- Messages
- 4,493
- Reaction score
- 15,418
- Points
- 523
u did it?thanks it helped
-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
Mathematics: Post your doubts here!
- Thread starter XPFMember
- Start date
- Messages
- 4,493
- Reaction score
- 15,418
- Points
- 523
its not tough.. its just that we had never seen this kind of questions ever beforei am soo glad i am not giving p3 till next year. s1 is killing me and i heard p3 is even tougher
- Messages
- 4,493
- Reaction score
- 15,418
- Points
- 523
depends on u... i loved p3 as compared to s1.
ii gave accelerated... and u wont guess.... with s2
got a b :/
and that also after only two months study of s2
ii gave accelerated... and u wont guess.... with s2
got a b :/
and that also after only two months study of s2
- Messages
- 1,601
- Reaction score
- 553
- Points
- 123
- Messages
- 1,601
- Reaction score
- 553
- Points
- 123
ummmm.........the work done isn't equal to zero
because the object moves through a distance with an applied force
- Messages
- 415
- Reaction score
- 252
- Points
- 53
can any one help me with mj 2010 varient 13 ... p1 ... q 8(ii) ??????
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_13.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_13.pdf
- Messages
- 1,601
- Reaction score
- 553
- Points
- 123
its okay i got it
no need to do it
- Messages
- 4,493
- Reaction score
- 15,418
- Points
- 523
ok here goes to find the limits of this curve y= sin cube 2x cos cube 2x we will put y=0 as on the x axis y=o
0= cos cube2x sin cube 2x o=sin cube 2x
sin cube 2x = 0 0 = cos cube 2x
x=0 2x=pi/2
x= pi/4
so the limits be pi/4 and 0
by substituting u= sin2x
du/dx = 2cos2x
du= 2cos2x dx
now substituting
(cos cube 2x)( u^3) (1/ 2cos2x) <----------- where this is du
one cos would be cut by the cos in the denominator
u^3 (cos square 2x)/2
remember that cos square x = 1-sin square x
so cos square x = 1-sin square x
so
cos square 2x = 1-sin square 2x
and sin 2x =u
so
cos square 2x= 1- u^2
then the total thing wud be
1/2 integral of u^3(1-u^2)
1/2 integral of u^3 - u^5
1/2 (( u^4)/4 - (u^6)/6)
now open u as sin2x
1/2 (( sin 2x^4)/4 - (sin 2x^6)/6)
1/2 (1/4- 1/6)
1/2 (1/12)
1/24
0= cos cube2x sin cube 2x o=sin cube 2x
sin cube 2x = 0 0 = cos cube 2x
x=0 2x=pi/2
x= pi/4
so the limits be pi/4 and 0
by substituting u= sin2x
du/dx = 2cos2x
du= 2cos2x dx
now substituting
(cos cube 2x)( u^3) (1/ 2cos2x) <----------- where this is du
one cos would be cut by the cos in the denominator
u^3 (cos square 2x)/2
remember that cos square x = 1-sin square x
so cos square x = 1-sin square x
so
cos square 2x = 1-sin square 2x
and sin 2x =u
so
cos square 2x= 1- u^2
then the total thing wud be
1/2 integral of u^3(1-u^2)
1/2 integral of u^3 - u^5
1/2 (( u^4)/4 - (u^6)/6)
now open u as sin2x
1/2 (( sin 2x^4)/4 - (sin 2x^6)/6)
1/2 (1/4- 1/6)
1/2 (1/12)
1/24
- Messages
- 4,493
- Reaction score
- 15,418
- Points
- 523
u did a bad thingits okay i got it
no need to do it
- Messages
- 4,493
- Reaction score
- 15,418
- Points
- 523
here are two hints that id give and i want you to do it urself
a) use the M1*M2 =-1 use this to take out the gradient of BD take midpoint
u have a grad u have coordinates make eq of line of BD
then to find b put x=0 as it is on the y axis
thenu have coords of B and M the midpoint
u can say that (coords of b + coords of c )/2 willgive you the coords of M the midpoint
here u can put midpoint and B to get the coords of C
- Messages
- 1,601
- Reaction score
- 553
- Points
- 123
- Messages
- 152
- Reaction score
- 82
- Points
- 38
part of itu did it?
- Messages
- 192
- Reaction score
- 103
- Points
- 53
I managed to get both M and A... but not C. Could anyone explain how to obtain the answer for C?
I'm reposting my query because it seems no one saw it and it got buried two pages back LOL.
- Messages
- 415
- Reaction score
- 252
- Points
- 53
thanks ... just by readind ur first sentence i remembered the concepthere are two hints that id give and i want you to do it urself
a) use the M1*M2 =-1 use this to take out the gradient of BD take midpoint
u have a grad u have coordinates make eq of line of BD
then to find b put x=0 as it is on the y axis
thenu have coords of B and M the midpoint
u can say that (coords of b + coords of c )/2 willgive you the coords of M the midpoint
here u can put midpoint and B to get the coords of C
i think ur a magician
- Messages
- 415
- Reaction score
- 252
- Points
- 53
check the guy who replyed to me ... i was having the same query !
- Messages
- 152
- Reaction score
- 82
- Points
- 38
please help me with 10(b)(i) and (ii). Link to question http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_33.pdf
- Messages
- 382
- Reaction score
- 315
- Points
- 73
I'm reposting this.. anyone?http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_61.pdf
Qn no. 7 - whole question.
Please help me solve the question above.. I'm totally confused!
- Messages
- 124
- Reaction score
- 49
- Points
- 38
Okay once you have the co ordinates of A and M, you have done the toughest part.
Since you havent given the co ordinates of A and M, im gonna take the numbers that come firrst in my head, A and M in this situation be (4,0) and (2,4)
Okay now Let C be (p,q)
Now we use Mid point formulas.
(4+p/2 = 2), (0+q = 4)
Find p and q.
we used the mid point formulas because M is the mid point of A and C. We have the points of A and M. Fill in the missing spaces.
yet again mid point = (x1+x2/2 , y1+y2/2) Use the co ordinates of A(and let C be p,q), which equals to co ordinates of M. Hope it helps!
- Messages
- 1,824
- Reaction score
- 949
- Points
- 123
Q7 i) Atleast one 3 .. means that it is saying P(X≥1) or P(X>0) n = 9
Where X is the number of 3s.
So for it to be zero 3s the probability will be (5/6) and 0 3s at all 9 throws would be P(X=0)=(5/6)^9
Now it asked at least 1 3 .. meaning there are no zeros .. so 1 - P(X=0) = 1- (5/6)^9 = 0.806 Answer
ii) When n fair dice are thrown, the probability of getting at least one 3 is greater than 0.9.
Find the smallest possible value of n.
This is a real tricky question .. we know that for one dice which is thrown 9 times the probability of getting atleast one 3 is 1-(5/6)^9
so for n number of throws the probability will be 1- (5/6)^n
Now the question says .. the probability of n number of throws is 0.9 so
0.9 = 1-(5/6)^n
-0.1=-(5/6)^n
(5/6)^n=0.1
Now To get the value of n we need to use ln on both sides (a little something from my add maths).. using ln would get the power to the bottom
n* ln (5/6)= ln(0.1)
n = ln (0.1) / ln(5/6)
n = 12.62
so atleast 13 throws need to be made for the probability to be 0.9.
b)
This was a real tricky question .. I remember my friend asked me solve it..
5 Green Balls 3 Yellow Balls .. Player wins when a yellow ball is drawn.. Julie draws the first ball .. Probability of Ronnie Winning..
The ways in which ronnie can win are as follows
J'R , J'R'J'R , J'R'J'R'J'R (then the green balls end so yellow ball would come eventually..) To summarize these details in terms of balls
GY , GGGY, GGGGGY
Total = 9, 5 green and 3 yellow.. so probabilities become
P(GY) = (5/8 * 3/7) = 15/56
P(GGGY) = (5/8 * 4/7 * 3/6 * 3/5) = 3/28
P(GGGGGY) = (5/8 * 4/7 * 3/6 * 2/5 * 1/4 * 3/3) = 1 / 56
Total P = 15/56 + 3/28 + 1/56 = 11/28 Answer
Hope that helps...
- Messages
- 4,493
- Reaction score
- 15,418
- Points
- 523
wow such long answers