Was just trying to help a sophomore with math today. Fortunately shes smarter than me, figured it out herself. So if 3x^2 = 4-3x, do i solve for '3x' first or 'x^2' first? Im just trying to recup what ive lost by not needing much math over the years...
Ah, powers. Well, with conventional maths, you are supposed to do it from greatest to smallest, so, x to the power of two would be first. But, with my method, we take the power and do that to 'the first number.' So, instead of making it x to the power of two, which would be x times x times x, we would say [3^2x]! This would mean we would say three times three times three, equalling [9x] = 4 - 3x. Then, we would say 4 = [6x]. Then we would say [x = 0.75]. So 9x = [0.75 * 9] = [6.85].
Now for calculus. Calculus is where we need to find the points for continuous change. The 'cheat way' of mine is to take the number next to the number next to the system or [F] where it might be [f][3] = ... and say that number to the power of one, subtracting one from it, would result in the answer for the whole equation. But, this does not hold up 'evidence.' The evidence must come from the continuous change being measured by an algorithm, where the algorithm is equal to [x] * [the factor] to that power for the amount of [x]. So, it would be [x] times as many factors as there are, where, the [x] would be measured by saying the factor of the measurement by the power of [x] to the power of one subtracted from itself.
No. Again, that's not how it works. Seriously, you don't understand grade school math, and still think you're able to TEACH high school math.
I don't teach grade school math. I could show you where you're wrong, but not in a way you'd understand. You need a teacher who is able to teach, like special education or something, to get any progress on your grade school math.
X in your case would be a number with decimals afterwards, which makes it hard for a beginner to understand. So here's one little simpler: 2X^2+2X=12 This one will be simpler, (X^2 means X squared). And yes, you're supposed to do the squared before multiplying. So if X in this case is 2, it would be 2 times four, and not four times four.
Here in the US, math is optional by 12th grade. One result is that in math US kids trail the kids in more than 35 other countries. We're going through a period where education is considered even less than just optional. And, we're doing it at a time when world competition is moving toward information, innovation, high tech. This year, China filed 4 times as many patents as were filed by US sources. We're betting on stupid. After all, how could brains possible beat stupid.
With continuous functions, we try to find the logarithm of the function, like powers. If we were to observe that we could merely power the function, in various places, we could easily find the continuous function of the sum. So, if we were to have a function where the continuation of the sum would see the whole sum continued as a following through lot of numbers, we will find that the sum will be continuous or discontinuous. The easiest way to find this, would be, to simply work out there being a positive or negative number involved, by, observing what we know about 'the angle.' If it has an end, in application, then it is discontinuous - it has petrol - and if it has no end it is continuous - it is an electric engine.
When we see something 'to the power of a n,' we realize that it must be a power of a prime number, yes? This would mean that we need to find the prime number the the symbol is multiplied by, of course. Simply put, this would equal [nxx] and that would mean that the power has been said to be [2x^n] or [1x^n] as there is no multiplication yet. ~ Remember, you can take x^3 and say it is [27x]! This is because it is one times by that power and power again, owing to the power number score. Then, we could see that this [n] would be better served being a real number, so we can work out what it is, yes? That would be made easier by saying that [1x^n] is equal to [n^1{x}] which is [n ^ 1], of course. This is because the [n] or prime remains 'just ahead of the curve' remaining a prime number by observing that [x * x * n = 3n] = non prime. This is because the [x * x * n] = [1x + 1x + 1n] = 3 must be prime, as, prime plus even numbers equals even numbers, yes? Also, prime times by three equals a number that can go into three, so, the answer is log times by three, or, square root or three or some junk.
With the previous equation, it would stand to reason that [to the power of n] would mean dividing by 3?
Again, you have to pass grade school math, before trying to teach away high school math. Because you understand neither.
In linear interpolation, there is a lot of confusion. It seems there is a plus minus a minus, meaning there is dead lock. then, there is something divided by itself, equalling one. That [y] is multiplied by [xb] and that equals [y]! So, it is logical to say that [2y] = [xb], meaning that if [x] = [y] then = [2x].
May I ask what this is practical for? And I have another question.. It is about addition and multiplication.. Simple addition and simple multiplication. If I were to say that I wanted to multiply myself (1), two times, that might be written as 1X2, right? And the answer to that would be 2. If I were to say that I wanted to multiply the 2 oranges in my hand twice, that might mean 2X2, right? And the answer would be 4. So here is the question.. Let me begin with the first set. I multiplied myself (1) two times and I ended up with 2.. However, if I took myself and multiplied it once, i would have 2. Me, the original and the newly 'me'. And if I multiplied myself twice, I would have 1 of me, the original, and 2 new 'mes'. And the total would be 3 and not 2. The second set is similar. If I took the two oranges and multiplied it once, i would have another 'set' of 2 oranges, making the total count 4 oranges. And if I took the two oranges and multiplied it twice, once and once again, I would have 2 new 'sets' of 2 oranges plus the original 2 oranges, which would make the total count 6. How does this work? Could it be that it could be stated like this? If i say 1x3, what I am actually saying is that I will have 3 sets of the 1. which would come out to be 3. And if I say 2X2, what i am actually saying is that I will have 2 sets of the 2, which would be 4. But is that really multiplication or addition? 1x2 = 1 set with an additional 1 set is equivalent to 1 set to 2 sets or 1X2. 2x2 = (1 set) of 2 with an additional (1 set) of 2 is equivalent to 1 set of 2 to 2 sets of 2s or 2X2... let's try higher numbers.. 3x3 3 mes multiplied 3 times (over) would result in 3 mes plus 9 mes making it 12 mes total.. or is it (1 set) of mes with an additional (2 sets) of mes results in the same product, which is 9. When a couple 'multiplies' by having children, a new 'product' is formed while keeping the originals. But if it has to something to do with the multiplication as in 'cellular multiplication', then it really isn't multiplying but dividing or splitting to form new products. In a cell multiplication, with one cell, the total number 'whole' might remain the same, as the original 1, with many new 'little cells' but the original cell would have been diminished from its own Whole to many little wholes. Yet the total volume of the combined might be the same as the total volume of the original 1. (just on a side note.. there are some out there that have such a desire to 'kill' that they need restraint to not do so... strange, isn't it? Human beings with cell multiplication occurring within their bodies, needing restraint so as to not kill)..
With inverse functions, you need to; Take the [x] on the right hand side, and, find the rest of the sum without the [x]. This is what equals [x], yes? Then, seeing as how it is two [x], you merely half the value of, in this case [1000] to find the value of five hundred, so, [x] = 500.
I think I have run out of functions and trigonometry to simply and improve, so let's move onto some more ground work for what you are probably learning maths for, engineering and technical things. Sometimes, it is wise to find the answer based on the most common symbol. Sometimes it would make sense to simply ignore the rest of the sum to find the most common symbol and how it relates to the rest of them afterwards? ~ This is highly experimental! So, let's say we have the equation; [f] x^4 = [5y^3 + 9 + 3a] - [3y + 6 / [2y] 5]? We could say, first of all, that x is next to a sum sign, that the equation is basically the sum of the sum itself, being x to the power of four, yes? This [f] could be eliminated, therefore, or, We could take [x^4] and observe that is not the answer, is it? This [f] next to it means that there is more to, that it is a function, yes? This would infer that [x^4] would be equal to [66 * 4] = 66 * 10 = 660 / 2 = [66 * 5] 330 - 66 = 264. This would mean that 264x = all that other junk, and, x is negated, so the answer is 264 degrees relating to the function. Seeing as how they are not equal unless the rest equals [x] and, I happened to suck all the junk out my butt hole, the answer of the equation on the right hand side of the equals sign must be 264 or it doesn't matter.
With trigonometry, you will find it is crucial to working out stress values of moving parts. This, although it uses triangles, is used for calculating circular parts. They used triangles because the best way to calculate a circle in a cross section - those funny graphs - is to use squares, and, two triangles can make a square with a line going through it. This will let you plot 'the angle' by using multiple triangles, of course. We can find the angle by taking the two 'side angles' or the outside of the circle, and, drawing a line through the circle, and, then delivering the value of the half way point through the circle set against the zero point of the cross section, yes? This we can add to the other methods; [1] Find the two points, and, 'go down' on the right hand point, and 'left' on the top point. This will leave you with two values that you can plot the bisecting angle point with, and, that is your angle! ~ You extend the points to meet up, and, draw your line through the circle thingy. [2] You observe that the forty five degree stress bearing area is actually [22.5] degrees on either side, and, work from there...
With calculus, it usually works on a cross section graph so you can visualize the sum you are doing. If you were to merely put the numbers in there, as in [2x] and [4y], where you add up all the amounts in front of the letters or symbols, you will be able to convert directly to cross section, then, convert directly to answer, yes? Of course, the final answer should be a single number, so, you merely times the x by the y!
With trigonometry I am still looking for the simplest way of doing it, so, here I go again! If we were to observe that trig is all about finding the angles of stress, or, values of stress on circles and triangles, that they would have gear coming down onto the 'part' like a record pin coming down onto a record, then the right way to work it out would be to find the point of stress deference and applying mass to it, and, counter mass, yes? Think of this; if you were to have a lot of building blocks, then you would have to fit them together for an engine or building, yes? That is what trigonometry is about, while many students simply learn the formulas and how to apply them, the real mission is to find out what they are doing, something they do not even ask about, something I doubt the teachers could explain for them too, honestly. If you get to the theory before the sum, then you understand better - it is advised that one always uses sketches to depict the problem to the class as then it will become something they will feel more comfortable with in an exam, or, in the real world! So, where are the stress points? This would come down to mass of the point, let's find it... let's say, if it is a triangle, it would be are of the opposing mass part against the area of the other opposing point, yes? This would lead to one degree, for the point, of the triangle, where the triangle would be [360] / 3, and that would come to [120], against the full [360], coming to about [3]. SO, the point would be the sum of three minus nine, so the universal number of trigonometry is three. How do we apply this... well, we will one day, let's get back tot he crux of this! We have a mass against a mass, a point versus a stress yielding area, yes? This means we need to find the area of mass and pressure exerted onto the point, and, the circle, or the circle to the circle. This would mean that, as trigonometry uses the angle forty five a lot, and, [360] / [45] = [8], then, correction, not [3] but [8], that the point would constitute [8] and multiples of it, depending on the size of the circle. But, with triangles, we need to observe that they are half circles, and, that they always have a pinnacle of [one hundred and eighty], yes? This means that, seeing as how the angle is always NOT a pinnacle meeting a circle, that the angle here would be [180] / [45] / [2], coming to exactly half the circle, SO... If we want to find the stress yielding ratios of trigonometry, we need to observe that for a circle, well, we have covered that, and with a triangle, it would be half of the circle's area affected, being twenty two and a half degrees in each direction, cut back into the gentle curve where the two points get a line drawn between them, and, then calculating the are of that semi circle. If I was to be honest, I would say that a triangle seems similar to a semi circle, and, that the one just has slightly more area than the other to yield stress.
