Mathematics: Post your doubts here!

[Sarosh Jameel]

View attachment 49875

PART ( I ) : Sum to infinity of progression P is a / ( 1 - r )
a= 2 r= 1/2
Sum to infinity of P is 2 / ( 1 - 1/2 ) = 4

Sum to infinity of progression Q is a / ( 1 - r )
a=3 r= 1/3
Sum to infinity of Q is 3 / ( 1 - 1/3 ) = 9/2

Arithmetic progression is 4 , 9/2 , 5

Sum to infinity of R is 5


PART ( II ) : Sum to infinity of progression R is a / ( 1 - r )

R = 5 , a =4 and r = ?

5 = 4 / ( 1 - r )

r= 1/5

sum of first 3 terms is a ( 1 - r^n ) / 1 - r
put the corresponding values of a , r and n

sum of first 3 terms is 4.96

 

[The Chill Master]

PART ( i ) : Differentiate y = ( 4x + 1 ) ^ 1/2
dy/dx = 2 ( 4x + 1 ) ^ -1/2
put x = 2 to get gradient of 1st tangent

PART ( I ) : Sum to infinity of progression P is a / ( 1 - r )
a= 2 r= 1/2
Sum to infinity of P is 2 / ( 1 - 1/2 ) = 4

Sum to infinity of progression Q is a / ( 1 - r )
a=3 r= 1/3
Sum to infinity of Q is 3 / ( 1 - 1/3 ) = 9/2

Arithmetic progression is 4 , 9/2 , 5

Sum to infinity of R is 5


PART ( II ) : Sum to infinity of progression R is a / ( 1 - r )

R = 5 , a =4 and r = ?

5 = 4 / ( 1 - r )

r= 1/5

sum of first 3 terms is a ( 1 - r^n ) / 1 - r
put the corresponding values of a , r and n

sum of first 3 terms is 4.96

gradient of tangent = 2 ( 4 * 2 + 1 )^ -1/2
gradient of tangent = 2/3

Differentiate y = ( 1/2) X^2 +1
dy/dx = x
put x = 2 in above equation to get gradient of 2nd tangent
gradient of 2nd tangent = 2

let theta be x
Tan x = ( M2 - M1 ) / 1 + M2M1
Tan x = ( 2 - 2/3 ) / 1 + 4/3
Tan x = 4/7
theta = 29.7 degree

Part ( ii ) : find area under the 1st curve using 2 and 0 as limits .
find area under 2nd curve using 2 and 0 as limits .
subtract both areas to get shaded area ... final answer is 1 unit square ..

Thanx a bunch i was rele worried hv exam tom thanx a bunch

 

[Sarosh Jameel]

PLZ SOMEONE HELP TO SOLVE THIS QUESTION !!!

 

[TimBluesWin]

PLZ SOMEONE HELP TO SOLVE THIS QUESTION !!!

--For 8A, I'm sure you know how to find the length of AB.
For length of CD, we need to find the length of OD first. OAD is a right angle triangle. Hence, cos a= OD/OA. So OD= OA cos alpha -> 4 cos a.
Hence CD is a*4 cos a=4a cos a.
For BD, just OB-OD, which is 4-4cos a, and do the same for AC.
Just add them up (4a+4a cos a +2(4-4cos a))
--8B. To find the area of ABCD, subtract area of OAB with area of OCD.
Area of OAB= 1/2 * pi/6 * (4)^2
Area of OCD= 1/2* pi/6 * (4-4cos (pi/6))^2

Tell me if there's something wrong; hope this helps!

 

[Sarosh Jameel]

THANKSS A BUNCH !!

 

[Haya Ahmed]

Hey all
Happy New Year 2015 )

 

[Robert Flockhart]

Please I need the working solutions for 04/M/J/05 Q.2

 

[Awesome12]

Why do you even need the answers?

Human stupidity just touched the highest jackpot.

 

[TimBluesWin]

Please I need the working solutions for 04/M/J/05 Q.2

Okay, since the 5N and 6N forces are neither horizontal nor vertical, we need to break down these forces into horizontal and vertical component.
Horizontal component of 5N: 5*cos 50
Vertical component of 5N: 5*sin 50
Horizontal component of 6N: 6*cos 30
Vertical component of 6N: 6*sin 30
Now that we have found the horizontal and vertical components of each forces, we need to calculate the resultant of the horizontal and vertical component.
Resultant horizontal component: 7+5*cos 50-6*cos 30=+5.0178
Resultant vertical component: 5*sin 50-6*sin 30=+0.8302
To find the resultant force; use the phytagoras theorem; sqrroot of (5.0178^2+0.8302^2)
To find the direction; just use tan inverse (0.8302/5.0178), and there you go the direction with respect to the horizontal component.

Hope this helps! Just point me out if there's something wrong, and sorry if my explanation is a bit lengthy...

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w03_qp_3.pdf

Q8 (i) part how to do it !! I got B=-1 but what about the rest

ZaqZainab

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w03_qp_3.pdf

Q8 (i) part how to do it !! I got B=-1 but what about the rest

ZaqZainab

I hope you get it

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w03_qp_3.pdf
q9 (i) and (ii)

ZaqZainab

 

[TimBluesWin]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w03_qp_3.pdf
q9 (i) and (ii)

ZaqZainab

Do you mind if I help you instead?

Hope my workings make you clear (Note: I use Marking Scheme to verify if my answers are correct)

 

[A*(a*)]

AS level topical MCQs + Theory upto 2014 including ALL variants!

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_qp_3.pdf

q4 (i)

 

[Starlight97]

somebody please help me solving these equations
1. 4sin4x=2
2. cot3x=4

solve these equations for 0<x<360

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf
Q 7(i)

 

[The Sarcastic Retard]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_3.pdf
Q 7(i)

integration by parts separable variable or what ever its called basically rearrange to this format:
(1/N) dN = Kcos(0.02t) dt
and integrate

 

[The Sarcastic Retard]

sorry its totally unmanageable to check threads as I have so many works in real life i hope u understand my problem else I would have helped u all

 

[The Sarcastic Retard]

somebody please help me solving these equations
1. 4sin4x=2
2. cot3x=4

solve these equations for 0<x<360

For 1

&

For 2