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You're right but they subtracted distance of P from distance of Q.. it has to be Sp-Sq=4.9
oh really ? I didnt seem to understand the method did the Q myself and posted what i thought was logical
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You're right but they subtracted distance of P from distance of Q.. it has to be Sp-Sq=4.9
how did you do it then?because P is further away and has covered more distance than q . it it were the other way around then we would get a negative distance which is impossible ( I am talking about distance not displacement )
BTW i did it another way
all i know is if you have 3 different forces on an object you can use lami
i'm not exactly sure why i got CP2X it just made more sense to me and after getting all the other angles i was sure it was right but it will make more sense if you turn it into a triangle (well for most people i know)
why t-2 and not t+2 as it is 2 seconds late?I took time taken by P to be 't' and time taken by Q to be (t-2). Then i formed the equations for the distance travelled by both the particles and subtracted distance of Q from distance of P. I got t=2.4. So the time taken by Q is t-2= 2.4-2= 0.4
why t-2 and not t+2 as it is 2 seconds late?
how did you do it then?
why t-2 and not t+2 as it is 2 seconds late?
ok thanks that really helped now i hope they get this on the exam its not as hard as it seemsSee, P started before Q. So at any instant, P has been travelling for more time Q has. So if time taken by P is t, the time taken by Q to reach that point will be less by 2, hence (t-2)
Q2)Dear, mathematicians I've hit rock bottom with two of the questions on this paper - Questions 2 and 8.The former I'm absolutely clueless how to do and the later, I'm struggling to substitute du for dx. Any help would be welcome. Thank you .
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf
the whole thing sounded gibberish to me....then i realized its paper 3Can any one please help me with this sum's part iv below please!(I could do till iii but got badly stuck with part iv)Q.The complex number −2 + i is denoted by u.(i) Given that u is a root of the equation x3 − 11x − k = 0, where k is real, find the value of k. [3](ii) Write down the other complex root of this equation. [1](iii) Find the modulus and argument of u. [2](iv) Sketch an Argand diagram showing the point representing u. Shade the region whose pointsrepresent the complex numbers z satisfying both the inequalities|z| < |z − 2| and 0 < arg(z − u) < 1/4π. Please help asap am at door! And please draw n upload the diagram here too...pleaaase i m really confused abt the diagram dat i should draw for part ivThank u! May u be blessed for helping!BESIDES here is the mark scheme below... to me it sounded completely gibberish so i couldnot work out the last part of the sum!(iv) Show point representing u in relatively correct position in an Argand diagram B1Show vertical line through z = 1 B1Show the correct half-lines from u of gradient zero and 1 B1Shade the relevant region B1 [4][SR: For parts (i) and (ii) allow the following alternative method:
State that the other complex root is –2 – i B1
State quadratic factor x2 + 4x + 5 B1
Divide cubic by 3-term quadratic, equate remainder to zero and solve for k, or, using
3-term quadratic, factorise cubic and obtain k M1
Obtain k = 20 A1]
AND Also :
Can you please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you
Link to Question: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf
Link to Mark Scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_ms_33.pdf
Anika Raisa said: ↑
Can you please help me with 9709/paper 33 year:October/November 2012 question 10 b (ii). Please help badly stuck with it! Thank you
Pleaase help!
First of all made equations for the dist covered by p and q :
Dist of p= o.5*1.75*(2+t)^2 (s=ut+0.5at^2) (u=0)
dist of Q = o.5*1.75*t^2
the difference was 4.9 => dist of P - Dist of Q
o.5*1.75*(2+t)^2 - o.5*1.75*t^2 = 4.9 (took o.5*1.75 common and divided it by 4.9 )
(2+t)^2 - t^2 =5.6
expanded the eq 4 +4t+t^2 -T^2 =5.6 ( T^2 is cancelled )
t =0.4
Haha...sorry i should have mentioned above!the whole thing sounded gibberish to me....then i realized its paper 3
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