Mathematics: Post your doubts here!

[Tabi Sheikh]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf
Please I need help with #6 (ii) and #7 , how do i draw the graph? If anyone could draw it it would really be helpful if you post a picture of the graph Thanks!

since work done here is the work done against gravity+ gain in potential energy + gain or loss in kinetic energy
to calculate gain or loss of kinetic energy we need final and initial velocity
now we are available with initial velocity = 6 ms^-1 =V1
power at bottom of the hill=P1
Force at the bottom of the hill =F1
now at top f the hill velcity =?
Power=P2=5*P1
Force=F2=3*F1
recall the formula P=Fv
take the ratio of both
(P1/5*P1)=(F1/3*F1)*(6/V2)
now we have
(5*6/3)=10=V2
now
work done against gravity+ gain in potential energy =945000
kinetic energy change = 0.5*1250*(10^2-6^2)=625*64=40000
therefore
945000+40000=985000=work done

 

[PANDA-]

since work done here is the work done against gravity+ gain in potential energy + gain or loss in kinetic energy
to calculate gain or loss of kinetic energy we need final and initial velocity
now we are available with initial velocity = 6 ms^-1 =V1
power at bottom of the hill=P1
Force at the bottom of the hill =F1
now at top f the hill velcity =?
Power=P2=5*P1
Force=F2=3*F1
recall the formula P=Fv
take the ratio of both
(P1/5*P1)=(F1/3*F1)*(6/V2)
now we have
(5*6/3)=10=V2
now
work done against gravity+ gain in potential energy =945000
kinetic energy change = 0.5*1250*(10^2-6^2)=625*64=40000
therefore
945000+40000=985000=work done

Alternative method... similar but I think it's easier

F2=3F1
P2=5P1

F2V2=5F1V1
Substitute F2=3F1 and V1=6
3F1V2=5F1(6)
3F1V2=30F1
F1V2=10F1
Cancel out F1
V2=10

I think you already got the height which is 50, using trigonometry, we'll use that to get P.E.

Work done by driving force <What we want> = Change in K.E. + Change in P.E. + Work done against resistance to motion
W = 1/2m(v^2-u^2) + mg(h2-h1) + Work done against resistance to motion

Work done against resistance to motion = Fd, and because F of the resistance to motion = 800, and d = 400, 800x400 =320000

W= 1/2(1250)(10^2-6^2) + 1250(10)(50) + 320000
W = 985000 J

 

[Sanis]

p = 2/5
q = (1-p) = 3/5
n =400

Variance = npq = 2/5 * 3/5 * 400
Variance = 96

thx alot man

 

[Scafalon40]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_33.pdf
Q 4 part (ii) please
This is worth only 3 marks, I'd prefer a method which take into account this fact
PhyZac could you help
or anyone else

 

[cute97]

heey guys can someone help me with this http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_43.pdf
the first question plzz and question 3 part 2
thanks in advance

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf
Q 4 part (ii) please
This is worth only 3 marks, I'd prefer a method which take into account this fact
PhyZac could you help
or anyone else

Always, when I see such a formula to be integrated, like cos square, i refer to the formula
cos2A = 2cos^2(A) -1

Now back to question, [let ! be integral sign]
!4 cos^2(3x)
4 ! cos^2 (3x)

refering to the identity

cos^2A = cos2A+1/2
cos^2(3x) = cos(6x)+1/2
so
4 ! cos^2(6x) + 1/2
4/2 ! cos^2(6x) +1
2 (sin 6x + x)

 

[Scafalon40]

Always, when I see such a formula to be integrated, like cos square, i refer to the formula
cos2A = 2cos^2(A) -1

Now back to question, [let ! be integral sign]
!4 cos^2(3x)
4 ! cos^2 (3x)

refering to the identity

cos^2A = cos2A+1/2
cos^2(3x) = cos(6x)+1/2
so
4 ! cos^2(6x) + 1/2
4/2 ! cos^2(6x) +1
2 (sin 6x + x)

Thanks!
Could you also explain why we need square roots in Q10 part ii b of the same paper please?

 

[PhyZac]

Thanks!
Could you also explain why we need square roots in Q10 part ii b of the same paper please?

Okay sorry for late reply, but i was solving it in order to understand.

Well it is because see lets take the example (z +2)
so one root is
z = -2

now if we take p(z^2)
then
(z^2 + 2)
z^2 = -2
z = √-2
z = +/- i √2

basically the square root of the answers! get it?

 

[19islandprincess96]

Can someone please show me how to solve question 6 of this paper?
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_61.pdf

 

[19islandprincess96]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_62.pdf

How do we do
1) Question 2?
2) 5 iii?
30 7 ii?

 

[littlecloud11]

Can someone please show me how to solve question 6 of this paper?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf

6i) P(<6) = 63/600 = .105
p(>6) - 1-1.05 =.895
z value for .895 = 1.253
for the P(>6)-

6-μ /σ = -1.253
μ - 1.253 σ = 6 ------1

P(>12) = 155/600 =.2583
P(<12) = 1-.2583 =.7417
z value for .7417 = .648
for P(<12) -

12- μ/σ = .648
rearrange:
μ + .648 σ =12 -----2

solve eq 1 and 2 simultaneously:

μ - 1.253 σ = 6
-μ - .648 σ =-12
-1.901 σ = -6
σ = 3.15

therefore μ = 9.95

ii) When the distribution of a variable Y, lies within +/-σ the Z value is +/-1
So the range for this is Z>1 and Z<-1
P (Z<-1 and Z>1)
For Z>1 = 1 -Ф1
For Z<-1 = Ф(-1) = 1 -Ф1
So,
=1 -Ф1 + Ф(-1)
=1 -Ф1 +1 -Ф1
=2 - 2* .8413
P=.3174
number of feathers =.317* 1000 = 3174

 

[Scafalon40]

Okay sorry for late reply, but i was solving it in order to understand.

Well it is because see lets take the example (z +2)
so one root is
z = -2

now if we take p(z^2)
then
(z^2 + 2)
z^2 = -2
z = √-2
z = +/- i √2

basically the square root of the answers! get it?

Yeah I think so, thanks
So, if we let y=z^2, we get the same equation as in part (i)
and the same solutions i.e y=-2 etc
but y=z^2
so z^2=-2
and z=sqrt(-2)
so z=+/- sqrt(2)
Thanks

 

[~`Heba`~ :)]

since work done here is the work done against gravity+ gain in potential energy + gain or loss in kinetic energy
to calculate gain or loss of kinetic energy we need final and initial velocity
now we are available with initial velocity = 6 ms^-1 =V1
power at bottom of the hill=P1
Force at the bottom of the hill =F1
now at top f the hill velcity =?
Power=P2=5*P1
Force=F2=3*F1
recall the formula P=Fv
take the ratio of both
(P1/5*P1)=(F1/3*F1)*(6/V2)
now we have
(5*6/3)=10=V2
now
work done against gravity+ gain in potential energy =945000
kinetic energy change = 0.5*1250*(10^2-6^2)=625*64=40000
therefore
945000+40000=985000=work done

Alternative method... similar but I think it's easier

F2=3F1
P2=5P1

F2V2=5F1V1
Substitute F2=3F1 and V1=6
3F1V2=5F1(6)
3F1V2=30F1
F1V2=10F1
Cancel out F1
V2=10

I think you already got the height which is 50, using trigonometry, we'll use that to get P.E.

Work done by driving force <What we want> = Change in K.E. + Change in P.E. + Work done against resistance to motion
W = 1/2m(v^2-u^2) + mg(h2-h1) + Work done against resistance to motion

Work done against resistance to motion = Fd, and because F of the resistance to motion = 800, and d = 400, 800x400 =320000

W= 1/2(1250)(10^2-6^2) + 1250(10)(50) + 320000
W = 985000 J

Thanks a lot!!
and please if you could do the graph part in #7 it would be great if you explain it to me

 

[Amaryllis]

Well from one you got
2(x-2)^2 + 3

now you have to know the possible values of y.

The x values are all real numbers ( as stated in the question)

Now what i did was, make the red bit of the equation zero , to do this, you have to make x as 2, now 2(2-2)^2 + 3 will equal 3, then i try a negative number, i choose -1, 21 came, so this bigger than 3, i tried a positive number as 1, and 5 came, so in all cases, a number bigger than three comes, so
y => 3, [i mean y is more than or equal to 3]

Check this, might help you, it has everything you need for statistic.
http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/S1/module.php

awesome ! thank you alot .. jazak allah 5air

 

[PANDA-]

Thanks a lot!!
and please if you could do the graph part in #7 it would be great if you explain it to me

Lol.... i don't think it's possible to show a graph here
Unless we draw it scan etc...

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_62.pdf

How do we do
1) Question 2?
2) 5 iii?
30 7 ii?

That is Question part 1 , the second part needs just the formula!

And for that question, Y value is 2 X values minus eachother.
[Two independent values of X are chosen at random. The random variable Y takes the value 0 if the
two values of X are the same. Otherwise the value of Y is the larger value of X minus the smaller
value of X.] this is from the paper....
anyway so the possible values of Y are
0 [when (2,2) or (4,4) or (6,6) are chosen]
2 [when (4,2) or (6,4) or (2,4) or (4,6) are chosen]
4 [when (6,2) or (2,6) are chosen]

So now we know the values, we have to find the probabilities
for 0, (0.5 x0.5) + (0.4 x 0.4) + (0.1 x 0.1) = 0.42 [P.S, i sub the probabilities of the numbers above from table.]
for 2, (0.4 x 0.5) + (0.1 x 0.4) + (0.5 x 0.4) + (0.4 x 0.1) = 0.48
for 4, (0.1 x 0.5) + (0.5 x 0.1) = 0.1

 

[Mayedah]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_11.pdf
question 7 part (ii) please anyone solve this !

 

[A star]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf
question 7 part (ii) please anyone solve this !

from solving first part you are able to get to sinx=0.5 . now if you compare both equations the only diff is n so we can say that
30/10=n hence n= 3 now for greatest solution find all solutions up to 3 x 360. then divide the highest with 3 you will get the answer

 

[Sadiqa Ali]

can anyone please provide me the worked solutions for m/j 2011 p3 questions??

 

[Esme]

This is from M/J 08 p4


For ii) can you explain why the speed at C is equal to the speed at A. Is it because both the points are at the same level? Even then, can someone provide the logic behind it? Maybe from a practical point of view?