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Do you have any trouble drawing the graphs? Draw the graphs and then find the points of intersection, which occur at x=-1 and x=14: So just find the integral of (x+2)1/2, evaluated from 14 to -1 and then subtract the area between y = (x+6)/5, the x-axis and the lines x=-1 and x=14 to find the...

Yeah you could do that, but also make sure to label the graph as y=6x3. Shouldn't be any problem.

Actually, if you draw the graph, it should become readily apparent: First find the area between y=2x, x=2 and the x-axis, easy enough just integrate 2x from x=2 to x=0. Then just find the absolute integral from x=2 to x=0 of -x^2 with respect to x, which will give you the area of the part below...

Woah, you serious tommykins? There are 8 topics in total (I think) lol. We've completed three topics so far.

10 units.

Draw the graph as you would normally, then reflect anything below the x-axis in the x-axis. This is because y≥0.

1. dy/dx = 3sec23x At x = pi/9; dy/dx = 3[1/cos2(pi/3)] = 12 2. dy/dx = -cos(pi-x), at x = pi/6; dy/dx = -cos(pi-pi/6) = &radic;3/2 &there4; Equation of tangent at (pi/6 , 1/2) is given by: y - 1/2 = &radic;3/2 (x - pi/6) <-- you can simplify that further if you wish :) 3. dy/dx = 6cos2x...

3. Let the number be represented by a+10b; a+10b=8(a+b) (1) a+10b-45=b+10a (2) From (1): 7a = 2b therefore a = 2/7 b From (2): 9a = 9b - 45 a = b - 5 b = a + 5 b = 2/7 b + 5 therefore 5/7 b = 5 => b = 7 and a = 2 Checking this: 72 is between 10 and 100, sum of its digits is 9, multiplied by 8...

Just so damn unmotivated. Not crazy ...yet :D

The HSC isn't something to judge someone's intelligence by, just so you know. You could be mentally brilliant but couldn't really give a shit about the HSC just cause you don't need a super-high UAI for whatever you want to do in uni. Just making a point :D.

Best to use a diagram if you're unsure. Oh, and Aerath, you were doing it incorrectly, not sure what you were thinking, maybe you should wait until you do integration properly in school (unless you've already done it? But it doesn't seem like you have :rolleyes:)?. Avenger6 is correct in his...

3) z6 + 1 = (z2)3 + 13 Now using the identity for sum of cubes: z6 + 1 = (z2 + 1)(z4 + z2 + 1)

Lol. 50% marks from doing 45-60 mins of Maths a night? In prelim? No way :rofl:

&int; [From 1 to 3] (x + 1/x)2 dx = 14 - 2/3

Roger Zelazny - Lord Demon

It's just ... :/

du/(8u3) = (1/8) . u-3 . du = (1/8).(-1/2) . u-2 + c = -1/(16u2) + c

Re: Important news for all year 11 students 4[a-2(b-c)-{a-(b-2)}] = 4[a-2b+2c-(a-b+2)] = 4[-b+2c-2] = 8c - 8 - 4b

Yes, circle geometry, after a bit of practice, becomes probably one of the easiest topics of extension maths as it is such a confined topic. You have a set of rules to learn, and then you only have to apply them. Just stick with it, and it will become a LOT easier.

Try: y=asin(x) <--the parentheses are important!