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A curve is defined by the parametric equations x = 3^(-t) + 1, y = 2 x 2^(t). Show that dy\dx = -2 x 3^(2t).

Write 3^(t) as an expression involving x : We can rewrite x = 3^(-t) + 1 as x - 1 = 3^(-t) ; equivalently, 3^(t) = (x-1)^(-1). Substitute this expression into y, to write y in terms of x: y = 2 x 3^(t) = ...

MK
Answered by Maleeha K. Maths tutor
3566 Views

expand and simplify (x+1)(x-1)

first begin by expanding the brackets using the CLAW method (shown using whiteboard diagram)multiply the first of each bracket together, we get x2multiply the first term of the first bracket an...

AT
Answered by Adam T. Maths tutor
9679 Views

root3 (root6 + root12) can be written as Aroot2 + B

root3 (root6 + root12) (root3 * root6 ) + (root3 +root12) = root18 + root36= (root9 + root2) + 6= 3root2 +6a= 3 b = 6

DM
Answered by Dorcus M. Maths tutor
3490 Views

Write down the expansion of (cosx + isinx)^3. Hence, by using De Moivre's theorem, find cos3x in terms of powers of cosx.

First, we expand by using binomial expansion (it is always helpful to draw Pascal's triangle to find each term's coefficients): 1 1
1 2 1
1 3 3 1
Since the function is raised to the...

HF
Answered by Hiro F. Maths tutor
12606 Views

v^2 = u^2 + 2as u = 12 a = –3 s = 18 (a) Work out a value of v. (b) Make s the subject of v^2 = u^2 + 2as

v^2 = 12^2+2(-3)(18) = 36. Therefore by square rooting v = 6.To make s the subject first minus u^2 from both sides to have v^2 -u^2 = 2as then divide 2a from both sides to have (v^2-u^2)/2a = s

HN
10385 Views

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