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????!!!!!

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

### 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

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

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

### 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

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=2.6

Taking log

log

1 = 0.2t log

t = 1 / 0.2log

= 11.18

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

Notice I used log

^{0.2t}you need to use logs.Taking log

_{10}of both sides giveslog

_{10}(10) = log_{10}(2.8)^{0.2t}1 = 0.2t log

_{10}(2.8) <-- using the 'fly' rule to bring 0.2t to the frontt = 1 / 0.2log

_{10}(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 log

_{10}here; you could have used any base (eg log_{e}, ln) but log_{10}meant the left hand side simplified to 1.