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Do you know what sort of marks he had?

I would probably change the uniform, gahh.

Either method is appropriate I guess, using the substitution u=cosecx+cotx works out easier than you would think.

int. sec3x.tanx.dx Let u=secx du/dx=secxtanx dx=du/secxtanx int. u2.du =u3/3 + c =sec3x/3 + c 2. int. cosecx.dx Multiply top and bottom by cosecx+cotx, then use u=cosecx+cotx. 3. int. sec2x = tanx + c

I go to Gosford High.

It's great :D Although if you want to do it, be prepared to put in a lot of time because even if you might understand everything, there is quite a lot of content to cover. But yeah, definitely give it a go if you love maths.

What maths are you doing in chemistry Aerath? I'm curious as I can't remember really doing any maths in prelim chem.

Thanks heaps

Hey I don't do music, but I'm just asking for a friend whether or not you need all the instruments in the song that you are playing for your HSC performance? I thought that you can use a backing track but apparently you will lose marks? Thanks

Fixed :D.

Where did the three extra marks come from? 16/30 * 100 = 53.333...%, so Aplus' calculation was correct. We know the highest mark, and the lowest mark, but that doesn't tell us anything about the average ...I think that's what Aerath was wondering about.

If F''(x)>0, then the point at x is a minimum. If F''(x)<0, then the point at x is a maximum. OR we can use the first derivative: If F'(x) changes from positive to negative through x, then (x,F(x)) is a maximum stationary point, conversely, if it changes from negative to positive, then (x,F(x))...

Well, oxygen actually has 7 different oxidation states. All we need to know: OS(O) in H2O2 = -1 [peroxides] OS(O) in H2O = -2 [oxides] OS(O) in O2 = 0 [of course :uhhuh:] OS - oxidation state. The other possible OS(O) are -1/2, -1/3, +1 and +2.

No problem :wave:

Produce AD to a point E, and join AB. Then: &ang;CAE = &ang;ABC (angle between tangent and a chord through the point of intersection is equal to the angle in the alternate segment) &ang;ABC = &ang;ADB (similarly) Hence &ang;CAE = &ang;ABC = &ang;ADB => &ang;CAE = &ang;ADB => AC||DB...

Differentiate the given equation for the area and solve dA/dx = 0 to find any maximum, minimum values of x. Use either the first derivative method or the second derivative to determine whether these values give a max. area or min. area, and substitute the maximum value into the equation for A to...

Damn, that's painful. :eek:

hehe :D

Uh there's no extension art course as far as I know.

y=e^-x(cos2x+sin2x) yex=cos2x+sin2x yex+y'ex=-2sin2x+2cos2x [differentiating both sides] ex(y+y')=-2sin2x+2cos2x ex(y+y')+ex(y'+y'')=-4cos2x-4sin2x [differentiating both sides] ex(y''+2y'+y)=-4(cos2x+sin2x) y''+2y'+y=-4y y''+2y'+5y=0 Quite a neat method.