Mathematics: Post your doubts here!

[Rutzaba]

guys my net is being damn slowwww so even if i do the explanation wont b able to upload it... will try though

 

[shezi1995]

please help me with question number 6(i). Link to the question is http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_4.pdf

First you have to find the displacement of P which is equal to that of Q as is given. Integrating (3t-0.3t^2) with limits 10 and 0 gives you the displacement. Then simply assume greatest speed to be v and find area under graph given i.e 0.5 *v*10 and equate it with the displacement found. I think you can do all the computations now.

 

[InternationationalGuy]

Guys. I have a question. Will we be provided a formula sheet for the papers. and can anyone give me one for Paper 1 and 4 cuz i dont have them.

 

[Rutzaba]

Guys. I have a question. Will we be provided a formula sheet for the papers. and can anyone give me one for Paper 1 and 4 cuz i dont have them.

u mean MF9?

 

[~`Heba`~ :)]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_43.pdf

#7 (ii) and (iii) urgently please

 

[unseen95]

First you have to find the displacement of P which is equal to that of Q as is given. Integrating (3t-0.3t^2) with limits 10 and 0 gives you the displacement. Then simply assume greatest speed to be v and find area under graph given i.e 0.5 *v*10 and equate it with the displacement found. I think you can do all the computations now.

thanks

 

[sma786]

Help please
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_6.pdf
Question 3(a), where did we get the -1.282 in the markscheme for the equation?
Markscheme here : (http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_ms_6.pdf)

 

[PhyZac]

Guys. I have a question. Will we be provided a formula sheet for the papers. and can anyone give me one for Paper 1 and 4 cuz i dont have them.

Yes we get it.
This is the syllabus, and go to page 31, You will see heading Pure Math, Mechanics, Statistic. Pure math has both 1 and 3. and Mech has both Mech1 & 2 same with Stat.
http://papers.xtremepapers.com/CIE/...d AS Level/Mathematics (9709)/9709_y13_sy.pdf

 

[PhyZac]

Help please
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_6.pdf
Question 3(a), where did we get the -1.282 in the markscheme for the equation?
Markscheme here : (http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_ms_6.pdf)

See, they said that P(X > 5.2) = 0.9
-Φ(z) = 0.9
z = 1.282 [so from the Normal distribution table]
-z= -1.282
Very simplified, if u dont get ask.

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_3.pdf
question 7 part 4
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_3.pdf
question 7 part 3
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_3.pdf
question 9 part 3
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_3.pdf
question 10 please please solve it somebody !
PhyZac Rutzaba or anyone else

Check the first link again. There is no part 4...

w05 qp 3
Q7 iii

Mark a point at (1, 2).
|z| = |z - 1 - 2i|
|z - (0 + 0i)| = |z - (1 + 2i)|
Construct a perpendicular bisector of (0, 0) and (1, 2).
----------------------------
w04 qp 3
Q9 iii

Put Q in m:
<2, 0, -1> = <-2, 2, 1> +t<-2, 1, 1>
From one equation, you get t = -2. Test the remaining two equations with this value. It should satisfy them.

PQ = <-2, -1, -3> (By putting s = 2)
Make a dot product with 'm' and see if the answer is zero.
-----------------------
w04 qp 3
Q10
i)
V = 1000h
dV/dh = 1000

dV/dt = 30 - k√h

dh/dt = dh/dV x dV/dt
dh/dt = (1/1000) (30 - k√h)

When h = 1, dh/dt = 0.02
0.02 = (1/1000) (30 - k)
k = 10

dh/dt = 0.01(3 - √h)

ii)
[(x - 3)/x] dx = 0.005 dt
(1 - 3/x) dx = 0.005 dt
x - 3ln|x| = 0.005t + c

When x = 3, t = 0
3 - 3ln3 = c

x - 3ln|x| = 0.005t + 3 - 3ln3
0.005t = x - 3 - 3ln|x| + 3ln3
t = 200(x - 3 + 3ln|3/x|)

iii)
x = 3 - √h
x = 3 - √4 = 1

t = 200(1 - 3 + 3ln|3/1|) = 259

 

[PANDA-]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_43.pdf

#7 (ii) and (iii) urgently please

I second this if someone could reply to this would be great.

 

[applepie1996]

Check the first link again. There is no part 4...

w05 qp 3
Q7 iii

Mark a point at (1, 2).
|z| = |z - 1 - 2i|
|z - (0 + 0i)| = |z - (1 + 2i)|
Construct a perpendicular bisector of (0, 0) and (1, 2).
----------------------------
w04 qp 3
Q9 iii

Put Q in m:
<2, 0, -1> = <-2, 2, 1> +t<-2, 1, 1>
From one equation, you get t = -2. Test the remaining two equations with this value. It should satisfy them.

PQ = <-2, -1, -3> (By putting s = 2)
Make a dot product with 'm' and see if the answer is zero.
-----------------------
w04 qp 3
Q10
i)
V = 1000h
dV/dh = 1000

dV/dt = 30 - k√h

dh/dt = dh/dV x dV/dt
dh/dt = (1/1000) (30 - k√h)

When h = 1, dh/dt = 0.02
0.02 = (1/1000) (30 - k)
k = 10

dh/dt = 0.01(3 - √h)

ii)
[(x - 3)/x] dx = 0.005 dt
(1 - 3/x) dx = 0.005 dt
x - 3ln|x| = 0.005t + c

When x = 3, t = 0
3 - 3ln3 = c

x - 3ln|x| = 0.005t + 3 - 3ln3
0.005t = x - 3 - 3ln|x| + 3ln3
t = 200(x - 3 + 3ln|3/x|)

iii)
x = 3 - √h
x = 3 - √4 = 1

t = 200(1 - 3 + 3ln|3/1|) = 259

thank you very very very much
i get it now
oh yeah sorry
i meant part 2

 

[mania _ manal]

Can some1 help me with octnov 2010 pp41 question 6 lst part

 

[sma786]

See, they said that P(X > 5.2) = 0.9
-Φ(z) = 0.9
z = 1.282 [so from the Normal distribution table]
-z= -1.282
Very simplified, if u dont get ask.

Thanku, i got it
can you send me a link of the normal dis. table please??

 

[PhyZac]

Thanku, i got it
can you send me a link of the normal dis. table please??

You see, this is the syllabus.
Go to page 31, you will find the formula booklet, now next few pages is the table.
http://papers.xtremepapers.com/CIE/...d AS Level/Mathematics (9709)/9709_y13_sy.pdf

 

[sma786]

You see, this is the syllabus.
Go to page 31, you will find the formula booklet, now next few pages is the table.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_y13_sy.pdf

Thanks, you're an angel (^_^)

 

[sma786]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_6.pdf
In question 6 (ii)
Why does the markscheme show 1-(7.5-8) and also (8/7.5) when the P to find is just 7 ?
Markscheme : http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_ms_6.pdf

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_6.pdf
In question 6 (ii)
Why does the markscheme show 1-(7.5-8) and also (8/7.5) when the P to find is just 7 ?
Markscheme : http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_ms_6.pdf

This is because this is an approximation, not normal.

So you add 0.5 or subtract 0.5 depending on the question. Since this is more than, you take 7.5

EDIT: Whenever you approximate, make sure to make continuity correction.

 

[cool Asviva]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w06_qp_6.pdf

question 6 part ii. please explain how 7p3 not 6p3?

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w06_qp_6.pdf

question 6 part ii. please explain how 7p3 not 6p3?

Solved this before. Here you go.

6 men 3 women
no women beside each other
thus
_M_M_M_M_M_M_
u see, i made blanks between men, and women can choose any of these blanks.
therefore
6! x 7P3 ( 6! because u rearrange the 6 men) (7P3, because out of these 7 blanks, 3 are permutated depeneding on where will 3 women stay)
answer = 151200