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=(r^1 + r^2 + r^3 + r^4) + (r^5 + r^6 + r^7 + r^8) = (r^1 + r^2 + r^3 + r^4) + r^4(r^1 + r^2 + r^3 + r^4) = (r^4 + 1)S4 = 17S4 "[S8 - S4] = r^4 S4 = 16S4 r^4 = 16, r = +-2" is probably all I would write in an exam if you divide terms 5-8 by r^4, you get terms 1-4

the first two lines are right, but I wouldn't bother writing it in the form you wrote it as in the 3rd line. [S8 - S4] = r^4 S4 = 16S4 is all I would do * r^1 + r^2 + r^3 + r^4 + r^5 + r^6 + r^7 + r^8 = r^1 + r^2 + r^3 + r^4 + r^4(r^1 + r^2 + r^3 + r^4) in case you still don't quite understand

Well terms 1-8 is 17 times larger than terms 1-4. So terms [1-8 - 1-4] must be 16 times larger. If r is the common ratio, then to get from a term to the term n after it you must multiply the number by r^n. So, term 5 is r^4 times greater than term 1 ... term 8 is r^4 times greater than term 4...

terms 5-8 are 16 times larger than terms 1-4, and also r^4 times larger. Therefore r^4 = 16, r = +-2

I would suggest not squaring both sides. Just solve for (4x-3y+2) = (3x+4y-7) and (4x-3y+2) = -(3x+4y-7), which gives you x-7y+9=0 and 7x+y-5=0.

the sum/product identities can be easily derived by adding the right angle-sum expansions eg. sin(A+B) + sin(A-B) = sinAcosB + sinBcosA + sinAcosB - sinBcosA = 2sinAcosB so you just find the correct A and B eg sin70 + sin10 [A = 40, B = 30] = 2sin40cos30 = sqrt.3 sin40 I used this type of...

made enough threads on this yet? -_-

49. is a difference of two squares, then converting 2sinXcosX into sin2X 51. Solve each pair normally, then sub in exact value of cos60 54. write cos40 as sin50. solve normally, then sub in exact value of sin30 61. let A + B = C, then get rid of the A's

have you typed that out correctly?

you just find "the integral from 1->3 of [(x^2 - 5x + 5) - (-x + 2)]" ie, ∫13 [x^2 - 4x + 3]dx giving a primitive of 1/3x^3 - 2x^2 + 3x (+C) so just sub in for 1 and 3 and you get 1+1/3

is it 1 + 1/3? which is (I think) the integral from 1->3 of [(x^2 - 5x + 5) - (-x + 2)]

10. I don't understand what question 9 means

Re: What UAI are you aiming for? well, for one, you could end up changing your mind about which uni course you want to do I suppose you'd probably be more likely to get >85 if your aim was 95, rather than 85 anyhow

if whether you wear a seat belt doesn't affect any other crash-related variables, then those statistics would suggest that you are 6 times more likely to die whilst not wearing a seat belt. However, even if you did have those statistics you couldn't make any accurate judgements about how much...

If you weren't good at/interested in those subjects then you shouldn't have chosen them. Just like how if you weren't good at/interested in English you shouldn't have... oh wait. imo if they're gonna make a subject compulsary, then, especially if it's not obvious, they should explain how it...

the minimum of a cos function is the negative of the maxmimum (-1 and 1 respectively), you if you change the signs you end up with the same thing anyway

the range of a sin/cos function is -1 -> 1 in this example it is 1 + 2(-1 -> 1) = -1 -> 3

we were only told the rules early in the year when drawing dot diagrams - we weren't expected to remember it then, so we'll be learning it again and yeah, looking in my textbook precipitation is only in the water module - we just haven't really done much specifically about water

basically just remember that the opposite angles of a cyclic quad' add to 180, and that the triangle made by two points between them is isosceles. for c)BRQ you have to use the rule that the angle from two points to the centre is twice the angle from two points to the circumference