Recent content by Villyer

In those steps you didn't distribute the cos back to the 1.

Have you attempted to use conservation of energy principles?

This isn't any different from those. How would you go about finding the domain of the example function you posted?

These do start with 41 and 41^2 instead of 1 and 1^2, but the formula they have (the ##\frac{x(x+1)}{2}##) is for the sequence that starts with 1. So that means that you can rewrite the formula given as a sigma notation from 1 to n and then simplify from there.

Close, but not quite the right formula. The condensed form that they gave is ##\frac{n^2 + n}{2}##, is that what you meant?

For the second one there, what is the sum of all numbers between 1 and n?

I amended the third line of the proof to incorporate that. Thank you!

Glad I could help :)

Look at the first two terms of your answer. Is there a common factor there that you can pull out?

Homework Statement If r is a rational function, use Exercise 57 to show that ##\mathop {\lim }\limits_{x \to a} \space r(x) = r(a)## for every number a in the domain of r. Exercise 57 in this book is: if p is a polynomial, show that ##\mathop {\lim }\limits_{x \to a} \space p(x) = p(a)##...

That is true if it is just the quadratic equation, but if you are dividing two numbers what happens if they are each positive? Each negative? One positive and one negative? You need to use the answers to those questions to determine the sign of the division of two functions.

I don't know if you want a mathematical method for doing a problem like this, but I would approach it by saying that if a person speaks only English, than they don't speak Spanish or Swahili. So there are 100 people, and at least 60 of them don't fit in the set of people who only speak English...

It does effect the acceleration of the mass system because as the mass drops, energy is transferred into the pulley as well, instead of remaining in the system.

The force of gravity on the hanging mass will not only accelerate the system of masses, but also cause the pulley to rotate.

Ah okay :) You said that you get the correct answer when you use the formula with a ##2g## in the denominator instead of just ##g##, and when you use the ##g## formula you get it wrong. But when you use the formula with just a ##g##, did you recalculate your ##V_o##?