Mark the centre of the circle as point O.

The angle at the centre of any circle is twice the angle at the edge, made by the same chord.

Bob

]]>hi point,

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

Bob

I like the simple, so I give a link, and the ambiguities can respond

]]>

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

Sleep well, and thanks for delaying your well-earned sleep for me!

]]>You mean delete this thread?

]]>list = 1,2,3,4,5

it is possible to combine only 1 with 5, and 2 with 4, symmetrically swapped with 3 as the center

the outcome should be:

1,2,3,4,5

1,4,3,2,5

2,1,3,5,4

2,5,3,1,4

5,2,3,4,1

5,4,3,2,1

4,1,3,5,2

4,5,3,1,2

(3 can also being left out, as it is the center and is not being picked / taken into account for the combinations)

trying to show you the relations by marking the numbers who are swapped together with either () or []:

(1),[2],3,[4],(5)

(1),[4],3,[2],(5)

[2],(1),3,(5),[4]

[2],(5),3,(1),[4]

(5),[2],3,[4],(1)

(5),[4],3,[2],(1)

[4],(1),3,(5),[2]

[4],(5),3,(1),[2]

or, to say it in different words, those 24 solutions / combinations of 1,2,4,5 (order important, repetition not allowed) are NOT VALID because 1,5 and 2,4 not symmetrical as a pair in relation to 3 (as explained above):

1,2,3,5,4

1,4,3,5,2

1,5,3,2,4

1,5,3,4,2

2,1,3,4,5

2,5,3,4,1

2,4,3,1,5

2,4,3,5,1

5,2,3,1,4

5,4,3,1,2

5,1,3,2,4

5,1,3,4,2

4,1,3,2,5

4,5,3,2,1

4,2,3,1,5

4,2,3,5,1

I would love to learn how to use your Haskell script, I tried a few programming tutorials, and it's rough to get into with no background!!(http://learnyouahaskell.com/chapters).

In this problem, is it possible to designate the separate 104 and 96's as 104a, 104b, 96a, 96b? Will the program allow that? It would be helpful even though I said it was unnecessary in a previous post. I am just exploring something, and the more I find out the more important the specific results become.

Thank you for the post Agnishom! If its easy and you have time can you post the results with the 104 and 96 designated as separate? If it's a pain and can't be done before 1500 EST 1/24/2015 I won't be able to use it, but if you can it would be incredible!

]]>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.

]]>