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Considering the equation T^2 = 4pi (l/g), when length is zero, T^2 would also be zero so I'd say that the graph would pass through the origin. But since your teacher disagrees with another teacher, it could well be more complicated.

1 b) To get P, Q and R, I think you just multiply A, B and C by cis45 respectively. So P = Acis45 = z cis45 = z (sqroot2/2 + sqroot2i/2) Q = Bcis45 = (z + iz) cis45 = sqroot2iz R = Ccis45 = iz cis45 = z (-sqroot2/2 + sqroot2i/2) But I'm not 100% sure about my method so...

b) When a projectile reaches its maximum height, v = 0. To find max. h, you can use either vy^2 = uy^2 + 2aysy or sy = uyt + 1/2ayt^2 However, both require knowledge of uy. So, using sy = uyt + 1/2ayt^2: 0 = 7.5uy - 4.9 x 7.5^2 7.5uy = 275.625 uy = 36.75ms-1 Now to find max. h...

That's cool! I think it's hilarious that Elvis is now considered a planet. :haha:

u = 12ms-1 ux = 12cos 45 and uy = 12cos 45 = 6sqroot2 Using sx = uxt 25 = 6sqroot2t t = 25/6sqroot2 Using sy = uyt + 1/2ayt^2 0 = 6sqroot2 x (25/6sqroot2) + 1/2 x (25/6sqroot2)^2 x a 0 = 25 + (625/144)a (625/144)a = -25 a = -5.76ms-1 (assuming up is positive) Therefore...

I think that's because you subbed in "cos^2a - 1" for sin^2a and I used "1 - cos^2a" (derived from cos^2a + sin^2a = 1). I also forgot that from sina = 0, a = 360 as well as 0.

I tried the first one: 3sina - 4sin^3a + 2sinacosa = sina 4sin^3a - 2sina - 2sinacosa = 0 2sina (2sin^2a - 1 - cosa) = 0 Therefore 2sina = 0, a = 0 Therefore (2sin^2a - 1 - cosa) = 0 2 (1 - cos^2a) - 1 - cosa = 0 2 - 2cos^2a - 1 - cosa = 0 2cos^2a + cosa - 1 = 0 (cosa + 1)(2cosa - 1) = 0...

This is how I did it: z^4 + 1 = (z^2 - 2zcos (pi/4) + 1)(z^2 -2zcos (3pi/4) + 1) Divide both sides by z^2 (for RHS, I just divided each bracket by z, which is the equivalent of dividing by z^2): z^2 + 1/z^2 = (z - 2cos (pi/4) + 1/z)(z - 2cos (3pi/4) + 1/z) Now group z + 1/z and change...

I'm pretty sure it's still better than IPT and also general maths. On "table A3", which shows the raw and scaled data for 2005, the highest mark in ancient was 50 (and it did not change with scaling) then the max in IPT was 49.5 and it dropped to 47.0 (these marks are per unit so double them...

In terms of your subjects, IPT doesn't usually scale that well whereas extension english and ancient history do carry better weightings. If you pick ancient history instead of IPT, you will be doing 14 units and you only need 12 so you may be creating an unecessary workload for yourself. As...

I'm currently doing 12 units but I had 14 at the start of yr12 and was having serious difficulty deciding what to drop. My choice was between 2 sciences and, in the end, I kept physics coz I'm doing extension maths and they're sort of interrelated so I thought maybe I could cut down on my...

Yep, my mistake, I didn't read the question properly. I hope I don't do that in a test! :evilfire:

Use De Moivre's Theorem: cos7@ + isin7@ = (cos@ + isin@) ^7 Expand RHS Gather real terms, which equal cos7@ And, using trig identities, eliminate sin@'s to get the answer in terms of cos

This question should work if you switch the complex numbers to polar form. This is what I get: z4 = -1 z = rcis theta, therefore, z4 = r4cis 4theta (De Moivre's Theorem) -1 = cis pi Equate the two: r4 cis 4theta = cis pi r4 = 1 r = 1 and 4theta = pi + 2pi.k (accounting for...