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Lessons in maths.

Maybe another thing to make maths easier is to observe that there are square roots out there that get used to fit things into other things, imagine a 'rivet' or 'link' in a wheel with those 'gears.' this will cycle around to create a circular effect and then turn the one gear to turn the other. for this we need geometry and square roots, on the outside of the circle, thereby making the circle the opposite of 360 degrees, reversing the angles to be right angles opposed to the right angle points, leaving us with four three quarter circles rather than four quarter circles - four quarters makes a complete circle, while if you were to turn the circle inside out, you would be left with 270 degrees times four, being 1080 degrees.

So, the outside of a circle is 1080 degrees. of course, using a new type of maths to calculate the outside of the circle for a lot of engineering would be preferred to making right angles on the outside of the circle or around the circle to find that degrees the gears are found at, rather finding lines that continue through the circle the gear is on to find a pattern for the 'gear shifts.'

So, we could, if maths is to be believed, change all relevant formulas to show that the old way of doing maths, far longer if you ask me, are converted to this form of maths. for example, trigonometry would now be much easier, yes?
 
BrettNortje said:
Maybe another thing to make maths easier is to observe that there are square roots out there that get used to fit things into other things, imagine a 'rivet' or 'link' in a wheel with those 'gears.' this will cycle around to create a circular effect and then turn the one gear to turn the other. for this we need geometry and square roots, on the outside of the circle, thereby making the circle the opposite of 360 degrees, reversing the angles to be right angles opposed to the right angle points, leaving us with four three quarter circles rather than four quarter circles - four quarters makes a complete circle, while if you were to turn the circle inside out, you would be left with 270 degrees times four, being 1080 degrees.

So, the outside of a circle is 1080 degrees. of course, using a new type of maths to calculate the outside of the circle for a lot of engineering would be preferred to making right angles on the outside of the circle or around the circle to find that degrees the gears are found at, rather finding lines that continue through the circle the gear is on to find a pattern for the 'gear shifts.'

So, we could, if maths is to be believed, change all relevant formulas to show that the old way of doing maths, far longer if you ask me, are converted to this form of maths. for example, trigonometry would now be much easier, yes?
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How much is four times three quarters?
 
zyzygy said:
How much is four times three quarters?
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Well, the original 'circle' is divided into four right angles, yes? this would mean, the rest of the four circles, or the outside, is four times three quarters going in the opposite angles.

Four times three quarters is three of course.
 
Now i would like to cover quotient and product rules, but product first. if we were to observe that the product of calculus is to find angles for, i think, the inside of the circle, while trigonometry is for the outside, we could use trigonometry rules with calculus rules, or vice versa, but let us find an approach first?

http://www.math.com/tables/derivatives/identities.htm said:
d/dx f(x)g(x) = f'(x)g(x) + f(x)g '(x) (product rule)
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So, we would take the [x]es first and try to find out what they are, more or less? this would come to [f'g' + fg * 4x]. or, we could stick to the left hand side of the equals sign, leaning towards 3x * [d / d =] 1 * x * [fg]. this would lead us to consider that d divided by dx equals one times x? this would come to [4x] times [fg].

Now, if we were to look at the right hand side of the equals sign, we would find that [4x] = four [x] points.

The theory to this would be that you can separate the [x]es from the rest of the sum to get the answer simpler, and then you can find the answers much faster. but, i remember that previously i found, quite to my surprise, that [f] = [x ^ x - x], so, [f] times [g] equals [4x], then maybe [2x] equals [f]?
 
BrettNortje said:
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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you lost me at maths LOL
 
BrettNortje said:
Well, the original 'circle' is divided into four right angles, yes? this would mean, the rest of the four circles, or the outside, is four times three quarters going in the opposite angles.

Four times three quarters is three of course.
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Nonsense. If you divide a circle into four equal parts you get four segments. What you mean by the outside of the circle is anybody's guess.
 
zyzygy said:
Nonsense. If you divide a circle into four equal parts you get four segments. What you mean by the outside of the circle is anybody's guess.
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Okay, you have two semi circles in one circle, and four half semi circles in a circle. all four half semi circles span from zero to ninety degrees, yes?

Now, if you reverse it so that it goes on the outside of the circle, then imagine a half semi circle spraying in all the other angles? that would be from 90 to 360 degrees, because it is the opposite of the inside of the circle. i believe the inside of the circle would be the diameter times four, coming to four lots of ninety degrees, or, if you were to be into classical mechanics, four right angles forming a square to measure the 'outer angles.'
 
Maths to me is all about getting values for questions, where you find the right angle or ratio and so forth to get the job done. Maths is practical in ways that include it in many disciplines for people, most notably engineering and science related things. i have covered a lot of engineering related maths, so now it is time to do some science type maths, yes?

Maths in science is about finding balances between things, or good 'exchange rates.' this would be where you measure how much oil and polymers you put into petrol, for example. there needs to be ratios showing how each works with the other and that is science, the maths is finding the non language based solutions to the goals of the 'quest.'

So, if you have a ratio, then you can easily convert that into decimals by finding the lowest common denominator, or, the nearest hundred denominator for the greater value. this would be like having 3:7 - the nearest ten denominator, as i call it, is 30:70 or 300:700. to get it into decimals, you need to say, it is [3x/7y]. this will leave you with two remainder one, yes? then, this would be 2 + 1 / 7 meaning that it could also be [1 + 1 + 1 / 7 =] [100 + 100 + 100 / 700] coming to, if we know our times tables, [700 / 100] * 3 = [21 = 21 + 21 + 21 + x =] [7 / 100 =] [70 + 30] = [7 * 3 =] [21 * 4 = 84 + x =] [%].

This means the answer is [4 * 3 =] 12 sevens for 84. then, you would say the remaining [16 = 7 * 2 + 2]. 12 + 2 sevens plus 2 is the answer in fractions. 14 remainder two, of course. then we would say [2 / 7 =] 3.1 so the answer is 14.31!

I hope you can see that these science cross overs are relevant?
 
https://en.wikipedia.org/wiki/Algebraic_statistics said:
Algebraic statistics is the use of algebra to advance statistics. Algebra has been useful for experimental design, parameter estimation, and hypothesis testing.

Traditionally, algebraic statistics has been associated with the design of experiments and multivariate analysis (especially time series). In recent years, the term "algebraic statistics" has been sometimes restricted, sometimes being used to label the use of algebraic geometry and commutative algebra in statistics.
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I think this comes down to [2pi = 1]. this is because there is three choices, being that [i = 0, 1, 2]. this means that if [2p * i = 1], and this is the main equation, then we might be better served to make our own?

Okay,

[Two of something] times by [something else] = [break even point]. to find what the break even point is, we need to double the [p] and then times that by the . this will give what we have broken even to. this stuff about [i = 0, 1, 2 ] is rather useless, as, it will be setting the math used into binary, basically, which would take a long time to work out.

So, i suppose that to find the algebraic statistics of a given set of values, would be to, first, take [value one] times by [value two] = [percent.] in this, the percent is stipulated at one, so that we can see how much a full volume of this statistics is equal to, raising the value of one to the numeric value using algebra.

This means, you merely need to take [value one] times by [another]. if the other is non existent, then the first remains at zero too, as the answer is [1]. basically, what i suggest to find out what the values are, if they meet the 'target,' you take the target and minus bad values from the good values.
 
This whole maths thing where everything revolves around the value one, and finding what one equals, is like finding a hundred percent of the other stuff, yes? this would mean that we merely need to reduce one into the parts of the sum, where if [x / 4t / y^7 =] 1, then you simply multiply it backwards, the opposite of bodmas!

This would see the one become part of the sum, instead of having the sum equal one, of course.
 
BrettNortje said:
Maths to me is all about getting values for questions, where you find the right angle or ratio and so forth to get the job done. Maths is practical in ways that include it in many disciplines for people, most notably engineering and science related things. i have covered a lot of engineering related maths, so now it is time to do some science type maths, yes?

Maths in science is about finding balances between things, or good 'exchange rates.' this would be where you measure how much oil and polymers you put into petrol, for example. there needs to be ratios showing how each works with the other and that is science, the maths is finding the non language based solutions to the goals of the 'quest.'

So, if you have a ratio, then you can easily convert that into decimals by finding the lowest common denominator, or, the nearest hundred denominator for the greater value. this would be like having 3:7 - the nearest ten denominator, as i call it, is 30:70 or 300:700. to get it into decimals, you need to say, it is [3x/7y]. this will leave you with two remainder one, yes? then, this would be 2 + 1 / 7 meaning that it could also be [1 + 1 + 1 / 7 =] [100 + 100 + 100 / 700] coming to, if we know our times tables, [700 / 100] * 3 = [21 = 21 + 21 + 21 + x =] [7 / 100 =] [70 + 30] = [7 * 3 =] [21 * 4 = 84 + x =] [%].

This means the answer is [4 * 3 =] 12 sevens for 84. then, you would say the remaining [16 = 7 * 2 + 2]. 12 + 2 sevens plus 2 is the answer in fractions. 14 remainder two, of course. then we would say [2 / 7 =] 3.1 so the answer is 14.31!

I hope you can see that these science cross overs are relevant?
Click to expand...

Relevant to what?
 
Casper said:
Exactly what I was thinking. Hey OP it is MATH, good grief.
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Most other english-speaking countries say "maths" because it's short for "mathematics."
 
Another theory i have been messing around with is that [x] / [y] equals [x] * [-y]? what does this mean, you may ask; well it is about the ease with which it is to multiply rather than divide, so we would have a better answer for [x] / [y] in [x] * [-y], yes?

This would mean that we could have [-xy] leaving a negative number as the answer. the obvious way forwards then is to simply make the negative number a positive number by leaving the minus sign out of it.
 
BrettNortje said:
Another theory i have been messing around with is that [x] / [y] equals [x] * [-y]? .
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It doesn't. For example 4 divided by two equals 2. 4 times minus two equals minus 8. Take a basic maths course. And it's not a theory.
 
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zyzygy said:
It doesn't. For example 4 divided by two equals 2. 4 times minus two equals minus 8. Take a basic maths course. And it's not a theory.
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I believe that would be [minus four] times by [minus 2] equals minus two, and;

me said:
the obvious way forwards then is to simply make the negative number a positive number by leaving the minus sign out of it.
Click to expand...
 
BrettNortje said:
I believe that would be [minus four] times by [minus 2] equals minus two, and;
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No. [x] * [-y]. There is no minus sign in front of the x. 4 x -2=-8. You don't know much about maths. And -4 x -2 = 8, not -2. You do need lessons.
 
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BrettNortje said:
Another theory i have been messing around with is that [x] / [y] equals [x] * [-y]? what does this mean, you may ask; well it is about the ease with which it is to multiply rather than divide, so we would have a better answer for [x] / [y] in [x] * [-y], yes?

This would mean that we could have [-xy] leaving a negative number as the answer. the obvious way forwards then is to simply make the negative number a positive number by leaving the minus sign out of it.
Click to expand...

And how would the rest of know that you have left out the minus sign?
 
BrettNortje said:
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?
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You are wrong. 1X^0 = 1*X^0=1*1=1
The exponent does NOT raise the number to the left of the variable, it raises the variable.

And e is a constant, not a variable, an roughly equal to 2.71828
 
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