Math Is Fun Forum

This sounds like your next question.  Do you want the bit left after you subtract the circle? This is correct if so.

The diameter = the length os side of the square.

Bob

I did try to find a good example of why my advice is right.  It'll take a bit of trial and error but an example is lurking out there for me.

Bob

Yes!!!

Bob

Yes, that answer is OK. But, a word of warning. If a calculation has, let's say, 2 stages and you round off after stage 1, using that result for stage 2, it could happen that the final answer is different from what you would get if you only round off once at the end of stage 2.

In UK exams, candidates are advised to maintain full accuracy until the end and then to round to a sensible number of figures based on the accuracy of the initial values and the precision of the model.

Some years ago I had a dispute with my gas supplier about the bill. Their calculation had four stages and they were rounding off after each stage. The final figure was 'wrong' by a considerable amount.

Bob

You have lost the (1/xy) term here.

Bob

hi paulb203

Now compare that with

If these represent the same circle then by subtracting

If this was just a random equation then there are lots of values for x, y, a and b that would work.

But it's for points lying on a circle so we don't have that much freedom.

Because x and y can be lots of values but everything else if fixed, the only way that can work is if a=b=0 thus making the x and y terms disappear.**  That means the original equation has r^2 = 68 and centre (0,0).

I've never seen a formal proof for the ** statement; just used it loads of times.  I now realise that's not fully satisfactory so I'm going to work on filling in the proper details of a proof.  (It's accepted in GCSE without proof) Meanwhile you can use what I've told you to do the problem.

LATER EDIT:

Here's the proof:

If you have {a circle, radius 68 centred on the origin} = statement A 
then you can use Pythag to get x^2 + y^2 = 68 {statement B} as the equation.

So A => B

Our problem is does B => A

So we have an equation x^2 + y^2 = 68.  Let's say (p,q) is a point satisfying this equation. ie p^2 + q^2 = 68.

As a negative squared gives the same result as the same number but positive squared this means that

(p,-q)   (-p,q) and (-p, -q) all also fit the equation.

Join (p,-q) to -p,q)  The midpoint of this line is ( (p-p)/2 , (q-q)/2 ) = (0,0).  Similarly (0,0) is the midpoint of the line joining (-p,-q) to (p,q). 

The distance from (0,0) to each of these points is the same.  So all four lie on a circle centred on the origin.  But this is true for all p and q fitting the equation. Thus the equation is for a circle, radius 68 centred on (0,0) 

Bob

Well sort of.  In this example

P(5 heads) + P(4 heads) + P(3 heads) + P(2 heads) + P(1 head) + P(no heads) = 1

It has to be 1 because one of those events must happen.

If we call  P(5 heads) + P(4 heads) + P(3 heads) + P(2 heads) + P(1 head) the same as P(some heads) then we have

P(some heads) + P(no heads) = 1 and that's what you are using.

Bob

Sorry again, my mistake.  But I'm thinking this is an odd model.

  y = 240 - 20(x-100)

So if the grower plants more trees the yield per tree drops (I'm thinking increased competition for resources).

But the model appears to have backward extrapolation: ie. we can set x to smaller values than 100, and get a bigger yield.

So what about x=1 ?  y = 240 -20 times -99 = 240 + 1980 = 2220. So if we have just one tree we get 2220 oranges on it.  Hhhmmm. I think the model needs to have a lower limit for the domain of x.

Bob

Yes, it does get complicated.  I think this is the diagram:

LB represents the (hill + lighthouse) and H is a point on the horizon.

You'll have to convert the feet into miles.

Angle LHO is 90 so you can use Pythag to get LH.

I'll leave the ship for a moment. Once we have a method for the plane it will be easy to adapt it for the ship so I'll attempt the plane next.

For this part you'll still only be able to see to the horizon but the plane P is above the Earth's surface, so we'll just be able to see it if LHP is a straight line.

The difficulty this time is that triangle LPO does not have a right angle.

LOP looks a bit like 90 in my diagram. It isn't!

I need to think about this part so I'll come back to this later.

LATER EDIT: Ok I have it, I think.

We know these distances:  LO, OH, AP and OP.  We can use trig on triangle LHO to calculate angle OLH = OLP as LHP is a straight
line

So in triangle LOP we have two sides and an angle. That's enough to use the cosine rule to find LP.

Do you know the formula for this?

Now replace P by S, the ship. Assume the ship's distance is as given and work backwards to get SA.

Bob

This will work but again you need to say P(no tails) = 1/8

Bob

Yes.

Bob

Correct.

B