Recent content by coderot

Okay, Used your second edit I produced the diagram attached. This got me the correct solution. Thankyou. :) However I had to use trigonometry to get there. I essentially worked out the other two angles (using trig) around A, added these to angle A, and then subtracted this from 360. It...

Homework Statement Three cylinders are placed in contact with one another with their axes parallel. The radii of the cylinders are 3, 4 and 5 cm. An elastic band is stretched around the three cylinders so that the plane of the band is perpendicular to the axes of the cylinder...

Ray, thanks for taking the time to respond. Most helpful. ehild. You've spotted another set of typos, thanks. Yes, your suggestion offered a far more direct way to the result. The discriminant approach for roots that are not real didn't occur to me at the time of tackling this; but it's...

Ray, been through the calculation, same result. Here is my working in full... f(x) \equiv k(x+2)^2-(x-1)(x-2) \le 12.5k(x^2+4x+4)-(x^2-3x+2) \le 12.5kx^2+4kx+4k-x^2+3x-2 \le 12.5(k-1)x^2+(4kx+3)x+4k-2 \le 12.5for coefficient of x^2 < 0, k < 1 note: not assuming this is the correct result but...

Ray, went down this route and got exactly the same inequality as before. I need to show this result algebraically. I've tried a number of different routes on this and I always get the same result. The only conculsion that I can draw is that my initial statementf(x) \equiv k(x+2)^2 -...

Sorry, I can't use the calculus technique for this. Yes, i know from the properties of quadratic functions that if the coefficient of x^2 is < 0 then f(x) is a max at x = -b/2a. Thanks for the reply.

1) Find the range of values of k such that the function f(x) \equiv k(x+2)^2-(x-1)(x-2) never exceeds 12.5. I've missed several stages of the computation because it is quite lengthy. I hope you get the flow of things. My attempt... (k-1)\left[ x + \left( \frac{4k+3}{k-1} \right)x +...

Thanks for your responses guys. You've cleared this one up for me. :)

That is what I thought. However there is another example of this in the accompanying exercises. Here is the table. I'm asked to state why this isn't a group. \begin{array}{c|ccc} * & 0 & 2 & 4 \\ \hline 0 & 0 & 2 & 4 \\ 2 & 2 & 0 & 2 \\ 4 & 4 & 2 & 0 \\ \end{array} Again the...

Hi, I'm having trouble understanding why the follow composition table for the set \left\{ a, b, c, d \right\} with operation * doesn't define a group. \begin{array}{c|cccc} * & a & b & c & d \\ \hline a & c & d & a & b \\ b & d & c & b & a \\ c & a & b & c & d \\ d & b & a &...