Mathematics: Post your doubts here!

[GorgeousEyes]

GorgeousEyes
8iii) Simple integration:
View attachment 24746

After that put in the value of x = -1 and find "c" i guess you could do that.
And sorry about the stupid drawing

when i tried it my self , i couldn't get c , i don't knw why :S

 

[GorgeousEyes]

please jun010 variant 12 :no.10 ii and no.11 iv , v
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_12.pdf

 

[Maz]

when i tried it my self , i couldn't get c , i don't knw why :S

4/15(3(-1) +4)^(5/2) - 3(-1)^2 -8(-1) +c = 5
4/15 -3 +8 +c = 5
c = 5 -8 +3 -4/15
c = -4/15

There you have it!

 

[Maz]

please jun010 variant 12 :no.10 ii and no.11 iv , v
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_12.pdf

If you put in the value of x, i.e. '0', in both dy/dx and the original equation you will get the gradient and coordinates of the y intercept.
and so put those in, in y = mx +c

 

[sma786]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 6, i .

 

[iKhaled]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 6, i .

use cosine rule

A^2 = B^2 + C^2 -2BC.COS A
A^2 = 8^2 + 8^2 -2(8)(8).cos 2.4
A = 14..9

 

[shafayat]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 6, i .

there are many ways to do this !
You can directly use the formula (2)(r)(sin(O/2) (i.e. 2 radius sin theta by two ) for any chord where theta is in radians

 

[shafayat]

help please! I am always confused with rates questions ! if you can please tell me the method of doing these questions

 

[PANDA-]

on it!

plus coordinates of B?

You didn't tell me if you understood my explanation for your Mechanics question

 

[sma786]

shafayat , iKhaled i found a much easier way, thanks

 

[GorgeousEyes]

4/15(3(-1) +4)^(5/2) - 3(-1)^2 -8(-1) +c = 5
4/15 -3 +8 +c = 5
c = 5 -8 +3 -4/15
c = -4/15

There you have it!

thanks !

 

[sma786]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 7 part i !

 

[falcon678]

You didn't tell me if you understood my explanation for your Mechanics question

dude i liked tht post? meaning i got it?

 

[falcon678]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 7 part i !

it simple....using the equation...find the sum for two diff terms.....use tht term n its sum in the original sum of terms eqation...u will get two equations...solve them simultaneously!
hope u got it?

 

[daredevil]

when scaler product less then 1 we sayy acute if greater then 1 then obtuse
other is simple integration

ok thanks.. igot the integration one by myself already but thanks all the samee
and yeah our teacher never told us this ..... thanks

 

[syed1995]

help please! I am always confused with rates questions ! if you can please tell me the method of doing these questions

dA/dt = DA/dr * dr/dt

where t is time
r is radius and
A is area ..
dA/dt is change in Area and
dr/dt is change in radius .. and
DA/dr is the diffrentiation of Area with respect to radius..

A= πr^2
DA/dr = 2πr
DA/dr at start = 2π*50 = 100π


DA/dt = DA/dr * dr/dt
DA/dt = 100π*3
DA/dt = 300π

The Area is increasing at a rate of 300π per second

 

[sma786]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 8, ii and iii , lol wtf is this?

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 8, ii and iii , lol wtf is this?

This is CIE making life difficult for us...

To get the vectors we will find the unit vector and multiply by the magnitude..

OA = Unit Vector .. 3i − 4k/√25

15/5(3i-4k)
=3(3i-4k)
=9i-12k

Unit Vector of OB = 2i + 3j − 6k / √(2^2)+3^2+(-6^2)
Unit Vector of OB = 2i+3j-6k/7

OB = 14/7(2i+3j-6k)
OB = 2(2i+3j-6k)
OB = 4i+6j-12k

 

[shafayat]

shafayat , iKhaled i found a much easier way, thanks

:O what much easier ? tell me also ..

 

[shafayat]

dA/dt = DA/dr * dr/dt

where t is time
r is radius and
A is area ..
dA/dt is change in Area and
dr/dt is change in radius .. and
DA/dr is the diffrentiation of Area with respect to radius..

A= πr^2
DA/dr = 2πr
DA/dr at start = 2π*50 = 100π


DA/dt = DA/dr * dr/dt
DA/dt = 100π*3
DA/dt = 300π

The Area is increasing at a rate of 300π per second

thanku , this helped me a lot !...