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As pointed out by ultra908, question asks for the point of intersection of 2 lines: L1: \binom x y = \binom 2 3 + \lambda \binom {-2} 3 \\ \\ L2: \binom x y = \binom 6 2 + \mu \binom 1 {-4} Rewriting these vector equations in their parametric form: L1: x = 2-2\lambda $ and $ y = 3 +...

Truth Tables are easy - for the basic ones at least. Interestingly, at the lowest level of IB Maths, called Maths Studies(SL), Logic is one of the its topics, covering: Propositions, Compound Propositions, Truth Tables & Logical Equivalence, Implication & Equivalence, Converse, Inverse &...

Why don't you wait till you've done a term or two of Yr 11 and see how you go. If you're coping fine, then you don't need a tutor. If you are not doing so great, or the maths is not being well-taught in your class, then you can consider getting a tutor.

Your problem is to find the point, P say, on the parabola y = f(x) where the normal passes. This normal has gradient -(1/4), so that the gradient of the tangent to the parabola at P is 4. You use this fact to find P, using f'(x) = 4. f'(x) = 4x-5 = 4 \implies x = \frac {9}{4} \implies y =...

I have not read any of the input above. But for what it is worth, for collision to occur: i) both projectiles must be capable of being in a same location ii) both projectiles must be at this location at the same time. ps: The said location is of course where the 2 trajectories intersect.

Thanks for pointing out the typos in my solution, CM-Tutor. I didn't even notice them.

Can you show examples of your "silly mistakes"???

Find someone in his/her 60s or older for advice on how to write. Speedy handwriting comes with doing running writing - but most people do not know how to join their letters. Also I find that once you are set in the way you write, it is very difficult, if not impossible, to unlearn your way of...

Every year I read about the blind trying to lead the blind of handwriting. So hilarious and so tragic.

radius of base of cone r = h x tan 60 = sqrt(3) h \therefore V = \frac {1}{3} \times \pi r^2 \times h = \frac {1}{3} \times \pi \times (h\sqrt 3)^2 \times h = \pi h^3 m^3/min\\ \\ \therefore \frac{dV}{dt} = \frac {dV}{dh} \times \frac {dh}{dt} = 3 \pi h^2 \frac {dh}{dt} = 3\pi \times 4^2...

I asked you to forward me those questions, but you have not. I may be able to help you, even if you cannot afford my rates.

Jasmine Send me shots of the questions and let me have a look. You can send to my mobile : 0490 780 347. I'll then let you know if I can be of help. Drongoski

25 - 30 questions in one lesson! And how long is the 1 lesson - 1hr, 2hr, 3 hr ?? Solution outline or full solution?

How about this way? sin\theta tan \theta + 2 sin\theta = 3 cos\theta\\ \\ \therefore sin\theta(tan\theta + 2) = 3 cos\theta\\ \\ \therefore \frac {sin\theta}{cos\theta}(tan \theta + 2) = 3 $ $ (cos\theta \neq 0)\\ \\ \therefore tan^2\theta +2tan\theta - 3 = 0\\ \\ \therefore (tan\theta +...

Volume = $ vol of cylinder $ - \pi \int ^* _* x^2dy \\ \\ = \pi \times 2^2\times ln4 - \pi\int ^{ln4} _0 (\frac{1}{2}e^y)^2dy \\ \\ =4\pi ln4 - \frac {\pi}{4} \int _0 ^{ln4} e^{2y} dy \\ \\ = 4\pi ln4 - \frac {\pi}{8} \left [e^{2y} \right ]^{ln4} _0\\ \\ = \frac{\pi}{8}(32ln4 - 15)

Reading widely and a variety of well-written material will definitely help you greatly in your future, even if it may, or may not, directly improve your performance in your HSC English. It is amazing such question even need to be posed.

Fully elaborated, and beautifully explained, above. Being a slow coach, and not being sure which TeX codes to use, I was entering a couple of lines at a time. Sorry: I did not see the missing part (now rectified) in line 3.

P(reject) = 0.03 + 0.03 = 0.06 P(accept) = 1 - P(reject) = 1 - 0.06 = 0.94 (0.94 + 0.06)^{10} = 1^{10} = 1 = \sum _{i=0} ^{10 }\binom {10} i 0.94^{10-i}0.06^i There are 11 terms in this binomial expansion, corresponding to i =0, 1, 2 . . . 10 P(none rejected) = P(when i=0) and P(1 rejected)...

The centre of the cube is M(0.5a,0.5a,0.5a). Find the angle between vectors MA: (0.5a,-0.5a,-0.5a)' and MC: (-0.5a,0.5a,-0.5a)'. You'll get the angle = 180 - inverse cosine of (1/3) = 109 deg 28 min. Above not a direct response to your question.

i remember doing a similar question in an IB Exam paper. can't find my old solution. I think the bonding angle is the same as that for methane: 109.5 degrees?