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Mathematics: Post your doubts here!

Anum96

Anum96

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holoholo said:
can you help find the midpoints ? please explain.
Click to expand...
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 = 11/2

Same with rest of the values.

(11+20)/2 = 31/2

(21+30)/2 =51/2

(31+40)/2=71/2

(41+60) =101/2
 
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***amd***

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holoholo said:
can you explain why it isn't 5.5 ?
Click to expand...
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.
and mid pt = (lower limit + upper limit) / 2 = (0 + 10) / 2 = 5
 
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***amd***

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Magnanimous!!! said:
Assalamalikum.
can anyone please help me in this question
may june 2104 paper33 question 5
Jazak Allah!!
Click to expand...
rearrange the eq u get
(2x + 1)^-0.5 dx = 1/2 *(1/ cos^2 O) dO (writing theta as O)
as cos O = 1/ sec O, we get
(2x + 1)^-0.5 dx = (1/2) * (sec^2 O) dO

integrate both sides
[(2x + 1)^(-0.5 + 1)] / [(-0.5 + 1)*(2)] = 0.5 * tan O
(2x + 1)^0.5 = 0.5 * tan O

just simplify and u'll get ur answer
 
qwertypoiu

qwertypoiu

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Maayee said:
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:)
Click to expand...

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)
 
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Maayee

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qwertypoiu said:
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)
Click to expand...
Thank you so much!!!:):):):)
 
The Sarcastic Retard

The Sarcastic Retard

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Anum96 said:
Now use any equation to equate. Lets take the first one.
x+2>2x+1
x - 2x > 1-2
-x>-1
x<1 :)

Or.
You can even take -x-2 and equate. You will get the same answer. But remember to change the sign of RHS as well.
Click to expand...
For the first time I saw this. :3
Anyways ty. Though I am not getting why we did like this. :p

Can u help me wid other doubts as well?

Thanks.
 
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