Alan

No, what you suggest is against the rules and would result in nonsense.

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Generally, the equation of a circle can be written as: 

(x - a)^2 + (x - b)^2 = r^2

where the centre coordinates are (a, b) and the radius is r.

Here a = 7, b = -4 and r = 4

You should attempt to use this to answer several of your other similar questions.

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CPhill has written 2 instead of -2 in the matrix. If you correct this and then repeat his process you should come up with one of the specified answers.

This is  looks pretty obvious when a graph is plotted:

I deleted my earlier entry because my reasoning was incorrect.  Here is (what I hope is) a correct proof:

lhs = (9/7)^n
rhs = n

When n = 9  lhs = 9.6...   rhs = 9   so lhs > rhs

When n increments by 1 the lhs increments by (9/7)^n(9/7 - 1) = (2/7)*(9/7)^n
The smallest value of this (for n >= 9) is 2.7...,  given when n = 9, and gets larger as n increases.

When n increments by 1 the rhs increments by n + 1 - n or just 1 for all n.

Hence the lhs starts larger than the rhs (where start is n = 9) and increases more rapidly than the rhs as n increases.

i.e. (9/7)^n > n for all n>=9

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1.  z(x,y,t) = k*x*y/t   where k is a constant

We know that 5 = k*2*(-3)/(1/2) or 5 = -12k  so k = -5/12  

Hence z(x,y,t) = -5xy/(12t)

2. When x = -12/(5z)^2  we have z = -5*(-12)/(5z)^2*y/(12t)  or  z = y/(5z^2*t)  or t = y/(5z^3)  

     Since z = 2/5  then t = y/(5*2^3/5^3)   or  t =  25y/8

3.  Since f(y) = t we must have f(y) = 25y/8

(check the algebra - I did this quickly!)

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Like:   \(mean=\frac{\int_0^{\pi/w}V(t)dt}{\int_0^{\pi/w}dt}\rightarrow\frac{w}{\pi}\int_0^{\pi/w}V(t)dt\)

See:

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The mean over the whole period is zero.  You might want to do the calculation over a half-period.

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Have a look at this Khan Academy video:  https://www.khanacademy.org/math/in-eighth-grade-math/squares-square-roots/square-roots/v/simplifying-radicals