Mathematics desk
< October 29	<< Sep \| October \| Nov >>	October 31 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

October 30

height of a plane...

This is a homework question, but both me and my math teacher weren't able to figure it out, so I thought I'd ask it here: You observe a plane traveling at 550 MPH toward your house. The angle of elevation of the place is 16 degrees. 1 minute later the angle of elevation of the place is 41 degrees. What is the height of the plane?

Thanks. 68.231.151.161 02:51, 30 October 2007 (UTC)[reply]

It's hard to show you without a diagram but here it goes

dist = 550MPH/60minutePH = 55/6 Miles (distance travel in a minute (60 secs))

Tan[16deg] = x/dist

x = dist * Tan[16deg]

x/Sin[25deg] = y/Sin[106deg] where 25deg = 41deg - 16deg and 106deg = 180deg - (90deg - 16deg)

y = x * Sin[106deg]/Sin[25deg]

Sin[41deg] = height/y

height = y * Sin[41deg]

202.168.50.40 03:57, 30 October 2007 (UTC)[reply]

An important point to mention is that the plane is moving parallel to the ground. If you take this to mean that the plane is ascending, there is no single answer. I tried to work through the above solution, but I couldn't get anywhere without an explanation of the symbols, so here's my version. You draw a diagram as follows. You are point O on the ground. From you emerge two line of sight at 16 and 41 degrees with the ground. The plane, which flies a height h, parallel with the ground intersects these lines of sight at two points. You draw two lines, straight down to the ground (ie. perpendicular to the ground) making points p1 and p2 (p2 being closest to you) on the ground. The distance between point p1 and p2 is equal to the distance d the plane travels in one minute and the distance from p2 to O we'll call x.

We can now state that h = (x + d) * t16 (I'll use t16 as shorthand for tan(16)) and h = x * t41.

Rewriting the second equation: x = h / t41

Substituting this in the first: h = (h/t41 + d) * t16

Solving for h: h = (d*t16)/(1-(t16/t41))

We know d, so we know h. I usually make mistakes with this sort of stuff, so please check it for yourself.risk 04:40, 30 October 2007 (UTC)[reply]

h = (dt41-t16)/(t41 - t16) if you prefer. risk 04:54, 30 October 2007 (UTC)[reply]

(edit conflict)

There is a bit simpler solution, if you use the Law of Sines twice you can find h=height. First off let me give you some instructions to create the diagram. Note that since we are solving for height it is reasonalbe to assume that the plane's height, h, is constant therefore draw to parallel lines. Next on the top line, Make a point A and point B (I did A on the right B on the left, but it doesn't really matter) point A is when the 1 minute time began, and point B is when that minute ended, and point C, which is where you are standing. Now draw the Triangle ABC. Next create a perpendicular line segment through A to the other parallel line. Mark that intersection point as D. Line segment AD has distance h which we are trying to solve for. Now Angle ACD is 16 degrees, Angle BCD is 41 degrees and Angle BCA is 25 degrees (or 41-16 degrees). Now since AB and CD are parallel we know that, Angle ACD=Angle BCA=16 degrees. So now to get the third angle, 180-(25+16)=180-41=139 degrees, so Angle CBA=139 degrees. Also since AD is perpendicular to AB and CD, the angles BAD and CDA are right angles (90 degrees). Now it's time to use the Law of Sines to find h. In order to use the Triangle ACD, to find h, we need to know how long AC is. So We convert 550 mph to feet using the fact the one minute passed between the two. 550 mph / 60 minutes/hour gives. 55/6 miles/minute, now convert mile to feet, 55/6 miles/minute * 5280 feet/mile gives 48,400 feet. But this distance is AB not AC, so AB=48,400 ft. Now we can use triangle ABC and the Law of Sines to find AC, then use Triangle ACD and the distance of AC (and again the law of sines) to find AD and thus h. Note that in the Law of Sines the angles are named by the capital letters A, B, and C and the corresponding OPPOSITE sides a, b, and c. (For example, angle BAC in Triangle ABC is called A and the side BC is called a and is the opposite side). The Law of Sines states that

${\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}$ .

So to find AC, now called b, we will solve, ${\frac {c}{\sin C}}={\frac {b}{\sin B}}$ for b

$b={\frac {c*\sin B}{\sin C}}$

So C=Angle BCA=25 degrees, and c=AB=48,400 ft B=Angle ABC=139 degrees. Thus

$b={\frac {48400*\sin 25}{\sin 139}}$

Now to find h using our answer to b.

We call the sides of triangle ACD, a, c, and d. b=AC=d $d={\frac {48400*\sin 25}{\sin 139}}$ And h=AD=c

The Law of Sines gives.

${\frac {c}{\sin C}}={\frac {d}{\sin D}}$

Solve for c (remember c=h).

$c={\frac {d*\sin C}{\sin D}}$

Note that D=90 degrees and sine of 90 degrees is 1 so that simplifies to

$c=d*\sin C$

So $h=d*\sin C$ and C=16 degrees

$h={\frac {48400\sin 25*\sin 16}{\sin 139}}$

Or approximatly

$h=8593.86999015$ feet

I highly encourage you to construct the diagram, otherwise most of my explanations won't make sense. Hope that helps A math-wiki 05:07, 30 October 2007 (UTC)[reply]

Doubling the cube???

I read the article on Doubling the cube (do a search and it will be on the top of the list) in which it claims that a compass and straightedge construction of ${\sqrt[{3}]{2}}$ is impossible but says that it is possible with ruler and compass. Looking at the construction the page gives, I am quiet capable of constructing that entire object with only a straightedge and compass.

[Cube Duplication] Artoftransformation 11:02, 3 November 2007 (UTC)[reply]

This is probably do to the fact that I am allowing myself to use the compass to carry a specific length (probably better called a radius, which is the 1 used to create point G I think) to a point it normally could not be constructed on. This be the case, then is their another axiomization of straigthedge and compass construction that would reflect the larger field this possibility creates. It is also different from compass and ruler constructions because I am not allowing myself to create special lengths or ratios off of a ruler. A math-wiki 09:47, 30 October 2007 (UTC)[reply]

You actually are. Your using a compass as a ruler! Artoftransformation 11:02, 3 November 2007 (UTC)[reply]

It sounds like you might be referring to what's called "compass and marked straightedge". The rules of compass and straightedge construction are quite arbitrary and there's no great significance to the fact that certain constructions are impossible within this framework (though the proof is interesting). I'm pretty sure that the ancient Greeks knew how to exactly trisect angles and duplicate cubes under different rules. -- BenRG 11:09, 30 October 2007 (UTC)[reply]

Indeed, the construction states: Now take the ruler and place it so that ... the distance GH is exactly 1. That is not possible with an unmarked straightedge. Putting markings on a straightedge is not in the repertoire of allowed actions. If you actually try to do this construction with a physical marked ruler, you may notice that it is not at all easy to position it in such a way that all constraints are met, since you have to make sure that three conditions of the form line-goes-through-point are met simultaneously, but you can only focus on one at a time, and any adjustment to the placement of the ruler will invalidate at least one of the other two conditions. This in contrast to the allowed actions. Maybe the rules are not that arbitrary. --Lambiam 12:05, 30 October 2007 (UTC)[reply]

I can actually carry this out without a marked straight edge, remember that a compass can keep a given length created earlier and use it on a different point than either one from which the measurement was taken, So I can set the compass to radius 1, and move it to the appropriate place to complete that construction. I would think that the rules for compass and straightedge should allow this since it is possible to do with said mechanical devices. A math-wiki 12:57, 30 October 2007 (UTC)[reply]

What is the appropriate place for the compass? Where will you place its center? At A? But then you can only find points at distance 1 from A, which is not what you want. Somewhere on DCF or on BCE? Where? The points of intersection are exactly those you are trying to find! Or maybe just play around with it until you've found a distance of 1, but then you don't know you are on a straight line from A! In short, this is impossible to do with a compass, and as Lambiam has mentioned, difficult to do with a marked ruler. -- Meni Rosenfeld (talk) 13:10, 30 October 2007 (UTC)[reply]

You aren't allowed to store a length in the compass. Sometimes the phrase "collapsing compass" is used to clarify this, meaning that the compass will collapse when you pick it up from the paper. There are ways to make a copy of a circle that can be used to move a given length from point to point in a construction, but in this case because you don't know the point H beforehand you cannot do so. - Rainwarrior 15:49, 30 October 2007 (UTC)[reply]

The collapsing compass is irrelevant. The postulates traditionally don't allow transferring lengths directly, but (I'm not sure if this is what you meant in the rest of the comment) you can prove that it is in fact possible. -- Meni Rosenfeld (talk) 16:09, 30 October 2007 (UTC)[reply]

That is what I meant, yes. I meant that you can construct a copy of a circle at an arbitrary known point, thus allowing a transfer of lengths. - Rainwarrior 17:06, 30 October 2007 (UTC)[reply]

My point is, if we are going to call it compass and straightedge construction, then the postulates should reflect the full capabilities of the mechanism for which it is named. It is very much possible for a real mechanical compass to hold a length that has been constructed elsewhere and use that length as a radius or to create a segment. I'm also curious if within this less specific set of postulates if it is possible to construct, say a heptagon. Which is impossible under the current interpretation of compass and straightedge constructions A math-wiki 16:39, 30 October 2007 (UTC)[reply]

As Meni pointed out above, a collapsing compass can in fact perform the functions of a non-collapsing compass. In fact, the compass can be dispensed with entirely if you start with one circle and its centre. Algebraist 16:48, 30 October 2007 (UTC)[reply]

Besides, didn't I explain in length why the doubling of the cube is not possible, even with "the full capabilities of the mechanism for which [the compass and straightedge are] named"? -- Meni Rosenfeld (talk) 16:52, 30 October 2007 (UTC)[reply]

I took a closer look at the construction, I am mistaken in my thinking. I thought G was fixed from memory when originally writting this, in fact it is not and neither is H, so it is clear now why a ruler is needed. I also found a nifty program for doing sketches which is completely free. I have since constructed all possible constructible n-gons through n=24. A math-wiki 17:00, 30 October 2007 (UTC)[reply]

(edit conflict) Compass and straightedge constructions is a standardized term with rules going back thousands of years. It's very practical for mathematicians to use the same terms with the same meanings. With the standard rules, you can not find G and H in Doubling the cube#Solutions. Whether you think you can find it "exactly" in "real" life with real tools is a matter of interpretation but I would say you cannot. Even if you have infinite accuracy, you would need to move the tools infinitely many times just to get the points as a limit (unless you hit the right spot by chance with probability zero). PrimeHunter 17:01, 30 October 2007 (UTC)[reply]

A dimension matrix

A matrix is two dimensions table with rows and columns. How does one call a three dimension table? How does one call the third dimension? Pierre de Lyon 17:29, 30 October 2007 (UTC)[reply]

These entities are often called arrays or tensors. I don't know off the top of my head what the third (and possibly additional) dimensions are called. Baccyak4H (Yak!) 17:38, 30 October 2007 (UTC)[reply]

As a programmer, I would refer to it as an n-dimensional array and discuss the nth index (so in this case, a 3-dimensional array with a first, second, and third index). At some point it just makes sense to abandon the attempt to fit the number of dimensions to a physical model. — Lomn 20:20, 30 October 2007 (UTC)[reply]

I have also heard the dimensions referred to as extants, although that has some particular technical context as well. Baccyak4H (Yak!) 20:27, 30 October 2007 (UTC)[reply]

extants or extents? —Tamfang 21:47, 30 October 2007 (UTC)[reply]

Round numbers

Can roundness of numbers be measured? The round extreme can be, of course, as powers of 10 are the most round numbers. But can the unround extreme be measured? I know intuitively that 2000 is less round than 1000, 1500 is less round than 2000, 1600 is less round than 1500, and so on. But how to make this into an objective mathematical formula? JIP | Talk 19:36, 30 October 2007 (UTC)[reply]

The first step to realize is that roundness depends entirely on the base of the number system. 256 may not be very round in base 10, but in base 2 it's 100000000, which is pretty darn round. Take base pi, and none of the natural numbers will be very round. If we take base ten as an assumption, you can easily sort numbers by roundness if you use exponential notation. By your example 2000 is rounder than 1500, because 2E3 (meaning 2 * 10^3) takes one number on the left and 1.5E3 takes two numbers. 2000 is less round than 1000, because 2E3 has a higher number on the left than 1E3. Finally, 10000 is less round than 10, because 1E4 has a lower number of the right than 1E1. So, if you sort first by the length of the decimal part of the exponential representation, if those are equal by the actual numbers of the decimal part (ie. 2 is less round than 1), and then by the size/length of the exponential part of the notation, you get a complete ordering by roundness.

As a final tangent, you may also want to express that 3333 is more round than 3141. The above system makes 3333 less round, but intuitively, you may think that 3333 is 'simpler' and therefore more round. This notion is described by Kolmogorov Complexity, which determines the minimal string needed to describe another string. This means that since 3333 can be described as 4x3 and 3141 has no such simpler description, 3333 is rounder. Kolmogorov complexity is very well defined, but is not computable, so you cannot sort by Kolmogorov complexity. risk 19:54, 30 October 2007 (UTC)[reply]

I just realized that my ordering makes 1400 more round than 1500 (because it's smaller) while the 5 in 1500 perhaps makes 1500 the rounder number. The more I think about it, the more it seems to me that there are several definitions of roundness. Usually people round number off so that mental calculation becomes easier. This means it really depends on the context, and on the strategies you use for calculation. For addition and multiplication, powers of ten are considered round, but for many computer programmers (like myself) powers of two are considered round numbers. Then again, if you're working with angle, then simple fractions of pi, or simple fractions of 360 may be more round. As a final attempt at sort of a generalization, I'd say that given some base (10, 2 pi) the ease with which the number can be expressed as a sum of powers of that base determines its roundness. That still doesn't explain 1500 over 1400, though. The reason 500 is rounder than 400 is that 500 is half of 1000, but then 1/2 isn't round in base ten, it's round in base 2. So it's about combinations in bases that we find to make computation easier. This is getting closer and closer to cognitive psychology and further away from mathematics. risk 20:23, 30 October 2007 (UTC)[reply]

For base-independence, you could define roundness as a decreasing function of the greatest prime factor. —Tamfang 21:46, 30 October 2007 (UTC)[reply]

The values of coins and bank notes are round numbers. They are powers of ten multiplied by powers of two. So round numbers are 1 2 4 8 16 32 64 and 1000 500 250 125. As the double of 64 is close to the half of 125, we get the following ascending sequence of round numbers: 1 1.25 1.6 2 2.5 3.2 4 5 6.4 8 10. This is approximately the powers of the tenth root of ten. So 10 times the base 10 logarithms of these numbers are 0 1 2 3 4 5 6 7 8 9 10 decibel. It is useful in calculations by hand when one-figure accuracy is sufficient, and for expressing big and small magnitudes. To answer the mathematical question above, consider 2^a·10^b more round than 2^c·10^d iff |a|<|c|. Bo Jacoby 11:30, 31 October 2007 (UTC).[reply]

What if a and/or c are non-integers? That would make, for example, 17389 rounder than 20000 because log₂1.7389 < log₂2. And leaving b and d as free variables makes the entire formula open to interpretation, for example in the above calculation, if we allow a and/or c to be non-integers. On the other hand, if we don't allow them to be non-integers, then the roundness of numbers having other prime factors than 2 and 5, for example: 3, 6, 7, 9, 11, 12, 13, 14, 15, 17, 18, 19, 21, 22... is undefined. JIP | Talk 17:52, 31 October 2007 (UTC)[reply]

Certainly, if there are no integer exponents a and b such that the given number x is equal to 2^a·10^b, then the number x is not round. The round number x determines the integer exponents a and b uniquely. Roundness may be defined for round numbers only. The countable set of round numbers is dense on the positive real semiaxis. So, for any positive real x some sequence of round numbers having decreasing roundness converges towards x. Bo Jacoby 19:57, 31 October 2007 (UTC).[reply]

I do not want a black-and-white division of numbers simply into "round" and "non-round". I want to define roundness as a quantifiable measure that can be applied to any real number, for example to be able to say "a is rounder than b but less round than c". JIP | Talk 16:48, 1 November 2007 (UTC)[reply]

Again if you ignore base, then another way is too take the number, add up the number of prime factors it has and divide the number of prime factors by the number itself. A math-wiki 19:33, 31 October 2007 (UTC)[reply]

One possibility is given by smooth number: measure how round a number is by the size of its largest prime divisor, or maybe better by its radical. E.g. 1000, 1600, and 2000 all have radical 10, but 1500 has radical 30, so would be less round by this measure. —David Eppstein 20:17, 31 October 2007 (UTC)[reply]

I don't think I understand your goal. Could you give more examples of what you consider round, and how round you think it is? Black Carrot 23:40, 1 November 2007 (UTC)[reply]

Down Under Factorization

Hello. When factoring polynomials, there is this "down under method". For example, to factor 10x² - 17xy + 3y², one must multiply the first and last coefficients and find two factors of that product that add up to the middle term's coefficient. 10 × 3 = 30. The two factors of 30 that add up to 17 are -2 and -15. The first term was taken with the variable square rooted and was subtracted by one of the two numbers. Similarly, the first term was taken with the variable square rooted and was subtracted by the other number. Each bracket was divided by its own common factor. In this case, 10x² - 17xy + 3y² = [(10x - 2y)/2][(10x - 15y)/5)] = (5x - y)(2x - 3y). How did the mathematician who invented this down under method prove that it always works? I would not have thought of all this myself. Thanks in advance. --Mayfare 22:13, 30 October 2007 (UTC)[reply]

I never even knew there was a real way to do it (except for those ginormous factorization formulas), I've always just eyeballed it o_O --⁪ffroth 23:45, 30 October 2007 (UTC)[reply]

By the way, the above factorization is wrong. I think you're looking for (5x+y)(2x+3y). That's the power of eyeballing :) --⁪ffroth 23:48, 30 October 2007 (UTC)[reply]

Possibly this was found by working "backwards", like this. Assume that the trinomial ax² + bx + c can be factored into (a₁x + c₁)(a₂x + c₂). By multiplying out the product and equating coefficients, we find a = a₁a₂, b = c₁a₂ + c₂a₁, and c = c₁c₂. Since ac = a₁a₂c₁c₂ = (c₁a₂)(c₂a₁), we see that in a trinomial that can be factored, the middle coefficient b is the sum of the pair of factors of some decomposition of the product ac of the outer coefficients. Initially we don't know all these numbers a₁ etc., but suppose we have succeeded in expressing b as a sum b₁ + b₂, where b₁b₂ = ac. Then we can put b₁ = c₁a₂ and b₂ = c₂a₁, so ax + b₁ = a₁a₂x + c₁a₂ = (a₁x + c₁)a₂, and likewise ax + b₂ = (a₂x + c₂)a₁. Dividing these by, respectively, a₂ and a₁, gives the factors of the trinomial. For a solution in integers, these divisors have to be common divisors of, respectively, a and c₁ and a and c₂, but not necessarily the greatest common divisors. --Lambiam 23:59, 30 October 2007 (UTC)[reply]

My bad, ffroth. I fixed the question. --Mayfare 01:54, 31 October 2007 (UTC)[reply]

A circle

If you move a point along X=T Y=Sqrt(1-X^2) does the point move at a constant speed? This is related to a question at WP:RD/C. Is there any way to tell mathematically whether a point on a parametric function moves at a constant speed? It doesn't seem to be related to any of the derivatives.. is there a function to express the speed of the particle? All I've had on parametric equations is the 1-day highschool lesson that just tells you what they are. --⁪ffroth 23:41, 30 October 2007 (UTC)[reply]

The answers are, respectively, and all preceded by "of course": no, yes, yes. Since X=T you can also write Y as a function of T,

Y={\sqrt {1-T^{2}}}

. The speed of the point is

{\sqrt {X'^{2}+Y'^{2}}}

. You can see that this is not constant - in fact, it is infinite at the start of the motion (when T=-1). -- Meni Rosenfeld (talk) 00:03, 31 October 2007 (UTC)[reply]

Oh, and you should read up on Differential geometry of curves, though I don't know if this particular article is good as an introduction. -- Meni Rosenfeld (talk) 00:10, 31 October 2007 (UTC)[reply]

The velocity of the point is the vector addition of the velocities in the x direction and the y direction. Since these are orthogonal, you can use the Pythagorean theorem to find the magnitude: v² = v_x² + v_y², where v_x = dx/dt and v_y = dy/dt. If v is constant, then so is v² and vice versa. Constant means that the derivative with respect to the time parameter t equals zero. Now d(v²)/dt = d(v_x² + v_y²)/dt = 2(v_xa_x + v_ya_y), where a_x = dv_x/dt and a_y = dv_y/dt are the accelerations along the x and y directions. This can be written as 2v·a, in which v and a are the velocity and acceleration as vectors, and the product is the dot product of two vectors. That product is zero when the vectors are orthogonal: you get constant (scalar) speed when the acceleration is orthogonal to the direction of movement.

Applying this to your parametric circle, if you work everything out, does not give you constant speed. This can be seen rather simply: v_x = 1 at all times, but v_y varies in time, and so then does the sum of their squares. To get constant speed, use x = cos(ωt + φ), y = ± sin(ωt + φ), of which the simplest form is x = cos t, y = sin t. --Lambiam 00:22, 31 October 2007 (UTC)[reply]

Algebra

Why does the government make us learn algebra in school when we will probably never use it. —Preceding unsigned comment added by 64.119.61.7 (talk) 23:53, 30 October 2007 (UTC)[reply]

It is extremely unlikely that you will never use Algebra in your entire life. -- Meni Rosenfeld (talk) 00:05, 31 October 2007 (UTC)[reply]

It's fun to learn. Not everything has to have a practical side. Theresa Knott | The otter sank 00:06, 31 October 2007 (UTC)[reply]

I suppose the question is motivated by the fact that for the OP, it is not fun. Too bad. But not everything worth doing is fun to do, either. -- Meni Rosenfeld (talk) 00:11, 31 October 2007 (UTC)[reply]

Mathematics has tons of applications too, and more importantly pure (theoretical) mathematics is responcible for having proven the theorems now used everyday to make computers, calculators, cars, machines and just about everything you can thing of. I would say math is far more valuble to you than english or history or even science is some regards. Mathematics is also good for the brain, as it requires significant use of your logic/reasoning center of your brain, which is in repid development until about the age of 25. It is unfortunate that the teaching methods used in the US education system are quiet ineffective. A math-wiki 00:20, 31 October 2007 (UTC)[reply]

You use algebra in real life. Here is an example

Someone: Bet you ten bucks that Mr Sun can do Y. Double your money or nothing!

You: $10 = X / 2

You: Therefore X = $10 * 2 = $20

Here is another one.

You and N of your friends are going to see a movie. The movie ticket is $9. Since the number of friends going is not known until tomorrow evening. The cost of the tickets are

TotalCost = $9 * (N+1)

See you used algebra already. 202.168.50.40 00:36, 31 October 2007 (UTC)[reply]

Does the betting example really count? Not sure what you're trying to do with the variable there- if you win you get two 10s, if you lose he gets two tens. I definitely wouldnt think of it in algebraic terms unless 10 * 2 = X counts as algebra.. --⁪ffroth 01:41, 31 October 2007 (UTC)[reply]

There's a lot of deeper math and science that builds on algebra. You can't be well-educated without knowing some good math and science. The government wants you to be well-educated. --Sean 14:21, 31 October 2007 (UTC)[reply]

My husband never uses algebra, and in fact gets very confused if I write down any math with letters in it. It can make it hard to communicate sometimes, but he seems to survive ok. I, on the other hand, can't get by a day without it. I design airplanes. He is a homemaker and soon a stay-at-home parent. It seems some jobs have much more use than others for algebra. moink 18:12, 31 October 2007 (UTC)[reply]

From "...if I write down any math..." I get the impression that he has, in fact, encountered situations in which knowledge of algebra could have been useful. From "...gets very confused..." I get the impression that he wasn't taught it all too well. This supports the idea that algebra should be taught in school. -- Meni Rosenfeld (talk) 20:03, 31 October 2007 (UTC)[reply]

"Situations in which knowledge of algebra could have been useful" tends to be when his wife says ridiculous things like "So if x is nudity, and f is attractiveness..." but yes, I agree that he was taught badly and that people should learn algebra in school. I will be working hard to make sure he doesn't transfer his algebra-phobia and curve-plotting-phobia to our child. moink 20:44, 31 October 2007 (UTC)[reply]

I probably shouldn't tell you this, but the reason the government makes you learn algebra has to do with the project to create an alien-human hybrid. It has to do with the fact that the faceless rebels are particularly bad at mixture problems, especially when they're infected with the black oil. They made me swear I'd never tell, but hey, with the end of the Mayan calendar just around the corner, what does it matter now? --Trovatore 20:56, 31 October 2007 (UTC)[reply]

What do you mean, "the government"? Do you really thing private-school kids don't have to learn algebra? Everybody learns algebra. On your question, though, I'd like to answer with two more relevant questions: Exactly what occupation are you planning for yourself? How much money do you plan to make? Many jobs require a lot of math behind the scenes, and basic finance requires a firm grasp of how numbers work. Let's say you get a credit card in a few years. Let's say that, like a lot of people, you let a measly $50 ride on the card for a year and a half. Can you make a guess, off the top of your head, how much that would cost you? Or look a few decades farther, to retirement - do you think you're competent to judge how much you need to save up now to keep yourself fed then? Counting in, of course, depression, inflation, bank failure, and stock failure, not to mention the outside chance that your company will go belly-up Enron style and take your savings with them. If you don't understand basic middle-school math, you have two choices: flail around helplessly, waste money, and miss many golden opportunities, or pay untrustworthy strangers to take control of your property and future. You need math more than any other class. Black Carrot 23:29, 1 November 2007 (UTC)[reply]