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Don't give up. Combinatorics isn't an easy topic for most people. You should spend this summer break working on any weak areas in your maths. If you want to do Engineering, you'd better be good in your Maths. Getting frustrated with your MX1 isn't going to help you. Tutoring centres are...

For b): I think x>= 0.5. So locus is the arc of the unit circle, centre O, with -\frac {\pi}{3} \leq arg z \leq \frac {\pi}{3}

a) x^4-x^2 + 1 = x^4 + 2x^2 + 1 - 3x^2 = (x^2 + 1)^2 - (\sqrt 3 x)^2 \\ \\ = (x^2 +1 + \sqrt 3x )(x^2 + 1 - \sqrt 3 x) b) x^6 + 1 = (x^2)^3 + 1^3 = (x^2 + 1)(x^4-x^2 +1) \\ \\ =(x^2 + 1)(x^2+\sqrt 3 x + 1)(x^2 - \sqrt 3 x + 1)

Where the place?

Is answer ??: \frac {47}{72} - \sqrt { \frac {\sqrt {119} + 12} {24}}

The general angle formulae for sin x = sin a, cos x = cos a and tan x = tan a are: \\ $sin x = $ sin\theta: x = n\pi +(-1)^n \theta \\ \\ $cos x = $ cos\theta: x = 2n\pi \pm \theta \\ \\ $tan x = $ tan \theta : x = n\pi + \theta\\ \\ n = 0, 1,2,3, . . . .

sin3x = cos x =sin(\frac{\pi}{2}-x)\\ \\ \therefore 3x = n\pi + (-1)^n (\frac {\pi}{2} - x) $ $(n = 0, 1, 2, . . .) \\ \\ $For n = 0, 1, 2, 3, 4, 5, 6 we get respectively: $ \\ \\ x = \frac {\pi}{8}, \frac{\pi}{4}, \frac {5 \pi}{8}, \frac {\5\pi}{4}, \frac {9\pi}{8}, \frac {9\pi}{4} $ (outside...

I think it would be highly impractical to teach 2 students with different texts simultaneously. Is it not possible to tutor each separately? Why are you being required to teach them together, unless they are from a same family?

The diagonals of a kite also intersect at 90 degrees; so should work. MABC here is both a kite(a kite does not normally have all 4 sides equal; one pair of sides is usually shorter than the other pair) and a rhombus.

For (ii) let the "centre" of the regular hexagon be M ( M is also the common mid-points of AD, BE & CF). All the triangles MAB, MBC, MCD, MDE, MEF & MFA are equilateral. The figure, MABC is made up of triangles MAB & MBC is a rhombus so their diagonals AC & MB are perpendicular to each other...

Q1. S_n = \frac {n}{2}(2a + (n-1)d) \\ \\ \text{ is simply the formula for sum of the 1st n terms of the the A.P.: a, a+d, a+2d, a+3d, . . . . }\\ \\ \text {whose n-th term is simply: } a_n = a+(n-1)d Q2. \text {The formula:} S_n = \frac {a(r^n -1)}{r-1} \text { is simply the formula...

Read up on Mechanics (in Physics). Maybe get friend to help you in Mechanics and further Maths. Or, if you can afford it, come to me, to fill you in in your Maths & Mechanics.

Not clear to me what course you are planning to pursue? Communication as in Media or Journalism(wonder if you need any maths) or in Communications Engineering(you'll need very advanced maths)?

You can see from the simple, real, quadratic equation: x^2+3x+5 = 0 \\ \\ \implies x = \frac {-3 \pm \sqrt{3^2 - 4\times 1\times 5}}{2\times 1} \\ \\ $its 2 complex roots are conjugates: $ \alpha = -\frac {3}{2} + \frac {\sqrt {11}}{2} i $ and $ \overline \alpha = -\frac{3}{2} - \frac...

Another way of looking at the situation is that the equation z^5 - 1 = 0 has 5 roots, a, b, c, d and e say. In this case z^5 - 1 = (z-a)(z-b)(z-c)(z-d)(z-e). For z^5 - 1 = 0, we have 5 roots, one real and the other 4 being complex. Since z^5 - 1 = 0 is a polynomial eqn with real coefficients...

Question incomplete? Here's a counterexample, with w & z satisfying the specifications. If |w| = 2|z|, then you can get the result. Or have I made a mistake somewhere?? $Let $ w = 3zcis \frac{\pi}{4} $, say.$\\ \\ \therefore w^4 + 16z^4 = (3zcis \frac{\pi}{4})^4 + 16z^4 = 81z^4cis(4\times...

Explanation. Please refer to your graphs of i) y = arctan(x) and ii) y = arccos(x) \text{For: }y = tan^{-1}x \text { x can be any real number. }\\ \\ \text {But in } y = tan^{-1}\sqrt{x^2 -1} $, $ \sqrt{x^2 -1} \geq 0 $ so that $ 0 \leq y <\frac {\pi}{2}\\ \\ \\$For: $ y = cos^{-1}(e^x) =...

But I still don't know if my answers were correct?

For last one: \text {Let: }sin^{-1} \frac {1}{3} = A \text { and } sin^{-1} \frac {1}{4} = B\\ \\ \text {Now draw 2 right-angled triangle, one with angle A, sides 1, 3 and } \sqrt 8 \text { and the other with angle B and sides 1, 4 and } \sqrt {15}\\ \\ \text {From these triangles, you can...

I'll explain it later. Please bear with me.