Mathematics: Post your doubts here!

[PhyZac]

No I meant how did you get
a+4d/a=a+14d/a+4d

Okay i will give you an example.
1,2,4,8,
now this is a geometric progression, of ratio 2.
and you see,
2/1 = 4/2 = 8/4
2 = 2 = 2

Always in geometric progression when you divide a number with the one before is equal when you divde a different number with the number before ( hope u get my point. )

Therefore
a, a+4d, a + 14d

a+4d/a=a+14d/a+4d

 

[PANDA-]

Okay i will give you an example.
1,2,4,8,
now this is a geometric progression, of ratio 2.
and you see,
2/1 = 4/2 = 8/4
2 = 2 = 2

Always in geometric progression when you divide a number with the one before is equal when you divde a different number with the number before ( hope u get my point. )

Therefore
a, a+4d, a + 14d

a+4d/a=a+14d/a+4d

Yup, got it, thanks a lot.

 

[~`Heba`~ :)]

ii) First you have to realise that boxes do not have to slide. We know the mass of A which is 200 kg and can calculate the vertical contact force which is equal to weight i.e 2000 N. Multiply the C. of friction i.e 2000*0.2 to get 400 N. It means that this is maximum force which the box B applies on box A to keep it going without sliding. So using F=ma , we get max a i.e. a=400/200=2 ms^-2.
iii)because you have the max a then you can use again F=ma to calculate the maximum resultant force on A as acceleration will be same for the boxes. resultant force=(200+250)*2=900. resultant=P-3150 .So P=3150 + 900 to get 4050.
I hope you have understood it.

thanks a lot!!

 

[~`Heba`~ :)]

http://papers.xtremepapers.com/CIE/...S 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!

 

[Maz]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_61.pdf
5 (ii) anyone please. Thnx

 

[megamind]

can anyone do these two questions on permutaion combination with working (and reasoning)

pleease!

 

[megamind]

hardest questions in the last five years
only a geenious could do it!

 

[syed1995]

edit

 

[syed1995]

can anyone do these two questions on permutaion combination with working (and reasoning)

pleease!

Link the paper.

 

[syed1995]

can anyone do these two questions on permutaion combination with working (and reasoning)

pleease!

1.) 6P4 (Simple 4 choosing from 6)

2.) 4!/2! (Repetition)

3.) 6P4 (There are 2 of everything .. if it has to be unique.. then there can be only 1 of each to choose from .. 12/2 = 6 .. 4 to choose from 6)

4.) This is a difficult one...

R B G O Y Bck

3 different..

If you remove 2R.. remaining will be 10 .. of which there are two of each.. so in total 5 colors to choose from.

5C2 for each..

2R _ _ *4!/2!

2B _ _ *4!/2!

2G _ _ *4!/2!

2O _ _ *4!/2!

2Y _ _ *4!/2!

2Bck _ _ *4!/2!

6*(5C2*4!/2!) (Cuz all of their combinations become 5C2*4!/2!)

6*(10*12) = 720 Ways

5) I don't understand this question .. ? What on earth are we finding here?

12P4 /2!*2!*2!*2!*2!

11880/10

1188.. lol which I think is more than what's possible..

I will do the other question later...

 

[PANDA-]

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!

I second this as well.

 

[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