-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
Search results
-
MATHS P3 HELP!
In part (i) u can cross the two direction vectors to get a vector perp. to both the lines. Then, u can use the fact that PQ shud b parallel to this perp. vector, to get the vector PQ which turns out to be -2i + 2j - k NB: This step is wrong. jux ignore it @ (i - j - 5k) + s(i - j - 4k) = (2i -... -
Hydrogen Bonding--chemistry
It is present on this page under the heading:- "Water as a "perfect" example of hydrogen bonding" -
help with math mechanics paper4 m/j 02 question!!!!!!please!
Re: help with math mechanics paper4 m/j 02 question!!!!!!ple p4 o/n 01 question 6 (ii) & (iii) for part (ii) When Q hits the ground, P is ready to start free-fall (in upward direction) with an initial velocity, say u. We can figure this u out using the fact that Q and P travel the same... -
help with math mechanics paper4 m/j 02 question!!!!!!please!
Re: help with math mechanics paper4 m/j 02 question!!!!!!ple In (ii), use the 1st equation of motion v(f) = v(i) + at = 0 + 2.5(2) = 5m/s (the velocity of A upwards b4... -
math question!
Forn n ≥ 2 we have:- 1 = cos^n(x) - sin^n(x) ≤ Abs [cos^n(x) - sin^n(x)] ≤ Abs [cos^n(x)] + Abs [sin^n(x)] ≤ cos^2(x) + sin^2(x) = 1 Hence sin^2(x) = Abs [sin^n(x)] and cos^2(x) = Abs [cos^n(x)], from which it follows that sin(x), cos(x) ∈ {1,0,-1} ---------> x ∈ pi*Z/2. By inspection one... -
OCT/NOV 2009 Paper 12
In this question, jux use the logic that when "the monomers are completely polymerised," they in fact are coverted to polymers wich comprise groups of 2 or more monomers (but not only 1 obviously, cos den it wd b a monomer :P) combined together. This contradicts with the statement 1 that implies... -
MATHS P3 HELP!
U shud try urself 1st den ask others. :x Anyways (wdout justifying my steps), 10.(a) (i - j - 4k) x (2i + 5j + 6k) = 14i -14j + 7k wich is // to 2i - 2j + k (i - j - 5k) + s(i - j - 4k) = (2i - 9j -14k) + t(2i + 5k + 6j) [(1 + s)i + (-1 - s)j + (-5 - 4s)k] - [(2 + 2t)i + (-9 + 5t)j + (-14 +... -
-
MATHS P3 HELP!
sec(x) - tan(x) = 0.5 1/cos(x) - sin(x)/cos(x) = 0.5 Multiplying throughout by cos(x) ---> 1 - sin(x) = 0.5cos(x) cos(x) = 2 - 2sin(x) Squaring both sides ---> cos^2(x) = 4 - 8sin(x) 4sin^2(x) 1 - sin^2(x) = 4 - 8sin(x) 4sin^2(x) Rearranging ---> 5sin^2(x) - 8sin(x) + 3 = 0 [sin(x) -... -
AS MATHS Help Needed P1
(1 + ax)^n = (1)^n + n(1)^(n - 1)*(ax) + (1/2)*n(n - 1)(1)^(n - 2)*(ax)^2 = 1 + 24x + 270x^2 Now, comparing coefficients ----> an = 24 .....[1] (1/2)*n(n - 1)*a^2 = 270 .....[2] Solving them simultaneously yields n = 16 and a = 3/2 = 1.5 -
Maths P1-Help Needed!!
yeah indeed it ain't in the syllabus, bt anyways 1)y = sec(0.5x) int[y dx] = int[sec(0.5x) dx] Let u = 0.5x ----> dx = 2du int[y dx] = 2*int[sec(u) du] Jux multiply by "sec(u) + tan(u)" above and bottom to get a function(as denominator) with its derivative(as numerator)... -
derivative.......
yar actually, m essentially studying the same syllabus as yours (i.e. CIE,& i guess most probably, dat's wt u r studying 2 :P) :!: 2 b precise, m in AS level n studying Phy(9072), Chem(9701), Maths(9709) and Further Maths(9231), and will b havin AS xmx in the cming May/June session -
-
-
derivative.......
In the above case, u all hav assumed that the question was to differentiate "ln[(x)^sin(x)]." Bt what if the question was to find the derivative of "[ln(x)]^[sin(x)]." In dat case, we could employ logarithmic differentiation of the form y = f(x)^g(x) ---> ln(y) = ln[f(x)^g(x)] = g(x)*ln[f(x)]... -
derivative.......
I guess u mean to ask "differentiate [ln(x)]^[sin(x)]....?" ain't it or do u mean "differentiate ln[(x)^sin(x)]....?" or sth else...??? -
MATHS P-2 AND P-3
I'll try the problem 2 1st:- The temp/deg C after t min shud be T(t = t min) = T(t = 0 min) * exp[(rate of change of T) * t)] = 85*exp(r*t) where; r = rate of change of T, t = time t Applying this to the given condition... -
-
Chemistry oct/nov 2010. (paper 12)
Or to put it more simple, We can use the idea of PV=K......and dalton's law of partial pressures which says that total pressure is equal to sum of partial pressures.......after connecting the both flasks 5 dm cube helium will spread throughout the two flasks and will occupy vol of 15 dm... -
Chemistry oct/nov 2010. (paper 12)
U can jux use the idea of ratios here: [5*12+10*6]/[5+10] = 8kPa