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Permit granted-_- May I take the glass and pour it on your head?
Can you please plug in the values given in the table for any one country cause I am a little confused...
Thank you so much for the time and effort
first part after getting dy/dx I used (a+b)(a-b) = a^2 - b^2 to get (1- x^2)^0.5question 9
Hope my doubts make you also benefit
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
part (i) i got till - 1 / {( 1-x) (1+x}^1/2 * 1 / (1+x)^2 , and its right but i cant simplify it
part (ii) didnt know how to solve it
for part (ii) they asked for max. value of gradientquestion 9
Hope my doubts make you also benefit
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
part (i) i got till - 1 / {( 1-x) (1+x}^1/2 * 1 / (1+x)^2 , and its right but i cant simplify it
part (ii) didnt know how to solve it
for part (ii) they asked for max. value of gradient
so I considered the equation for gradient as m and took dm/dx
tip for these kinda sums
when in calculus you are asked to find maximum or minimum value of something, first differntiate it and then equate that to 0
you get one of the variables this way and then use it to find any other information the question indicates
And yes! your doubt helps me a lot
I am quite in a messy situation regarding chem so I don't really pay the required attention to maths but solving your doubts helps me practice the critical sums
so Thank YOU
Thanks!The formula to calculate median (50th
percentile or any other percentile) without the
cumulative frequency curve is :
lower class boundary +
((50n/100 - cf (till prev. class)/ f (of that
interval))
where n is the total number of observed items
or maximum cumulative frequency. First
calculate the corresponding class by checking
in which class the n/2th value lies then
perform the relative ascribed function
For country A
median lies where 300/2 = 150th frequency lies which lies in the interval 20 < x < 35 (or x< 35) and it's corresponding frequency is 159 - 68 = 91
applying in formula
20 + (150 - 68) / 91 = 20.9 (different from your answer*)
similar method of median for B
but in this 2 marks question, all this working is not required, we just need to simply state that median frequency of A lies in interval 20<x<35 while median frequency of B lies in interval 50<x<70 hence median for B is greater than median of A
Thank you so much!when line and curve intersects
equation of curve = equation of line
x^2 - 2x = kx - 4
x^2 +(-k-2)x + 4 = 0
a=1, b= -k-2, c=4
given, line intersects curve at 2 distinct points
therefore, b^2 -4ac >0
after solving the inequality, take the 2 critical values of x
write the possible ranges
find the value of individual term in that range
eg- (k+6) is -ve when k<-6
then multiply the signs for each possible range
eg- (k+6)(k-2) = -ve multiplied by -ve = +ve
our initial inequality was greater than 0
thus the range is k<-6 or k>2
Anyone can help with question 10iii and 7iii http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf
A2 level ??For question # 7iii) Given: u = 1 + 2i/1-3i
Taking argument on both sides we have; arg(u)=arg(1 + 2i/1-3i)
or,arg(u)=arg(1+2i)-arg(1-3i)
[From 7i: arg(u)=tan-1(-.5/.5)--->3pi/4] or, 3pi/4=tan-1(2)-tan-1(-3)
Therefore, 3pi/4=tan-1(2)+tan-1(3) shown
Thanks aloot can u help me in ths too http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_63.pdf Q2 ii)
For 7i) General Point of line: x=s y=1-2s and z=1+sHi
Please help me with number 7 vector.
Thank you!
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