Mathematics: Post your doubts here!

[psychiatrist]

http://onlineexamhelp.com/wp-content/uploads/2014/08/9709_s14_qp_13.pdf
Q7
How AB vector is 4i - 2j + 4k ??
I got -2i - 2j + 4k
Anyone???

Q11
I got A and C
My approach to find B and D was I found en of BD then use the substituting method to find the coordinates, but answer is coming wrong, why?

 

[nehaoscar]

How to do this??

 

[Sarosh Jameel]

what is the dy/dx for this ???

 

[eliyeap]

what is the dy/dx for this ???

dy/dx= 1-2sinx

 

[eliyeap]

View attachment 57180
How to do this??

um, thetha=a, alpha=b (because its easier to type)
3sina+2cosa=Rsin(a+b)
=Rsinacosb+Rcosasinb (compare with original eqn)
Rcosb=3 R^2cos^2= 9
Rsinb=2 R^2sin^2= 4
Add up: R^2= 9+4=13
R=3.51
sinb/cosb= 2/3
b=33.7
3.51sin(a+33.7)=1
a+33.7=16.55, 163.44
a=129.7

 

[eliyeap]

http://onlineexamhelp.com/wp-content/uploads/2014/08/9709_s14_qp_13.pdf
Q7
How AB vector is 4i - 2j + 4k ??
I got -2i - 2j + 4k
Anyone???

Q11
I got A and C
My approach to find B and D was I found en of BD then use the substituting method to find the coordinates, but answer is coming wrong, why?

Q7 AB = AO+OB
= (-2i-j-3k)+(-j+7k) = -2i-2j+4k

 

[musiclover gurl]

Eqn of Line : y = -3/4x + 12
Area = x * y = 12x - 3/4*x^2

I'm sorry for the late reply but thank you so much! I got it. Thanks for replying

 

[eliyeap]

http://www.cie.org.uk/images/157318-november-2012-question-paper-12.pdf
Q3(i)

Area= (40*60) - 1/2(x)(60) - 1/2(2x)(40-x) - 1/2(40)(60-2x)
Simplify and you will get the answer.

 

[nehaoscar]

um, thetha=a, alpha=b (because its easier to type)
3sina+2cosa=Rsin(a+b)
=Rsinacosb+Rcosasinb (compare with original eqn)
Rcosb=3 R^2cos^2= 9
Rsinb=2 R^2sin^2= 4
Add up: R^2= 9+4=13
R=3.51
sinb/cosb= 2/3
b=33.7
3.51sin(a+33.7)=1
a+33.7=16.55, 163.44
a=129.7

Thankyou! But the answer of R is slightly different
It says R=3.61 which is sqaure root of 13
so the angle comes out to be 130.2 but close enough!
For the a+33.7=16.55, 163.44 part
the question says that the range of a is : 0 < a < 180
So for : a+33.7 do you take the range as 33.7 < (a+33.7) < 213.7 ?

 

[eliyeap]

Thankyou! But the answer of R is slightly different
It says R=3.61 which is sqaure root of 13
so the angle comes out to be 130.2 but close enough!
For the a+33.7=16.55, 163.44 part
the question says that the range of a is : 0 < a < 180
So for : a+33.7 do you take the range as 33.7 < (a+33.7) < 213.7 ?

Ah, i typed down the wrong value for R. (sorry!) But the method is correct, so just work it out again
And yes we take the initial range as 33.7 < (a+33.7) < 213.7 when we first find the inverse of sin

 

[Sarosh Jameel]

PLZ help me with part (iii)...

 

[Saad the Paki]

Hey can some body help me with Q2!
It seems so easy yet I'm unable to do it

 

[The Sarcastic Retard]

Hey can some body help me with Q2!
It seems so easy yet I'm unable to do it

2(i)
4x^2 - 12x
4(x^2 - 3x)
4((x-3/2)^2 - 9/4)
4(x - 3/2)^2 - 9 = 1 * ( 2 * (x - 3/2))^2) - 9
-------------------> (2x - 3)^2 - 9
(ii)
4x^2 - 12x - 7 > 0
2x(2x - 7)+1(2x - 7) > 0
(2x+1)(2x-7)>0
x<-1/2 and x>7/2

 

[psychiatrist]

Q7 AB = AO+OB
= (-2i-j-3k)+(-j+7k) = -2i-2j+4k

So the question is wrong... right?
As according to this answer we get cos^-1(1/sqrt(6)) but we have to get cos^-1(1/3)

 

[psychiatrist]

If I get more than 65 in P1 and more than 25 in S1 What will be my grade??
Like, total is 90 if I score 65 and 25. So 90/125 * 100 is 72% will I get B or C? :/

 

[Saad the Paki]

2(i)
4x^2 - 12x
4(x^2 - 3x)
4((x-3/2)^2 - 9/4)
4(x - 3/2)^2 - 9 = 1 * ( 2 * (x - 3/2))^2) - 9
-------------------> (2x - 3)^2 - 9
(ii)
4x^2 - 12x - 7 > 0
2x(2x - 7)+1(2x - 7) > 0
(2x+1)(2x-7)>0
x<-1/2 and x>7/2

Thanks for answering. ..
Although I didn't get this part "1 * ( 2*" ... where'd the 4 go? (Pardon my stupidness)

 

[The Sarcastic Retard]

L

Thanks for answering. ..
Although I didn't get this part "1 * ( 2*" ... where'd the 4 go? (Pardon my stupidness)

Like* according tto me there is no trick to it, u just need ur logic.

 

[eliyeap]

So the question is wrong... right?
As according to this answer we get cos^-1(1/sqrt(6)) but we have to get cos^-1(1/3)

I think the question is wrong?

 

[eliyeap]

PLZ help me with part (iii)...

y'= -e^p + e^p(4-p) = y2 - y1/ x2 - x1 ( gradient formula)
= (4-p)e^p - 0/ p - 0
-p^2+4p-4=0
p=2

 

[awesomaholic101]

Can someone please solve this question?