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Thanx.. i got Q7(i), but cud u plz xplain how exactly u used the completing the squares method in 10(iii)..n 10(iv) the ans in ms is 3+root(9-x), theres no minus0ct-nov 08
7i) first you have to find the perimeter of both the square and circle then add both and equate to 80cm.( because 80cm is the total length)
so,
Perimeter for square = 4x , for circle = 2(pi)r hence,
4x + 2pi*r = 80 divide everything by 2 and make r subject of formula
==> r = (40 -2x)/pi
now find area, area of square = x^2 , area of circle = pi*r^2
so, A = x^2 + pi*r^2, if u substitute r into this equation u will find ur answer
10 (iii) here u use the completion of square method:
h(x) = 6x - x^2
= -(x^2 - 6x)
= -[x^2 - 6x + (-3)^2 - (-3)^2]
= -[(x - 3)^2 -9]
= 9 - (x - 3)^2
(iv) here u have to find inverse of h:
h(x) = 9 - (x-3)^2
let y = 9- (x-3)^2
(x-3)^2 = 9 -y, taking square root both sides
x - 3 = +/ - sq.root of(9-y)
x= 3 +/- sq.root of(9-y) hence
inverse of h = 3 +/- sq.root(9 -x)
hope u got it
oh so q9(i) is on intergration?...i havent yet learnt intergration..sry 4 the trouble0ct/nov 07
m not so sure of q4
9i) here since they are asking u to find equation of the curve u have to integrate dy/dx
if u integrate dy/dx u'll get:
y= 4x - (x^2 / 2) + c
u have a point (2,9), substitute this point and find the value of c:
===> 9 = 4(2) - (2^2 / 2) + c
c = 3
then substitute c in ur equation and u'll have ur answer :
y = 4x - (x^2 / 2) + 3
ii) here u will have to find the gradient of normal first then u can find the equation, to find grad. of normal first find grad of tangent (i.e dy/dx) at x=2
so,
dy/dx at x=2 will be = 2
now to find grad of norm use m1*m2 = -1
u can let m1= 2 (grad of tangent) so ur m2 = -1/2
now since u have the gradient of normal and a point (2,9) u can find the equation using:
y - y1 = m(x -x1) and u'll get
y = 10 - x/2
hope u understood
..n sry it was supposed 2 be q9(iii), not q9(ii)..so if u cud plz ans dat
not a prob...sure just waitoh so q9(i) is on intergration?...i havent yet learnt intergration..sry 4 the trouble
ok to find co-ordinates of Q u have to solve the equation of the curve and the equation of the normal simultaneously. if u do that one of ur co-ordinates will be for P and the other one for Q.oh so q9(i) is on intergration?...i havent yet learnt intergration..sry 4 the trouble
if u dont mind cud u plz do it..coz thats wat i did but wasnt getting the ans n that other question in the other post q10(iii)ok to find co-ordinates of Q u have to solve the equation of the curve and the equation of the normal simultaneously. if u do that one of ur co-ordinates will be for P and the other one for Q.
hope u understood
sure just wait
eqn of curve:
k thanx
you need to know that only those innings get counted in which a batsmen bats and gets out.
The answer for this one is absolutely correct and Igot how u did it!!thanks a ton friend!!I forgot to take -9....This question is for a cricket fan
65 Total matches..
Innings he got to bat .. 65-16 = 49
Innings in which he got out = 49 - 9 = 40
His old average was 47 .. so he scored 47*40 runs = 1880 runs
he got out on 129 in his next inning (41st Out) .. (1880+129)/41 = 49
His new average is 49.
Check the answer
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