A.Maths Qn replied by escadaraindrops @ Thu, 01 May 2008 00:33:23 +0800

escadaraindrops — Thu, 01 May 2008 00:33:23 +0800

this is linar law right? refer to the textbook, there is a similar example

A.Maths Qn replied by wishboy @ Tue, 29 Apr 2008 00:35:18 +0800

wishboy — Tue, 29 Apr 2008 00:35:18 +0800

oic....
thanks

A.Maths Qn replied by weewee @ Tue, 29 Apr 2008 00:27:32 +0800

weewee — Tue, 29 Apr 2008 00:27:32 +0800

Originally posted by wishboy:

wad is curvy equality?

x ≈ y means x is approximately equal to y.

A.Maths Qn replied by wishboy @ Mon, 28 Apr 2008 22:55:40 +0800

wishboy — Mon, 28 Apr 2008 22:55:40 +0800

Originally posted by weewee:

(ii) By inspection sub in x = 0.02. The answer is an approximated solution so use curvy equality.

wad is curvy equality?

A.Maths Qn replied by weewee @ Mon, 28 Apr 2008 22:22:51 +0800

weewee — Mon, 28 Apr 2008 22:22:51 +0800

(1 + ax + x²)^10

= ((1 + ax) + x²)^10

= (1 + ax)^10 + 10(1+ax)^9 (x^2) + ....all the rest will be x with powers above 2

= 1 + 10ax + 45a^2x^2 + 10(1 + ...)x^2 + ....

= 1 + 10ax + (45a^2 + 10)x^2 + ...

therefore by comparing of coefficients, 10a=20=>a=2

45a^2+10=190

(ii) By inspection sub in x = 0.02. The answer is an approximated solution so use curvy equality.

A.Maths Qn replied by wishboy @ Mon, 28 Apr 2008 21:40:49 +0800

wishboy — Mon, 28 Apr 2008 21:40:49 +0800

need to expand out the whole thing?

A.Maths Qn replied by macshaobao @ Mon, 28 Apr 2008 03:30:22 +0800

macshaobao — Mon, 28 Apr 2008 03:30:22 +0800

how to expand (1 + ax + x²)^10 ?

EDIT: ok i got it =D

ok the qn was

Given that the expansion of (1 + ax + x²)^10 up to the x² term is given by (1 + 20x + bx² + ...),

i) find the values of a and b.

ii) hence, without using the calculator, find the value of (1.0404)^10, giving your answer to 3 decimal places.

i found tat a = 2, b = 190 for part (i)

how do i do part (ii)?

Given that your answer to part one is right, i.e., a=2,b=190,

then (1 + ax + x²)^10 is also (1 + 2x + x²)^10 ,

so u sub x =0.02 into (1 + 20x + bx² + ...) to get the value of (1.0404)^10... goddit?

A.Maths Qn replied by wishboy @ Sun, 27 Apr 2008 19:34:57 +0800

wishboy — Sun, 27 Apr 2008 19:34:57 +0800

how to expand (1 + ax + x²)^10 ?

EDIT: ok i got it =D

ok the qn was

Given that the expansion of (1 + ax + x²)^10 up to the x² term is given by (1 + 20x + bx² + ...),

i) find the values of a and b.

ii) hence, without using the calculator, find the value of (1.0404)^10, giving your answer to 3 decimal places.

i found tat a = 2, b = 190 for part (i)

how do i do part (ii)?

A.Maths Qn replied by wishboy @ Mon, 21 Apr 2008 23:24:29 +0800

wishboy — Mon, 21 Apr 2008 23:24:29 +0800

thx alot

A.Maths Qn replied by eagle @ Mon, 21 Apr 2008 23:01:03 +0800

eagle — Mon, 21 Apr 2008 23:01:03 +0800

Originally posted by wishboy:

another qn

solve the equation

2 sin² x = 3 cos x , for 0 <= x <= 5pi

2 sin² x = 2(1-cos² x)

so equation becomes

2 - 2cos² x = 3 cos x
2cos² x + 3cos x - 2 = 0
(2cos x - 1) (cos x + 2) = 0
cos x = 1/2 or cos x = -2 (reject)
so x = pi/3, 5pi/3, 7pi/3, 11pi/3, 13pi/3

A.Maths Qn replied by wishboy @ Mon, 21 Apr 2008 22:55:45 +0800

wishboy — Mon, 21 Apr 2008 22:55:45 +0800

another qn

solve the equation

2 sin² x = 3 cos x , for 0 <= x <= 5pi

A.Maths Qn replied by wishboy @ Mon, 21 Apr 2008 20:17:00 +0800

wishboy — Mon, 21 Apr 2008 20:17:00 +0800

Now Y is actually ln y and X is actually ln x

ok i understand liao

previously i subbed the coords into the x/y in ln x/ln y instead of the whole ln x/ln y

tyvm!

A.Maths Qn replied by weewee @ Mon, 21 Apr 2008 20:10:35 +0800

weewee — Mon, 21 Apr 2008 20:10:35 +0800

Originally posted by wishboy:

need help with this qn

A straight line is in the form Y=mX+c

here gradient m = (3-1)/(4-1)=2/3

Hence Y=(2/3)X + c

Sub in X=1, Y=1, we get 1=2/3+c so c=1/3

Therefore, Y=(2/3)X + 1/3

Now Y is actually ln y and X is actually ln x

ln y = (2/3) ln x +1/3

y = exp(2/3 ln x +1/3)

y = x exp(2/3)exp(1/3)

y=x exp(1)

Not 100% sure.

A.Maths Qn replied by ohnoez! @ Mon, 21 Apr 2008 20:03:41 +0800

ohnoez! — Mon, 21 Apr 2008 20:03:41 +0800

Y=mX + c

Y= ln y, X = ln x

ln y = m ln x + c

sub (1,1)

1 = m(1) + c

m = (3-1)/(4-1) = 2/3

c = 1/3

ln y = 2/3ln x + 1/3

y = e^(2/3 ln x +1/3)???

A.Maths Qn replied by wishboy @ Mon, 21 Apr 2008 19:53:42 +0800

wishboy — Mon, 21 Apr 2008 19:53:42 +0800

The variables x and y are related in such a way that, when ln y is plotted against ln x, a straight line is obtained which passes through the points (1, 1) and (4, 3).

Express y in terms of x.

need help with this qn

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