Mathematics: Post your doubts here!

[aksameerkhan27]

O/N-P3-2005-(7)
M/J-P3-2007-(8)
M/J-P3-2008-(5)
Plz help in complex numbers. Tips/hints in solving a complex number question. To find locus of a complex numbers.Plz urgent.

 

[rizzu78692]

heyy.. i need help with one question:
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
question 10 part (1)
please reply ASAP THANKS !

 

[2pac]

see did you get the answer from part (i) its 13sin(x+67.38)
so you do 13sin(x+67.38)=11
sin(x+67.38)=11/13
angle=57.38
2x+67.38=122.2 from 180-57.38
2x=122.2-67.38
x=27.4
other one is 360+57.38=417.8
x=175.2

okay,could you explain why we do the steps you mentioned.I have underlined the things I didn't understand.If you know any link which is useful for this particular topic,please let me know.

 

[SUMMERLOV35]

okay,could you explain why we do the steps you mentioned.I have underlined the things I didn't understand.If you know any link which is useful for this particular topic,please let me know.

well i don't have any links but my teacher told me this:

to find sin we do
1. sin x=k
2. sin inverse k=x
3.sin( 180-x)=sinx
and sin(x+360) or sin(x-360)
so these were the steps i followed

for tan
1. tanx =k
2.tan inverse (x+180) or (x-180)

for cos
1. cos x=k
you should also know that anything negative becomes positive with cos
2. cos(-x)=cos x
3. cos(x+360) or (x-360)

 

[2pac]

well i don't have any links but my teacher told me this:

to find sin we do
1. sin x=k
2. sin inverse k=x
3.sin( 180-x)=sinx
and sin(x+360) or sin(x-360)
so these were the steps i followed

for tan
1. tanx =k
2.tan inverse (x+180) or (x-180)

for cos
1. cos x=k
you should also know that anything negative becomes positive with cos
2. cos(-x)=cos x
3. cos(x+360) or (x-360)

Thanks a lot.

 

[aaakhtar19]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_42.pdf
Need help in q 4(ii) and 5(ii).
Please explain the method etc.

THX ALOT.

4(ii)
Friction = u*R
4cos(30)=u*[10-4sin(30)]
u=[4cos(30)/8]=0.433

5(ii)
As B continues to move for 0.3 sec , surely after those 0.3 sec B's velocity would be zero
so V=U+at
O=U -10*0.3
U=3
"3" is the speed with which A hits teh floor
So,
2as=V^2-U^2
2*2*s=3^2-O^2
s=2.25

Hope u got them

 

[Rabb94]

Waalaikumassalam wr wb!

Check this: http://www.xtremepapers.com/communi...st-your-doubts-here.9599/page-100#post-210258
a member has solved this

thanx...

 

[2pac]

I have one doubt regarding the sketching of argand diagrams.I had downloaded a pdf file which explained drawing the argument and other related stuff,in that I couldn't understand the last topic.It was about least value of argument and the complex was |z-3i|=pie/4.Normally we would plot it as 0,3 but the diagram in the file shows 0.-3.Just wondering if that's a mistake or is it supposed to be like that.Please let me know soon.
thanks

 

[2pac]

O/N-P3-2005-(7)
M/J-P3-2007-(8)
M/J-P3-2008-(5)
Plz help in complex numbers. Tips/hints in solving a complex number question. To find locus of a complex numbers.Plz urgent.

Hey a member of this forum was kind enough to spare his/her time by making some very useful notes for Argand diagrams.I've attached the file and I am sure it will help you clear your complex number doubts.

 

[aaakhtar19]

M/J 2011 P32 question 10(iii) link
Anyone help please. Thx

Alhamdulillah I git the answer right but i think its too Lengthy

 

[Zishi]

I have one doubt regarding the sketching of argand diagrams.I had downloaded a pdf file which explained drawing the argument and other related stuff,in that I couldn't understand the last topic.It was about least value of argument and the complex was |z-3i|=pie/4.Normally we would plot it as 0,3 but the diagram in the file shows 0.-3.Just wondering if that's a mistake or is it supposed to be like that.Please let me know soon.
thanks

That's a mistake for sure.

 

[2pac]

That's a mistake for sure.

so I guess the points will be never negative right?

 

[qassim]

this is question from mj 08 , question 7(i). I don't understand how 1 - x/(x+1)(x+3) becomes x/(x+1)(x+3). can someone please explain in detail it would be greatly appreciated. thanks

 

[allysaleemally]

Please could anyone help me in oct/nov 2011 m1 v3 question 6, I really need your help!
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_43.pdf
marksheet: http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_43.pdf


Thanks a million!

 

[RGBM211]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_43.pdf
question 1 part iii) finding the total distance
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_43.pdf

 

[Zishi]

so I guess the points will be never negative right?

No, the points can be negative, e.g if you had arg(z+3i)=x rad, then you had to use (0,-3). . .

 

[RGBM211]

someone please answer this 1 mark question atleast

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_43.pdf
question 1 part iii) finding the total distance
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_43.pdf​

 

[Zishi]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_43.pdf
question 1 part iii) finding the total distance
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_43.pdf

It's equal to Area of Trapezium + Area of Triangle + Area of Triangle, as the total distance is the area under velocity time graph. You'd be getting negative area for second triangle, but you have to add that into the the areas of trapezium and first triangle, as you have to find the distance(which is a scalar quantity)

 

[RGBM211]

It's equal to Area of Trapezium + Area of Triangle + Area of Triangle, as the total distance is the area under velocity time graph. You'd be getting negative area for second triangle, but you have to add that into the the areas of trapezium and first triangle, as you have to find the distance(which is a scalar quantity)

i am not asking about part ii) i am asking about part iii where the answer is 86.5 the one yu mentioned i used that for part ii)and got the correct answer pleaseee helllp!!

 

[Zishi]

i am not asking about part ii) i am asking about part iii where the answer is 86.5 the one yu mentioned i used that for part ii)and got the correct answer pleaseee helllp!!

I have also told you the answer to part (iii). Read what I said again. In part (ii), you're finding the displacement but in part (iii) you need to find the total distance.