Mathematics: Post your doubts here!

[qwertypoiu]

also i always dont get exact answer due to rounding in p1 when to round a side or an angle to get exact answer

You can show long decimals as rounded in your working, but you should keep the exact value on your calculator. The only time you should round a number is in your FINAL answer. Or else you'll suffer from premature approximation. Your final answer is in 3 sig figures. If you must round in between, I suggest 5 sig to be safe

 

[mohmed ahmed soliman]

You can show long decimals as rounded in your working, but you should keep the exact value on your calculator. The only time you should round a number is in your FINAL answer. Or else you'll suffer from premature approximation. Your final answer is in 3 sig figures. If you must round in between, I suggest 5 sig to be safe

and angles calculation

 

[Rizwan Javed]

and angles calculation

A similar scheme should be adapted for the angles calculation also as is explained by qwertypoiu . The only change is that your final answer should be rounded off correct to 1 decimal place.

 

[mohmed ahmed soliman]

A similar scheme should be adapted for the angles calculation also as is explained by qwertypoiu . The only change is that your final answer should be rounded off correct to 1 decimal place.

may u help me in any of above posts

 

[Rizwan Javed]

may u help me in any of above posts

View attachment 59922 help in II june 2013 p12

For this expansion you're only interested in finding the term in x^2. So to make calculations simple, find the terms in x^2, and x^0 only in the expansion of (2x -1/2x)^6 .
Then multiply out those two terms with the remaining expression (1 + x^2), considering only terms in x^2 after, and ignoring the term in x^4.

Hope you'll be able to solve the rest by yourself.

 

[Eugene99]

View attachment 59922 help in II june 2013 p12

This can more easily be solved by finding a general term first,
General term T(r+1) = 6Cr (2x)^6-r (-1/2x)^r
collect only powers of x now: (2x^6-r) (1/2x^r)
sums up to x^6-r/x^r =x^6-2r
Now, since we want to find the coefficient of x ^2
x^2=x^6-2r
only collect the powers (as there is the same index x) : 6-2r=2
2r=4 and r=2
Now, we simply put that into the formula we just made above, for the general term, (we just found the value of r):
T(r+1) = 6Cr (2x)^6-r (-1/2x)^r
6C2 (2x)^6-2 (-1/2x)^2
15(16x^4)(1/4x^2) = 60x^2
co-efficient of x^2 is 60

 

[Eugene99]

View attachment 59923 View attachment 59924 help

This you have to imagine in mind first or just sketch out so that you have an idea what the question means. when you draw out, you will see that the point (-1,3) and the point R if joined together will form a line exactly perpendicular to the line in which R is reflected.

Since we are given the equation of the line in which the reflection takes place: 3y+2x=33 ---(i)
we can easily find the gradient of this, by making y the subject: y=33-2x/3 , you might be able to see already from the equation that gradient of this line is -2/3
since the product of the gradients of perpendicular lines = -1
the gradient of perpendicular line, on which R or the point (-1,3) exists is: m ( -2/3)=-1, which becomes 3/2
Now we just got the gradient of the perpendicular line and we already know that it also passes through (-1,3) equation of that perpendicular line can be easily made:
y-3=3/2 (x+1)
y-3=(3x+3)/2
y=(3x+3)/2 +3 ---(ii)
Now solving the both perpendicular equations simultneously will give us the coordinates, which is the intersection point of both lines
It will get us (3,9)
but still we haven't got the reflected point, we are exactly halfway through that, which in other words is the mid-point of both points ( the (-1,3) and the R)
assume R(x,y)
use the mid-point formula (we have the mid-point (3,9) and the point (-1,3):
x-1/2 , y+3/2 = 3, 9
seperate them
x-1/2 =3 and y+3/2 =9
here we get the required coordinates x =7 and y=15

 

[farhan141]

For any staitonary point, dy/dx = o as dy/dx is a gradient, on maximum point or minimum point, change in y is 0 thus dy/dx should be zero

Ohh gotcha, thanks a lot.

 

[mohmed ahmed soliman]

No. of ways of arranging 3 cards without restriction (also found in iii) = 9C3 x 3!
= 504
( Two cards out of 3 are green and pink together) = 7C1 x 1C1 x 1C1 x 2P2 x 2P2
= 28
504-28= 476

ه
i didnt understand they said how many araangement of 3 cards in part III contain a pink card why u wrote ( Two cards out of 3 are green and pink together) they didnt mention anything about greeen in the question part v is clear now but my doubt is in IV

 

[Layla..]

Can anyone please solve Q 5 part iii
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s15_qp_33.pdf
Thank you

 

[The Sarcastic Retard]

Can anyone please solve Q 5 part iii
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s15_qp_33.pdf
Thank you

Two coordinates of P and Q
common will be (0,0) and other of P (acos^4t,0) and other of Q (0,asin^4t)
Use As level distance between two coordinates formula that is sqrt of ((x2 - x1)^2 + (y2 - y1)^2)
You will get asin^2t + bsin^2t = OQ + OP = a

 

[mohmed ahmed soliman]

Correct answer legit obtained ?? in MS what does this mean

 

[darks]

helo

 

[mohhefssssssssss]

Robert uses his calculator to generate 5 random integers between 1 and 9 inclusive.
(i) Find the probability that at least 2 of the 5 integers are less than or equal to 4. [3]
Robert now generates n random integers between 1 and 9 inclusive. The random variable X is the
number of these nintegers which are less than or equal to a certain integer k between 1 and 9 inclusive.
It is given that the mean of X is 96 and the variance of X is 32.
(ii) Find the values of n and k.

Some1 tell me how to get the value of k plz rest I know.
k will be equal 6

 

[mohmed ahmed soliman]

help in III

 

[mohmed ahmed soliman]

help in II

 

[mohmed ahmed soliman]

help in I

 

[The Sarcastic Retard]

Robert uses his calculator to generate 5 random integers between 1 and 9 inclusive.
(i) Find the probability that at least 2 of the 5 integers are less than or equal to 4. [3]
Robert now generates n random integers between 1 and 9 inclusive. The random variable X is the
number of these nintegers which are less than or equal to a certain integer k between 1 and 9 inclusive.
It is given that the mean of X is 96 and the variance of X is 32.
(ii) Find the values of n and k.

Some1 tell me how to get the value of k plz rest I know.
k will be equal 6

https://in.answers.yahoo.com/question/index?qid=20140529123109AAtHhJS

 

[Ruman Wajih]

Please anybody give full solution to this question

 

[i_try9621]

View attachment 59941 View attachment 59942
help in III

• extend the whole equation like this: 1^5 + 5C1 x (x + 3x^2) + ...... + 5C4 x (x + 3x^2)^4 + 5C5 x (x + 3x^2)^5
• 5C4 gives you 5 and 5C5 gives you just 1 .
• you will only get the coefficient of x^8 from the last two. Therefore,
• (5 x 81) + ( 1 x 270) = 675 which is the coefficient of x^8