Search results

Re: I dare someone to give me worked solutions for these 5 complex number qs (not har a)i) (√3 + i) / (1+i) = 2 (√3/2 + i/2) / [ (1/√2)(1/√2 + i/√2) = 2√2 (cis pi/6) / cis (pi/4) = 2√2 cis -pi/12 Mod z = 2√2 Arg z = -pi / 12 ii) So to be real, arg z must = 0 or pi Therefore, n = 12, to make...

Yeah, it's pretty bad... Because there's no tutorials or labs, the first day I have one 1 hour lecture and then on Tuesday, I have two 1 hour lectures, but with a 5 hour break inbetween O_O

First of all you need to find the points of intersection x^2 = x + 2 x^2 - x - 2 = 0 (x-2)(x+1)=0 x = -1, 2 Now to find the volume about the x axis, its V = pi S y^2 dx But here, since we're dealing with 2 graphs, we need to do the upper graph minus the lower graph. If we draw a diagram then...

If you differentiate again, then you will find that w is not a root of multiplicity 3. Also, w cannot be a root of multiplicity >2 as complex roots are always in conjugate pairs. So therefore one of the factors is (x-w)^2 Since w is a double root, then the conjugate of w is also a double root...

P(x) = 3x^5 + 2x^4 + x^3 - 6x^2 - 5x - 4 Therefore, P'(x) = 15x^4 + 8x^3 + 3x^2 - 12x - 5 P'(w) = 15w^4 + 8w^3 + 3w^2 - 12w - 5 = 15w + 8 + 3w^2 - 12w - 5 since w^4 = w^3 * w, where w^3 = 1 = 3 + 3w^2 +3w = 3 (1+w+w^2) = 0 , since 1+w+w^2 =0 Therefore, w is a repeated root of multiplicity 2...

Because you need to find the area between the curve and the y axis, you need to make the subject of the formula x. So y = 3 / (x-2) xy -2y = 3 xy = 3 + 2y x = (3+2y) / y = 2 + 3/y Now since we need 4 sub intervals, it means we will be taking the values of y; 1, 1.5, 2, 2.5, 3. Now S (1=>3) 2...

Re: TUTOR 2u/3u/4u Maths | 99.90 ATAR | 3u 99 4u 97| State Rank for Jap | Western Syd edited: change of a few details

There is actually a very easy way to do this question once knowing the centre So we have the equation x^2 + y^2 + 6x - 2y -15 = 0 which becomes (x+3)^2 + (y-1)^2 = 15+9+1 (x+3)^2 + (y-1)^2 = 25 So the centre is (-3,1) and radius 5, which is what you got. Now knowing that the centre is (-3,1)...

I'm guessing this is from Fitzpatrick 2U? Read this thread. http://community.boredofstudies.org/13/mathematics-extension-1/239452/locus-question.html

Was that you AKILA? xP But anyway, please remember that for optometry, UNSW offers many bonus points for subjects, making it quite easy to get 99.95 ATAR with bonus points. This then means that the UMAT will become a discriminator.

Hrm.. If it simplifies to that, there's probably values for A and B somewhere. Maybe it relates to a previous question or something? Sorry I can't check myself because I don't have the Fitzpatrick 2U book.

Oh my bad. Should be 1+4(2)^2 instead .. XD

Okay so notice how the length of the arc is S (a=>b) [1+{f'(x)}^2]^(1/2) Since we're concerned with the function y = x^2 f'(x) = 2x [f'(x)]^2 = 4x^2 So the function we are concerned with for Simpsons rule will become [1+4x^2]^(1/2). Now Simpsons rule is A = h / 3 [ y0 + 4y1 + 2y2 + 4y3 + ...

Basically you just need to allocate A and B positions. So let A = (x1,y1) B = (x2,y2) P is a variable so (x,y) Therefore, using the distance formula (PA)^2 = (x-x1)^2+(y-y1)^2 (PB)^2 = (x-x2)^2+(y-y1)^2 Then the difference between these equals 5 So [(x-x1)^2+(y-y1)^2]-[(x-x2)^2+(y-y1)^2]...

You can do the reverse chain rule with any integral, as long as the derivative of the function is there. But with this integral no. If it was say, S x (9-x^2)^(1/2) dx instead, with the x at the front, then yes you could do the reverse chain rule, because the derivative of 9-x^2 is -2x, but...

+1 I mean, if you can achieve the answer in like 2 lines, rather than ~6 wouldn't you rather just take the shorter path? Unless the question specifically asks you to solve it by integration. Where did you find this question btw? I'm pretty sure that if they didn't give you the substitution...

If you want working for the integral. It's easier to change the limits to 0 and 3, and multiply the answer by 2 Let x = 3sin theta dx = 3cos theta d(theta) So, S (9-x^2)^(1/2) dx = S (9-9sin^2 theta)^(1/2) 3 cos theta = S [9(1-sin^2 theta)]^(1/2) 3 cos theta = S 3 (cos^2 theta)^(1/2) 3 cos...

With this type of integral, try to think of what the graph is. It's a positive semi circle with centre (0,0) and radius 3. This means that the integral is equal to the area of a semi circle radius 3. Therefore, 9pi/2. :) edit: beaten XD

Okay, since we know that the diameter of the moon is approximately the length of the arc with the centre as the observation point, we use the formula l = r theta, where l is the arc length, r is the radius and theta is the angle subtended at the centre Now the radius = 400 000 km theta must be...

Okay so I can't type in Latex, and if I try to type it out here it'll be messy but.. As you can see in the answer you need to achieve, there is no 'y', so you just need to make the substitution y1 = x1^2 / 4a and y2 = x2^2 / 4a Once you do this, simplify the expression and it should come out...