Mathematics: Post your doubts here!

[Saly Elkady]

now multiply the brackets.. do it and reply me... multiply carefully no mistakes should be done

to minimiz efforts, just multiply that way that u get coff of x and x squared (if u dont get this, then do multiplication normally like i mentioned above) and reply me.

Okay

 

[Saly Elkady]

now multiply the brackets.. do it and reply me... multiply carefully no mistakes should be done

to minimiz efforts, just multiply that way that u get coff of x and x squared (if u dont get this, then do multiplication normally like i mentioned above) and reply me.

There's no n value
What r we going to multiply?

 

[Thought blocker]

There's no n value
What r we going to multiply?

now wait till i upload solution... i dont think u getting me at all xD

 

[Saly Elkady]

now wait till i upload solution... i dont think u getting me at all xD

Probably that's better xD

 

[Thought blocker]

The thig u dont get ask me without heaitation. Dont thank if u dont mean it

 

[Saly Elkady]

The thig u dont get ask me without heaitation. Dont thank if u dont mean it

Ohhhhh okay thanks for real this time xDDDDD

 

[Thought blocker]

Ohhhhh okay thanks for real this time xDDDDD

Curious. Where were u stuck in this TOUGH question? xD

 

[Saly Elkady]

Curious. Where were u stuck in this TOUGH question? xD

Where was 3x/2-2 xD
Because I actually wrote it in this form -2x+3nx/2
(sight problem xD)

 

[Thought blocker]

Where was 3x/2-2 xD
Because I actually wrote it in this form -2x+3nx/2
(sight problem xD)

hahaha ok

 

[Holmes]

guide please.!!

 

[shingiechingz@8]

View attachment 62047
guide please.!!

b(i) well first thing is first u need to find the value of r
if a=1 then ar=1/3tan^2(theta) so r=1/3tan^2(theta)
if the GP is convergent then -1<r<1 so -1 < 1/3tan^2(theta) < 1
multiply both sides by 3 u get -3 < tan^2(theta) < 3
so tan^2(theta) > -3 or tan^2(theta) < 3
ignore tan^2(theta) > -3 and u are left with tan^2(theta)<3
so tan(theta) < sqrt(3) or tan(theta) < -sqrt(3)
so 0<theta<pi/3
(ii) sum to infinity(S) = a/(1-r)
so S=1/(1-1/3tan^2(pi/6)) = 9+sqrt(3)/8

 

[student2]

3rd part plss??

 

[krishnapatelzz]

View attachment 62060

3rd part plss??

which year?

 

[student2]

which year?

s15_qp_33

 

[Thought blocker]

View attachment 62060

3rd part plss??

-pi/two and arctan (0.5)?
You will get two points.
(i) (2,1)
(ii) (0, -something)

arg(i) = arctan (0.5)
arg(ii) = arctan(-something/0) = -pi/two

 

[student2]

-pi/two and arctan (0.5)?
You will get two points.
(i) (2,1)
(ii) (0, -something)

arg(i) = arctan (0.5)
arg(ii) = arctan(-something/0) = -pi/two

sry but wht is arctan??

 

[Thought blocker]

sry but wht is arctan??

Come on man. You have the google.

 

[Simran72]

Hi do you have notes or an explanation for normal distribution (for P6)?

 

[Thought blocker]

Hi do you have notes or an explanation for normal distribution (for P6)?

No

 

[areeba240]

how to find the negative angle? plzz help