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hi point,

Welcome to the forum.

I am always a bit wary when someone posts a link like this. You never quite know where it may lead. Thankfully yours is genuine.

But you could post your work directly using LaTex http://www.mathisfunforum.com/viewtopic.php?id=4397

This will make it easier for people to follow, and, by copying out sections, they can add, and comment more easily.

Bob

hi demha

Q1. {3[SQRT(3)] + 7}[SQRT(3) - 5]

If this was (a+b).(c+d) you'd have to multiply each term from the first bracket with each from the second:

= ac + ad + bc + bd.

If you do this with the question you'll get a term with root 3 x root 3 = 3, and other terms. It should simplify to something of the form (p + q.root 3)

Q4. This is just a substitute and evaluate.

Q5 Yes, do what you say. To solve for 'r' you need to 'un-do' the formula ... divide by 4/3, and by pi, and then find the cube root of that.

Hope that helps,

Bob

That's why I suggested a simple example first. When I find a bit of maths hard, I look for simple cases first so I can slide into the hard stuff in small steps.

Try it with simple functions and just two points, letting the second move gradually closer to the first. If you set up the formulas on a spreadsheet it will be easy to change 1.1 into 1.01 into 1.001 ...........

Then change the fixed pint t=1 into t=2 and repeat. Or change the formulas into harder ones.

Think I'd better sign off now and wash away the literal garden mud and metaphorical middle lane hogging driver anguish. Good night. See you tomorrow.

Bob

With my simple functions

x = 2t so t = (x/2)

Therefore y = (x/2)^2

It would be somewhat harder with the functions you chose.

Bob

Just read your last post. What I had in mind was something like:

x = 2t

y = t^2

Take t = 1 for one point and t = 1.1 for a second.

Compute delta t, delta x and delta y.

Then change t = 1.1 into t = 1.01 so you're letting delta t tend to zero.

It's easy enough to eliminate the t from the above so you have y as a function of x. That way you can calculate lim (delta y / delta x) and compare results.

Bob

but I wouldn't want you to stay up on my account!

I've just driven 70 miles round the M25 so my brain is wide awake, even though I'm theoretically tired having spent the day doing jobs for my Mum. So this is probably a good way to unwind.

Bob

Oh drat, you're not.

Never mind I'm warmed up to it now so I'll carry on.

delta x etc is used for a small but not yet zero amount in the x direction.

So you can start with delta y over delta x, and divide top and bottom by delta t to get

If you let delta t tend to zero then you'd be dividing by zero so you need the small but not yet zero amounts.

Then, is it valid to let the amounts all tend to zero so you get

The rigorous proof of that step is the one that is left out. You could probably convince yourself it is OK by making up an example, but I suggest easier formulas for x and y in terms of t.

Bob

Firstly, the functions of t.

Choose a value of t and work out x and y, using your formulas. Plot the point (x,y)

Now choose another value of t and repeat.

You'll get a series of points on the graph. Those points display that x is a function of y and vice versa.

I'll post this and see if you're still with me.

Bob

hi Au101,

Yes, i'll explain. It's a bit past my bedtime, but I'll do it now if you stay online.

Bob

hi MaaMood93

Welcome to the forum.

You could:

(1) Become a forum member.

(2) Look at the teaching pages on http://www.mathsisfun.com/ There's enough there to keep you busy for a while.

(3) Post in the Help Me section of you want more help with something.

Bob

ps. To graduate to genius level you might consider a brain transplant.

hi Agnishom,

Firstly, consider a thin disc of the cone and work out its M of I.

Then sum up all such moments along the axis of the cone, integrating wrt distance from base to vertex.

Bob

Well done! That's my answer to.

Bob

hi Au101,

I thought it was true because these seem like the rules for multiplying powers.

Anyway, thanks for two proofs. They both look good to me! Well done!

Bob

hi SPARKS_CHAN

These can all be done using this property of circles http://www.mathisfunforum.com/viewtopic.php?id=17799 post 6

You will also need isosceles triangles and angles along a straight line = 180.

Q1. Mark the centre of the circle as point C. Join K, M, Q, and R to C with radius lines.

Then, for example, MCK = 130, and MRK = 65.

Keep working around the diagram using all the above rules to work out every angle.

Q2. Similarly, call the centre of the small circle point C. Join AC and BC. Once again use the angle properties of a circle.

Q3. Same again. I used the property to work out AOE and COF, and gradually progressed round the diagram working out every angle.

Hope that's enough of a hint. Post back if you need more.

Bob

hi Au101,

That all looks good to me except I would like a bit more to explain this line:

Bob

He says that (->0)/(->0) is indeterminate. Shouldn't it be 1 always? .

I would say no to both. Look here:

http://www.mathsisfun.com/calculus/slop … point.html

Bob

hi

Well I don't mind that proof, but I doubt it would pass the number theory rigorousness test.

Here's what may be wrong with it:

(0/0)= k => (0/0).x = kx ok so far.

But now you want (0/0).x = (0.x)/0 Can you switch about the brackets like this?

Let's continue.

So (0.x)/0 = kx => 0/0 = kx => kx = k => x = 1 which is a contradiction.

So one of your steps / assumptions is wrong. Maybe it is that 0/0 is indeterminate. I'd accept that.

Have you done any calculus yet ?

The differential calculus is founded in the idea that we can, for certain functions, find a value for 0/0 .

So a lot of mathematicians are going to be very unhappy if 0/0 is undefined.

Bob

hi Ajay Mohan

I don't think the Khan Academy guy is assuming that y/y = 1. He is saying that division is the opposite process to multiplication and then exploring a 'what if' to show that special rules are required for division by zero. The trouble is, we all know that his assumption is going to lead to a false conclusion; when you assume something is true that is actually false all sorts of strange results can happen. That's why mathematicians shy away from allowing division by 0. The problem arises because the additive identity (zero) causes all multiplications by it to be zero. It's a property of the algebra of 'field' theory and you cannot escape it.

In the end it comes down to this: in any mathematical theory you can make up any axioms you like. They don't have to make sense. But they do have to be consistent. You cannot be able to prove, with your axioms, that 23 = 47 for example. If you allow division by zero that is easy to prove:

23 x 0 = 47 x 0 (= 0)

divide by zero

23 = 47

Whoops.

I'm confused about whether you are then offering an improved proof. If you are, please post just that again and I'll give it some thought.

Bob

hi Ajay Mohan

If a times b = c, then you can define division as c/b = a.

But a problem occurs when b = 0.

c/0 = a requires that a times 0 = c and this has no solutions in real numbers. One way to avoid this problem is to say that division by 0 is excluded from the definition.

And as a times 0 = 0 and d times 0 = 0, this appears to suggest that 0/0 = a = d, where a and d are not necessarily the same. So 0/0 is declared to be indeterminate.

That's how I think of it.

Bob

Q2.

If a number a/b has the value k plus or minus delta then

So figure out what K delta and b are and you have the limits for a + b.

Bob

hi

Q1. The nth term is given by

and the sum by

So if the nth is also the last, use these two equations to write S in terms of r

So now consider the graph of r against S.

If you are unsure what this looks like, here it is:

http://www.mathsisfun.com/data/function … x-1)/(x-1)

Bob