Mathematics: Post your doubts here!

[PANDA-]

I know its messed up but i think u can get it..

Yes I do get the tan(pi-x)... But not the tan(0.5pi-x)

 

[MustafaMotani]

Yes I do get the tan(pi-x)... But not the tan(0.5pi-x)

sin x = cos (90-x)
cos x = sin (90-x)

tan(0.5pi-x) is same as tan (90-x)

tan (90-x) = sin(90-x)/cos(90-x)
= cos x/ sin x
= 1/tanx ..

hope it helps

 

[Hamza1996]

its chem bro..

HERE U GO

 

[Hamza1996]

its chem bro..

THE VALUE OF C IS NT COMNG RITE WEN M SOLVING IT!!!!

 

[MustafaMotani]

HERE U GO

just integrate dy/dx
u shall get
y= 4(3x+4)^2.5/15 -3x^2 - 8x + C

to get value of c just insert (-1,5) in the places of x and y respectively

 

[MustafaMotani]

mine is coming -4/15
wats urs..?

 

[Manobilly]

THE VALUE OF C IS NT COMNG RITE WEN M SOLVING IT!!!!

The c value is -4/15

 

[MustafaMotani]

The c value is -4/15

yup thats what i got ... u might have been integratimg incorrectly... refer to my last post i have integrated

 

[sma786]

for these type of equations use this formulae
NCr (a)n-r br equate their powers like -1=5-r-r this means r=3 solve it with this value

i dint get your formula

 

[PANDA-]

sin x = cos (90-x)
cos x = sin (90-x)

tan(0.5pi-x) is same as tan (90-x)

tan (90-x) = sin(90-x)/cos(90-x)
= cos x/ sin x
= 1/tanx ..

hope it helps

Yes! Got it. Thanks a lot.

 

[sma786]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
What sin formula did they use in Q9 part iii ?!

 

[MustafaMotani]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
What sin formula did they use in Q9 part iii ?!

I think u shall get AB and BC by cos rule

 

[ZainH]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
What sin formula did they use in Q9 part iii ?!

You can't use sin since it's not a right angle triangle.

For finding side AB, you know that angle BOA is 60, and two of your sides are 8. It has to be an equilateral triangle meaning AB is also 8.
For finding BC you just use cos rule using triangle OBC. If you want I can solve it :c


Also, I've got finally got question :3 A star daredevil asd
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3.

 

[ShreeyaBeatz]

9709/11/M/J/12 Question no 7(b)(i)
how do we solve that "convergent"progression??

 

[asd]

You can't use sin since it's not a right angle triangle.

For finding side AB, you know that angle BOA is 60, and two of your sides are 8. It has to be an equilateral triangle meaning AB is also 8.
For finding BC you just use cos rule using triangle OBC. If you want I can solve it :c


Also, I've got finally got question :3 A star daredevil asd
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3.

Oh, dude I just did it yest. D
P(a,0)
Q(0,b)
now the distance PQ= [a^2 + b^2]^1/2 = 45^1/2
a^2+b^2=45.
substituting (a,0), y=-1/2 (x + a)
2y=a-x ---> (i)
Now substituting (b,0), y-b=-1/2(x-o)
2y= 2b-x ---> (ii)
a-x=2b-x
a=2b
Now plug in the value of a in the distance equation to get the values.

 

[ZainH]

9709/11/M/J/12 Question no 7(b)(i)
how do we solve that "convergent"progression??

For arithmetic progressions, a = 1st term, a+d = 2nd term, a+2d = 3rd term , and so on.
d = the difference between consecutive terms.

The first term here is 1 and the second term here is cos²x.
SO, a=1 and a+d= cos²x

We can find d from this, as we have two equations.
Since a = 1 we can write the 2nd equation as 1+d = cos²x , which gives d as cos²x-1

Now, we use the formula for sum of a number of terms in an arithmetic progression which is Sn= n/2(2a + (n-1) d ) you should learn this formula.
Since there asking for the sum of the first ten terms we take n as 10.

S10= 10/2 (2(1) + (10 -1) cos²x-1 )
= 5 (2 + (9) cos²x-1 )

If you know identities you should recognize cos²x-1 is the same as -sin²x
So we replace that.

= 5 ( 2+ (9)(-sin²x)
= 10-45sin²x

a= 10 , b = 45.
Hope you understood.

 

[Sarah22]

3a where dot product =0
10a) idk

Oops sorry.. Not the 3(a), its 3(ii)

 

[ShreeyaBeatz]

For arithmetic progressions, a = 1st term, a+d = 2nd term, a+2d = 3rd term , and so on.
d = the difference between consecutive terms.

The first term here is 1 and the second term here is cos²x.
SO, a=1 and a+d= cos²x

We can find d from this, as we have two equations.
Since a = 1 we can write the 2nd equation as 1+d = cos²x , which gives d as cos²x-1

Now, we use the formula for sum of a number of terms in an arithmetic progression which is Sn= n/2(2a + (n-1) d ) you should learn this formula.
Since there asking for the sum of the first ten terms we take n as 10.

S10= 10/2 (2(1) + (10 -1) cos²x-1 )
= 5 (2 + (9) cos²x-1 )

If you know identities you should recognize cos²x-1 is the same as -sin²x
So we replace that.

= 5 ( 2+ (9)(-sin²x)
= 10-45sin²x

a= 10 , b = 45.
Hope you understood.

I meant Qno.7 (b)(i)

 

[daredevil]

You can't use sin since it's not a right angle triangle.

For finding side AB, you know that angle BOA is 60, and two of your sides are 8. It has to be an equilateral triangle meaning AB is also 8.
For finding BC you just use cos rule using triangle OBC. If you want I can solve it :c


Also, I've got finally got question :3 A star daredevil asd
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3.

already solved it in the previous threads -_-
seriously i'm thinking i've done every difficult question ... thankuuuuu xpf 8D

 

[ShreeyaBeatz]

u did (a)