NOVEMBER 2014 P11CAN YOU TELL ME THE ANSWER OF THIS QUESTION?????
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NOVEMBER 2014 P11CAN YOU TELL ME THE ANSWER OF THIS QUESTION?????
Can somebody upload GCE A level Mathematics Challenging Drill Questions Yellowreef written by Thomas Bond, Chris Hughes??
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mmm....i thot that so aswell but they might just be loading as u scrolll down just wait for a while nd then check againSome pages are blocked...
v = 4t - 1/16*t^3
So you got that answer or I should explain it to you?NOVEMBER 2014 P11
thnk u so muchv = 4t - 1/16*t^3
s = 2*t^2 - 1/64*t^4 + c
we know that when t=0, s=0 as well:
0 = 0 - 0 + c
c = 0
s = 2*t^2 - 1/64*t^4
Distance PQ is 64m, so we just have to find the value of t for which s=64:
64 = 2*t^2 - 1/64*t^4
t = +/-8
t = 8 (reject negative)
ii)
a = 4 - 3/16*t^2
a>0
4 - 3/16*t^2 > 0
and you do the rest
mmm explain ??So you got that answer or I should explain it to you?
That integration question.mmm explain ??
Plus the thing I was talking about 90 is odd, I know it did not made any sense, I meant to say, ignore the zero and see the previous number(s) is/are even or odd. In case of 90, "9" is a odd number. In case of 180, "18" is a odd number.mmm explain ??
yes can u plzzzzzzzzzzz explain it???That integration question.
To find the shaded area over here using integration is subtracting the integrals.
writing = ....thnk uTo find the shaded area over here using integration is subtracting the integrals.
Sorry for bad writing.
Please explain question 7 in the given paper below.http://studyguide.pk/Past Papers/CIE/International A And AS Level/9709 -Mathematics/9709_s04_qp_4.pdf
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