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Mathematics: Post your doubts here!

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Q. For the GP 4,6,9.... find the value of n if the nth term is the first term greater than 100.
Help me plzz
a = 4
r = 1.5
nth term = 100
a [1.5^(n - 1)] = 100
4[1.5^(n - 1)] = 100
1.5^(n - 1) = 100/4
1.5^(n - 1) = 25
TAking ln of both sides
(n - 1) ln1.5 = ln25
n - 1 = ln25 / ln1.5
n - 1 = 7.94
n = 8.94
Round it off
n = 9
 
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Oct/Nov 2011 - Question Paper 11:
Q3. (i) Sketch, on a single diagram, the graphs of y = cos 2q and y = 1
2 for 0 ≤ q ≤ 2p. [3]
(ii) Write down the number of roots of the equation 2 cos 2q − 1 = 0 in the interval 0 ≤ q ≤ 2p. [1]
(iii) Deduce the number of roots of the equation 2 cos 2q − 1 = 0 in the interval 10p ≤ q ≤ 20p. [1]

Can somebody please solve? Thanks :D
 
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Please remove the old link and put this one in place of Sketching Diagrams in the first post.
 

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The parametric equations of a curve are
x = a cos3 t, y = a sin3 t,
where a is a positive constant and 0 < t < 1
2
p.
(i) Express
dy
dx
in terms of t. [3]
(ii) Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t.
 
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The parametric equations of a curve are
x = a cos3 t, y = a sin3 t,
where a is a positive constant and 0 < t < 1
2
p.
(i) Express
dy
dx
in terms of t. [3]
(ii) Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t.

(i):

x = a cos^3 t
dx = -3a cos^2 t sin t

y = a sin^3 t
dy = 3a sin^2 t cos t

dy/dx = (3a sin^2 t cos t)/(-3a cos^2 t sin t)
dy/dx = - sin t / cos t

(ii):

x = a cos^3 t
y = a sin^3 t
m = - sin t / cos t

y - y1 = m ( x - x1)
y - a sin^3 t = (- sin t / cos t) ( x - a cos^3 t)
y cos t - a sin^3 t cos t = - x sin t + a cos^3 t sin t
x sin t + y cos t = a cos^3 t sin t + a sin^3 t cos t
x sin t + y cos t = a cos t sin t ( cos^2 t + sin^2 t)

Recall the identity 'cos^2 t + sin^2 t = 1'.

x sin t + y cos t = a cos t sin t ( cos^2 t + sin^2 t)
x sin t + y cos t = a cos t sin t ( 1 )
x sin t + y cos t = a cos t sin t

Therefore, the equation of the tangent to the curve at the point with parameter t is 'x sin t + y cos t = a cos t sin t '.
 
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I don't think that they've asked us to explain why is the angle twice the angle of arg z; In Q7(iii) N03 P3, they've asked us to find the greatest value of arg s for points on this locus. Are you referring to the same question?
yes the question is same......and i know they havent asked us to xplain this........but then for us to calculate the angle the diagram should be very accurate......or is there any other method.. using calculation
 
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Excuse me... I need help in MJ 2011 P61 question 5.. Asap bcoz tmrw stats mock
(a)
Let mean = m
standard deviation = s

3m = 7s^2
m = (7/3) s^2 ...... eq(1)

P(X > 2m) = 0.1016
P(X < 2m) = 1 - 0.1016
P(X < 2m) = 0.8984
(2m - m) / s = phi inverse (0.8984)
m / s = phi inverse (0.8984)
Read the table in reverse to get phi inverse
m /s = 1.272
m = 1.272s ..... eq(2)

Solving eq(1) and eq(2) simultaneously
(7/3)s^2 = 1.272s
(7/3)s = 1.272
s = 1.272 * (3/7)
s = 0.545

Substitute value of s in (2)
m = 1.272(0.545)
m = 0.693

So, mean = 0.693 and standard deviation = 0.545

(b)
P(X < a+33) = 0.75
P(Z < (a+33-33)/root21) = 0.75
a/root21 = phi inverse (0.75)
Read the table in reverse to get phi inverse
a/root21 = 0.674
a = 3.09
:)
 
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If the function " y=x^3+ax^2+3x -1" is always increasing,find the range of possible values of a.?
Plz help ..
 
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A circular cylinder has a diameter of 40 cm and is being filled with water at the rate of 1.5 litres per second. At what rate is the water level rising?
 
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A circular cylinder has a diameter of 40 cm and is being filled with water at the rate of 1.5 litres per second. At what rate is the water level rising?
dV/dt = dV/dr * dr/dt (dV/dt will be taken in cm^3 because the radius is given in cm^3)
vol of cylinder = pie * r^2 * h
1500 = pie * 20^2 * h
h = 15/(4pie)
V = pie * r^2 * 15/(4pie)
pie cancels out
dV/dr = 15r/2
1500 = 15*20/2 * dr/dt
dr/dt = 10 cm/s
 
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dV/dt = dV/dr * dr/dt (dV/dt will be taken in cm^3 because the radius is given in cm^3)
vol of cylinder = pie * r^2 * h
1500 = pie * 20^2 * h
h = 15/(4pie)
V = pie * r^2 * 15/(4pie)
pie cancels out
dV/dr = 15r/2
1500 = 15*20/2 * dr/dt
dr/dt = 10 cm/s
question asks for rate of change of water level means height..my ans is 1.19 but the book says 4.77
 
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can anybody give me the link to download 'list of formulae(mf9) ?? please !
 
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