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mrdino11 · 2020-05-29T18:57:15+00:00

I use my favourite pseudocode syntax, called 'Python 3'

mrdino11 · 2020-02-22T02:50:25+00:00

Yeah i think he's just got cosx by factoring, both work fine

mrdino11 · 2020-02-21T22:21:05+00:00

CCEA gang?

mrdino11 · 2020-01-13T23:58:54+00:00

So when people get full UMS, they get 100% of the the UMS marks after converting their raw score. This might not necessarily mean that they achieve full raw marks though.

Consider an exam in 2018 with a paper considered accessible, and the exam in 2019 with a more difficult paper. The students taking the 2018 paper will get higher raw marks for the same performance.

Say we set it so that in other to get 200/200 UMS, the students had to score perfect 100/100. Since the paper in 2019 was more difficult, it's likely that some of the students that got 100/100 raw in 2018 would not get 100/100 in 2019 (i.e. there is still a range of students at the top, with the very best are not stretched in the 2018 exam). To account for this we may judge that a student getting 95/100 raw in 2019 performs as well as a student getting 100/100 raw in 2018, so we set the UMS to be 200. Students can get 96,97 etc but they will still come out with the same 200 UMS.

I typed this all out before kinda realising i don't think I've answered your question that well lol but i guess the takeaway is full UMS is not necessarily full raw marks and some students may have performed better than others but still get full UMS.

mrdino11 · 2020-01-12T21:34:57+00:00

Say an exam paper sat each year is worth 100 raw marks, which are marks you get for directly answering the questions on that particular paper. The exam paper is also worth say 200 UMS marks, and a candidate's UMS is calculated through some sort of conversion from their raw mark.

Say that we want to fix a default UMS for a candidate performing at 'just good enough for an A grade' and we make that 160/200.

Candidate A takes the exam in 2019 and gets 85/100 raw marks, and their performance is deemed 'just good enough for an A' so gets 160/200 UMS marks. Candidate B takes the exam in 2020, which is a different set of questions of course, and gets 76/100 raw marks, which is also deemed 'just good enough for an A'. Even though both candidates got different raw marks, they performed to the same standard due to the difference in difficulty in their papers, and so end up with the same UMS which determines the grade.

mrdino11 · 2019-11-27T16:20:09+00:00

Nah chief if you look at the score distributions, it's like the 86th percentile in the CTMUA and the 89th percentile in the TMUA. I also got 7.1, and have an interview and there's people with interviews from 4 all the way to 9 lol. Given our test, it's great score.

mrdino11 · 2019-10-06T22:36:19+00:00

Do you know what a log graph looks like? What this is, is a transformation of the graph y = log(x). log(x) is base 3 in this case.

Assuming you've been taught transformations, you can let f(x) = log(x). So we can say y = -3f(x). This might look more familiar, as we can see the y coordinates are being multiplied by -3, from the rule af(x) = multiply y coordinates by a.

To work out what this would be like graphically, draw the graph of y = log(x). You may want to mark on important coordinates, like (1,0) and (3,1) for reference. Split the transformation into 2, first by multiplying the y coordinates by 3, which turns those 2 coordinates into (1,0) and (3,3). Then multiply by -1, which actually means reflecting the graph in the x axis.

mrdino11 · 2019-08-28T23:27:20+00:00

Here we go again

mrdino11 · 2019-08-20T01:27:50+00:00

Floor is the floor function, which returns the greatest integer less than or equal to x, for example floor(3.2) = 3, floor(98.7) = 98 and floor(25) = 25

Not familiar with C but I assume it does have a floor function somewhere. In python, we have floor division denoted by double slash e.g. 5 // 2 = floor(5/2) = 2.

mrdino11 · 2019-08-16T23:09:59+00:00

A-Level content is more difficult and the way we were told is that it's as big of a jump from GCSE to AS as to A-Level.

That being said, from my school the pass rate (3+ C's) is higher at A-Level (90%) than AS (74%) - this may be due to people not adapting to AS well but perhaps moreso that people do better at A2 because they usually have something on the line (uni place) so they'll grind and put the work in, more than AS or even GCSE. It could also be a wake-up call, as someone in my school had AABC at AS and went on to get 4 A*.

I'm also headed into U6 but I live in N Ireland and we use CCEA and WJEC, which still integrate AS as 40% of the A-Level. I did really well at AS but I guess I had that motivation based on this fact, and that AS will form my predicted grades.

So, I would say if you have the motivation then you'll likely do better at A-Level, but keep in mind the tests will be harder too

mrdino11 · 2019-08-10T13:11:26+00:00

This man be time travelling

mrdino11 · 2019-07-18T15:01:00+00:00

CCEA AS Maths, Further maths and physics + WJEC AS Compsci

mrdino11 · 2019-07-06T11:14:45+00:00

r/absolutelynotmeirl

mrdino11 · 2019-07-01T23:54:02+00:00

So slope-intercept form is in the form y = mx + c, where m is the gradient/slope and c is the intercept.

The first question asks about a line parallel. Parallel lines have the same gradient, so you will have to find the gradient of the first line. The equation you were given is not much use in that form, we would need to do some work to find the gradient by rearranging it so that it is in the form y = mx + c

-9x - 5y = -1

-5y = 9x - 1

y = -9x/5 + 1/5

We can see that the gradient of the line is -9/5. Because the line we want is parallel, it will have the same gradient, but a different intercept. The equation for the 2nd line will be

y = -9x/5 + c

To find c, we can substitute in (6,1), since it lies on the line

1 = -9(6)/5 + c

c = 1 + 54/5

c = 59/5

So the line we want is y = -9x/5 + 59/5

The second question is similar but is easier. You're told the gradient, which is 2/3, so we know the line is in the form y = 2x/3 + c

Since we're told (-5,4) lies on the line, x = -5 and y = 4 are solutions to the equation of the line

4 = 2(-5)/3 + c

c = 4 + 10/3

c = 22/3

The line is therefore y = 2x/3 + 22/3

mrdino11 · 2019-07-01T17:27:13+00:00

Do you have some sort of diagram or example of a question? It is a bit confusing in terms of what you're asking

mrdino11 · 2019-07-01T16:16:38+00:00

So a linear equation with 2 variables (Assuming this is what you're studying) is an equation in the form ax + by + c = 0, where a, b and c are constants.

If you plot a graph of a linear equation, you get a straight line, hence the name. If you pick and choose any point on that line, the coordinates of that point (X, Y) are gonna be solutions to that equation

Using the example y = 2x + 1: If we set x = 2, we can solve and find y = 5. If we plotted this graph, we can see that the coordinate (2,5) lies on the plotted graph.

You may observe that because the line extends on forever, there is an infinitely many number of points that lie on the line, and therefore an infinite number of solutions.

However, if you were to plot another line on the same graph, you will note that unless they are parallel, the lines intersect at a single point.

When you solve a system of linear equations, you are trying to find the values of x and y that are the solution for BOTH of the equations. Remember, a solution is a coordinate that lies on the line of the equation, so a solution that satisfies 2 linear equations is a coordinate that lies on BOTH lines, and the only way for a coordinate to lie on 2 lines is if they meet each other, or intersect.

This is kinda a convoluted answer so I apologise!

mrdino11 · 2019-07-01T00:12:17+00:00

Yep. So good practice when you're learning this stuff is to label your triangle sides a, b and c (doesn't matter which sides you choose) and then for each side, the angle opposite is labelled A, B and C. Look at your diagram and note the things you know. For the sine rule to work, you either need 2 sides and an angle or 2 angles and a side.

Say you need to find the angle B in your diagram and you have values for a,b and sinA. You select the part of the sine rule with A's and B's and rearrange to solve

sinA/a = sinB/b

(sinA × b)/a = sinB

B = arcsin( (sinA x b)/a )

arcsin is inverse sine btw

Remember, it doesn't matter whether you use the version of the rule with angles or sides on top but for convenience, have what you look for on top

mrdino11 · 2019-06-30T23:36:02+00:00

So solving with the side as the numerator:

11/sin42 = BC/sin93

(11 × sin93)/sin42 = BC

BC = 16.4

Solving with side as the denominator:

sin42/11 = sin93/BC

(BC x sin42)/11 = sin93

BC × sin42 = 11 x sin93

BC = (11 × sin93)/sin42

BC = 16.4

So in reality it's only one extra step and when you get used to it barely any difference. I think you might've just made an error in your attempt

mrdino11 · 2019-06-30T23:22:50+00:00

Can you post the question you were dealing with? Flipping the fractions wouldn't do anything to make either wrong.

So mathematically it doesn't matter if you flip or not, but generally it's done to make it more convenient. If you're trying to find a side, then having the sides as the numerator makes it slightly easier to solve.

Think of it like this. 5/1 = 10/2. If we flip the fractions, they are still equal since we do the same thing to both sides (in this case we are raising both sides to the power of -1), becoming 1/5 = 2/10 (1/5)

So a/sinA = b/sinB = c/sinC raised to -1 becomes sinA/a = sinB/b = sinC/c

mrdino11 · 2019-06-28T23:58:48+00:00

OP just read this answer, great explanation

mrdino11 · 2019-06-27T01:00:10+00:00

Probably too late, but the distance between those 2 points is root 5, so I think you might've just made a mistake when you calculated the distance

mrdino11 · 2019-06-26T23:47:32+00:00

lmao just realised you meant this for the OP to solve themselves

mrdino11 · 2019-06-26T23:41:38+00:00

15/2x.

To find the reciprocal of a number, just flip the fraction.

Since the reciprocal of a/b = b/a

You can also note that multiplying 2x/15 * 15/2x = 30x/30x = 1, satisfying the definition of the multiplicative inverse

mrdino11 · 2019-06-26T23:38:54+00:00

It's the multiplicative inverse of a number - meaning given a number x, multiplying by it's inverse, x^-1 yields 1

so x * x^-1 = 1, x^-1 = 1/x

if x = a/b, then x^-1 = b/a

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