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Lessons in maths.

BrettNortje

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Yes, this! this is like the funnest thing to do in the world, as, it makes you feel really clever when you get it right, and if you don't, well, it is maths! i have decided to start with maths theory for those of you that feel you cannot do it, i will guarantee that you will understand if you can read, write and count, okay?

Now, maths is about finding answers. these answers are numeric, all the time. the hight of maths is technical things, so, that is why you need to use numbers or symbols to represent the factors of the sum. this means, if you understand the method of getting the answer, you will have it eventually or be wrong. i am going to show you how to simplify methods of finding the answer, and make sure you answer right.

If you were to observe the maths you learned in lower school, you will say it was much easier. it was the same thing. it is just now that new concepts far easier than typical primary school maths have been presented. remember how hard was learn your division? that is the same as seeing all those things on top of each other, the division with symbols - it is just a fancy x! remember how hard it was to do multiplication? powers are multiplication. remember how hard it was to do square roots this is the opposite of powers!

So, you can do it. all you need to do is concentrate, and, then fill the values in, slowly. once you have all the values, you will have a much easier sum, but, seeing as how they use the 'same values in the symbols,' it is just repetition.

Let's say you need to divide [x] by [3y]? the answer in it's simplest form is [x / 3y], yes? division comes from needing to fit something into something else, yes? imagine if you were to try to fit a 'cog' or 'swivel' into a engine? this would be where you would imagine them not being negative numbers, as the 'spokes' all fit onto something that will work, yes? how can it be negative? if you were to attribute each formula rather to a picture in your mind rather than to a few symbols on a page, it will be much easier, of course. all you got to learn for now is all the pictures that relate to each formula, then you will find it much easier.
 

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maths....lol
 

BrettNortje

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Okay, so we understand how to do arithmetic, and, that is all you need as a base to do harder things in maths. in maths, there are problems and answers, but, often there is a formula to help you get the answers for the sum. these are like abbreviations in english, of course.

If you want to work out a quadratic, which sometimes might scare people, as most formulas contain quadratics, i will teach you how to simplify it. basically, the thing on top or on the left gets divided by the thing on the right or at the bottom. this is typical division mind you, so it stands to reason given the right symbols it could be done quite quickly, yes?

Then, you need to look for similar symbols - where else is [x], how many powers does it have? remember too that [x] is [1x] so [x] is also [1x^0], got it?

Okay, we have found all our [x]es,and need to find out how many we can 'cross out.' this reminds me of a album called "totally crossed out," by criss cross, but that is another story. so, if you see a power symbol with at the top of a quadratic, you may cross it out if it has a similar [x] that is divided by the same symbol.

But, now i want to learn something new! if we were to use 'binary' with the symbols, we could convert this language in 'lesser binary,' which would be 'circuitry,' where if you have a lot of symbols, you simply add them all up, like [x + x] = [2x], except a lot harder, so;

So, let;s take a harder equation? x = [2w^2 / 1r] * [1e^3 / 1w] would look like this with my new idea;

[8w / 1r] * [3e / 1w] = [8 / 1] * [3 / 1] = 24 units.

This is because, the power multiplies by the number on the left of the letter, as we have a numerical value for [w], being [2w], [1e^3] is one times by 3, so it is three, yes?

Am i wrong?
 

BrettNortje

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But, this is not complete, of course.

How would x = 24? i think it is answered by multiples of 24, of course.

This means that it is [w + r + e] / 24?
 

RetiredNSmilin

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well, there is a fallacy there.

x divided by 3y's ANSWER is NOT [x / 3y]. That is NOT an answer. It is only a formula used to determine an answer.

You will not get a real world answer until you plug in real world numbers into it and come out with a real world ANSWER.

[x / 3y] is merely a formula to get an answer, and NOT an answer unto itself.

Which is one of the main reasons mathematician's babble amongst themselves in their native tongues, and come up with strange theories only they understand.

While the rest of us remove the clog in the gears and continue on with life.

Never...never...ask an engineer to troubleshoot anything.
 
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catif001

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well, there is a fallacy there.

x divided by 3y's ANSWER is NOT [x / 3y]. That is NOT an answer. It is only a formula used to determine an answer.

You will not get a real world answer until you plug in real world numbers into it and come out with a real world ANSWER.

[x / 3y] is merely a formula to get an answer, and NOT an answer unto itself.

Which is one of the main reasons mathematician's babble amongst themselves in their native tongues, and come up with strange theories only they understand.

While the rest of us remove the clog in the gears and continue on with life.

Never...never...ask an engineer to troubleshoot anything.
I don't know if you realized, but mathematics is a requisite for physical understanding and for you to dismiss mathematics as babbling in a native tongue. Well perhaps, you never realized you were using a tongue to dismiss the value of tongues, described vaguely enough by yourself. So, in essence, your anti-intellectualism rant is just an indicator of how mad you are at yourself for not attempting to understand its value hard enough. This thread was not conceived on the basis of arrogance, as far as I can tell and you don't need to vent your unwanted self-esteem issues here.
 

BrettNortje

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In maths, there are also symbols that seem to be repeated. simply crossing some of them out would help a lot, no?

So, you got your question full of symbols, and we know x is reserved for the unknown factors, yes? if we were to observe that some symbols are divided by something that something else is divided by, we could easily make the connection between the two canceling each other out often, or, the sums doing the same thing, just like being long winded, of course.

Then, we could also say that in a division sum, there could be much to cancel. let's take this sum;

x = [r / v] * [s / v] + [v / g^2]

This would be where we can say that half [r] multiplied by half equals [v]? therefore, [r] * equals [v], yes? moving on along the equation, we could say that [4g] * [v] = [r + s] = [x], so the answer would be [x] = [r + s] = [v] / [4g]?

Is this the answer, and if so, did it help you?
 

BrettNortje

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Then we could also say that [r + s] / [4g] = [x]. as we just did previously, yes?

Then, we can say [2r] + [2s] = [x], as v doubled equals doubled r and s?
 
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BrettNortje

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Then, we could say that [4r] + [4s] = [x] as [2v] doubled equals [4g]. this means [g] = [r + s] / [4].

And,

[4v] = [1g] = [x]
 

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Then, we could say that [4r] + [4s] = [x] as [2v] doubled equals [4g]. this means [g] = [r + s] / [4].

And,

[4v] = [1g] = [x]

Are you doing your homework on DP and hoping someone will double check it you?

Sorry, my school days are 35 years in the past, and I try not to tax my brain quite so hard anymore
 

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Then we could also say that [r + s] / [4g] = [x]. as we just did previously, yes?

Then, we can say [2r] + [2s] = [x], as v doubled equals doubled r and s?

Who cares?
 

BrettNortje

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More to do with maths being made easier would be to observe what division really is. division is subtracting an amount from something - the number you divide by. so, you could say that division is actually like taking the value into the bottom number, or, the divider, yes?

Let me give you an example;

30 / 12 = 12 + 12 + 6, as 6 is 0.5 of twelve, the bottom number. but, how do we do this with symbols?

If it were that someone is getting divided into something else, we need to find out what is greater than or less than the concerned values, of course. this could be as simple as observing that [x / y] is as simple as [y] fitting into [x], yes? this could also mean that [y] * [x / y] = [x] of course. so, we could say that [1] * [x / 1] = [x] and this makes [x] ^ [x] / [y] * [y], of course.

So, as many times as x goes into y there will be the answer, of course. the trick is to find more repeats of the [x] or [y] values. let's expand on the sum? [z] = [x / y]? this would mean that we need to say that [z * y = x], of course.

As soon s you have a value, it will be much easier to work out. this value could be as simple as any set number anywhere, which will replace the symbol.

This would be like having [r] = [2x / y]. this would mean that [y = 2 x], so [r = 2 / 1] so r would be 1.
 
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zyzygy

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More to do with maths being made easier would be to observe what division really is. division is subtracting an amount from something - the number you divide by. so, you could say that division is actually like taking the value into the bottom number, or, the divider, yes?

Let me give you an example;

30 / 12 = 12 + 12 + 6, as 6 is 0.5 of twelve, the bottom number. but, how do we do this with symbols?

If it were that someone is getting divided into something else, we need to find out what is greater than or less than the concerned values, of course. this could be as simple as observing that [x / y] is as simple as [y] fitting into [x], yes? this could also mean that [y] * [x / y] = [x] of course. so, we could say that [1] * [x / 1] = [x] and this makes [x] ^ [x] / [y] * [y], of course.

So, as many times as x goes into y there will be the answer, of course. the trick is to find more repeats of the [x] or [y] values. let's expand on the sum? [z] = [x / y]? this would mean that we need to say that [z * y = x], of course.

As soon s you have a value, it will be much easier to work out. this value could be as simple as any set number anywhere, which will replace the symbol.

This would be like having [r] = [2x / y]. this would mean that [y = 2 x], so [r = 2 / 1] so r would be 1.
I just divide by long division or use a calculator.
 

BrettNortje

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Another thing to remember when doing maths is that there can be many answers, but only one is right. this is why maths is a science, i figure, as there is a right and wrong answer for each task or question. of course, if there were more than one, then it would be an art, yes?

So, if there is only one answer, there must be a clear way to calculate it, and a good way to ask it. the problem with maths today is students sit down to a lecture of values on a board, and, then the answer is asked for. of course, it would be easier if there was a picture for the question, yes? this would make the question a practical working example, with, measurements for the exact angles the question is asking for. this would make eduction so much easier, as, the students would get a practical and theoretical aspect of the task at hand, of course.

Now, how do we make each question a picture for the students? well, with calculus, which is a stumbling block for many students, we could draw a cross section and analyze where the dots go, of course. this would be where we draw the angles on the cross section where we put each symbol on each side of the cross section. then, we find the pattern of the cross section's angle by calculating each value given to us - easy, yes? this will make it much easier!
 

BrettNortje

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Then, we could say that each symbol has a power to it, placing the one in front of it, and multiplying it by itself that many times. this will evenly split each dot over the graph, of course. trigonometry works on this principle too.

Then, we could measure each interval, and then we could show 'ratios.' ratios can be converted into simple math by... well let's take an example?

34:15. this is a ratio, yes?

We take the decimal value from the ratio by finding the value of the ratio, by taking the smaller one down to one, yes? this would be two remainder four, yes? two remainder four is added together, and you got six, so the answer is six, yes?
 

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Then, we could say that each symbol has a power to it, placing the one in front of it, and multiplying it by itself that many times. this will evenly split each dot over the graph, of course. trigonometry works on this principle too.

Then, we could measure each interval, and then we could show 'ratios.' ratios can be converted into simple math by... well let's take an example?

34:15. this is a ratio, yes?

We take the decimal value from the ratio by finding the value of the ratio, by taking the smaller one down to one, yes? this would be two remainder four, yes? two remainder four is added together, and you got six, so the answer is six, yes?

Is this yes thing a form of Tourette's?
 

BrettNortje

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I find that nearly all maths in university, which is hopefully where you plan on going with maths if you do it, comes down to a few simple sums from high school. basically, combining algebra with quadratics will get you your answer, and, the rule of bodmas will make it simpler. if you have those things, in a real life scenario, you will have values for those sums, where most of the [x]s any [y]s will have a proper value, then it is down to powers and so forth, of course, if you were to observe that there is also those stupid sum of things that remind me of proper engineering, then it includes a bit of addition too.

So, what can we do with this to make it easier? these are most of the sums you will face in university, and learning that cone stuff is only used in rocket science, of course. if you want to make life easier on yourself, you will observe first that some things are easy to get the values for. this is because they are repeated with other things, often canceling themselves out due to repeated division and multiplication - if you were to see an [x] * then a [x] / then it would be safe to say that [x] and , you would have to find the relationship between them. if there is no indicated relationship, the sum would follow that they are multiplied by each other, leading to a [xb] and [x over b], basically. this would leave you with a chance to make a new quadratic, where would be over [x] and [x over b], leaving you with [xb / bx] i think? this would leave you with [x / b] and divided by [x], leaving you with [2] + [x - b]?
 

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But, what we are dealing with is a big sum with multiple little sums. this means that there will surely be more entries of [x] and and that we are more likely to deal with [x] + [x/b] and [b / x], for example. what if all of them come to one problem?

This would mean that there would be three [x] and three entries. the first is to add them, the second is to multiply them and the third is to divide them, yes? this means we would have [x] + {[x] } {[x] / } or so? this would mean [2b + 3x / 1b]? this would mean, if we were to use the numbers as a guide, [1 / 3 =] 3x + 2b?

I hope you can see it is easy to find your way around these simple mini formulas to scythe through the long sums out there, simply by eliminating whole clusters of compressed sums with the same symbols?
 

BrettNortje

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I think the answer we are looking for is that 3x / 1b = [b = 33.3 of x.]
 
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