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Yeah my bad, thanks Trebla.

Consider: y = x - \frac{1}{x} For the inverse, swap x and y: x = y - \frac{1}{y} xy = y^{2} - 1 y^{2} - xy - 1 = 0 , which is a quadratic in y. Using the quadratic formula: y = \frac{x \pm \sqrt{x^{2} + 4}}{2} But now we need to decide whether to take the plus or the minus. Try...

[ tex ] y = 2x^{2} + 5x - 3 [ /tex ] write that^ without spaces around 'tex' and '/tex' and you get the following: y = 2x^{2} + 5x - 3 Quote me to see

this lol no idea what it is

1. \int \frac{x^{2}e^{2x^{2}}-1}{x}dx (is that what you were after?) = \int xe^{2x^{2}}dx - \int \frac{1}{x}dx You can use a substitution for the 1st. (eg let u = 2x2) edit: btw answer is \frac{1}{4}(e^{2x^{2}}-4lnx) + c

But you don't need "to write all that crap down", I just did so above to make it clear what I was doing. I'm not saying you can't "just factorise" it, I was merely explaining another method that someone might find useful.

Assuming they have to be arranged 4 on bow, 4 on stroke, no. of ways = (4.3.2)*(4.3)*3! = 1728 I did it by placing each person individually: Firstly seat the 3 who must row on bow side: 1st one has 4 seats to choose from 2nd one has 3 seats to choose from 3rd one has 2 seats to choose from...

Not really... only takes 2 steps: 1) solve the quadratic, which shouldn't take more than a few seconds (giving x = a, b) 2) then write (x-a)(x-b) = 0

Another way to factorise a quadratic is to find its roots first. 4x^{2} + x - 18 = 0 using the quadratic formula: x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} x = \frac{-1 \pm \sqrt{(-1)^{2}-4(4)(-18)}}{2(4)} x = \frac{-1 \pm 17}{8} x = 2, \frac{-9}{4} Now we have the roots we can "go...

Sorry guys I've already bought it.

for 2007 Q10b part (i) at P, N_{1}= \frac{L_{1}}{x^{2}}, N_{2}= \frac{L_{2}}{(m-x)^{2}} N = N_{1} + N_{2} N = \frac{L_{1}}{x^{2}} + \frac{L_{2}}{(m-x)^{2}} part (ii) now we want to minimise the noise level N, so we differentiate (with respect to x) remember when you are doing this that L1...

The first term is: T1 = a The second term is: T2 = ar The last term is: Tn = arn-1 The "last term but one" (the second last term) is: Tn-1 = arn-2 So to answer the question, the product of the first term and the last term: T1*Tn = (a)*(arn-1) = a2rn-1 and the product of the second term and...

Are you sure it asked for integration by parts? Because it rolls out quite easily using the substitution t = tan(x/2)

Definitely not 3unit, it's actually a very simple 4unit mechanics question. We are given that: x = 6t - t^2 Therefore: v = 6 - 2t a = -2 which implies that the force along the direction that the particle is travelling is -2 (force = mass*acceleration, but let's assume mass = 1, so...

Isn't the integral of sec x = ln(sec x + tan x)? You could differentiate ln(sec x + tan x) to check that its derivative is indeed sec x, just like tacoqym did for the other one. Another way to look at the integral of sec x is to multiply top and bottom by (sec x + tan x), giving: [sec x(sec x...

rearrange (3) giving b = 7/a2 sub into (2) giving: a2 + 2a*(7/a2) = 15 a3 - 15a + 14 = 0 (a - 1)(a2 + a - 14) = 0 (found the root a = 1 and then factored by inspection... don't know if this is 2unit?) Therefore a = 1 (disregarding the solutions to the quadratic because they are irrational) And...