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z = 2y + iy z² = 4y² + 4iy² + i²y² = 3y² + 4iy² z² - 4i = 3y² +4iy² - 4i = 3y² +4i(y² - 1) y² = 1 :. y = +/- 1 (subbing back in to z = 2y + iy)...

a) Find an expression for sin^2x in terms of cos2x. rearrange the double angle formula for cos2x to get sin^2x = (1-cos2x)/2 b) V = pi times the integral of (sin^2x) between x = 0 and x = pi/4 Sub in the result from part a so you can integrate it. You have to be bothered doing part a...

Rats, shouldve gone to S4 to test it out. Back to the drawing board..

T₁ = 6 T₂ = 66 T₃ = 666 S₁ = 6 S₂ = 72 = 6 (11 + 1) S₃ = 738 = 6(11² +2) S_n = 6(11^n-1 +n - 1) My 3u paper was pretty straight forward, 3D trig question 7

How many marks was this one worth? 3?

I did the long method (adding up 0 + 1 + 2 + 3 + 4 days of rain) and got 0.903744.

Alot of people I know do not have tutors and still do considerably well. From memory, the dux of my school a couple of years ago didnt have a tutor (or so his sister tells me). I myself have never had or will ever want a tutor and am currently 3rd (out of about 50) and 2nd (out of 7) in ext 1...

I did okay, not as well as id hoped for physics and chem though Physics - 2nd /30 (80%) Chemistry - 4th /25 (83%) Ext 1 maths - 4th /50 (84%) Ext 2 maths - 3rd /7 (55%) Still waiting on SOR I and english though

I dont have a scanner

I just got my half yearlies back today. 55% (which was 3rd out of 7) and theres no way im dropping ext 2 maths. It may seem bad but its actually really good considering I know the calibre of the other students in my grade. Pretty much everyone received an E4 last year in 3unit maths so yeah...

Your given that the final term is 6p + 15 and you know that the final term is also a + (n-1)d. Sub in the values for a, n and d then equate them. for 15. Theres nothing wrong with your working for this question. You found the value of n and substituting that value serves as a means of...

For the first one t₅ = a + 4d = m t₁₁ = a + 10d = n n-m = 6d d= [n-m]/6 m+n = 2a + 14[n-m]/6 6m + 6n = 12a + 14n -14m 20m - 8n = 12a 5m - 2n = 3a a = [5m - 2n]/3 t₇ = a + 6d = [5m - 2n]/3 + n-m = [5m - 2n + 3n -3m]/3 = [2m + n]/3 For the second...

TRY to get off your ass and do some work :p Lets see.... cut down on tv and net. Try to start assessments early rather than doing it the day before its due. Start making notes for subjects, do homework... Screw this! Hard work takes effort, I'll just keep doing what I'm doing and pray...

No way Jose, exams are the best! I absolutely HATE hand in assessments namely because I dont think it actually tests how much a student knows. You can pretty much get away with plagiarism (like some politicians) and the teachers dont even care. Its like, heres the question. You have 2 weeks to...

Grr... I hate those people. They sit in class not even paying attention, do assignments on the day before its due and still end up with some pretty decent marks :rolleyes: :p

sin2@ = cos2@ tan2@ = 1 1st and 3rd quadrants 2@ = pi/4, pi + pi/4 = 5pi/4 therefore @ = pi/8, 5pi/8 you have to draw the two curves to see which one is the upper curve. Alternatively, sub in a value between the points of intersection to see which has the greater y value. The area is...

Maybe this will be easier to see.

Nice solution. Sum and products are part of 3unit but no one bothers to learn them (unless you do 4unit)

Yeah, its 540 and b is 48?

Let me give it a go tan@=k tan# sin@/cos@ = ksin#/cos# sin@cos# = ksin#cos@ (k-1)sin(@+#) = (k-1)(sin@cos# + cos@sin#) = (k-1)(ksin#cos@ + cos@sin#) = (k-1)(k+1)(sin#cos@) = (k+1)(ksin#cos@-sin#cos@) = (k+1)(sin@cos# -sin#cos@) = (k+1)(sin(@-#) hmm.... that isnt 3 lines.