-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
Recent content by imanmalik
-
What if I told you that...Meh I'm out of ideas. Go jump off a building or something as I think...
What if I told you that...Meh I'm out of ideas. Go jump off a building or something as I think of something to say.- imanmalik
- Profile post
-
-
-
DUDE! <3 I am so happy for you ;_;! HUG...DO IT!
DUDE! <3 I am so happy for you ;_;! HUG...DO IT!- imanmalik
- Profile post comment
-
<3333333333333333333333333333333 so proud of you <3
<3333333333333333333333333333333 so proud of you <3- imanmalik
- Profile post
-
Hello everyone, a Chinese Girl new here!
Please. Kill yourself. Dōitashimashite is Japanese not Chinese.- imanmalik
- Post #50
- Forum: Welcome Lounge
-
-
-
-
Hey, thanks for the follow :) Nice to meet you :D
Hey, thanks for the follow :) Nice to meet you :D- imanmalik
- Profile post
-
Mathematics: Post your doubts here!
Thank you so much. Our maths book is damn hard. CIE not Edexcel! :/ -
Mathematics: Post your doubts here!
Yeah I worked backwards and got that. But that's by working backwards! If we didn't know the equation was (|4+z|+|4-z|) (|4+z| - |4-z|) >=48 how would we get |4+z|+|4-z|≧ 8 from |4+z| - |4-z| = 6 :/ ? That's why I'm confused :( -
-
Mathematics: Post your doubts here!
P is a point in an argand diagram corresponding to a complex number z which satisfies |4+z| - |4-z| = 6. Prove that |4+z|^2 - |4-z|^2 ≧ 48 and deduce that Re z ≧ 3. The only trouble I'm having here is how do you get |4+z|+|4-z|≧ 8 -
registration forms for may/june 2013.
Everybody at our school's registered and all. So the registration form's been out for aaaaaaaaaaaaaaaaaaaaages