• 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!

Messages
2,703
Reaction score
3,939
Points
273
Messages
2,703
Reaction score
3,939
Points
273
P(X<6)=0.105 Let (6-mean)/sd be a
P(Z<a)=0.105
a is negative (im not sure how you've been taught to determine whether a is -ve or +ve. What we do is that if both the sign is less than, i.e <, and the probability is less than 0.5, or if both are more than, then a is -ve)
1- Fi(a)= 0.105
Fi(a)= 0.895
a= -1.253
thanks alot aleezay :)
 
Messages
96
Reaction score
27
Points
28
Screen Shot 2013-04-16 at 11.57.14 PM.png
HELP ME with the (ii) part,
i got (i) some what could you tell me whether i am right!? dx/dt= -3acos^2 t *sin^2 t and dy/dt = 3a sin^2 t* cos t ! now what to do?
 
Messages
66
Reaction score
20
Points
18
I = ⌡(tanⁿ⁺² x + tanⁿ x) dx
I = ⌡[tanⁿ x(tan² x + 1)] dx
u = tan x
dx = du/sec² x

I = ⌡[tanⁿ x(tan² x + 1)] du/sec² x
I = ⌡uⁿ du
I = uⁿ⁺¹/(n + 1)

u = tan x
tan(π/4) = 1
tan(0) = 0

I = 1/(n + 1)
Thank you SO much! May you get loads of A* and great success in this life and in the hereafter!
 
Messages
4
Reaction score
0
Points
1
need help! JazakAllah.

3. The first three terms in the expansion of (1 − 2x)^2 (1 + ax)^6, in ascending powers of x, are 1 − x + bx2. Find the values of the constants a and b. (9709/13, May/June 2012)
 
Messages
2,703
Reaction score
3,939
Points
273
Thank you for your reply..

Yeah, but that's the domain for the inverse.. not the function. I'm confused :(
when we inverse a function its domain becomes the range and range the domain of the inverse BUT the defining range remains constant . take it like this that only and only for value of x greater then 0 is the curve following the formulae given so it will only have a valid inverse function after zero
 

Tkp

Messages
1,660
Reaction score
11,026
Points
523
need help! JazakAllah.

3. The first three terms in the expansion of (1 − 2x)^2 (1 + ax)^6, in ascending powers of x, are 1 − x + bx2. Find the values of the constants a and b. (9709/13, May/June 2012)

(1-4x+4x2)(1+6c1*ax+6c2*(ax)2)
(1-4x+4 x2)(1+6ax+15a2x2)
For a u need the coefficient of x
6a-4=-1
A=.5
For b
4+15a2-24a=b(put a =.5)
B=-17/4
 
Messages
373
Reaction score
125
Points
53
When a function as such is written, the 'a' is the amplitude of the wave i.e. its maximum height/displacement from the mean position. Hence, a = 9 -3 =6.
For a normal sine curve, b = 1. However, if we compare a normal curve with the given, we find that this one completes two wave cycles in a time period of 2 pi. Therefore, b = 2.
'c' is the value of how high or low the mean position of the wave is when compared to the line y = 0. Here the graph is raised +3 respective of the line y = 0, hence c = 3.
Thank you for your help as it really was useful.
 
Messages
821
Reaction score
231
Points
53
Hey can anyone help me with this
A(a,-3); B(-3,1);C(2,-9) are collinear,Find A....can someone plz show me how to solve this sum plz asap!!:)
 
Messages
128
Reaction score
112
Points
53
when we inverse a function its domain becomes the range and range the domain of the inverse BUT the defining range remains constant . take it like this that only and only for value of x greater then 0 is the curve following the formulae given so it will only have a valid inverse function after zero
Oh.. alright.. thanks :)
 
Top