Question 24.
In a flock of 1000 chickens, the number P infected with a disease at time t years is given by
P=1000/(1+ce^-1000t) where c is a constant
i.Show that eventually, all the chickens will be infected.
ii.Suppose that when time t=0, exactly one chicken was infected...
x=cos2t - sin2t
= √2/√2 (cos2t - sin2t) (multiply top and bottom by √2)
= √2 (1/√2 x cos2t - 1/√2 x sin2t)
= √2 (cos(pi/4)xcos2t - sin(pi/4)xsin2t) since cos(pi/4)=1/√2 and sin(pi/4)=1/√2
= √2 (cos(2t+pi/4)) using the trig identity cos(a+b)=cosaxcosb-sinaxsinb
= √2...
You know that this is a temperature question so that it relates to Newton's law of cooling which states that the cooling rate of a body is proportional to the difference between the temperature of a body and that of surrounding medium.
i.e dT/dt = -k(T-M)
where T is the temperature at...
A container contains 20 lollies of 3 types (5 black, 5 white and 10 clear). Ben Bob and Peter decides to choose 5 lollies each, Ben would only choose the lollies that has an opposite colour to what peter chooses ( i.e. Ben would only choose white if Peter chooses black and vice versa ) unless...
7c) Prove that for any positive integer m,
[2m+1]/[2m+4] <= sqr.root [(3m-2)/(3m+1)]
Hence deduce that (1/2)(5/8)(7/10).....(2n+1/2n+4) <= 1/sqrroot(3n+1)
also
8bi) Show that
r+1/r-1 = 1 + 2/r + 2/r^2 + 2/r^3 + ..... + 2/r^n + 2/r^(n+1)
ii) show that
sum of...
did u approximate n?....
if 4n+1=1001
n=250 (for the mid inequality)
but 4(250)+3 is not equal to 10^-3 (as in the right inequality)
......unless i did something wrong....
yeh i got that part sorry didnt mention that to u but thx anyway.....u kno how to do parts iii) and iv)?.......even part ii), i can do it but im not sure if im correct.
1997 4u maths
6a) The serues 1-x^2+x^4-.....+x^4n has 2n+1 terms
i) Explain why
1-x^2+x^4-.....+x^4n = [1+x^(4n+2)]/[1+x^2]
ii) Hence show that
1/[1+x^2] <= 1-x^2+x^4-......+x^4n <= 1/[1+x^2] + x^(4n+2)
iii) Hence show that, if 0<=y<=1, then
inv.tan y <= y - (y^3)/3 +...
i dont have a diagram here but if u do u'd prolly get it,
Tcos30 = mg
Tsin 30 = mv^2/r (since v=rw, where w = omega)
= mrw^2
divide these equations to eliminate T,
therefore,
tan 30 = rw^2/g
now, sin 30 = r/1 = 1/2
so, tan 30 = 1/2 x w^2/9.8
1/sqr...