Skywave

Suppose I have a total of n resistors (all of the same value). I wish to use all n resistors in a variety of series, parallel and series / parallel arrangements, to produce a range of resultant resistance values. What is the total number of arrangements that can be produced, duplicate values of resultant resistance allowed? What is the total number of combinations that can be produced if duplicate values of resultant resistance are not allowed?

I'm not so much interested in the final values: I'm really interested in your method to solve these two questions.

The very essence of Science is measurement. But how can a scientist be sure that by the very act of his measurement, he is not changing the very object or phenomenon that he / she is trying to measure? If he can't be sure, what compromises must he accept which are not irreconcilable with his Scientific philosophy?

(X^3 + X^3) - (X² + X²) - X = XXX

(2x - 1).(x + 2) / (2x + 1).(x - 2) is +ve.

Using Euclidean principles of plane geometry only, prove that (AB + BC) > (AD + DC).

(Yes, it looks very obvious to the eye, but can you prove that with appropriate mathematical rigour?▲ denotes 'triangle'.)

This reference from the BBC News gives the full story:

http://www.bbc.co.uk/news/world-asia-pacific-19190...

It was in the mathematics section. A perfectly formed question. No spelling mistakes, made perfect sense and could not be construed as violating the Rules in any way whatsoever. So why - after it had been 'live' for 3 minutes - was it deleted? (Known, valid reasons please, not guesses - thanks).

Ellipse 1: x²/a² + y²/b² = 1; ellipse 2: x²/b² + y²/a² = 1. What is the size of the area that is common to these two ellipses? Please show full workings. Thank you.

1. Find the equation of the normal to the hyperbola xy = c² at the point P (ct, c/t).

2. Find the co-ordinates of the point P' in which the normal at P cuts the hyperbola again.

3. Write down the co-ordinates of the point P'' in which the normal at P' cuts the hyperbola again

4. Find the equation of the locus of the mid-point of PP'

Note: This Q. is from my GCE 'A' level exam. in Mathematics, 1969. I was allocated 20 minutes for this Q. ~ so it can't be that difficult. Care to try it? (I've retained my solutions).

including the determination of the maxima and minima values of the function & the points of inflexion, if any of these values exist.

Over the centuries, various eminent mathematicians have sought in vain for a formula to find the nth. Prime Number - all to no success. So can the inverse be proved? That is, it is possible to prove mathematically - or otherwise - that a formula for the nth. Prime Number cannot be found or that such a formula cannot exist? (Hopefully, the answer is 'yes', and with a proof here provided - then I can stop wasting my precious time!)

Thank you.

Such as ad.yieldmanager - amongst others.

And why does Yahoo! Answers insist on doing this? Some other Web Forums that I use do not do this.

Express (x + 1) / [x(x - 1)²] in partial fractions, and hence expand this function in a series of ascending powers of x stating with x^(-1), as far as the term in x².

State the general term of the expansion and the set of values of x for which the expansion is valid.

I obtained:

1/x - 1/(x - 1);

x^(-1) - 3 - 7x - 11x² - . . . . . . . .; x<=1

Am I correct?

When a certain quadratic function in x is divided by (x - 1), (x - 2), (x - 3) the remainders are r, 2r, 4r respectively. Find, in terms of r, the remainder when the expression is divided by (x - 4).

( I obtained the answer 7r. Is this correct?)

I thought I knew this; Yahoo have told me to go to my Q., click the Action Bar . . . . etc. But when I go to my Q., the Action Bar isn't 'clickable'!

What's going on?

Here is a question from my GCE "A" level maths exam. of Summer 1969! I've put it here to check if my exam. answers were correct. {Also, you may like the challenge!}

(a) If α, ß are the roots of the equation ax² - bx + c = 0, form the quadratic equation whose roots are α + 1/α, ß + 1/ß.

(b) Find the range of values of x for which (x - 4) < x(x - 4) <= 5

(N.B. the symbol <= means less than or equal to)

.

For (a) I obtained: acx² - b(a + c)x + (b² - 2ac + c²) = 0;

for (b) I obtained: -1 <= x < +1 and +4 < x <= +5

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I've never met anyone who 'sits on the fence' on this Q. I'm sure that there will be some pretty sharp and deep feeling answers to this! So - let's be hearing from you . . . . .