Amaths question help?
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The binomial expansion of (1+px)^n, where n>0, in ascending powers of x is
1 - 12x + 28p²x² + qx³+ ...
Find n, p, and q
I got np = 12
nC2 = 28
n!/n-2! = 56, how can i solve this? or my path is wrong? -
(1+px)^n
= 1 + npx + [n(n-1)/2]p²x² + [n(n-1)(n-2)/6]p³x³+ ...
= 1 - 12x + 28p²x² + qx³+ ...
By comparing,
np = -12 ----(1)
n(n-1)(n-2) = 6q ----(2)
n²-n= 56
(n-8)(n+7)=0
n=8 or -7 (rej. as n>0)
sub n=8 into (1): p= -1.5
sub n=8 into (2): 8×7×6×(-27/8)=6q, q=56×-(27/8)=-189 -
g(x) = |x-3|-4
Find the inverse of g(x) -
[quote]Originally posted by ^tamago^:
(1+px)^n
= 1 + npx + [n(n-1)/2]p²x² + [n(n-1)(n-2)/6]p³x³+ ...
= 1 - 12x + 28p²x² + qx³+ ...
By comparing,
np = -12 ----(1)
n(n-1)(n-2) = 6q ----(2)
n²-n= 56
(n-8)(n+7)=0
n=8 or -7 (rej. as n>0)
sub n=8 into (1): p= -1.5
sub n=8 into (2): 8×7×6×(-27/8)=6q, q=56×-(27/8)=-189[quote]
a small mistake here in bold.
should be [(n)(n-1)(n-2)/6]p^3 = 6q
It didn't really affect the rest of the workings except the
final ans of q should be
q = 8(7)(6)/6(-1.5)^3
= -189 -
In this case, g(x) has no inverse function if the domain of g(x) is from -infinity to infinity.Originally posted by AndrewPKYap:g(x) = |x-3|-4
Find the inverse of g(x)
Reason: It is not a one-one function. -
p³ was reflected inOriginally posted by ^tamago^:
(1+px)^n
= 1 + npx + [n(n-1)/2]p²x² + [n(n-1)(n-2)/6]p³x³+ ...
= 1 - 12x + 28p²x² + qx³+ ...
By comparing,
np = -12 ----(1)n(n-1)(n-2) = 6q ----(2)
n²-n= 56
(n-8)(n+7)=0
n=8 or -7 (rej. as n>0)
sub n=8 into (1): p= -1.5
sub n=8 into (2): 8×7×6×(-27/8)=6q, q=56×-(27/8)=-189[quote]
[quote]Originally posted by tanjun:
a small mistake here in bold.
should be [(n)(n-1)(n-2)/6]p^3 = 6q
It didn't really affect the rest of the workings except the
final ans of q should be
q = 8(7)(6)(-1.5)^3
= -1134
sub n=8 into (2): 8×7×6×(-27/8)=6q
n(n-1)(n-2)p³ = 6q
so it remains as -189.
paiseh it's very hard to keep track of all the vars on the small screen. my error correction and checking techniques just didn't come in at this part other than to check that the answer at the end matches.
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tammy ~
i know chiu ish sexpert in amaths. get A1 sumore -
collect a lot of rust liao, need to buah gu you alrd.Originally posted by spadeTwo:tammy ~
i know chiu ish sexpert in amaths. get A1 sumore 
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gu you not good lahOriginally posted by ^tamago^:collect a lot of rust liao, need to buah gu you alrd.
i want kaya
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lol. I made a mistake too. forget to divide by 6 for the ans.Originally posted by ^tamago^:p³ was reflected in
sub n=8 into (2): 8×7×6×(-27/8)=6q
n(n-1)(n-2)p³ = 6q
so it remains as -189.
paiseh it's very hard to keep track of all the vars on the small screen. my error correction and checking techniques just didn't come in at this part other than to check that the answer at the end matches. -
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A bloody motor boat travels in a stright line across a river which flows at 3m/s btwn straight parallel banks, 200m apart. the bloody boat, has speed 6m/s in still water, travels directly frm point a to point b, 150 downstream of a, on the opposite bank. find the time taken to travel frm a to b.
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differential eqn wor, and have to find out the flow profile of the river. and how long the boat isOriginally posted by 77th Cloud:A bloody motor boat travels in a stright line across a river which flows at 3m/s btwn straight parallel banks, 200m apart. the bloody boat, has speed 6m/s in still water, travels directly frm point a to point b, 150 downstream of a, on the opposite bank. find the time taken to travel frm a to b. -
huh? what do u mean? this question is found on the ten year series.
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common angle between the velo diag. and dist diag is angle ABX
angle ABX = tan^-1( 200/150)
= 53.1 deg
using sine rule,
3sin53.1 / 6 = sin angle YAB
angle YAB = sin^-1 (0.4)
= 23.6 deg
therefore, angle AYB = 180-23.6-53.1= 103.3 deg
using cos rule, resultant velocity AB = ( 6^2 + 3^2 -2(6)(3)cos103.3)^1/2
resultant velocity AB = 7.30m/s
distance AB = (150^2+200^2)^1/2 (pythagoras theorem)
= 250
time taken = distance AB / resultant speed (also indicated by AB) = 250/7.30
= 34.25084106s
= 34s (nearest sec)
hope this helps...