Grade twelve maths help.

Discussion in 'Science' started by Brett Nortje, Feb 14, 2017.

  1. Brett Nortje

    Brett Nortje Well-Known Member

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    Let's start with functions - yes, i feel like doing this again. you never know, i may come up with something even better than previously!

    Functions are used to find degrees in a circle, like all useful maths, as then you can measure the angles that, for example, the screw tip needs to be 'cut at.' of course, this could be made easier with parabola, but we will get there later... for now, functions!

    As a polynomial function, a function is expressed as;

    Now, if we were to observe that [x^k equals x^k], and [x^1 = x^1], then we are left with very little to do, yes? this would mean that we need only f = b + 2a, yes?
     
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  2. Brett Nortje

    Brett Nortje Well-Known Member

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    This is where you need to reverse some things in maths. basically, it comes down to reversing pressure or angles to find some circles in maths, as people in maths like circles, like some cult of old men that draw circles... or something... anyway, here is the formula of inverse functions;

    So, we need to, observe [1g 2f 1x] = [0.33x - 5], as i see it, this is because every symbol without a number in front of it has a one in front of it, yes?

    ~ this is one thing i will never forget about what my maths teacher taught me!

    Now, if we were to observe that, then we could plan the sum on the right hand side to be [4.66], so [x = 4.66].
     
  3. btthegreat

    btthegreat Well-Known Member

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    This is chaos, and it will never work and it is just plain unnatural. Integration was a forced policy instituted in our public and private schools . The Numbers need to stay on their side of the tracks and intermarry as they see fit. Letters should stay in their own neighborhood and stop pushing themselves where they are not wanted. Teachers need to stop being instigators for 'change'. Segregation today. Segregation tomorrow. Segregation forever
     
  4. Brett Nortje

    Brett Nortje Well-Known Member

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    Quadratic functions. this is used nearly always, practically, for graphs. these graphs need to find the starting point, the curve, and the end point. this means that you need only, find these three angles to find the degrees you want with your protractor to measure the angles at any given point.

    Of course, you will need to have values for the points, and, they should be part of the circle that you 'have information of.' this would mean that you need to measure to your 'start and end point,' then use a thin pencil line to the point you find with your protractor to be the middle point. you can find the middle point by crossing two equal angles from each start to end point, and, then you can have a straight line from one point to the middle point, to the end point.

    Then, you need to use your compass to make a semi circle from points one and two to the middle point. this will no doubt leave as few degrees as possible - the most direct route, of course. then, you have your quadratic function!

    To put this to paper, you need to use your protractor to find the middle point, being [z], and the coordinates of the 'length points' to be [x] and the 'breadth points' being [y]. this will reveal the values that you are vexed with with the functions symbols, of course. if you were to observe that the function value is equal to z = one fortieth, or {[x] + [-x] = 4.0} you could multiply the [z * x / 10].

    - - - Updated - - -

    You are very funny. kids might grow to like this stuff from these efforts.
     
  5. btthegreat

    btthegreat Well-Known Member

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    Seriously, I applaud your offerings that are well above my comprehension. One of the reasons it is above my comprehension, was that I was never exposed to math as a pleasure, a passion or joy. The biggest problem with mathematics as a discipline in this country is an image problem, not a content problem.

    I have not seen a thread like this before. In truth you presented a series of ideas in a language I just don't speak very well. I can't read music either, but I am smart enough to know incredible ideas blossom in notes and such. You are bilingual and I am not. Teach it with my admiration.
     
  6. Brett Nortje

    Brett Nortje Well-Known Member

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    This trigonometry is all about triangles inside circles. why they want to know the angles inside the bloody circles is beyond me. i suspect it is some form of stress handling on the circle by any combination of factors, but, it is part of maths all the same, constituting at least a fifth of your final grade twelve maths exam mark, i suppose. so, here goes;

    The best way to approach this, triangle mess, is to say that there is a right angle to this. if you were to observe that there are two angles from the outside in, and one line connecting the two points once they reach length and breadth of the 'triangle.' but, how do we find this angle?

    If we were to observe that there are two triangles inside the circle, then we could find that they both contain right angles, yes? these right angles lead from the angle you are at to the 'length and breadth original depth' -0 so they both get measured until they are zero, yes? this means that they need to make a right angle that compliments each of them. this is the centre line, and, then you can find the zero points on the graph of [y] and [y], and easily work the rest out.
     
  7. btthegreat

    btthegreat Well-Known Member

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    Your problem, of course, may be finding your audience on a political forum chat website. I hope you do this elsewhere as well.
     
  8. Brett Nortje

    Brett Nortje Well-Known Member

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    So, now we need to find the lengths of the x and y inside the circle, yes? this can be done by multiplying x by y to find the area, and halving this total to find the area of each cosine and sine thing respectively.

    To find the length of [x] or [y] you need to take the angle or degree that you are working with, and then measure it in right angles to the [y] and [x] graph. then, you need to take 360 degrees and find remember that each degree equals a 'unit.' seeing as how most of the time you will be working with right angles, each degree will be equal to the [radius * 360] = [area].

    To justify this, let's see if we can make three equal triangles out of the whole right 'angle curve?' a triangle has 180 degrees in total, so, observing that there is a right angle in these triangles, then we can say that [180 - 90 = 90], leaving two forty five degree angles to make the circle. this means that it might change the angles but the two triangles are equal to 360 degrees, yes? 360 degree curved circle equals, at eight inner right angles, 720 degrees, and the outside of a circle is 1080 degrees if you use the outside 'like a blazing sun.' this means that 360 degrees is left over, so, seeing as how the 'square area' is equal to 2 / 3 or 'two thirds of the angle,' then you need to multiply your angle by two and divide it by three to find the area that is used, and, from there, the length of [x] and [y] is simply [x] / [y] or [y] divided by [x].
     
  9. Brett Nortje

    Brett Nortje Well-Known Member

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    I have always found these two are quite similar, as they both divide or multiply into something. they say every positive number has two square roots, and every logarithm is the total divided by the base, yes?

    So, with square roots, the answer is always that number divided by the square, so it is reversing the square. let's say we have [square root] of x? this would come down to [x^2 / 3x] , as that would say that, if real numbers were given, and x equals [x = 9]; [9 * 9] = 81 / [3x = 27] = 3. this means that if you multiply [x] by itself, and then use 3x to find the answer to divide by. let's try it with something else? [x = 8]; [8 * 8] = 64 / [3x = 28] = [56 + 8 = 64] = 4.

    This means that [x^2 / x] = the square root!

    With a logarithm, you simply do it backwards.
     
  10. Brett Nortje

    Brett Nortje Well-Known Member

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    So, it is where we combine functions to get a new value for [k] and [l], as the definition describes.

    So, it is all about finding k, l and x. it is easy to see that by crossing out things to the power of zero, they will be zero, yes? let's look at this again without all the zeroes and cross some others out too?

    1 = [2a^1] x + k^2 [4a^k-2] x^k.

    Now we can see that it is a lot shorter, but is it short enough? let's combine [2a^1] * [4a^k-2] = [16a^k-2]. this would be where all the a is added together, minus the parts that amounted to zero, yes? what is left is [1 = x + k^2 * x^k] = [1 = x + 4k * kx].

    Adding it all up will leave us with [1k * 16a * 2x]. find this not too easy, i suggest you start with the infinity - it is basically zero as it has no value, yes? i mean, what good is a line without an end point - it is a dot, if that!

    So, you need to simply add all the symbols together, remembering that a symbol with a power is merely a multiplier.

    Now, to find the rational answer, you need to reverse the powers of k and the symbols of a! this would mean that you need to use k... with a as the power. this would mean that it is much easier, and, all you need to do is use my system of having powers equal; times two, times two, or, times three, times three for the value by the number in front of the symbol, of course, remembering that every symbol without a number is one. with my formula, [k = 1k] in the end, and the rest also goes into the super value answer of one, so it would be 1 = 1k = [16a] / [2x] = [8a / 1x] so k = x = 8a.
     
  11. Brett Nortje

    Brett Nortje Well-Known Member

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    So, we need to estimate the prime numbers given for this theorem. this would mean that we take the base and find the estimated prime number for it to be given value to. this means we need to use [n] * [log] ... [product] - in other words we merely take the [base], and multiply it by the [prime] up until it reaches the [product], and then estimate.

    The objective of this exercise or formula is to find the prime number that fits between the [base] and the [product], of course. this would be where you take the base = [p + p + 1] ... product.

    Then, we need to find the time and distance covered by [x]. this would be where we take the [distance] * [time] / [n].
     
  12. Brett Nortje

    Brett Nortje Well-Known Member

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    Now let's do some "conic sections?" this is where one works out the angles that need to be given on 'a cone' so that we can work out what to put where on the cone, or, how much stress it can handle in certain points.

    I think the key here is to find that the 360 degree base of the cone becomes a one degree point when it 'reaches the middle.' 120 degrees and 60 degrees symbolize the true angle of the cone, from base to 'tip,' as this is the one and a half of 90 and 45 degrees into the angle, of course.

    For every degree that passes, there is about three 'measurements of length' from one degree to the next.
     
  13. Latherty

    Latherty Well-Known Member

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    Just out of curiosity, are you mentally unhinged?
     
  14. Brett Nortje

    Brett Nortje Well-Known Member

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    Let me put some of my ideas into a summary of sorts, a grand theory for the 'grouping of symbols?'

    Basically, what you want to do is observe something like, especially in college with 'real maths;' [a^3] / [b^2 * 4y] * [b^1 / 6a]? this can be summed up as having; [6a] / [4b * 4y] * [2b / 6a]. the theory so far is that you multiply the 'letters' by the powers to find the multiplier, yes?

    Then, [12a ~ 6b ~ 4y] = halved = [6a ~ 3b ~ 2y]. we are only looking for values of symbols at ratios, so, [2a = 1.5b = 1y], basically. this would mean that [x = 2a = 1.5b = 1y] of course.
     
  15. Brett Nortje

    Brett Nortje Well-Known Member

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    With calculus, you need to find the values of the circle, as calculus is all about circles, and, the angles we work with are all about the points of operation on the circle. this means that, first of all, there is a one to two thirds mathematical difference between angles on the inside, being one third, and angles on the outside, being two thirds of the angles. this means, to get a complete angle, sending it into the circle and without, you need to work out the thirds problem to get a comprehensive answer.

    So, back to calculus. if you were to observe that the whole point of calculus is to get a 'cross section' that shows the angles that the thing must be cut or grooved at, then we can easily forego the calculations about the angles, and, dividing the total area by the angles could come down to checking the values you do know, and comparing them to one another. this would mean, instead of finding the calculations according to the formula, you take the values you have, and divide them by three and multiply them by two, if on the outside of the circle, to find the correct values, or, by one third to find the inside angles.

    ~ But wait a minute, it must be harder than that? i mean, how can we just multiply them by two thirds and one third respectively? let's look at a typical calculus formula?

    This would mean that 4xy - y^3 is a constant, or, is used on both sides of the equals, and, that leaves other things to add up to various numbers, which would be;

    This means that;

    This means 2x + y = 3a-d.
     
  16. Brett Nortje

    Brett Nortje Well-Known Member

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    Did i forget to mention square roots are the opposite of logarithms?
     
  17. primate

    primate Well-Known Member Past Donor

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    Interesting. Not certain whether math is considered as science but fire away.
     
  18. Brett Nortje

    Brett Nortje Well-Known Member

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    With sigma notation, the series of the numbers is supposed to be simplified, where, the sequence of sums, usually addition or such, will be summarised by setting the sum up by it being a 'expression' of a "bodmas" sign or such, where it is multiply, divide, add and s forth, and, then show the sum to work around this 'symbol.'

    Let us take the sum;

    ]

    This would be where "100" at the top is somehow related to "i=1" at the bottom, yes? then, the formula is associated with the on the right. so, somehow this means that 100 / [i = 1] * i, yes?

    ~ Man, this is one weird little funny big e sign!

    So, what is [x + x] = 100 where the x = "i," as we have dealt with before, it means there is a imaginary number, yes? this means that, as [x] is an imaginary number, this number or symbol represents a 'vague concept,' which has a real value, in this case fifty, of course.

    Now, if we were to observe that these sums will always balance to be at the top of the 'funny big e thing,' or, "sigma," we could still add up all the related symbols in the sum, to find, in this case, there is '2i,' yes? in this case, [2i = 100] = [1i = 50], of course.
     
  19. DennisTate

    DennisTate Well-Known Member Past Donor

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    On that note I would indeed be honoured to see your reply to:


    http://www.politicalforum.com/index.php?threads/geology-physics-high-tides-question.463012/

    Geology, physics, high tides question.

     
  20. Brett Nortje

    Brett Nortje Well-Known Member

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    Hyperbola is where we 'cut cones into pieces.' This is where we want to examine, in engineering the stress delivered onto screws or parts of machines as if they were under pressure from various angles, of course.

    The right way to do this has been studied for centuries, but no I come along and tell you there is an easier way! Forget about that rubbish they teach you with pages of formulas - this will work better, and, give the right answer too!

    If we were to observe that the cone would be shaped as if it were circular, first we need to understand the cone. The cone is a 'screw' or mechanism inside the machine that needs to be understood by children or people wanting to renter the maths fields wanting to learn the understanding of what they will be used for - what good is it to learn these diagrams without knowing the applications - this is like learning a recipe without a bowl or fruit, just theory, yes?

    So, if we were to observe the cone, it is a circle becoming a smaller circle or zero. This is where the circle gets to remove a degree or ratio from each of it's points considering the other sides they all need to be 'equally removed,' yes? So, if it comes to be that 'a circle' being studied will become bigger or smaller, then it would stand to reason that they would be equally shaped, or, will they? If you were to remove something from something else, it will fall or not work, and, that is why the ratios need to be similar, of course.

    Then, if we were to observe this;

    [x] = indentation [a] multiplied by indentation minus [c] where [c] is the half way point between the two circles, [c] equals a right angle, which is ninety degrees, as well as the others, also equalling ninety degrees each, so, it is [90 * 3] for the total shape, minus ninety, leaving you with one eighty, which is a straight line with that angle we just made up, but please remember it.

    Then, we need to merely measure [90 * 3 = 270] with the other circle inside the 'sphere' equalling where the mere addition or subtraction of either outer point would result in a [90] degree point between the two. Then, the three are added together and divided by four, to leave the unseen of the [360] degrees for the whole circle, where the known point is taken to represent the entirety of the whole square, triangle or whatever the angles are - every cross section has [360] degrees, yes? This means, you take the first angle, the other known angle, and simply add ninety and divide by four to find the answer to 'the full circles' or cross section's measurements or angles.

    Now, to find the right angle at any given point, we need only take the circumference or like round the outside of the cone shape measured into the tape measure, then, scale up to the highest point or greatest point, and, add that to the lowest point, plus ninety, divided by four.
     
  21. Otern

    Otern Active Member

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    Uhm, do you understand what the sigma is used for?
     
  22. Brett Nortje

    Brett Nortje Well-Known Member

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    This would be like the another way?
     
  23. Brett Nortje

    Brett Nortje Well-Known Member

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    With trigonometric functions, we need only, to find "tan," take [90] degrees plus [90] degrees, plus cos [x]. or, maybe it is ninety degrees minus cosine?
     
  24. Brett Nortje

    Brett Nortje Well-Known Member

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    With nth roots, we need to find the prime numbers that use the roots. If we are to understand that, we need to understand what a root is, and, hopefully, find an easier way of doing roots, yes?

    Roots are where we find the lowest number that the prime will go into. This means we need to use [x] for the root number, as i has no divisors. This means we need to take it as it is, and, then observe that roots are the opposite of powers, yes? If you were to multiply something by itself it would lead to an algorithm, where, the jump would be signified by the root being the number in question, of course.

    So, if you were to multiply 4 by 4 by 4, you would have an algorithm of four, by three, of course. this would mean the algorithm would be three or four, I am not sure which, but, it would be like a power formula, of course.

    So, with roots, we merely reverse this process. with primes though, they do not go into anything, so merely multiplying them by two would lead to the algorithm. This would be where we take, for example, [seven] and get fourteen, then divide [fourteen] by [seven] for the answer!

    So, we would multiply it by two and divide it by itself, of course. This would lead to the square root of the prime number.
     
  25. Otern

    Otern Active Member

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    Please, take some grade school math before posting further.

    You don't understand anything.
     

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