Recent content by besulzbach

I wan't further explanation on this. Well, we can say that l_{1} = l_{2} and then use l_{1}^{2} to account for both variables. But, when you say that a third variable is not necessary you are claiming what exactly? Do you see a way to get a quadratic equation for this problem? I don't...

Yes, of course. Thank you. I'm not a native English speaker, I messed it up. What exactly was done? We got l, w and h in terms of one variable. Just getting {2, 2, 15} through "thinking" may not satisfy some tutors/mentors/teacher/evil cyborgs. I don't know if it would be possible to solve...

Exactly. Notice that if you try to "proof" it mathematically you could do the following: l_{1} \cdot l_{2} \cdot l_{3} = 60 l_{1} \cdot (l_{1} + 13) \cdot l_{3} = 60 l_{1} \cdot (l_{1} + 13) \cdot l_{1} = 60 l_{1}^{2} \cdot (l_{1} + 13) = 60 l_{1}^{3} + 13l_{1}^2 = 60...

My point is that you are going to want to set some equations and solve them, and you can't (logically) equal l to a comparison. (Tell me if this makes sense for you) A box (parallelepiped) has 3 different edge lengths (or less), so... You already know that l1 = l2 [SIZE="4"]+ 13 Now try...

Got everything clarified, thanks to everyone who answered my questions.

I understand it. So, if we suppose they are irrational*, then we can say that sin(ex) + sin(πx) IS aperiodic, right? And, of course, that e/π is also irrational.

I made a mistake there, I know the period of \sin{2\pi x} + \sin{\pi x} Which two periods you were talking about? f(x) = \sin{2\pi x} P_{f}=\frac{2\pi} {2\pi}=1 g(x) = \sin{\pi x} P_{g}=\frac{2\pi} {\pi}=2 h(x) = f(x) + g(x) = \sin{2\pi x} + \sin{\pi x} P_{h}=LCM[2, 1]=2 What...

It may be only me, but what the hell did you wrote there? if length is 13cm longer than the width we could ADD 13cm to the width to get the length, id est: l = 13cm + w Maybe I look pesky now, but > is absolutely (at least according to my math studies) different from +. Tanya Sharma...

What I got from this topic (up to this moment): \sin{(\mathrm{e} x)} + \sin{(\pi x)} is aperiodic. \sin{(2\pi x)} + \sin{(\pi x)} has a definable period. What I would still like to know: How do I get the period of \sin{(2\pi x)} + \sin{(\pi x)} with arbitrary precision? I don't want...

You do not need to apologize, as I told you. I just found you quotation unexpected, nothing else. I guess that his textbook does NOT provide the constants that I got from Wolfram|Alpha, so I posted them. If that is really the case (something only the author of this topic can say), he should...

That is the plot of sin(ex) + sin(πx) period of sin(ex) = 2π/e period of sin(πx) = 2π/π = 2 Therefore the period of sin(ex) + sin(πx) would be the LCM of 2π/e and 2. I think that the this LCM is 2π/e. But the graphics show that I'm absolutely wrong.

What? So: LCM[222, 444] = LCM[222, 222 * 2] = 2? (as the 222 would cancel out?) LCM of 222 and 444 is clearly 444. As the GCD is 222.

Thanks, but, would the LCM of 2e and e exist? They are incommensurate, but, nonetheless, it seems really straightforward to stipulate that LCM. If that is possible, sin(ex) + sin(2ex) would be considered periodic with a period of 2pi/e? Thanks in anticipate, I have no background whatsoever...

Consider f(x) = \sin{(ex)} and g(x) = \sin{(\pi x)} Determine the period of f(x) + g(x)? Is it possible? Would LCM[\frac{2\pi}{e}, 2] = \frac{2\pi}{e}?Homework Equations e = \lim_{h \to 0} (1 + h)^\frac{1}{h} \approx 2.718281828 The period of the sum of the functions would be the LCM of their...

Write the units (ALWAYS). my volume is 27000 cm^3 Right. my Density would have to be 1 g/cm^3 x 27000 You are assuming that the ICE density IS 1g/cm^3 which is incorrect. my final mass is of my ice cube is 729000 correct? No. The final mass is the volume times the density, as your density is...