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b) Draw a parallel line, parallel to both EF and CB through C. The angle 'Y' will split up into two angles. Using your alternate angles (parallel lines), then the two angles are alpha and beta, which add up to angle 'Y' Hence Y = alpha + beta d) The converse of (b). Split up angle ADE into two...

Re: 2015 permutation X2 marathon used for a question like how many ways can you arrange 8 boys and 4 girls so that no girl is next to each other arrange boys first (looks like this): B B B B B B B B = 8! then you "insert" the girls inbetween the spaces so that at least one B separates two G's...

Re: HSC 2015 4U Marathon New question: question b http://imgur.com/XPZ7Ass http://imgur.com/bqCRBsv Also, would anyone know what paper this is from?

I've found a way that's shorter but i'm not sure if Mathematics students learn this. $ Convert $ cos^{2}\theta $ into $ \frac{1+cos2\theta}{2} (\sqrt3 + 1)(\frac{1+cos2\theta}{2}) - 1 - ( \sqrt3 - 1)sin\theta cos\theta = 0 (\sqrt3 + 1)(1+cos2\theta) - 2 - 2(\sqrt3 - 1)sin\theta...

Set the girls first So it looks like this: G G G G G There are 6 possible spots for a boy to be seated such that a girl separates two boys (denoted by a star): * G * G * G * G * G * There are 6 spots possible for boy 1, 5 possible for boy 2, 4 for boy 3 = 6x5x4 = 6P3 Arranging the girls...

If I interpreted your question correctly, then you could do this. \frac{T_{r+1}}{T_r} > 1 $ for greatest coefficient $ $ Simplifying, we'll get $ r<15 $ So for $ T_{r+1} > T_r , r<15 $ For $ r = 14, T_{15} > T_{14} r = 13, T_{14} > T_{13} ... r = 1, T_{2} > T_{1}...

Wouldnt it be n^(n-2) instead? n choices for 1 person, n choices for 2nd, n for 3rd, ... , n for (n-1)th person = n^(n-2) dunno

And just checked you're right until the very last line. 98b = 70a dividing: a/b = 98/70 So, a:b = 7:5

I'll check your working out but what I just did was: let angle O = B Area OCD = 1/2 a^2 sinB Area OAB = 1/2 (a+b)sinB Area OCD / Area OAB = 49/144 so: a^2/(a+b)^2 = 49/144 Square rooting, a/(a+b) = 7/12 12a = 7a+7b 5a = 7b a/b = 7/5 So, a:b = 7:5

Don't think so. the cats are indistinct

What Integrand said.l But A similar question could be How many ways can you place 'n' cats into 'n-2' cages?

What do you have trouble with specifically?

The one I'm thinking of right now is to convert the cos^(2)x into 1/2 [cos2x + 1] and then convert 7sinxcosx into 7/2 sin2x (double angle). Then expand and collect al the trigonometric functions to one side and constants on the other. You'll get something like cos2x + 7sin2x = 5 Then transform...

Have you tried integratong both sides of part i and letting v = 0 for maximum height x?

Re: MX2 2015 Integration Marathon I tried using the substituion x = u^(15/4) Which will then lead to u^4/(1+u^3) requiring long division and then a fractional decomposition. But then it didnt give me an arctan?

Re: HSC 2015 3U Marathon Oops didn't see it asked for locus If the line connecting P and Q is a focal chord (passes through the focus S(0,a)) then gradient PS = SQ When you equate this, you will find a relationship between p,q which is pq = -1 (what InteGrand said) so intersection =...

It wouldn't be. It's right angled if the two lines are perpedicular ie. their gradients multiplied equal -1 The second part made you find the perpendicular distance (the 'length' of the line when you draw it from B connecting to line AC at a right angle) which will then split the triangle into...

Re: HSC 2015 3U Marathon dy/dx = x/2a ==> dy/dx (at x = 2ap) = p Hence, tangent at P : y - ap^2 = p (x-2ap) y = px - ap^2 Similarly, tangent at Q: y = qx - aq^2 Subtracting, px - qx - ap^2 + aq^2 = 0 (p-q)x = a(p-q)(p+q) x = a(p+q) Substituting x = a(p+q) into one of the equations: y = apq...

I think it's because a thicker wire would provide less resistance, and so it would reduce the energy loss due to heat