Mathematics: Post your doubts here!

[iKhaled]

hello guys!!

i need help in Maths P1, oct/nov 08

Q9.iii)

please help me , i didn't get it at all

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_11.pdf

 

[anonymous123]

hello guys!!

i need help in Maths P1, oct/nov 08

Q9.iii)

please help me , i didn't get it at all

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_11.pdf

AoA wr wb
create eq for the differential of the curve and put points p and q in it
u get m(p) = 3/2 and m(q) = 3/4
now use the rule
tan theta = m
theta(p) comes out to be 56.3 and theta(q) 36.9
the acute angle bw thm is 56.3 - 36.9 = 19.4

 

[2pac]

hi guys need help in 9709 nov 2009 p31 q2.
Thanks

 

[DragonCub]

hi guys need help in 9709 nov 2009 p31 q2.
Thanks

3^(x+2) = 9 (3^x)
9 (3^x) = 3^x + 9
8 (3^x) = 9
3^x = 9/8 =1.125
x = ...

 

[iKhaled]

AoA wr wb
create eq for the differential of the curve and put points p and q in it
u get m(p) = 3/2 and m(q) = 3/4
now use the rule
tan theta = m
theta(p) comes out to be 56.3 and theta(q) 36.9
the acute angle bw thm is 56.3 - 36.9 = 19.4

thank you so much and sorry for the wrong link..opps :$

 

[Esme]

Whenever there are odd number of possibilities you always do n+1 hence it will be 13+1=14. Then take 75% of it which is equal to 10.5. As the answer is not a whole number, you have to take the 10th and 11th number and then divide it by two:
(49+51)/2 =50
Hope it helped

Ok. just corect me if im wrong: whenevr we hav odd number of possibilities to find the median n the quartiles we do (n+1) ??
for even numbers we do not ?
Because my stats book says dat to find the median we alwys do n+1. it doesnt matter if its odd or even.

 

[Esme]

AoA wr wb
9709_s10_11
Q1 explanation needed plz..

i) tan (pi-x) means that the angle lies in the second quadrant. And an is only +ve in frst and 3rd quadrant. So in the second quadrant tan will be negative which gives us -k.
P.S. im assuming you have the angles in the four quadrants .( All Silver Tea Cups )
ii) draw a right-angle triangle. one angle will be 1/2pi he other will be x. Now as tan x = k, the opp side to angle x will be k and the adjacent side will be 1. The third angle in the triangle will obviously be 1/2pi -x. Therefore tan(1/2pi-x) is opp/adj. the opp side to this angle is 1 and the adjacent side is k. this gives us 1/k.

 

[Esme]

iii) similarly draw another right-angle triangle. One angle shall be x n the other 1/2 pi. Again name the sides and then find the hyp which will be square root of 1+k^2.
As sin is opp/hyp.Sin x will be k/root of 1+k^2.

i knw its a lil complicated. but i hope this helped

 

[anonymous123]

got it..Jazakallah

 

[Rampag3r]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_61.pdf

QUESTION 5 (B).

 

[2pac]

3^(x+2) = 9 (3^x)
9 (3^x) = 3^x + 9
8 (3^x) = 9
3^x = 9/8 =1.125
x = ...

Thank you for sparing your precious time for helping me out.
Really appreciate it.I have tons of more doubts,will be very grateful for your help in the future(1-2days time)lol
Thanks again

 

[DragonCub]

Thank you for sparing your precious time for helping me out.
Really appreciate it.I have tons of more doubts,will be very grateful for your help in the future(1-2days time)lol
Thanks again

No problem. I shall be ready to help you solve the doubts.

 

[DragonCub]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s11_qp_61.pdf

QUESTION 5 (B).

 

[2pac]

No problem. I shall be ready to help you solve the doubts.

haha ok then I couldn't understand 9709 q6 nov 2005 p3.I got the first step but then the square root seems to complicate it.
thanks

 

[Rampag3r]

View attachment 7301

THANK YOU, finally got it.

Here's another one, similar in some regards. I can't solve this either. http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_63.pdf Question number 7(b).

 

[confused123]

Q. 9 ii part and iii part.
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w05_qp_1.pdf

 

[smzimran]

Q. 9 ii part and iii part.
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w05_qp_1.pdf

AOA,
(ii)
xy =12 ---(1)
2x + y = k ---(2)
From (1),
y = k -2x ---(3)
Subs. (3) in (1)

x (k - 2x) = 12
kx - 2x^2 = 12
2x^2 -kx + 12 = 0
Since there is no intersection, roots of this equation must not be real
b^2 -4ac < 0
(-k)^2 - 4(2)(12) < 0
k^2 - 96 < 0
(k + √96) (k - √96) < 0
- √96 < k < + √96

(iii)
y = 12/x
dy/dx = - 12 / x^2
At P(2,6)
dy/dx = m = -12 / (2)^2
m = -12 / 4
m = -3
Using the formula m = tanθ1
θ1 = tan-1 (-3) = -71.6◦

the line is 2x + y = 10
rearrange in the form y = mx + c
y = -2x + 10
gradient is -2
m =tanθ2
θ2 = tan-1 (-2) = -63.4◦

Angle required = θ2 - θ1
= ( -63.4◦) - ( -71.6◦)
= 8.2◦

 

[DragonCub]

haha ok then I couldn't understand 9709 q6 nov 2005 p3.I got the first step but then the square root seems to complicate it.
thanks

The square root is not that terrible. It is just used to compensate certain squares.
Also to solve this question I think you need to master the trigonometric identities.

 

[DragonCub]

THANK YOU, finally got it.

Here's another one, similar in some regards. I can't solve this either. http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w10_qp_63.pdf Question number 7(b).

 

[Rabib_69]

Hey guys! need help with these two questions! Both are from O/N 11 33..

The (iii) of number 9 and the last part of the 2nd sum..