Search results

Assuming suggested answer is correct, here's my effort: \int \frac{1} {\sqrt{u^2 - a^2}}du = ln (u + \sqrt {u^2 - a^2}) + C \\ \\ sin 2x = 2sinxcosx = (sin x + cos x)^2 - 1 \\ \\ -d(sin x + cos x) = (cos x - sin x)dx \\ \\ \therefore I = \int \frac {sin x - cos x}{\sqrt {sin 2x}} dx = -\int...

Is this a possible answer?? -ln[(sin x + cos x) + \sqrt{(sin x + cos x)^2 - 1}] + C

Another way to look at it: \frac{49\pi}{30} = 2\pi - \frac{11\pi}{30} $ and $ \frac {41\pi}{30} = \pi + \frac {11\pi}{30}\\ \\ So: sin \frac {49\pi}{30} $ and $ sin \frac {41\pi}{30} have the same related angles (11pi/30) and they are both negative, being in the 3rd & 4th quadrants resp...

\therefore tan^2x + tan x -6 < 0\\ \\ \therefore (tan x + 3)(tan x - 2) < 0 \\ \\ \therefore -3 < tan x < 2

a = 1 or 3?

Yes. Walter Rudin was a popular textbook or reference for Analysis in the 1960s and 1970s. Schaum Series was very popular in those days and I've a good dozen plus volumes on my shelves.

Is calculator in degree or radian mode? Maybe this is irrelevant.

7500 is only a simple constant multiple; so product rule not required. That is: \frac{d(7500h(t))}{dt} = 7500\frac {dh(t)}{dt}

\therefore \frac {dg(t)}{dt} = g(t) \times \frac {d(0.02t + 0.5cos(...))}{dt} \\ \\ =g(t)\times (0.02 - 0.05sin(...) \times \frac{\pi}{6})\\ \\ = 7500(0.02 - \frac{\pi}{12}sin(\frac{\pi}{6}t-4\pi))e^{0.02t + 0.5cos(\frac{\pi}{6}t - 4\pi)}

What if dumbcurry were mentally deficient, or has a wobbly foundation in basic maths. Do you think, in this case, he'll be able to pick up a copy of Cambridge and do as you suggested? That approach may suit somebody bright like you. But we just don't know much about dumbcurry. Appropriate advice...

If you decide you want to change to Standard, then it's best you do so ASAP.

dumbcurry: maybe you need good guidance. You may have worked hard; but if you have been approaching it the wrong way, you'd not do well. No guarantee you will do well in Standard either if you drop. Luckily you are at beginning of Yr 11. You still have some time to seek remedy. Suggest you find...

For (a) hypothesis step: \sum _{r=1} ^k (r+1)\times 2^r = k\times2^{k+1} \\ \\ \therefore \sum _{r=1}^{k+1} = k\times2^{k+1} + (k+1 + 1)\times 2^{k+1} = 2(k+1) \times 2^{k+1} = (k+1)\times 2^{k+1 +1} So if true for n=k, true also for n=k+1 . . . blah blah blah

1st continuous; second(Free Throws) discrete. Not being well versed with Basketball: you can be awarded 1, 2 or 3 free throws. So you can, if awarded 3 throws, you can score 0, 1, 2 or 3: so the percentage can be 0/3, 1/3, 2/3 or 3/3 [x 100%] - only 4 (a finite and countable number, you can...

Can you show us the questions from Baulko?

f(x) is concave down, if f''(x) < 0, assuming a < 1 < b. (f(a)+f(b))/2 is the value of the mid-point of the line segment joining the points A(a,f(a)) and B(b,f(b)), which is below the point M((a+b)/2, f((a+b)/2)).

As suggested by synthesisFR by Induction: a_n =2^n -1 $ is true for $ n = 1 $ as $ 2^1 - 1 = 1 = a_1\\ \\ $Assume true for n = k $ (k \geq 1)\\ \\ $i.e. $ a_k = 2^k - 1 \\ \\ \therefore a_{k+1} = 2a_k + 1 = 2(2^k - 1) + 1 = 2^{k+1} - 1 So, if the formula holds for n = k, it holds also for n...

Got the same answer as in your Method-2, using almost the same approach. Didn't post as I was not sure if my answer was correct.

I'm no expert in physics. Here you have a simple relationship connecting 4 quantities: E, h, c $ and $ \lambda From elementary algebra, if you know any 3 of these 4, you can find the 4th one. Here you want to find lambda, the wavelength; since you know the other 3, you should be able to find...

The poster most likely no longer active on Bored.