Recent content by Bohrok

Oh I know how to solve simultaneous equations; I just wanted to see if my work seemed alright :smile: Thanks! If no one else chimes in, I'll take it that they agree

Homework Statement In the following figure, the boy has a mass of 40kg, and the platform he is sitting on has a mass of 15kg. If the force of the board on the boy is 180N; a) Find the acceleration of the boy b) Find the reading of the scale...

One more possible route to take is\frac{e^x}{e^x-1} = \frac{e^x-1+1}{e^x-1} = 1 + \frac{1}{e^x-1} However, That's not true in general for the product of a function and its inverse. What is true is that f(f-1(x)) = f-1(f(x)) = x

ln(a) + ln(b) ≠ ln(a+b), but ln(a) + ln(b) = ln(ab). Try working with that.

It's usually understood what the independent variable is in a given differential equation from the context; often it's either x or t when dealing with one independent variable.

If you use L'Hopital's rule, you'd have to differentiate with respect to h and not a (you always differentiate with respect to the variable under lim with l'Hopital's rule):\lim_{h\to 0}\frac{a^h-1}{h} = \lim_{h\to 0}\frac{\frac{d}{dh}a^h-1}{\frac{d}{dh}h} = \lim_{h\to 0}\frac{d}{dh}a^hwhich...

This isn't true; take a look at log/exponential rules.

Thanks, that worked perfectly for me :smile:

Just try doing the long division again and you'll get a quadratic that you can factor; no need for any other tool to get the other factors.

It's been years since I last programmed in C++ and I want to brush up on it over the next couple weeks. Problem is I have no idea what's out there that I could use for programming/debugging C++ programs; the compiler I used before came on a CD with the textbook, but now I'm completely on my own...

Start with e^{i\theta} = \cos\theta + i\sin\theta and e^{-i\theta} = \cos\theta - i\sin\theta Then you add the equations and gete^{i\theta} + e^{-i\theta} = 2\cos\theta \Longrightarrow \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}

I'm surprised no one mentioned the other way without using Euler! z^4 + 16 = z^4 + 8z^2 + 16 - 8z^2 = (z^2 + 4)^2 - 8z^2 This is a difference of squares that will factor and give you not just one but both real quadratic factors.

What substitutions did you try?

A better simplification would be 2(√2 - 1)

You can also use the Rational Root Theorem to find the possible rational roots of a polynomial.