zelda
Member
Does anyone have the solutions to this trial? or is the "safety/security" period not over yet? If it is, can anyone be kind enough to post it up? thanks!
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2004 CSSA 2u Trial (1 Viewer)
- Thread starter zelda
- Start date
zelda
Member
that only has the independent one
zelda
Member
omg, did u even read what i asked?? *sigh*
no hard feelings
no hard feelings
rainonmay
floating around
are there any solutions available for the 2004 CSSA trial??
For my own practice, and for your benefit
Some might be wrong... I don't like working on the comp... hard to check answers =/
9.)
V = x(20-2x)(50-2x)
= 4x(10-x)(25-x)
= 4x(250-3x+x^2)
= 4x^3 -140x^2 +1000x, 0<x<10
V' = 12x^2 -280x + 1000 = 0 when x = 4.4
V"= 24x - 280, <0 :. max at x= 4.4
being the only local maximum in domain, volume is maximised when x = 4.4
V(max)= 2828.56 cm^3
M2 = (1+12/100)500
=560
M20 = (1+12/100)^19*500
=4306.38
S20 = 500((1.12)^20-1)/0.12
= 36026.22
10.)
dv/dt = k
:. v = lnt(k)
=kt + c1
x= kt^2/2 + ct + 1
sub values
2=k+c+1
9=4k+2c+1
k=3, c=-2
:. x = 3t^2-2t+1
v=0
i.e. 3t-2=0, t=2/3s
in ACD
h/b=cosB
in BCD
h/a=cosA
hence h = bcosB = acosA
area ACD = h.AD/2
but AD/b = sinB
so area = hbsinB/2
= abcosAsinB/2, since h=acosA
area BCD = h.BD/2
but BD/a = sinA
so area = hasinA/2
=absinAcosB/2 since h = bcosB
area ABC = absin<ABC/2
=absin(A+B)/2
since area ACD + area BCD = area ABC
so abcosAsinB/2 + absinAcosB/2 = absin(A+B)/2
and hence sin(A+B) = sinAcosB + cosAsinB
Some might be wrong... I don't like working on the comp... hard to check answers =/
9.)
V = x(20-2x)(50-2x)
= 4x(10-x)(25-x)
= 4x(250-3x+x^2)
= 4x^3 -140x^2 +1000x, 0<x<10
V' = 12x^2 -280x + 1000 = 0 when x = 4.4
V"= 24x - 280, <0 :. max at x= 4.4
being the only local maximum in domain, volume is maximised when x = 4.4
V(max)= 2828.56 cm^3
M2 = (1+12/100)500
=560
M20 = (1+12/100)^19*500
=4306.38
S20 = 500((1.12)^20-1)/0.12
= 36026.22
10.)
dv/dt = k
:. v = lnt(k)
=kt + c1
x= kt^2/2 + ct + 1
sub values
2=k+c+1
9=4k+2c+1
k=3, c=-2
:. x = 3t^2-2t+1
v=0
i.e. 3t-2=0, t=2/3s
in ACD
h/b=cosB
in BCD
h/a=cosA
hence h = bcosB = acosA
area ACD = h.AD/2
but AD/b = sinB
so area = hbsinB/2
= abcosAsinB/2, since h=acosA
area BCD = h.BD/2
but BD/a = sinA
so area = hasinA/2
=absinAcosB/2 since h = bcosB
area ABC = absin<ABC/2
=absin(A+B)/2
since area ACD + area BCD = area ABC
so abcosAsinB/2 + absinAcosB/2 = absin(A+B)/2
and hence sin(A+B) = sinAcosB + cosAsinB
flower_farie
Hot Pink Mitten
Answers
Has anyone put up the solutions for the 2004 CSSA Trials?
Has anyone put up the solutions for the 2004 CSSA Trials?