Are you A'Level student?Hahaha!
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Are you A'Level student?Hahaha!
Was Gave my papers this novemberAre you A'Level student?
oh nice...I'll Inshallah take my exams in June of 2016 (maths, chem and Bio)Was Gave my papers this november
Wbu?
can u tell me the answer.?
im getting 42.67 for the first part. and 0.5 for the second
the answer is same for both the parts since you are to find the area of the same region in two (almost) different ways. For the first part, i guess u both have already done it, but fr the second one, i guess u guys are getting confused by the 'horizontal components'. All u need to do is, rotate your x-y plane by 90 degrees, now your x-axis will be your y axis, and your y axis will be your x axis. I guess I made it more confusing. Just replace the y with x, and x with y. and u get the equation x = sq rt (16-y).......( as given in second part).Accurate! very nice indeed. You got it right for the first part! but not for the second though, it's the same as first both parts have answer 42.67
You can take c or k. Its the same thingcan anyone show me how to do Question 4 of 9709_s14_qp_31?
It is about differential equations- I don't know why the answer has k in the answer instead of just +c.
http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_31/
Oh damn. I did exact same thing. Miscalculation of the power. I must've subtracted 1 instead of adding it. *facepalm* Moreover I was too lazy to redo it.the answer is same for both the parts since you are to find the area of the same region in two (almost) different ways. For the first part, i guess u both have already done it, but fr the second one, i guess u guys are getting confused by the 'horizontal components'. All u need to do is, rotate your x-y plane by 90 degrees, now your x-axis will be your y axis, and your y axis will be your x axis. I guess I made it more confusing. Just replace the y with x, and x with y. and u get the equation x = sq rt (16-y).......( as given in second part).
integrate this eq wrt y. u get (-2 (16-y)^2/3)/3. Apply the limits y = 16 (when x = 0) and y = 0 on this and u get 128/3 = 42.67
hope i made it clear enough.
Oh wait. That k is a constant neha. They have just written k instead of the answer. I got 2ln(2 plus e^3x) So k is basically 2. Dont get confused. If you want me to solve it then I will.... ?can anyone show me how to do Question 4 of 9709_s14_qp_31?
It is about differential equations- I don't know why the answer has k in the answer instead of just +c.
http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_31/
yes could you solve it please?Oh wait. That k is a constant neha. They have just written k instead of the answer. I got 2ln(2 plus e^3x) So k is basically 2. Dont get confused. If you want me to solve it then I will.... ?
Sure. 5 minutesyes could you solve it please?
I'd understood that though thank you!the answer is same for both the parts since you are to find the area of the same region in two (almost) different ways. For the first part, i guess u both have already done it, but fr the second one, i guess u guys are getting confused by the 'horizontal components'. All u need to do is, rotate your x-y plane by 90 degrees, now your x-axis will be your y axis, and your y axis will be your x axis. I guess I made it more confusing. Just replace the y with x, and x with y. and u get the equation x = sq rt (16-y).......( as given in second part).
integrate this eq wrt y. u get (-2 (16-y)^2/3)/3. Apply the limits y = 16 (when x = 0) and y = 0 on this and u get 128/3 = 42.67
hope i made it clear enough.
Dy/dx = 6ye^3x/(2+e^3x)yes could you solve it please?
Thankyou soooo much! I was taking the 6y to the other side in the first step so it was becoming all complicated and i wasn't getting the answerDy/dx = 6ye^3x/(2+e^3x)
Dy/y = 6e^3x/(2+e^3x)dx
ʃ dy/y = ʃ 6e^3x/(2+e^3x)dx
lny = 6 ʃ e^3x/(2+e^3x)dx
lny = 6 (1/3)ln(2+e^3x) + C
Now you have y = 36 and x = 0. Use them to find C.
ln36 = 2ln(2+e^3(0)) + C
ln36 = 2 ln(2 +1) + C
ln36 = 2ln3 + C
ln 36 = ln3^2 + C
ln 36 =ln9 + C
ln36 – ln9 = C
ln(36/9) = C
C = ln4
Re-write.
lny = 2ln(2 + e^3x) + ln4
lny = ln(2 + e^3x)^2 + ln4
lny = ln[4((2 + e^3x)^2]
ln cancels out.
y= [4((2 + e^3x)^2] Ans.
ii) Maximum point. Put dy/dx = 0View attachment 57707
Part (ii) please
Hello? When y=0 , (x-2)^2 = -3, and that means no real roots for that eq. 2 values for x must be in terms of i(x-2)^2 + 3
I am asked to sketch this graph. All that is ok. I found its minimum point and y-intercept and sketched. But what is confusing is that when I equated the equation(that is, y=0) , I got two answers for x.
But how is that possible when the minimum point of the graph is (2,3) and it doesn't touch the x-axis at all.
Help please?
tell me ur ans to iView attachment 57728
Part (ii) please
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