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Q3 - mt = -1/mn so gradient is -1 The point of intersection between curve and tangent is (3,-1) - solve them simultaneously as I have started above. so y = mx + b -1 = -3 + b b = 2 y = -x + 2 as required.

Q1 looks good except for x = 2 you had y = 4/3 instead of -4/3. Also you did not find point of inflexion, only x value. (y value will be -2/3) Q2 - basically - you need to show it touches the curve once. x - 4 = x2 - 5x + 5 x2 - 6x + 9 = 0 b2 - 4ac = 36 - 36 = 0 so one solution...

On a circular table everyone is to the right of you.

So 4! arranging one of each type of shirt, then 3! arrangements of the next three shirts (would be 5P3 but each shirt has two places it can not go), then fit the last shirt in one of 6 places (two places it can not go in)? 4! x 3! x 6 = 864 Hi Aaron; ready for the test next week?

a) 4! = 24 b) 4! = 24 c) 3! = 6 d) 2! x 2 = 4 e) 4! - 3! x 2 = 24 - 12 = 12 Looks pretty good to me.

d) Total possibilities = 4P1 + 4P2 + 4P3 + 4P4 = 64 Only four digit numbers can be greater than 3000. First consider numbers begining with 3 (if you try to do them all together you run into problems with the odd part). 3! = 6 4 digit numbers will begin with 3. Now either list them or...

So you can have a group of 1 and group of 5, 2 & 4 or 3 & 3? This means the total number is (6C1 + 6C2 + 6C3 + 6C4 + 6C5)/2 - it is divided by two as each group of two corresponds to a group of four, each group of one corresponds to a group of five and each group of three is repeated. So...

This is a binomial question that you may not have covered yet, however I will answer it anyway. Probability of no tag is 0.403 Probability of at least one tag is 0.597 So probability none on three of the days = (0.403)3 x (0.597)4 (chance of getting a tag on four days) times 7C3 (the...

You are selecting 28 fish in total. So the answer is 80C28/100C28 = 5.82 x 10-4 EDIT: oops missed the replacement part - pLuvia is correct. And I actually think my method is more in full - each repetition is still a possibility - even if the ones you don't count cancel out in the end. What do...

I think you need to use the product rule as you are not adding x a constant number of times, but a variable amount of times. You must take into acount that as x changes, so does the number of additions. In other words you have x times x. u = x u' = 1 v = x v' = 1 d/dx = x + x = 2x...

He's just done it a different way. He's considered what order you draw them in, then subtracted the combinations that are the same but in a different order. I like my way better :P

3 diamonds - 13C3 (ways of choosing three of the thirteen diamonds) 2 diamonds - 13C2 * 39 (ways of choosing two diamonds and one other card) Total ways - 52C3 (13C2 * 39 + 13C3)/52C3 = 64/425

Be a bit careful when squaring both sides - what if x is negative?

Well part one is simply sin(pi/3) = rt3/2 tan(pi/3)/2 = rt3/2 therefore x = pi/3 is a solution. Here is a tan graph source: www.mathsrevision.net/gcse/sincostan.gif For 1/2 tanx just half every y value. sin x=1/2 tan x 2sin x -sinx/cosx = 0 2sinxcosx - sinx = 0...

I agree with that and it is dangerous to act on simply faith. No one has really responded properly to give any evidence for God, so you can not simply just assume he exists, based on what your parents and others have told you. The majority of "believers" have simply been indoctrined from a very...

d) Choose the letter Y. This leaves 7 letters of which four must be chosen. The y can then be put in 5 different places so the answer is 7P4 * 5 = 4200 e) Choose I and A (the only two vowels). These are two letters of the word and will be next to each other. Then you have 6 letters of...

No, sin (-pi/4) = -1/rt2 sin (-pi/2) = -1

Ok, so you have four people on each side, right. Three of them must be on one side, so you are selecting three of the four ie how many ways are there to arrange three people in four seats? 4P3 Two of the four must be on the other side, so how can you arrange these two people into four seats...

cos^2θsin^2θ+cos^4θ = cos2θ (1 - cos2θ) + cos4θ = cos2θ - cos4θ + cos4θ = cos2θ

cos^2θsin2θ+cos^4θ = cos2θ * 2cosθsinθ + cos4θ = 2cos3θ sinθ + cos4θ = cos3θ (2sinθ + cosθ) Not sure if that's the answer they're looking for - its hard to tell when it just says simplify.