Impossible system of equations - Mathematics - 420chan samsung galaxy s4 http://www.mddesigncontractor.com/ [url=http://www.mddesigncontractor.com/]samsung galaxy s4[/url]

Impossible system of equations - Mathematics - 420chan 传奇私服 http://www.txingzuo.com/

Impossible system of equations - Mathematics - 420chan 传奇私服 http://www.txingzuo.com/ [url=http://www.txingzuo.com/]传奇私服[/url]

Impossible system of equations - Mathematics - 420chan Samsung Galaxy S4 http://www.mddesigncontractor.com/ [url=http://www.mddesigncontractor.com/]Samsung Galaxy S4[/url]

God damn I hate spambots

No. If you can't prove something then you have no reason to believe it.

Say there are facts you can't prove are true but are true nonetheless. That doesn't give you reason to believe these facts since you don't know whether they are true.

Yes, if you looked at the idea of god as an unprovable truth or falsity. However, as soon as you regard God as innately and rationally undecidable, Bertrand's paradox is sidestepped without any help from Godel.

>>11893

You don't need (and cannot expect) a rationally coherent argument to prove something that is by nature unprovable. People use methods other than reason to decide whether or not something is true, and that's where ineffable faith comes in to the picture. Judicious use of reason points away from and outside of itself, as Godel found.

Basically, I don't understand if the first C and M are variables or if they are just indicating what the numbers that come before them are. Also, I don't know what to do with the second C and M or the D.

mu = Q * r
Q as mentioned above is the charge of one of the poles, so Q = 1.6*10⁻¹⁹ C
(C stands for Coulomb and is the unity of charge.)
r is the distance between the two poles, so r = 100 pm = 100 * 10⁻¹² m = 10⁻¹⁰ m
(m stands for meter and is the unity of distance, p stands for pico and is a 10⁻¹² factor.)
So mu = 1.6 * 10⁻¹⁹ * 10⁻¹⁰ Cm = 1.6 * 10⁻²⁹ Cm

The next step is changing the unity from Cm to D.
Since 1 D = 3.33*10⁻³⁰ Cm, you get 1 Cm = 1/(3.336*10⁻³⁰) D.
Finally,
mu = (1.6*10⁻²⁹)*(1 Cm) = (1.6*10⁻²⁹)/(3.336*10⁻³⁰) D = 1.6/3.336 * 10 D ≃ 0.48 * 10 D = 4.8 D

I hope this is clear enough. I'll be around if you have any more questions.

>>11262

Thank you! That helped a ton!

*feeds you a biscuit*

DICKS EVERYWHERE

>>11828

The breaking of space-time fabric like you would put your finger on your shirt. Your finger presses inward, subsequently "bending" the cloth fabric inward.

>>11831
You know "inward motion" as you describe it is just "outward motion" from a different frame of reference? That was already handled by physics a long long time ago. Congratulations on discovering the "number line" they teach that in kindergarten here, to help visualize basic arithmetic and coming to the groundbreaking discovery that A-B= A+ (-B)

>>11836
No need to be cheeky here. Yes I know the part about perspective, but I am saying let the fabric be fixed so that the viewer has only one reference.

>>11838
I think hes being sarcastic because this thread is a shit show of people who dont actually know mathematics or sufficiently advanced physics to know they're wrong

>>11896
A few people here are qualified. I would say most arent; that includes me

>>11806
I guess it could be artifacts like you suspect, but I don't know how Google Earth coded the surface, so I have no way of verifying this.

I can't really help you on starting math again, since I haven't taken a break yet. But concerning your cannabis question, I gotta say I know some guys who do math only when they are high, and they have great results. I tried once and I did well too, but maybe it was only because I was doing some pretty abstract stuff, had there been any serious calculation to do and I doubt it would have gone this well. I guess I'd recommend you to maybe take a little cannabis break for beginning, then try working high and decide from there on.

Good luck.

This post is quite the coincidence. I am the OP of the antarctic artifact thread, and I too am attempting to learn math after a long period of none for a university program I'm switching into. Khan Academy is working very well for me so far and I would highly recommend it. The skill levels is encompasses are very broad, so if anything has you stumped you can easily look back at previous concepts to see why it isn't working. Best of luck to you!

I have a ti-89 emulator for my smart phone for anything that needs to be done on the go which is complicated... because I'm out of school that basically never happens.

Tbh, if it's not something super complicated that I need matlab/r/eviews for, I just use wolfram alpha these days. I'm pretty much always near a computer anyhow (and they have a mobile app).

tl;dr there's not really a good reason to use a calculator these days other than maybe the fact that some of them are solar powered.

>>11254
seconded, in the modern age wolfram is the way to go

I liked my TI-83+ for a long time, had it for about 7 years. Did most stuff I wanted it to and I never needed a calculator much during my degree anyways.

Lately I've studying for actuary exams and they only allow a select few calculators so I've been using a TI BA-2 Plus pro. Basically just a glorified scientific calculator.

>>11230
fucking truth dude
I actually bought another casio fx-115es when mine broke, because its really fucking easy to use and i love it for geometry/algebra

i've got a TI-89, which I use for all heavy-duty calculations when I'm on the go and don't have any other computer available.

Besides that, I've got this rockin' scientific calculator that I've had for like 10 years now. Still works perfectly, and if I need something fast I use this instead of the more clunky 89.

>>11847
er, actually sorry, not the sum, the product

>>11847

Thanks for your answer, but I still don't get it. Isn't what you just said what I did?

I proved that it was true for n = 2, I assumed it was true for arbitrary k, then for k +1. And the operation should be first second term (assuming k) * first second term (assuming k +1) is true for second second term.

>>11850
No, you do not assume that it is true for both k and k+1 , you have to SHOW that if it is true for some k, then it necessarily is true for k+1. Start with your assumption that it is true for k, multiply both sides by the term that would make the left half equal to the sum for k+1, (which would just be (1- (1/(k+1))) right?). Then use arithmetic to show that the right half equals ((k+1)+1)/(2(k+1)). That it is how you use induction.

>>11846
Logical errors aside, the last line should have a multiplication sign between the two major expressions:
[(k+1)/(2k)] * [1-1/(k+1)^2].

Then you will get
[(k+1)/(2k)] * [1-1/(k+1)^2] = (k+1)/(2k) - [(k+1)/(2k)] * [1/(k+1)^2] by distribution
= (k+1)/(2k) - (k+1)/[(2k)*(k+1)^2] multiplication of fractions
= (k+1)/(2k) - 1/[(2k)*(k+1)] eliminating like terms in second expression
= (k+1)^2/[(2k)*(k+1)] - 1/[(2k)*(k+1)] finding the common denominator
= [(k+1)^2 - 1]/[(2k)*(k+1)] adding fractions
= (k^2 + 2k + 1 - 1)/[(2k)*(k+1)] FOIL on (k+1)^2
= (k^2 + 2k)/[(2k)*(k+1)] definition of additive inverse
= [k*(k+2)]/[(2k)*(k+1)] distributive property ("pulling out" the common term)
= (k+2)/[2*(k+1)] eliminating the common multiple k from numerator and denominator
= (k+2)/(2k + 2) distributing the 2

Comment too long. Click here to view the full text.

It's critical to prove from one side of the equation as well.
In particular, look at step 1.
You should show that 1-1/(2^2) = 3/4
and then show that (2+1)/(2*2) = 3/4

They shouldn't be shown at the same time, on opposite side of an equation. Writing it like that assumes that they are already equal and you should get used to writing it properly, it'll come up repeatedly.

>>11458
You have 12 players and you need to choose 3 of them so 12!/(3!(12-3)!)

https://en.wikipedia.org/wiki/Combination

>>11461
And that's 220, if you don't know binomials.

>>11463
and thats 11011100 in binary, if you don't know the decimal system

>>11548

1101110

>>11549
Disregard this. I don't even.

Why would acceleration due to gravity vary?

>>11842
well, if you have a particle falling from a significant distance away from a gravitational attractor, because gravity is an inverse square law, the field is not constant. you can only ignore the variance of the gravitational field at different radius' for changes in height much much less than the radius of the earth.

As per the OP, I would say the best way to solve this problem would be to solve for the velocity as a position of function using energy, and then integrate.

>>11841
acceleration is derivative of velocity, and displacement is integral of velocity. So the change in displacement is the velocity. Right?

>>11861
In this case acceleration is a function of position, however

What do you like? What are your hobbies?

If you like game theory, then why don't you simply learn about game theory?
I suggest you read about the minimax algorithm, with alpha/beta pruning and shit like that. It's between game theory and computer science so you might like it.
You could make a cool game and a cool AI to go with it. Like an online othello or something like that. The game itself is more /prog/, while the AI is more /math/.

If you prefer algebra and calculus then you could make a 3D game with pixel/vertex shaders. 3D engine are full of matrices and quaternion everywhere. A shader is basically a program to manipulate functions. Even a simple animation is math, it's just a function of time to your screen.

>>11869
I don't think my programming skills are quite good enough yet to make games, the closest I could get is to use a previous OpenGL program I made in class that just displayed moving particles that formed a pixelated pac-man, and get it to use input to determine movement of those particles and such. Not much math involved there, unless I were to get into gravity and stuff like that.
I think I might just pick at my textbook.

For useful/interesting mathematics to learn,
Linear Algebra is practically mandatory to do anything interesting or large scale in CompE so go as deep as you can down that rabbit hole.
Abstract Algebra and Group Theory appear everywhere including CompE and it's one of the most interesting areas of mathematics so definitely take a look if you have time.
(Non-Euclidean) Geometry especially projective geometry appears a lot in Comp Vision, Robotics, Comp Graphics so it would be very beneficial to read through college level geometry book like Brannan before you run into those topics and quickly get lost.

Complex Analysis, Fourier Analysis, PDEs, and Functional Analysis are also important but trying to learn them by yourself might be a little too difficult at your level...

>Computer Engineering
>Java

Why the fuck are you wasting your life learning Java? Learn C++/Python/Matlab/ARM/Verilog, something useful for your profession...

subtracting a vector is the same as adding a vector of the same magnitude but opposite direction. If the vector that was to be subtracted was at magnitude 3 at 90 degrees, it would be the same as adding a vector of magnitude 3 at 270 degreees.

Strictly algebraically: You can just subtract the components of B from the components of A.

Geometrically: A - B means, assuming you understand how to add vectors, that you add A and -B. If it makes it easier for you to think about, -B is just B in the opposite direction, so you can draw the negatives of each vector and simply treat them all like addition. I hope we helped. I know I said the same as the other guy, but different wording helps for different people. nb since redundancy.