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[Jim Groovester]

-> 327 = 256 + 64 + 4 + 2 + 1
-> 101000111

I'll just add on to this so I can feel smart and say that the resulting digits in binary come from this breakdown:

(256 * 1) + (128 * 0) + (64 * 1) + (32 * 0) + (16 * 0) + (8 * 0) + (4 * 1) + (2 * 1) + (1 * 1)
= (2⁸ * 1) + (2⁷ * 0) + (2⁶ * 1) + (2⁵ * 0) + (2⁴ * 0) + (2³ * 0) + (2² * 1) + (2¹ * 1) + (2⁰ * 1)

Then you just take the 1s and 0s from the second line and that's how you get the binary number.

-> 327 = 256 + 64 + 4 + 2 + 1
-> 101000111

I find bases pretty easy to grasp, but I'm not following your trick here.  How does 256+64+4+2+1 turn into 101000111 in one step?

I think this answers your question.

[MaximumZero]

Okay, here's another stupid question that I forgot to ask in class. When writing numbers above 256 in binary, (Say, 9054, or something like that), do I just write a big string of 1s and 0s, or do I break it into bytes?

Depends. In math, no. You just use a giant string of 0s and 1s. In electronics there might or might not be a reason to write it as multiple bytes(like, say, if you're using an 8-bit system.)

So, if I'm writing in binary for Win7 for some reason, the string can go up to 64 places, but on the NES it only goes to 8? And for that purpose, math is an infinite bit system?

...am I reading this semi-correct?

[dragnar]

Pretty much. Though it would probably be a good idea to get in the habit of separating the bytes anyway, just so it's easier to read.

[MaximumZero]

Schweet. I should be able to finish my homework no problem now. Thanks, all! ^_^

[Jim Groovester]

You can count to 1023 on your fingers?... D: *head esplode*

It's surprisingly easy. You have ten fingers. When they are extended, they count as 1s. When retracted, they are 0s.

00000 00000 = 0
00000 00001 = 1
00000 00010 = 2
00000 00011 = 3
.
.
.
11111 11111 = 1023

I was bored in high school one day and did this for fun and managed to get all my friends to do it too. The funnest number when counting on your fingers is binary is 132.

00100 00100

819 is pretty cool too.

11001 10011

[ToonyMan]

If your fingers could somehow do hexadecimal you could go up to 1,099,511,627,775.

FFFFF FFFFF = 1,099,511,627,775

[Heron TSG]

I have a friend that can finger-count in binary. I never got the hang of it, though.

[Solifuge]

Bio exam tomorrow, and suddenly Scientific Notation! Though it's floating somewhere in my cerebral soup, I forgot all my old tricks for doing math with exponents.

ex. 4.24 x 10⁶ / 2 x 10^-4

It's mostly for doing metric conversions and suchlike. I seem to recall something about adding/subtracting the exponents, but my old shortcuts have suddenly returned in a big jumbled heap, to bite me in the brainstem. HALP!

[Urist Imiknorris]

Multiplying/dividing powers? Add/subtract the exponents.
Raising powers to powers? Multiply the exponents.

[ToonyMan]

If it's positive you move the decimal to the right whatever the number says

5.24 x 10³ would be 5,240.00

If it's negative you move the decimal to the left whatever the number says

5.24 x 10^-3 would be 0.00524


Okay for adding and subtracting exponents you do what Urist said.  I need to go to sleep now before I do something more stupid.

[Solifuge]

Just to make sure I'm not borking it, in the example:

(4.24 x 10⁶) / (2 x 10^-4)

is equivalent to

(4.24 / 2) x 10^(6-[-4])

[Vector]

Just to make sure I'm not borking it, in the example:

4.24 x 10⁶ / 2 x 10^-4

is equivalent to

(4.24 / 2) x 10^(6-[-4])

Yeah.  I sort of do it like this:

4.24 x 10⁶ / 2 x 10^-4
(4.24 * 10^10)/2 = 2.12 * 10^10

So, I think of the division bar as a sort of... converter.  Moving a 10 from top to bottom or vice versa changes the sign of the exponent.  Then when you multiply two exponented numbers with the same base, you just add together the power.

The rest of it can be dealt with as normal.

[Blargityblarg]

Well, this is a mighty convenient thread. From my Calculus homework:

Sketch the graph of the function y=2x^3-12x^2 showing the x and y intercepts and giving the co-ordinates of all stationary points. State the nature of each stationary point show your work blah blah.

Okay, so the y-int is 0, easy enough. X-int is when y=0, so 2x^3=12x^2. I've only come up with 0 and 6 as possible values of x for that, if I had batteries for my gfx calculator I'd be more certain.

Now, the derivative is y'=6x^2-24x, so the stationary points are when that equals zero, i.e. when 6x^2=24x. 0 works, as does 4, and I think that's it. We'll get the y-values of those points, so it turns out that (0,0) and (4,-64) are my stationary points.To find out the nature of those points, we take an adjacent x-value from either side of each and find its gradient, so for (0,0) we'll take -1 (m=30) and 1 (m=-18). So, (0,0) is a local maximum. Now, for (4,-64), we'll take 3 (m=-18) and 5 (m=30), so it's a local minimum.

My question is thus: With my x-intercepts at x=0 and x=6, does it work to have a turning point between those points but *not* in the centre (i.e. x=3), or have I done something wrong?

[Heron TSG]

Question for other people who know aught of cack-yool-uss:

Let's say we have a graph that's formula is (1)/(X-2). The graph is discontinuous at X=2, because you divide by zero. Why is the limit as X=>2 'no limit' instead of 'limit is undefined'? What is a limit?

[Vector]

Blargh: take a second derivative and find out =)

Let's say we have a graph that's formula is (1)/(X-2). The graph is discontinuous at X=2, because you divide by zero. Why is the limit as X=>2 'no limit' instead of 'limit is undefined'? What is a limit?

Meh.  That's an artifact of shitty schooling standards.  A limit is (in essence) whatever value a function approaches from one direction or another.  Say, as x approaches infinity, y approaches zero or similar.

The formal definition is significantly more complicated, but the basic idea is that you have a sequence of terms and a limit n exists if, for every number m you can think of, there is a member of the sequence past which all terms will be within a distance m of n.

They say there is "no limit" because in that instance, they have defined limits to be finite-valued.  The limit is defined: it's positive infinity.  It's just that some people think of infinity as a number, and others think of it as an abstract concept.