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Your answer doesn't work for the second equation: 3(3)-(-2) = 9 + 2 = 11 \neq 7.

\frac{1+\sqrt{1+\frac{\sqrt{3}}2}}{\sqrt{1+\frac{\sqrt{3}}2}} \cdot \frac 22 = \frac{2+\sqrt{4+2\sqrt3}}{\sqrt{4+2\sqrt3}} = \frac{3 + \sqrt3}{\sqrt 3 + 1} = \frac{\sqrt 3 (\sqrt3+1)}{\sqrt 3 + 1} = {\sqrt 3}

ii) The line through AP has equation y = t(x+1) . (can you see why?) Solve this simultaneously with the equation for the unit circle to obtain a quadratic in x : (1+t^2)x^2+(2t^2)x+(t^2-1)=0 This equation tells us where the line intersects the circle. You could just throw the quadratic...

For k > 0, \, n \in \mathbb{Z}^+, let I_n = \int x^k (\log x)^n \, dx. Prove that (k+1) I_n = {x^{k+1}(\log x)^n} - {n}I_{n-1}. Given that \lim_{x \to 0} x^k (\log x)^n = 0, Show that \int_0^1 x^k (\log x)^n \, dx= (-1)^n \frac{n!}{(k+1)^{n+1}}, \frac 1 e \int_0^e (\log x)^n \, dx =...

Construct lines to the centre from each intersection point. This forms sectors of isosceles triangles. Because equal arcs of a circle subtend equal angles at the centre, you can find all the angles in each sector. Then solving for x should be easy. I believe the answer should be 60^\circ.

The standard formula for volumes of solids with similar cross-sections is V = \pi \int_a^b r^2\, dh (in this case, r = y, \,\,h=x) This can be thought of as approximating the volume of the solid with several cylindrical slices, and then making these slices thinner and thinner. The volume of...

Not really, my team was great and it was really nice working with them. It was mostly due to: - our robot's parts being poor quality and some failing on us in final testing - bad marks on individually submitted reports, no feedback was given (having them squished in the last two weeks of term...

======================================================== T1 COMP1511 Programming Fundamentals.........95 HD T1 ENGG1000 Engineering Design...............64 PS T1 MATH1141 Higher Mathematics 1A............99 HD ======================================================== Term WAM...

getting a 99 in math1141 was a really pleasant surprise... was expecting somewhere around 80-90.

Note that before we do anything else, b must be nonzero. Also, to apply the identity we require 1 + \cos ax \neq 0 \iff ax \neq \pm\pi, \, \pm 3\pi ... One method is \begin{aligned} & \frac{1 - \cos ax}{bx} \\ &= \frac{\sin^2ax}{bx(1+\cos ax)} \\ &= \frac{4...

Let P(x) = ax^3+bx^2+cx+d . Because P is odd, P(x) = -P(-x) and therefore b = d = 0 . P(x) = ax^3+cx . By the remainder theorem, P(-4) = 0 and P(3) = 21 . Performing substitutions, we obtain \begin{cases}16a + c &= 0 \\ 9a + c &= 7 \end{cases} \implies a= -1,\, c =16

Here's another solution: \begin{aligned} \frac{y-x}{x+y} &= \cos \theta \\ 1 - \left(\frac{y-x}{x+y}\right)^2 &=\left(\frac{y-x}{2}\right)^2 \\ (x+y)^2- (y-x)^2 &= \frac 1 4 (y-x)^2(x+y)^2 \\ 16xy &= (x^2-y^2)^2 \end{aligned} i.e. (x^2-y^2)^2-16xy=0 It's also clear to see solving this...

Firstly, (2-x)^n = \sum_{k=0}^n \binom{n}{k} 2^{n-k} (-x)^k Set k = 2, 3, giving the coefficients of x^2 and x^3 : \binom{n}{2} 2^{n-2} (-1)^2 and \binom{n}{3} 2^{n-3} (-1)^3 Now all you need to do is to solve \binom{n}{2} 2^{n-2} (-1)^2 = - \left(\binom{n}{3}...

I = \int^{\pi/3}_{\pi/6} \frac{\tan^4x+\cot^4x}{1+\tan^{\frac{2019}{2020}}x} \,dx First we note that (\tan x)^{-1} = \cot x for x\in [\pi/3, \, \pi/6] . Using \int_a^{b} f(x) \, dx =\int_a^{b} f(a+b-x) \, dx , \tan(\pi/2 - x) = \cot x , and \cot(\pi/2 - x) = \tan x...

Re: HSC 2018 MX2 Integration Marathon Hint: \sqrt{x^2} = |x|

i.e. "count all the rows before, and add a new row". The algorithm could look like: BEGIN total = 0 FOR row = 1 TO 10 STEP 1 total = total + row Display total NEXT i END Specifically, look at this line: total = total + row In total + row, total...

your pseudocode is logically correct (if you intended the "count+1" to be outside of the IF statement) but you have some syntax errors. BEGIN count = 0 present = 0 GET advent WHILE advent has not ended IF advent[count] = "Present" THEN present =...

Re: HSC 2018 MX2 Marathon The equation of the tangent at the hyperbola is given by y-a\tan\theta = \csc \theta(x-a\sec\theta) Because the shortest distance between a point and a line is the perpendicular distance, M is the intersection between the tangent line and the line perpendicular...

Sure, feel free to PM me. (the same applies to anyone else reading this who wants help)