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winter diet calc 2002 questions 8 9b 10a and 11a!!!

hssc11045

Posted 19 May 2004 - 07:59 PM

god this is def a nightmare paper.....questions 1-7 no problems but after tht am jus stuck!

question 8----bad day me finks...the integration jus doesnt seem 2 match the answer!!

question 9b....o..k size when observation began was 950 therefore multiplied by 10 = 9500 => i got 9500 = 950 X (2.6)^0.2t.....but am lost from there.


question 10a.........jus confused for what to use for the a,b,c of the discriminant

question 11a) I was pretty sure i worked out how 2 do this one in my head but it does not work properly for me on paper..............i thought I would do integration of F(x) between zero and negative 5 then once that is sussed i would add 30 to tht answer.....then would do integration of G(x) between zero and 5 and take this value away from thirty........then add the 2 values together from either side of the x-axis and weeeeeeeeelaaaaaa......but i did not get the correct answer!!!


any help please????!!!!!

Allan

Posted 19 May 2004 - 08:13 PM

for question 8 I would say make sure ur calculator is in radians

any differentiation or integration involving sin or cos must be worked in radians

if that doesn't work ill write down the working...

Allan

Posted 19 May 2004 - 08:36 PM

for 10...

subsituting the line into the circle you get:

x² - (k - 2x)² - 2x - 4 = 0
x² + (k - 2x)(k - 2x) - 2x - 4 = 0
x² + k² - 4kx + 4x² - 2x - 4 = 0
5x² - 4xk - 2x - 4 + k² = 0
5x² - (4k + 2)x - 4 + k² = 0

so a = 5, b = (-4k - 2), c = (-4 + k²)

(-4k - 2)² - 4 x 5 x (-4 + k²) = 0
(-4k - 2)(-4k - 2) - 20(-4 + k²) = 0

leading to:

4k² - 16k - 84 = 0
4(k² - 4k - 21) = 0
4(k + 3)(k - 7)
so k = 7

hssc11045

Posted 19 May 2004 - 09:27 PM

cheers 10 was very helpful...... think it was what i used for a common factor b4 applying discriminant.......went on 2 do part b successfully tho so thanks for clearing that up!!!!

for 8 tho a fink it wud b helpful if u put ur working up....for some reason i never knew tht my calc had 2 b in radians for tht again thanks for pointing that out

AndyW

Posted 19 May 2004 - 09:36 PM

Questions 8:

After integration you should get:

1/3 sin 3x + 3 cos(1/3x + 1)

Substitute to get:

[ 1/3 sin 3 + 3 cos 4/3] - [1/3 sin 0 + 3 cos 1]
= 0.047 + 0.706 - 0 - 1.621
= -0.868

Dave

Posted 20 May 2004 - 08:33 AM

for question 11 the first line is:

int(f(x)-(-6)) dx between -5 and 0 + int(g(x)-(6)) dx between 0 and 5

however i also can't get the right answer i'm dancing about the answer giving in the book

fro question 9b

you then get 10=2.8^0.2t

Then i remember doin it but can't remember how i don't think you go guess and check but the answer does say about 12

Tell you if i work it out

George

Posted 20 May 2004 - 09:16 AM

To solve 10=2.60.2t you need to use logs.

Taking log10 of both sides gives

log10 (10) = log10 (2.8)0.2t
1 = 0.2t log10 (2.8) <-- using the 'fly' rule to bring 0.2t to the front
t = 1 / 0.2log10 (2.8)
= 11.18

Rounding up gives t = 12, so the colony will take 12 hours to multiply in size by 10.

Notice I used log10 here; you could have used any base (eg loge, ln) but log10 meant the left hand side simplified to 1.

Dave

Posted 20 May 2004 - 11:37 AM

yeah can't believe i forgot how to do that but anyway

superstar

Posted 20 May 2004 - 01:08 PM

Ok this paper is a nightmare. Paper 1 was fine but a paper 2... the first page or qus were ok but it went VERY downhill after tht!!
Lets hope it is not as hard as that tomoz!!

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