Mathematics: Post your doubts here!

[Bilal Khan]

View attachment 53653 Bilal Khan

Good hai , was testing you .

 

[Bilal Khan]

9709 May 14 variant 33 q7 part b .
Can anyone solve it please?

 

[TheJDOG]

Solve the equation
2( ln5 − e^−2x) = 1,
giving your answer correct to 3 significant figures.

Bilal, my answer is wrong because you put the eqaution here wrong no? Kitkat did it like this, 2ln(5-e^-2x)=1

 

[Bilal Khan]

Bilal, my answer is wrong because you put the eqaution here wrong no? Kitkat did it like this, 2ln(5-e^-2x)=1

i put the right equation man .

 

[TheJDOG]

9709 May 14 variant 33 q7 part b .
Can anyone solve it please?

There you go.

 

[TheJDOG]

i put the right equation man .

Ok, as you like

 

[trao11]

Can someone please help with M/J 2012 QP 31 question 3 (ii) (b)?

Thanks!

 

[trao11]

And Question 4 (ii) M/J 2012 QP 31. Thank you!

 

[TheJDOG]

And Question 4 (ii) M/J 2012 QP 31. Thank you!

Here's 4 ii)

 

[trao11]

Here's 4 ii)

Thank you!

 

[sara kamal]

http://maxpapers.com/wp-content/uploads/2012/11/9709_s12_qp_62.pdf
please explain q6 part 5

 

[TheJDOG]

http://onlineexamhelp.com/wp-content/uploads/2012/05/9709_s07_qp_3.pdf

I need help for question 9(ii). I don't get what the examiner report says about finding unit vector k??

To find the acute angle, you need to find the normals of the two planes and find the angle between them, from i) we know the normal of the plane ABC, but plane OAB has a normal of (0,0,1) because it's normal is perfectly vertical, it's direction is 1k
Then scalar product: Costheta= (N1.N2)/(|N1||N2|)

 

[Haya Ahmed]

How to solve Q7 (iii) without calculator!?
http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_s12_qp_32.pdf

 

[Farah Ismail]

Do you have S1 formula sheet?

 

[Debonny1]

If the equation of a curve= 0 the will the equation of the tangent also equal 0?? CAN SOMEONE PLS ANSWER THIS

 

[Farah Ismail]

If the equation of a curve= 0 the will the equation of the tangent also equal 0?? CAN SOMEONE PLS ANSWER THIS

If you equate the equation of the curve to zero you get the solution of the curve
and if you equate the derivate of the equation to zero, you get the stationary point.
If you have a point on the curve, you can use the coordinates of that point and the gradient of the curve at that point to find the equation of the tangent to the curve at that point.

 

[Haya Ahmed]

Hardest P3 to solve?

 

[Debonny1]

If you equate the equation of the curve to zero you get the solution of the curve
and if you equate the derivate of the equation to zero, you get the stationary point.
If you have a point on the curve, you can use the coordinates of that point and the gradient of the curve at that point to find the equation of the tangent to the curve at that point.

Yea I know.. But will the equation of the tangent also be equal to 0? I saw some qs in which they did that

 

[Charlotte20102013]

HI. Can someone help me with this please?
QP: http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_w12_qp_31.pdf
9709/31/ON/12 Q7 (ii). When I equate the denominator to 0, shouldn't I get both x=0 and 3y^3-1=0 ? Why does the mark scheme solve for only 3y^3-1=0?
MS: http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_w12_ms_31.pdf
Thank you!

 

[tiki-taka]

HI. Can someone help me with this please?
QP: http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_w12_qp_31.pdf
9709/31/ON/12 Q7 (ii). When I equate the denominator to 0, shouldn't I get both x=0 and 3y^3-1=0 ? Why does the mark scheme solve for only 3y^3-1=0?
MS: http://onlineexamhelp.com/wp-content/uploads/2013/11/9709_w12_ms_31.pdf
Thank you!

you have to put the y value into the main equation ln(xy) − y ^3 = 1. to find x