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Look The first data value says x is greater than(>) 0. Just greater than means we don’t have to include 0. However x is less than equal to(≤) 10. Therefore 10 is included. Hence, (1+10) /2 = 11/2can you help find the midpoints ? please explain.
Okay. Wait. I'll confirm it.marking scheme says that the midpoints are 5, 15...
i guess it will still be 2.8+3.0 / 2, no? coz there might be values between 2.9 and 3 in the data
so I was right theremarking scheme says that the midpoints are 5, 15...
Hahaha. I was about to say the same
what is the lower limit? ">0" ? i.e. Zero is NOT included, but 0.000000001 is. so we just round off the min possible number in that group to Zero, which eventually becomes our lower limit.can you explain why it isn't 5.5 ?
rearrange the eq u getAssalamalikum.
can anyone please help me in this question
may june 2104 paper33 question 5
Jazak Allah!!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w13_qp_31.pdf
Pls can someone help me with question 4, thank you![]()
(x+2)^2 > (2x+1)^2
Answer is coming incorrect.(x+2)^2 > (2x+1)^2
Use your graph for final answer.
Thank you so much!!!y = e^-t * sint
dy/dt = (e^-t) * (cost) + (-e^-t) * (sint) = e^-t *(cost - sint)
x = e^-t * cost
dx/dt = (e^-t) * (-sint) + (-e^-t) * (cost) = -e^-t *(sint + cost)
dy/dx = dy/dt ÷ dx/dt = [e^-t *(cost - sint)] / [-e^-t *(sint + cost)] = (sint - cost) / (sint+cost)
We want tan in there somewhere, so divide everything by cos:
= (tant - 1) / (tant + 1)
remember that tan(pi/4) = 1, above expression becomes:
= (tant - tan(pi/4)) / (1 + tant*tan(pi/4))
Looking familiar? If not, time to revise the tangent sum rule thingy!
= tan(t - pi/4)
Page 28 of:Does anyone have or know a link for A-level math data booklet
thanks
x < 1Whats the answer?![]()
Yes.You have to sketch three graphs here.
1. x + 1
2. -x-1 (because of absolute value)
3. 2x + 1
First two intersect at x = 1
Are u getting the same?
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