid formula for how frequencies interact.


So I will paraphrase my story. Basically, I have a jar of mushroom vodka currently fermenting in the corner of my room.
To keep the jar cool I have a small fan running.
The fan despite being cheap is a fast and efficient beast with a powerful motor.
I have verified that the motor's vibration is vibrating precisely at 459 HZ G#

So because the alleged effects of frequencies on our brains I have to assume that the binaural effect created by my room fan in conjunction with remarkably wide room resonance
is going to be doing something to my brain, and Maybe this is an opportunity to test the so called theory in analog manner.
I'll post a diagram to show exactly how this room effect is achieved. I tried to get an audio recording of it, but now luck.. maybe if I had a set of sensitive binaural mics.
Comment too long. Click here to view the full text.
1 posts omitted. Click Reply to view.
William Paddlenutch - Tue, 22 Jul 2014 14:47:48 EST ID:Gw2IN3ba No.14228 Ignore Report Quick Reply
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I seriously hope you're trolling OP. Also, wtf is mushroom vodka?

This is now a timecube thread.


Nigel Murdville - Tue, 22 Jul 2014 16:02:10 EST ID:gU9kqsUV No.14229 Ignore Report Quick Reply
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what am i reading here
William Paddlenutch - Tue, 22 Jul 2014 19:09:07 EST ID:Gw2IN3ba No.14231 Ignore Report Quick Reply
1406070547200.jpg -(128255B / 125.25KB, 1280x716) Thumbnail displayed, click image for full size.
>what am i reading here

A paradigm shift that will rock your world to its cubic core. You are beginning to understand that everything you were ever taught before the realization of cubic time amounts to nothing!
Molly Worthingbury - Wed, 23 Jul 2014 02:31:55 EST ID:v2qInvsx No.14235 Ignore Report Quick Reply
Hugh Turveyhall - Thu, 24 Jul 2014 03:02:04 EST ID:lx6v5XWF No.14244 Ignore Report Quick Reply
"a set of sensitive binaural mics."

Do you mean "two microphones"?
Or, if the motor's vibration at "precisely" 459hz doesn't change, "one microphone"? You do realize that "how frequencies interact" is addition, right?

is math real by Ernest Nopperbury - Tue, 03 Jun 2014 16:57:49 EST ID:Uq/TU4om No.14060 Ignore Report Reply Quick Reply
File: 1401829069682.gif -(1375B / 1.34KB, 60x60) Thumbnail displayed, click image for full size. 1375
do discrete systems even exist outside of conceptual renderings?
I guess our number system came out of counting things by orders of ten, using fingers to represent whatever was being counted, but the map is not the territory and information doesn't translate into number of fingers so easily.. In a digital world, things stick to the program, but I can't really tell if reality is quantifiable, if you know what I mean.
2 posts omitted. Click Reply to view.
Matilda Wankinnure - Sat, 05 Jul 2014 22:10:50 EST ID:kFCSHfPa No.14174 Ignore Report Quick Reply
>do discrete systems even exist outside of conceptual renderings?

/pss/ fag here. I think we can take a few different tacks with this.

First: how is it that we can conceive of discrete systems if none exist in nature? For that matter, what, exactly, is going on when we conceive of a discrete system? Does said system exist in our brains somehow?

Secondly: mathematics isn't all discrete. There are mathematical continua. Couldn't those be real?
Martha Drummleman - Wed, 16 Jul 2014 02:37:20 EST ID:jEbtLayo No.14216 Ignore Report Quick Reply
Do our thoughts exist physically? Surely our thoughts are some kind of representation of the physical world. But are the two the same thing?
Charlotte Grimridge - Thu, 17 Jul 2014 20:23:26 EST ID:Dn4zhXKX No.14218 Ignore Report Quick Reply
There are quite a few Pythagoreans who talk about pi as an algorithm, rather than a number, and who think discrete math is true to reality and continuous math is conceited. Doron Zeilberger, for one.

Seems to go hand in hand with digital physics and universe/brain-as-computer doctrine. Personally, I don't care about whatever's the most empirical metaphysics; thought and experience are continuous.
Archie Wazzletere - Sat, 19 Jul 2014 04:01:08 EST ID:d7wWJNnk No.14219 Ignore Report Quick Reply
In mathematics there are structures and theories. Structures are things like the natural numbers or real numbers, and theories are the collection of statements deduced from axioms using deduction rules. Mathematics attempts to describe these structures using theories, but due to the incompleteness theorems we know that we will never be able to know all the true facts about, say, the natural numbers. But it's clear that natural numbers exist (as counting objects) and the continuum exists as well (as distances between objects). The theories on the other hand, that attempt to describe these structures, are purely human constructs that optimistically try to approach the truth about the structures.

I like to think of mathematics sort of like a biography about a real person. The subject really exists as a corporeal thing (structures), but the biography is a human construct that tries to tell the true facts about the subject. It should be clear that a biography could never tell every little fact about a person, but if it's a really great person they're worth writing about even if you can't tell it all! What makes the biography of mathematics special is that it is a biography of the most fundamental concepts of existence.

In short, both discrete and continuous things are real. Our descriptions of them in mathematics cannot encompass all the truth about them, but what they are about is fundamentally real.


I've heard of these people being called constructivists rather than pythagoreans. Or finitists, like zeilberger. They don't deny the possibility of continuity, the rationals are continuous and none of these guys would deny the existence of fractions. What they stand against is the notion of a "completed" infinity, the idea of being able to produce any infinite ordinal is a good point but I think is an over-reaction to a misperception of infinity in the public.
Shit Smallman - Thu, 24 Jul 2014 02:50:21 EST ID:VbY4tW0M No.14243 Ignore Report Quick Reply
Math, at least pure math, consists entirely of induction and rationalism. Any math proof you read is a priori which is fucking insane.

Simple Algebra Question by Barnaby Hadgeville - Tue, 15 Jul 2014 15:01:32 EST ID:qCDKBx4v No.14213 Ignore Report Reply Quick Reply
File: 1405450892751.jpg -(4806B / 4.69KB, 328x70) Thumbnail displayed, click image for full size. 4806
Can someone explain in a very verbose way how the first term = the second term in this image?
Shit Nicklewater - Tue, 15 Jul 2014 15:26:08 EST ID:Fj/YvlCk No.14214 Ignore Report Quick Reply
Yesh. Remember that a root is the same as a (fractional) power. That is:
(a+b)^2 =/= a^2+b^2 but
(ab)^2 = a^2b^2

You can factor out 4 below: sqrt( 4 * (-x^2-x) ) which is sqrt(4)*sqrt(-x^2-x)
And sqrt(4) = 2 which cancels out the 2 above.

Goddamn I wish this website supported latex.
Jarvis Munningman - Tue, 15 Jul 2014 20:33:17 EST ID:qCDKBx4v No.14215 Ignore Report Quick Reply
Thank you sir. I think I'll just keep this 1 thread for all my basic questions.
Hugh Turveyhall - Thu, 24 Jul 2014 02:48:19 EST ID:lx6v5XWF No.14242 Ignore Report Quick Reply
(a^m * b)^1/m = (a^m)^1/m * b^1/m = a * b^1/m

Business Math by ConfusedSoul - Wed, 23 Jul 2014 00:12:37 EST ID:BfGCwHN9 No.14233 Ignore Report Reply Quick Reply
File: 1406088757916.png -(303224B / 296.12KB, 500x500) Thumbnail displayed, click image for full size. 303224
I, for the life of my cannot figure this solution out. This is the last question on my study guide and I am totally stumped:
On April 1st, the unpaid balance in an account was $174. A payment of $70 was made on April 8th. On April 26 a $26.00 purchase was made. The interest rate per month was 1.75% per month of the average daily balance. Find the finance charge and new balance at the end of April.
2 posts and 1 images omitted. Click Reply to view.
Whitey Billingshaw - Wed, 23 Jul 2014 10:19:31 EST ID:2aEkBtv5 No.14237 Ignore Report Quick Reply

Yes? No?
Fucking Ponderseg - Wed, 23 Jul 2014 10:29:38 EST ID:GT+uOqbr No.14238 Ignore Report Quick Reply
"Business Math" is just story problems from 3rd grade. Math with a narrative should not be considered a separate type of math. Whoever thought of this needs to die.
Molly Worthingbury - Wed, 23 Jul 2014 17:23:12 EST ID:v2qInvsx No.14239 Ignore Report Quick Reply
Sorry I was a dick. I'm assuming the "finance charge" is the interest charged at the end of the month.

To find the interest, you need to know the average daily balance. There were 8 days at $174, 17 days at $104, and 4 days at $78. That's an average of $115.73. So the finance charge is .0175*115.73. And the balance is 104-the finance charge.
Molly Worthingbury - Wed, 23 Jul 2014 17:30:25 EST ID:v2qInvsx No.14240 Ignore Report Quick Reply
Oh shit, nevermind. It's a credit card.

There were 8 days at $174, 17 days at $104, and 4 days at $130. That's an average of $123. So the finance charge is .0175*123. And the balance is 130 plus the finance charge.
Hugh Turveyhall - Thu, 24 Jul 2014 02:38:26 EST ID:lx6v5XWF No.14241 Ignore Report Quick Reply
Break it down into smaller problems.
First, find the average daily balance.

So the initial balance is $174, and this balance lasts for 7 days.
On the 8th, $70 is paid off, and so the new balance of $174 - $70 lasts for 18 days.
On the 26th, a $26 purchase is made, so the new balance is $174 - $70 + $26, and this lasts through the end of the month.

Thirty days hath September, April, June, and November, so that balance lasts for 5 days.

So the average daily balance is:
[ 7 * 174 + 18 * (174 - 70) + 5 * (174 - 70 + 26) ] / 30

"Cent" means 100 (century, centurion, 100 cents in a dollar), and "per" means "divide", so "per cent" means "divide this by 100"

( 1.75/100 )
Comment too long. Click here to view the full text.

Get X Y by Matilda Daggleville - Mon, 21 Jul 2014 21:33:34 EST ID:Bbuuretd No.14226 Ignore Report Reply Quick Reply
File: 1405992814795.png -(15867B / 15.50KB, 439x305) Thumbnail displayed, click image for full size. 15867
If i have a angle to an object and i know the distance to it. how can i know its coordinates?
made in geogebra
William Paddlenutch - Tue, 22 Jul 2014 04:30:42 EST ID:Gw2IN3ba No.14227 Ignore Report Quick Reply
(r*cos(theta), r*sin(theta)) where r is the distance and theta is the angle

Nathaniel Woshshaw - Tue, 22 Jul 2014 18:40:14 EST ID:yaOOvKGH No.14230 Ignore Report Quick Reply
That's assuming the observer is in (0,0) of course.
James Gerrynut - Tue, 22 Jul 2014 23:26:00 EST ID:Vq7cxA7H No.14232 Ignore Report Quick Reply
thanks m8

i literally just ate my taxes by Thomas Chunninghore - Sun, 20 Jul 2014 23:50:21 EST ID:Q0uL8y2L No.14225 Ignore Report Reply Quick Reply
File: 1405914621916.jpg -(358185B / 349.79KB, 1280x1024) Thumbnail displayed, click image for full size. 358185

Pre-Calc question by Emma Docklegold - Sat, 19 Jul 2014 21:37:36 EST ID:sTqzkRzm No.14220 Ignore Report Reply Quick Reply
File: 1405820256520.gif -(1083B / 1.06KB, 26x27) Thumbnail displayed, click image for full size. 1083
I'm having a problem with a strange question, my textbook doesn't exactly give an explanation on how to do this one, but I'm hoping you can make a little bit more sense for me.

5.) Maximizing Revenue: The price p (in dollars) and the quantity x sold of a certain product obey the demand equation:

p= -5x +100 0 < p ≤ 20

a.) express the revenue R as a function of x

the back of the book says the answer is:

-(1/5)x^2 + 20x

but I'm not sure how to get this answer, I know I'm supposed to express it as x * p, but i don't know how the coefficients got so small
Barnaby Niggerson - Sun, 20 Jul 2014 01:19:45 EST ID:xvgqavvT No.14221 Ignore Report Quick Reply
Exact solution

Plus you can graph it on google to see if it makes sense/visualize it
Google "graph -(1/5)x^2 + 20x"
Jack Pemmleham - Sun, 20 Jul 2014 21:23:55 EST ID:sTqzkRzm No.14224 Ignore Report Quick Reply
Wow, thank you so much, I'm sure to study well with this. Also I didn't know google had a graphing function, cool!

Multiply Roots by Matilda Croblingson - Wed, 02 Jul 2014 00:05:33 EST ID:qCDKBx4v No.14159 Ignore Report Reply Quick Reply
File: 1404273933097.jpg -(109411B / 106.85KB, 921x523) Thumbnail displayed, click image for full size. 109411
What's the answer to #3?
Cyril Sunningsted - Wed, 02 Jul 2014 11:32:24 EST ID:MTIV7/tU No.14161 Ignore Report Quick Reply
Well x^n * x^m = (x * x * ... * x) * (x * x * ... * x) = x * x * ... x = x^(m+n), where in the first equality, the parenthesis on the left has n times x, and the one on the right has it m times.
David Buzzhall - Wed, 02 Jul 2014 14:08:42 EST ID:Gw2IN3ba No.14162 Ignore Report Quick Reply
This. I'd just like to add that this reasoning only works for integer exponents. Not that it matters in this case, since the problem is only asking for a reason and not a proof. Take a look at this:

Nicholas Gosslegold - Wed, 02 Jul 2014 16:13:08 EST ID:e1p9nP+4 No.14163 Ignore Report Quick Reply
For non-integer exponents, the actual definition of x^a involves the exponential and logarithmic functions and the identity in the general case follows from the identity in the special case of these two functions.

The way exp and log are usually defined involves integration (one defines log first, then exp as the inverse of log) and the identity for them can be checked from the defining integral for log.

Things get fun (and much more complicated) when complex and imaginary exponents are assumed, then the additivity of the the exponents no longer true, but one instead has to deal with branching behavior. Of course, to understand that, you would have to know a little bit about Riemann surfaces and you'd be touching some extremely deep mathematics
Martha Garrywire - Mon, 07 Jul 2014 16:31:05 EST ID:qCDKBx4v No.14181 Ignore Report Quick Reply
Thanks everyone. This came from some Harvard or Stanford or Princeton test from way back in the day.
I don't even remember how to do polynomial division.
William Fimmermark - Mon, 14 Jul 2014 15:01:57 EST ID:K+W3fm6W No.14212 Ignore Report Quick Reply
Exponentiation is a special case of multiplication, so by multiplying exponents you increase the number of groups to be multiplied.

SICP by Jack Trotdock - Fri, 11 Jul 2014 08:14:52 EST ID:kDfCm/bw No.14204 Ignore Report Reply Quick Reply
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Hi, i enountered something weird whilst studying the answer to an exercise on SICP (computer science). The anwer declared

(1-square root of 5)/2 <1
is the same as
(square-root of 5-1)/2 <1

can someone explain? Thanks
Jack Trotdock - Fri, 11 Jul 2014 09:46:10 EST ID:kDfCm/bw No.14205 Ignore Report Quick Reply
This is the exercise
and you can find the point i am talking about by ctrl-f'ing "That is. let's show:"
Martin Haddleberk - Fri, 11 Jul 2014 21:20:43 EST ID:Gw2IN3ba No.14209 Ignore Report Quick Reply
1405128043153.gif -(988801B / 965.63KB, 500x281) Thumbnail displayed, click image for full size.
>(1-square root of 5)/2 <1
>is the same as
>(square-root of 5-1)/2 <1

You forgot to include the absolute value notation. |x| = |-x| for real x is the reason why the two expressions mean the same thing.

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