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Hi there, I'm no expert on math but I am curious about the implications of 0.999... = 1. I've read that this statement is true (please...

So if Av = \lambda v and Bv = \mu v what can we say about B(Av)?

Noted. 📝

So if Av = \lambda v and Bv = \mu v what can we say about B(Av)?

BBB, while replying to a post, please use the "reply" button of the post - then the OP cannot be erased.

The definition as I recall ... \displaystyle \lim_{x \to a} f(x) = f(a) \iff f(x) \text{ is continuous at } x=a Your attached graph is...

Did you delete your original question?

x^2 * y^2 + \frac{x^4}{y} = [x^2 * y^2] * \frac{y}{y} + \frac{x^4}{y} = \frac {[x^2 * y^3] + x^4}{y}

Did you delete your original question?

v is called an eigenvector of A if there exists a scalar lambda such that Av= lambda*(v) and v=/=0.

Take a more concrete example. \frac{6-2\cdot 2^{-1}}{5} Say you want to get rid of the 2^{-1}. How would you do it? Check your answer...

I thought the y^-1 on the numerator could just become 1/y so the y just goes to the denominator.. so do I have to always multiply the...

That should be 380. And we can find more by adding 360 to 160 too right?

The definition as I recall ... \displaystyle \lim_{x \to a} f(x) = f(a) \iff f(x) \text{ is continuous at } x=a Your attached graph is...

If f(a) is undefined, then how can the limit(x->a)f(x) = f(a) ? How can it be equal to something undefined?