Recent content by meiso

It is significant because an angle formed by a chord and a tangent to a circle is half the arc it intercepts. Angles 2 and 4 are inscribed angles that intercept the same arc, so they are congruent.

When a post is written in response to a user asking for homework help, I would assume that the original poster is going to read the response in its entirety (which, from his response, it seems he did), so I don't think about whether a certain item will be hard to "find" while I'm writing it.

Exactly right. And, sorry, I thought I was clear enough when I said,

No problem. And, actually, you can set u = tan(x) + 2 to make things even easier. 2 is just a constant, so the derivative of tan(x) is the same as the derivative of tan(x) + 2.

The simplest way to solve that indefinite integral is to realize that \frac{d}{dx}tan x = sec^{2}x , and try a u-substitution from there...

An inverse function can be found by switching the positions of the x and y variables of a function and (to put it in true function form) solving for y. For example, if we have f(x) = y = 3x -1 and we want to find the inverse of f, which we can call g, these are the steps to take: 1...

For your first solution, you are correct, but if when you say "format" you are referring to a simpler form, you may wish to use the identity cos(x - \pi) = -cos(x) to simplify the equation.

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Try multiplying both sides of the equation by r and substitute directly from the conversion equations. Remember, r = sqrt(x^2 + y^2). I got an answer different than yours when I did this. By the way, was the shape of your graph a limaçon?

There is most definitely a solution to this equation. Take giant_bog's advice in using that log expansion rule.

Ok, first of all, I am assume that \omega, m, and y are constants. After we dispose of the denominator, we have the numerator equal to zero. Notice in the first line of the handwritten image you posted, the expression \pi\omega^{5}m^{2}y is a common factor of both terms on either side of...

I believe your professor is actually wrong on this one. Let R = 1.005 The answer he/she gave you is actually the value that results from \frac{10(1-R^{120)}}{1-R}. This is equal to 1638.79 rounded to two decimal places. But this expression would correspond to the sum of the geometric...

Very good point! Thank you for pointing that out. Good luck with the results of your exam.

I guess the answer to that question would depend on the scope of the course/ personality of the professor. Only you could answer that, but I guess it's remotely possible. I'm betting it was a mistake. I even tested it for other odd degree root behavior at x = 1 up to the 15th power just in...

If the polynomial definitely has those roots and that y-intercept, then it can't have a max at (3,32). Were all of the points (roots, intercept, max) labeled on the test paper, or did you have to estimate them from grid lines??