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Soln Q33. Solving 2lnx=ln(5+4x) Question 34 Express 0.3494949... (recurring 49) as a fraction in simplest form.

show that the equation what?

note: first principles is strictly extension-1 Question 30 If f'(x) = cot(x) + x and f(pi/2) = 0, find an expression for f(x).

Solution to Q27- Question 28 Find a primitive for tan{2-1x}.

it's perfectly legal to square both sides with absolute values x > 4 why the hell would you square both sides here? the reason you can square both sides with absolute values is because absolute values are defined as the square root of the square and it allows the inequality to be...

close DeanM, but remember a parabola can't have two focusses 4x = 8y - y2 you missed the negative sign. so y - 8y = -4x (y - 4)2 = -4x + 16 (y - 4)2 = -4(x - 4) focal length 1; concave left hence the focus is (3,4) Anyway: Soln. Q23 Question 24 The sum of the first n terms of...

you must remember that (a + b)2 = a2 + 2ab + b2 also, when a + b = c, squaring both sides gives (a + b)2 = c2, not a2 + b2 = c2 ((x - 2) + (x + 2))2 = (x2 - 4x + 4) + 2(x2 - 4) + (x2 + 4x + 4) 2x2 + 8 + 2x2 - 8 > 36 x2 > 9 x < -3, x > 3

according to cambridge AA suffices, and according to keypad from an old post set out like so: from cambridge: in triangles TCN and RAY 1. angle TCN = angle RAY = 90 degrees (given) 2. angle CNT = angle AYR (given) so ΔTCN ||| ΔRAY (AA). also you can use SSS, SAS, RHS if the sides...

Q17 Answer Question 18 Find the focus of the parabola given by 4x = 8y - y2.

kickstart: Question 11 Consider the geometric series: 1 + lnx + (lnx)2 + ... Find the values of x for which the limiting sum of this series exists.

I got a different answer to Q7 I think you made an error finding the y co-ordinate: Answer to Question 8: Question 9 Use the trapezoidal rule with three function values to find an approximate value of the definite integral of (x+1)-1 from 2 to 4.

20x/5x = 4 anyway post up a new question

test for n = 1 1 = 1/2(1 + 1) (true) therefore the statement is true for n = 1 assume true for n = k i.e. 1+2+3+...+k=k/2(k+1) proving statement for n = k+1 i.e. 1+2+3+...+k+(k+1) = (k+1)/2((k+1)+1) k/2(k+1) + (k+1) = (k+1)(k+2)/2 LHS=(k2 + k)/2 + k + 1 = (k2 + k + 2k + 2)/2 =...

Inspired by other boards because they worked well, I thought it'd be interesting to start a Mathematics Marathon here... although it might not work as well because only a few people come here. Here's how it goes: You answer the previous question and post a new one. You don't necessarily have...

if you're ever unsure about rounding off to significant figures, let your calculator do it. assuming you're using my calculator:... press mode until you get to 'Sci', then it'll say Sci 0~9, press 3 for 3 significant figures. then press [shift] [ENG] until it's *1000.

1. if x = 2 is a root, then f(2) = 0 i.e. f(x) = (x - 2)(x - k) = 0 x2 - kx - 2x + 2k = 0 x2 - x(k + 2) + 2k = 0 2k = -6, so k = -3 x2 + x - 6 = 0 m = 1/2 2. let f(x) = (x - k)(x - {k + 1}) (x - k)(x - k - 1) = 0 x2 - kx - x - kx + k2 + k = 0 x2 - x(2k + 1) + (k2 + k) = 0 k2 + k...

1. if y = x + c is a tangent to x2 + y2 = 4, solve simultaneously: x2 + x2 + 2xc + c2 = 4 2x2 + 2xc + c2 - 4 = 0 for a tangent there is only one solution i.e. discriminant {b2 - 4ac} = 0 4c2 - 8(c2 - 4) = 0 -4c2 + 32 = 0 c2 = 8 c = +-2Sqrt(2) y = x +- 2Sqrt(2) x2 + y2 = 4 x2 + x2 +-...

This is an extension 1 question and uses the compound angle formula taught in the extension 1 course.

Um. No. You didn't take into consideration if MISS wasn't first. And you're supposed to divide from repetition not multiply because they're not distinct. By your logic, arranging the word "MMM" = 3 * 2 * 1 = 6, but really there is only one way because they're not distinct. KFunk is right...

y² = 2x and y = Sqrt{2x} are different. y² = 2x isn't a function, and y = Sqrt{2x} doesn't have any turning points. Whether (0,0) is considered a turning point in the relation y² = 2x is beyond the scope of the course.