NationStates Jolt Archive

I've just done an equation/maths question, which really isn't all that complicated but I want to know what answers other people come up with, so.

(1 - 1.1 x 460 x 5 Pi d squared
--------------
4
------------------------------
40 x 1000 x 169) x 169




A note D = 12., and the 5 Pi sqaured is divided by 4 not the entire thing.

For this the answer I got was 161.8

(1 - 1.1 x 460 x 5 Pi d squared
--------------
4
------------------------------
40 x 1000 x 169)

(1 - (1.1 x 460 x ((5 πd^2)/4))) / (40 x 1,000 x 169)

1. Let d = 12. Then d^2 = 144.
2. 1 - (1.1 x 460 x [5 π x 144] /4) = 1 - (1.1 x 460 x 5 π x 36).
3. 1 - (1.1 x 460 x 5 π x 36) = 1 - (91,080 π).
4. 40 x 1,000 x 169 = 6,760,000.
5. 1 - (91,080 π) = (1 - [91,080 x 3.1415926]) = 1 - (286,136.25) = -286,135.25
6. -286,135.25 / 6,760,000 = -0.0423277

Dammit If i could decipher my notes on how I did it I'd post them...

I make it 3 oranges.

1. Let d = 12. Then d^2 = 144.
2. 1 - (1.1 x 460 x [5 π x 144] /4) = 1 - (1.1 x 460 x 5 π x 36).
3. 1 - (1.1 x 460 x 5 π x 36) = 1 - (91,080 π).
4. 40 x 1,000 x 169 = 6,760,000.
5. 1 - (91,080 π) = (1 - [91,080 x 3.1415926]) = 1 - (286,136.25) = -286,135.25
6. -286,135.25 / 6,760,000 = 0.0423277

left out the negative at the end,

also, stop trying to get people to do your homework. Although I am interested in how you do math (not maths) in europe if you got 161.8 :P (maybe its the extra "s"... must investigate further.)

I can't read that. Can someone rewrite the equation in a way that makes sense?

Or should we not bother, if this is just homework and NSG does have a rule that says "no homework"?

And besides, if there is only one variable in a simple polynomial and you're given its value, where's the problem? Isn't that what calculators are for?

And Kahanistan, your answer should have a negative sign in front of it.

Uhh... homework?

Go 'way.

left out the negative at the end...

Fix'd. I made that mistake all the time in high school and all the way through university.

NL: I tried to rewrite it in a more easily translatable format; I can only assume I got it right.

Fix'd. I made that mistake all the time in high school and all the way through university.

NL: I tried to rewrite it in a more easily translatable format; I can only assume I got it right.

I always make that mistake too. No biggie, esp since you showed your work- thats partial credit.

NL: I tried to rewrite it in a more easily translatable format; I can only assume I got it right.
Fair enough. It's f(d) = (-2,526π/2,704,000)d² where d=12.

If the teachers had any sense in them, that's how they'd write it, and they'd see how stupid a question it is to ask.

Which makes this an opportunity to make this thread into something that doesn't need to be locked actually...how would you teach mathematics to people?

left out the negative at the end,

also, stop trying to get people to do your homework. Although I am interested in how you do math (not maths) in europe if you got 161.8 :P (maybe its the extra "s"... must investigate further.)

Not my homework, already done the equation and got it right, just seeing what other people got.

I always make that mistake too. No biggie, esp since you showed your work- thats partial credit.
You know, I used to tell my students that there was no partial credit in real world problems that people are paying you to solve... Turns out that there's as much partial credit and as many do-overs as your employer is willing to allow. But remember that each mistake costs money. And that you still need to produce what you were paid for in the end.

Fair enough. It's f(d) = (-2,526π/2,704,000)d² where d=12.

If the teachers had any sense in them, that's how they'd write it, and they'd see how stupid a question it is to ask.

Which makes this an opportunity to make this thread into something that doesn't need to be locked actually...how would you teach mathematics to people?
We don't know what the original form was, unless we can get it scanned and posted. At some point in school, though, you do need to teach a student to solve a problem in manageable parts. That means applying the rules you know to make a complicated problem more simple.

At some point in school, though, you do need to teach a student to solve a problem in manageable parts. That means applying the rules you know to make a complicated problem more simple.
Well yeah, but you can do that a lot better in algebra. In this case, one can solve it by simply putting numbers into a half-decent calculator, which would have brackets, fractions and all the other fancy stuff you could hope for. If you ask a student to solve the problem at home, that's precisely what he or she will do.

And besides, algebra can look more complicated too, so it'll scare the crap out of some people, teaching them how non-scary even bad-looking functions usually are.

It's the easy ones you've gotta look out for. ;)

EDIT: Oh, and here's a favourite of mine, posted for no reason at all. The problem with it is that it only applies to a select group of people just good enough to understand the symbols but not quite good enough to see the problem. But when it works, it's priceless:

1 = sqrt[(1²)] = sqrt[(-1²)] = sqrt[(-1) x (-1)] = sqrt[(-1)] x sqrt[(-1)] = i x i = i² = -1

Well yeah, but you can do that a lot better in algebra. In this case, one can solve it by simply putting numbers into a half-decent calculator, which would have brackets, fractions and all the other fancy stuff you could hope for. If you ask a student to solve the problem at home, that's precisely what he or she will do.

And besides, algebra can look more complicated too, so it'll scare the crap out of some people, teaching them how non-scary even bad-looking functions usually are.

It's the easy ones you've gotta look out for. ;)

I'd like to see what the original problem was -- we may be looking at H's substitutions. But I agree that just stringing numbers together, separated by parentheses and punctuated with divisions, is a pointless exercise in calculating.

Wow all your posts are making this sem far more complicated than it actually was...

EDIT: Oh, and here's a favourite of mine, posted for no reason at all. The problem with it is that it only applies to a select group of people just good enough to understand the symbols but not quite good enough to see the problem. But when it works, it's priceless:

1 = sqrt[(1²)] = sqrt[(-1²)] = sqrt[(-1) x (-1)] = sqrt[(-1)] x sqrt[(-1)] = i x i = i² = -1

I don't see that many know-it-alls on a regular basis anymore, but this would be a good test.

Always make sure you can estimate your answer to get a feeling if it's right.

40x1000*169 is a little over 6 million.
1.1*460 is about 500. 500*5= 2500. Pi*12^2 is also about 500, so that whole product is 1,250,000

From your format I don't have a clue what to divide through what.

This kind of calculation could easily follow from an engineering problem, so it's not all that unrealistic.

Wow all your posts are making this sem far more complicated than it actually was...I'm fairly confident that the way you posted it makes it seem more complicated than it actually was. The way you posted it there is no way it would yield the answer you got.

To work out the equation:

1) For the numerator - 1.1 x 460 x 113.097 (this being the result of 5 Pi D squared divided by four) = 57227.082
2) For the Demoninator - 40 x 1000 x 169 = 6760000
3) divide the numerator by the Demoninator
4) Take the resulting number away from one and then times the whole thing by 169.

Which should, if i gave the right numbers to you, equal 161.8.

I'm fairly confident that the way you posted it makes it seem more complicated than it actually was. The way you posted it there is no way it would yield the answer you got.

Yeah...

1 = sqrt[(1²)] = sqrt[(-1²)] = sqrt[(-1) x (-1)] = sqrt[(-1)] x sqrt[(-1)] = i x i = i² = -1
On the risk of failing: sqrt(-1) doesn't really exist as such, which makes the series of equations sqrt[(-1) x (-1)] = sqrt[(-1)] x sqrt[(-1)] = i x i flawed.

A favourite of mine:

step 1. a number N can be written as the sum of N 1's:
N=1+1+1+....
step 2. multiply on both sides with N:
N^2=N+N+N+....
step 3. ake the derivative towards N on both sides. d(N^2)/dN = 2*N; dN/dN=1.
2*N=1+1+1+...
step 4. As we saw in step 1, 1+1+1+... = N, so
2*N=N
step 5. devide by N:
2=1.

Wow all your posts are making this sem far more complicated than it actually was...
It's not complicated.

Still, I just saw that I actually made a mistake in mine, which was quoted. Hence, rather than sneakily edit it, I have to acknowledge the fact that I've thoroughly embarrassed myself. ;)

It's f(d) = (1/676,000) - (2,530π/2,704,000)d² where d=12

Which, the trusty calculator says, is about -0.4233.

But that's assuming Kahanistan was right in his rewriting of the thing.

It's not complicated.

Still, I just saw that I actually made a mistake in mine, which was quoted. Hence, rather than sneakily edit it, I have to acknowledge the fact that I've thoroughly embarrassed myself. ;)

It's f(d) = (1/676,000) - (2,530π/2,704,000)d² where d=12

Which, the trusty calculator says, is about -0.4233.

But that's assuming Kahanistan was right in his rewriting of the thing.

I remember getting a number similar to that at one point, so you've made a mistake in calculating it.

I remember getting a number similar to that at one point, so you've made a mistake in calculating it.
I'm reasonably sure I didn't. I may not know much, but I know how to use a calculator. :P

Tell you what: you post the original in MS Excel formula format, and we can see whether we actually both did the same problem.

To work out the equation:

1) For the numerator - 1.1 x 460 x 113.097 (this being the result of 5 Pi D squared divided by four) = 57227.082
2) For the Demoninator - 40 x 1000 x 169 = 6760000
3) divide the numerator by the Demoninator
4) Take the resulting number away from one and then times the whole thing by 169.

Which should, if i gave the right numbers to you, equal 161.8.

5 *Pi*144/4=565.5 not 113.1. You forgot the factor 5.

Kahanistan didn't include the final multiplication with 169; I guess the whole thing is thus
169*(1-(1.1*460*5*Pi*(D^2)/4))/(40*1000*169) which is -7.15

Well yeah, but you can do that a lot better in algebra. In this case, one can solve it by simply putting numbers into a half-decent calculator, which would have brackets, fractions and all the other fancy stuff you could hope for. If you ask a student to solve the problem at home, that's precisely what he or she will do.

And besides, algebra can look more complicated too, so it'll scare the crap out of some people, teaching them how non-scary even bad-looking functions usually are.

It's the easy ones you've gotta look out for. ;)

EDIT: Oh, and here's a favourite of mine, posted for no reason at all. The problem with it is that it only applies to a select group of people just good enough to understand the symbols but not quite good enough to see the problem. But when it works, it's priceless:

1 = sqrt[(1²)] = sqrt[(-1²)] = sqrt[(-1) x (-1)] = sqrt[(-1)] x sqrt[(-1)] = i x i = i² = -1

Order of operations. Square 1 first, then subtract it from 0 giving you -1. What do I win?

On the risk of failing: sqrt(-1) doesn't really exist as such, which makes the series of equations sqrt[(-1) x (-1)] = sqrt[(-1)] x sqrt[(-1)] = i x i flawed.
sqrt(-1) exists for the purposes of this question as the imaginary number i. The problem lies with the breaking up of the square roots. sqrt[ab] = sqrt[a] x sqrt[b] only holds for positive real numbers.

And wiki also points out the obvious, which I hadn't actually thought of, which is that really as soon as you introduce square roots you already have two solutions, and you're screwing things up by jumping from one to the other without telling anyone. Both 1 and -1 are numbers that made the equation hold, but that doesn't make them the same.

Order of operations. Square 1 first, then subtract it from 0 giving you -1. What do I win?
Wait, what? Why are you subtracting things?

sqrt(-1) exists for the purposes of this question as the imaginary number i. The problem lies with the breaking up of the square roots. sqrt[ab] = sqrt[a] x sqrt[b] only holds for positive real numbers.

And wiki also points out the obvious, which I hadn't actually thought of, which is that really as soon as you introduce square roots you already have two solutions, and you're screwing things up by jumping from one to the other without telling anyone. Both 1 and -1 are numbers that made the equation hold, but that doesn't make them the same.
This is the non-trivial answer. In other words, the fact that you have gained the ability to deal with the square root of a negative number doesn't mean that the associative property still holds.

sqrt(-1) exists for the purposes of this question as the imaginary number i. The problem lies with the breaking up of the square roots. sqrt[ab] = sqrt[a] x sqrt[b] only holds for positive real numbers.

And wiki also points out the obvious, which I hadn't actually thought of, which is that really as soon as you introduce square roots you already have two solutions, and you're screwing things up by jumping from one to the other without telling anyone. Both 1 and -1 are numbers that made the equation hold, but that doesn't make them the same.
I just remember always getting told off by my math professors when writing something like sqrt(-1). i is defined as such that i*i=-1, not as the sqrt(-1). Which, I guess, amounts to the same thing as the splitting of the root not allowed for negative numbers.

But yes, sqrt(1) has indeed 1 and -1 as solution, hadn't thought of that either...