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the terry lee one is good because it has the complete worked solutions but still better the have another textbook ie the cambridge aswell

from what i know the caffeine content of a no-doz pill is the same as a cup of coffee.

The method for all three questions is the same: For (1): Rearranging 2x - y + 1 = 0: y = 2x + 1 Gradient = 2 diff. y = x^2: dy/dx = 2x therefore: 2x = 2 x =1 ie at x=1 the tangent is parralel to 2x- y +1 = 0

oh no posts are out of order...

let the parabola be y =ax^2 + bx + c Sub (0,-2): -2 = C Sub (1,0): 0 = a + b -2 2 = a + b ____________(1) Sub (3,-8): -8 = 9a + 3b -2 -6 = 9a + 3b -2 = 3a + b____________(2) (2)-(1): -4 = 2a -2 = a b = 4 therefore parabola is y = -2x^2 +4x -2

HOBr is Hypobromous acid. I agree with your tutors equation. Also the first equation does have water: C6H10 + Br2 <b>(aq)</b> --> C6 H10 Br2 Ill take a stab in saying that the top layer is 1,2-dibromocyclohexane since it would have a density lower than water.

ie Prove: a + ar + ar^2 .......+ar^(n-1) = a(1-r^n)/(1-r) Test n =1: LHS = a RHS = a(1-r)/(1-r) = a Assume n=k: a + ar + ar^2 .......+ar^(k-1) = a(1-r^k)/(1-r) Prove n=k+1: a + ar + ar^2 .......+ar^(k-1) + ar^k = a(1-r^k)/(1-r) +ar^k =a(1-r^k)/(1-r) +ar^n(1-r)/(1-r) =a(1-r^k + r^k -...

Chem: Teacher: "And the electron comes over here and goes. oooooh thats a bit alright!" --------------------------- Me: Can we make rainbow fizz sir? Teacher: No Me: Can we make rainbow fizz sir? Teacher: No Me: Can we make rainbow fizz sir? Teacher: Oh alright but ony...

i agree, except with lincoln they was already been specified in 01 i thnk it was

anyone know the name of the 3rd piece? i think it was by blood sweat and tears. i want that annoying tpt one :D

supermarket jobs suck! lifting, stacking, facing...its soooooo boring go for rodger david!

because of the double bond, there is four electrons close together, negative charges repel, thus it is more reactive

Re: Music one 1 F**K yeah!!! when the cheesy drum came in, in the piece with the trumpet our whole class pissed themeslves.

holyfuckilovechilli

alternatively you could consider the: area = are of rectangle - the area between the parabola and the x-axis ie: Area = 4a*a - Int(between -2a and 2a)[(x^2)/4a] = 4a^2 - (4a^2)/3 = (8a^2)/3

first you need the subsitution u=x - π/2 du = dx and also to recognise that: cosx = cos(u + π/2) = cos(u)cos(π/2) - sin(u)sin(π/2) = - sin(u) then get you to first part, then since its odd and the intergral is between a and -a (in this case pi/2) its equal zero.

-gradient = acceleration -area under the curve = displacement

ooops lol sry my bad: x'' = -4.3cos2t = -12cost2t = -4(x-1)

x=1+3cos2t x'= -2.3sin2t = -6sin2t x'' = -4cos2t = -4x from: x'= -6sin2t |sin2t|<=1 therefore max speed when |sin2t|=1; x' = -6 therefore max speed is 6m/s

Maybe meant to be: Prove 5^n >= 4^n + 3^n for n >= 2 Test n=2; LHS = 25 RHS = 19 + 9 = 25 Assume for n=k; 5^k >= 4^k + 3^k Prove for n = k + 1; 5^(k+1) >= (4^k+3^k)*5 >= 5*4^k + 5*3^k >= 4*4^k + 3*3^k...