Search results

Thanks. :) Something I need to remember. :)

Thanks. :)

Hello All... :) Could someone please assist me with the following questions? 1. Use a calculator to find, to 2 d.p where appropriate the approximate value of: a) lim h--> 0 \left (\frac{10^h - 1}{h} \right ) 2. Find the value of k, correct to 3 d.p if: ln k = 1.9. Thanks for your help...

Got the next part as 9.... Sorry ! How about finding the area between the curves y = 2x^2 and y = x^2 -13x + 30 I've found the point of intersection is 2 4/13 is this right?!?! Then do the integral from 0 to 2 4/13 with y = 2x^2 and add the integral from 2 4/13 to 3 (Think this is right.)

You're right... First part = 6.5??

Yes. (roughly looks like that). :)

Yeah ok... I got 9!! Will have to check...

Yup, that's exactly what I meant... :)

No they're not given.... If you draw the two equations given... It's not inside the parabola... It's to the left of the parabola... To the y axis and the bottom is the x axis... Can you picture it??

Hello All If I have the line y = 5x + 4 and the parabola y = (x - 4)^2 How do I find the area between the x axis (on the bottom), the y axis (on the LH side), y = 5x + 4 (on the top) and the parabola on the RH side?? Hope this makes sense! Thanks in advance. :)

Of course... Thanks heaps. :)

Is anyone able to help me please? I'm unsure now! :)

Thanks for your help... :) I've tried that again but I'm not getting their answer of 3/4... So the height = 1 as you said... Then for x = 2, y = 1/2, x = 3, y = 1/3 x = 4, y = 1/4 So I've subbed it in as 2/2 (1/2 + 1/4 + 2(1/3) = 17/12 What am I doing wrong??

Hello All. :) Could someone please show me the steps for the following question? 'Use one application of the trapezoidal rule to approximate: \int^{4}_{2} \frac{1}{x} dx Thanks in advance. :)

Any 2013 Year 12 Mathematics students that could share their tips to success?? :)

Yup, whoops... I forgot to multiply the RHS!

Thanks for your help Carrot. :)

1. So equation to normal is: 2x + 2y - 5 = 0 which is arranged to y = -x + 5/2 This correct so far??

Thanks for your help... :) I'm not quite getting the right answer... Could you please explain further??

$The normal to the curve y $ = \frac{1}{8}x^2 $ at P $ (2, \frac{1}{2} ) $ meets the curve again at Q. Find the co-ordinates of Q.$ Sure ...:) I might have clicked 'edit' for a moment.