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Yes , though your answer is wrong.Is it only like this or is there more to it?
Good hai , was testing you .
Bilal, my answer is wrong because you put the eqaution here wrong no? Kitkat did it like this, 2ln(5-e^-2x)=1Solve the equation
2( ln5 − e^−2x) = 1,
giving your answer correct to 3 significant figures.
i put the right equation man .Bilal, my answer is wrong because you put the eqaution here wrong no? Kitkat did it like this, 2ln(5-e^-2x)=1
There you go.9709 May 14 variant 33 q7 part b .
Can anyone solve it please?
Ok, as you likei put the right equation man .
Here's 4 ii)And Question 4 (ii) M/J 2012 QP 31. Thank you!
Here's 4 ii)
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 1khttp://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??
If you equate the equation of the curve to zero you get the solution of the curveIf the equation of a curve= 0 the will the equation of the tangent also equal 0?? CAN SOMEONE PLS ANSWER THIS
Yea I know.. But will the equation of the tangent also be equal to 0? I saw some qs in which they did thatIf 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.
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