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Thanks for answering. Clearly, I'm not understanding something! for i: We are finding the volume of the solid formed when rotating the shaded region above, about the x axis. Why are you integrating from 1 to 0, instead of 1 to -2? for ii: I have a similar question. Why are we only looking at...

Q20. c) i: I got wrong, and I disagree with the working out of the solutions book. ii: I couldn't solve this, and I disagree with the working out of the solutions book. Just posting this here to see what you guys can do with it :) Probably an oversight by me; but nonetheless, your help is...

oh fek. yeah. thanks

For my work below; the book provides 1+2e^(-x) as the answer. If I simply re-write the initial problem with -dx on the RHS, and y-1 in the denominator of LHS I will get the same answer. But why doesn't it work otherwise? Or it does and I can't see why? Where is my working out wrong? (mind...

dx = d(sqrt(x))^2 , fine. Why does d(sqrt(x))^2 = 2*sqrt(x)*d*sqrt(x) ? Looking at your note: As gorgeous as your apples may be. The integral of 1/x is not my source of confusion with your solution. Why does dx^2 = 2xdx ? o_O --- Edit, I got it: dx^2 = d/dx * x^2 * dx = 2x dx d(sqrt(x))^2 =...

This reasoning mirrors my own thoughts, so thankyou for the clarity. This helped!

Hi guys, I'm stuck on something trivial. letting x = (u-4)^2 I get to the correct solution in terms of u: 2u - 8ln|u| + c However, my confusion begins when changing the answer back to be in terms of x. Since x = (u-4)^2 , u = x^(1/2)+4 OR u =-x^(1/2)+4 So shouldn't there be two solutions in...

Thanks legend 😁

Anyone else find this topic very tedious and frustrating? Anyway, I’m tired and can’t find how to solve q11. d). Could anyone help out pls

Hi guys, I have a question from this chapter (Proof). Please help, if you can. And if you have questions of your own from this chapter of the cambridge book, pls post them here and I'll try help as well. 2A Q13. Following directly the explanation of the question, it is clear that we have a)...

This was really intuitive and easy to follow. Can do the original question now thx to this. Good stuff!

yeah, nah. I've done all combinatorics questions in the cambridge y11 ext 1 book and this isn't like anything I've seen before. Sometime in future, I'll have a look at what stars and bars is. Thx always, Luukas

@Luukas.2 . Reckon you could give this a crack if you have time pls?

Do you have the final answer?

with a bit of algerba you can rewrite the equation as: y-2 = 1 / [2(x-2)] y = 1/x is the starting graph y-2 = 1/x moves the graph up 2 y-2 = 1 / 2x is a horizontal dilation by 1/2. y-2 = 1 / [2(x-2)] moves the graph right 2. You can put the 4 equations above into Geo Gebra, to see exactly how...

I did part a) the same way as Lukkas above. b) 6 possible ways to set out E and I: E _ _ _ I _ _ _ E _ _ _ I _ _ _ E _ _ _ I Note that E and I can be written in reverse order so that there are 3x2 = 6 possible ways. The remaining three ways are also shown below: I _ _ _ E _ _ _ I _ _ _...

It's been a while since I did this topic. Dunno if I'm right, but hoping to help you at least. As Liam said, since there must be at least 4 balls of each colour; 4x5=20, gives us 20 balls fixed in place. Hence, we have to find the remaining combinations. Since 1 colour can be no more than...

It does make sense, and I'm thankful for the clarity. I struggle with these subtleties, but I think they are very important and try and have them sink in whenever I come across them.

Interesting... Is the equation Arg(z-i) = Arg(z) - pi ? I had thought +pi or -pi didn't matter because they both ended up at the same place. Since the Complex plane is restricted between -pi and pi, I thought a result of 3pi/2 simply means the same as -pi/2. I simply saw +pi as a rotation...