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okay got it, thanks for all the help guys

But isn't friction similar to air resistance? Since they are both forces that oppose the movement of the pendulum?

but isnt air resistance an error in the system?

I try and refrain from using the pirate bay you have all these people going HOW DO I OPEN THIS RAR FILE?!?!?!! I DONWLOADED WINRAR BUT ITS NOT WORKING HELP!?!!!?!!!!!1111111 :burn:

My School's prac trial is coming up and I was wandering if air resistance is a valid response for why your experimental value is different from your theoretical value for motion type practicals e.g. determining a value for g using a pendulum because i remember one time i was marked incorrect for...

possibly a bunch of trig identities might simplify to a nicer term?

3. (sinx - cos)/(x-pi/4) = rt2sin (x - pi/4)/(x-pi/4) using the small angle theory sin (x-pi/4)/ x-pi/4 -> 1 as x -> pi/4 therefore its equal to rt2

1. ii) i just used long division you get the polynomial x^2 -4x + 1 using the GQF i get x = 2+/- 3^1/2 and (2+root3) x (2-root3) = 1

Hey Michael, it's Jeffrey how's the study going?

upon subbing u = sinx you get u/(2-u^2)^1/2 then you use the reverse chain rule e.g. (intergral) f'(x). (f(x))^n = (f(x))^n+1 therefore you get -(2-u^2)^1/2 if you have trouble understanding differentiate (2-u^2)^1/2 you'll end up the the question but with a negative sign sub sinx back in...

no takers?

Im sorry but for part i) this just had to be said, lol wut?

1st q just sub in x as 22.5 therefore tan^2 (22.5) = </abn>1-cos45/1+cos45 = (1 - 1/root2)/(1 + 1/root 2) = (root2 -1)/(root2 +1) = (root2 -1)^2 (rationalizing the denominator therefore tan22.5 = root 2 -1 <abn bc="2a," ca="b," ab="c," bn="CN=d<br"> </abn>

Thanks man I get it now

2) xy = 100 therefore y= 100/x - eq'n 1 2x + 2y = P(perimeter) - eq'n 2 sub 1 into 2 2x + 200/x = P therefore P' = 2 - 200/x^2 let P' = 0 for stat pts therefore 0 = 2 - 200/x^2 therefore 200 = 2x^2 therefore x = 10 P" = 400/x^3 when x = 10 P" > 0 therefore x = 10 is a minimum sub...

Some Math in Focus Q's i've completely forgotten how to find the volume of a solid of revolution formed between 2 lines 1. Find the volume of the solid of revolution formed if the area enclosed between y= x^2 and y= (x-2)^2 is rotated about the x axis 2. The area bounded by the curve y = x^2...

Not sure if this is right but for q 3 if P(x) = (x-2)(x-1)Q(x) then x = 2 and x = 1 are factors hence you sub those values in and you'll end up with all the terms cancelling out Question 4 using remainder theorem when x= -1 remainder is -11 therefore 0 + 0 + b = -11 therefore b = -11...

I'm in year 11 at the moment and also an accelerated student so it's hard for me to take year 11 seriously. I haven't really noticed much of a change in difficulty more of a change in structure since schools try to prepare you for the HSC e.g. in science courses, classes are more based on...

Whoa. I've never seen a question like this before, is this harder 3U?

I would try and help but i have no idea what the question is :confused: what does the 1st expression have to do with the sigma notation also why is there a power next to the sigma n ? I've never seen this kind of notation before :bomb:<==== my brain when I saw the question