Alan

Correction!  arg(z) is the angle formed by tan^-1(b/a)

i.e. tan^-1(imaginary component/real component)

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Here are a couple of hints:

3) What is d(sin⁷x)/dx ?

4) (a² - x²)^5/2 → a⁵(1 - (x/a)²)^5/2 Let sin y = x/a

arg(z) and |z| are the same thing, namely the absolute value of the complex number. If 

z = a + ib then arg(z) = |z| = sqrt(a² + b²)

Another approach is to use similar triangles to get the height of the cylinder:

height of cylinder/(base radius of cone - radius of cylinder) = height of cone/base radius of cone

or

h/(4 - r) = 12/4

h = 3(4 - r) → 12 - 3r

V = pi*r^2*h → pi*(12r^2 - 3r^3)

or. V = 12pi*r^2 - 3pi*r^3

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Do this by equating ratios, ensuring you use consistent units.

oj/100 = 100/4000. (There are 1000 ml in a litre)

oj = 2.5 ml

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Inside distance = 2 π r meters. Where r is inner radius of the track.

Outside distance = 2 π(r+5) meters

Hence extra distance = 10 π meters

Time taken for extra distance =  π seconds

Horse's speed = distance/time = 10 π/ π → 10 meters/second

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My third line above would be better written as:

 (4h^(2/3))^3 →  4^3*(h^(2/3))^3 → 64(h^(2/3))^3

Note that (a*b)^3 = (a^3)*(b^3)

With a = 4 and b = h^(2/3) we have

 (4h^(2/3))^3 → 64 (h^(2/3))^3

Also: (h^b)^c = h^(b*c)

with b = 2/3 and c = 3 we have b*c = 2, so (h^(2/3))^3 → h^2

Hence (4h^(2/3))^3 → 64h^2

Just noticed that Tiggsy has given the same solution.

These two solutions may be obtained by letting y = 2^sqrt(5x^2 - 8x) so the equation becomes:

8y^2 + 4 = 33y

(8y - 1)(y - 4) = 0

y = 1/8, or y = 4

y = 2^(-3) or y = 2^2

We can't have a square root being negative in the real number domain, so we must have:

sqrt(5x^2 - 8x) = 2

5x^2 - 8x  = 4

5x^2 - 8x - 4 = 0

(5x + 2)(x - 2) = 0

x = -2/5 or x = 2

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