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5cos30ºsin225º = 5rt(3)/2 * -1/rt(2) = -5rt3/2rt2 = -5rt6/4

Soulsearcher, doesn't work for y = 0.

The range of the function is y > 0 You don't need to make x the subject, although that is one way to solve it. You can not take the log of a negative number and 1/0 is undefined. Other ways are to draw it as a graph - exponential graph asymptoting at y = 0, x approaching infinity. Or...

Rama, where are you getting the quadratic from?

y = (3x + 25)/4 x2 + (3x +25)2/16 = 25 16x2 + 9x2 + 150x + 625 = 400 25x2 + 150x + 225 = 0 x2 + 6x + 9 = 0 x = -3 y = 4 There is only one pt of intersection with the circle, hence tangent. (-3,4)

Haha three posts in a minute all giving slightly differnet solutions. airie, that is not the problem, as cos x is only zero at 90, 270 which are not solutions in this case. If cosx is zero, then sinx will still not be zero, so the equation is not satisfied.

sin^3x + 2 sin^2x cosx + sinx cos^2x = 0 tan3x + 2tan2x + tanx = 0 tanx(tan2x + 2 tanx + 1) = 0 tanx(tanx + 1)2 = 0 I think your problem comes here - you are neglecting the possibility that tan x = 0. The solutions are tanx = -1 OR tanx = 0 So you get 135, 315 for the first one and 0, 180...

You don't need t, you've missed the simple solution! rt3 sinx = -cosx -1/rt3 = sinx/cosx tanx = -1/rt3 x = 150, 330

Shouldn't it be x^2 - 10x + 25 + 5 - 25 (you have added 25, therefore you have to subtract 25) So y = (x-5)^2 - 20 or y + 20 = (x-5)^2

cos^3x = (1-(sinx)^2)(cosx) = cosx - cosx(sinx)^2 u = sin x du = cosxdx So Integral of cosx(sinx)^2 dx = u^2 du So the integral of (cosx)^3 = sinx - u^3/3 = sinx - ((sinx)^3)/3 Sorry for the poor setting out, but the important part is turning (cosx cubed) into (cosx times...

(x+2)/3 - (2x-1)/5 = (5x + 10 - 6x + 3)/15 = (13 - x)/15 Do you mean (x+2)/3 - (2x-1)/5 = 5 Then the answer will be -62 (13+62)/15 = 75/15 = 5

2. displacement north = 120km displacement west = 400km total displacement = sqrt(400^2+120^2) = sqrt174400 = 40rt109 velocity = 40rt109/7 km/h (can't be bothered to make it m/s) tan angle = 400/120 73 degrees 360-73 = 287 so velocity is 40rt109/7km/h @ 287 degrees

From 100 to 200 multiples of 9 are 108, 117 ... 198 AS with a = 108 d = 9 198 = 108 + 9(n-1) 10 = n - 1 n = 11 S11 = 11/2[216 + 9 x 10] = 1683

3) Integers that are multiples of 6 are 6,12,18...96 Arithmetic series with a = 6 d = 6 96 is the final term 96 = a + d(n-1) 96 = 6 + 6(n-1) 15 = n- 1 n = 16 S16 = 8[12 + 6 x 15] = 816 Sum of integers from one to 100 = 1 + 100 + 2 + 99...50 + 51 = 50 x 101 = 5050 So sum...

2) If x= (2ab) / (a+b) show that 1/a, 1/x, 1/b are in arithmetic progression. If true 1/x - 1/a = 1/b - 1/x (a+b)/2ab - 1/a = 1/b - (a+b)/2ab (a+b - 2b) = 2a - (a+b) a-b = a-b

1) Sn = n/2[2a + d(n-1)] S3 = 24 S6 = 24 + 51 = 75 24 = 3/2[2a + 2d] 16 = 2a + 2d 25 = 2a + 5d 3d = 9 d = 3 a = 5

Oops - I solved it by looking at it then thought i had better do some working. And then when I checked it by substituting the values into the original equation it was right. I think the reason I still got the right answer is that y = 1 and 1/y should = x/4 not y = x/4 but this doesnt matter...

xy = 4 y = x/4 Substitute into second equn 2x - x/4 - 7 = 0 8x - x = 28 x = 4 y = 1 2x^2 -3x + 1= 0 (2x - 1) (x - 1) = 0

a = 1/10 + n ar = 1/4 + n ar^2 = 9/20 + n n = a - (1/10) ar - a = 3/20 r = (3/20 + a)/a a[(3/20 + a)/a]^2 = 9/20 + a - 1/10 9/400 + 3a/10 + a^2 = 9a/20 + a^2 - a/10 9 + 120a = 180a - 40a 20a = 9 a = 9/20 n = 7/20 r = 4/3 So GP is 9/20, 12/20, 16/20 = 9/20, 3/5...

6. a = -1 Vertex at (2,0) So directrix y = 3 8. kx^2- 8x+k b^2 - 4ac < 0 and a is negative 64 - 4k^2 < 0 k^2 > 16 k < -4