EvoN1020v

Yeah thanks Atheist and dave, I got it. I know why it's positive 51, and not negative 51. Meanwhile, I have another questions that I got confronted with. [math]lim_{x \rightarrow 1} \frac{1}{(x-1)^2}[/math] I assumed that the limit is not defined because there's a vertical asymptote at x=1. A teacher had this answer of [math]+ \infty[/math]. I thought this was wrong, because it would be correct if the limit goes [math]x \rightarrow 1^+[/math] or [math]x \rightarrow 1^-[/math]. What do you think?

Also, your sister could get a lot of money from the prenup.

I need a tiny bit favour from you guys. Given that [math]f(x)=\frac{x-1}{(x+2)^{50}}[/math]. The solution is [math]f'(x)=\frac{52 - 49x}{(x+2)^{-51}}[/math]. Am I correct? Because the teacher had this solution: [math]f'(x)=\frac{52-49x}{(x+2)^{51}}[/math]. Who is right?

The system will be inconsistent if they are not linearly independent. Also you have to remember that identity matrix is always a square matrix.

I'm thinking perhaps "factorial"? i.e. 5! = 5x4x3x2x1 n(n-1)(n-2)(n-3)...1 But it probably doesn't fit with your pattern because you are adding the terms. It looks like if a person is starting at the bottom of a mountain, and it's going uphill. Then at the top of the peak, it goes downhill back to the bottom. Anyways.

Wow, what an honour.

Does it really matter? You can just put your threads in the Envoirnmental Science forum.

What is "OU courses"? Mind explaining that to me please.

LU Decomposition is a mulitple of lower triangle and upper triangle matrices, so it does not produce the reduced matrix.

I second that. Gaussian or Gaussian-Jordan Elimination is the only way.

What's happening here?

Hyperbola have the general equation of [math]\frac{x^2}{a^2}- \frac{y^2}{b^2}=1[/math] OR [math]\frac{y^2}{b^2}- \frac {x^2}{a^2}=1[/math], while an ellispie have the general equation of [math]\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1[/math].

Yeah, sorry the domain should be [math](2,\infty)[/math]. Thanks Ragib.

Never underestimiate animals. You never know what they might do.

It's vital to have your life balanced with school and friends. I go to university during the week, and I go out and see my friends/girlfriend on the weekends. It has been going well for me so far.

You perhaps can assume that [math]15x+42[/math] is the equation for the length of line BD.

Completing the square is when you half the coefficient of the term x, and then add it to the both sides. i.e. [math]2x^2-4x+6=0[/math]. You factor it by 2. [math]2(x^2-2x+3)=0[/math] [math]2(x^2-2x)=-3[/math] You half the second term by -2 and add it to both sides. [math]2(x^2-2x+1)=-3+1[/math] So you get: [math]2(x-1)^2+2[/math]. That's the answer. I have checked it, and it's right.

The OP in the first post said "average intelligence", while Einstein is way above average intelligence.

I recommend that student to get courses in college or university.

Upon graphing those two functions together, I don't see any loop?

With for example [math]\frac{1}{\sqrt{x-2}}[/math], you have to be careful. You can't have a negative number in the square root, so the domain is [math][2,\infty][/math]. Also it will have an asymptote at y=0. [math]lim_{x \rightarrow \infty}f(x)= 0[/math].

Median in another word, is midpoint. In the first diagram, the line BD touches the midpoint of line AC.

Computers always use algorithms to generate random numbers. I emphasize that you understand the definition of an algorithm!!

If the scientists continue with the clonings, they will try to make a "perfect world", whereas disability associations/organizations wouldn't be welcome anymore. Everyone who is perfect will compete to get the control of the world, because there is nothing else wrong to handle with. I'm deaf and blind myself, and the world wouldn't be an exciting place without disability people. I know I'm deaf and blind but I am still taking Mechanical Engineering at University of New Brunswick. There's nothing that can stop me, and cloning "perfect" people is insulting to the disability community all over the world. What do you guys think?

Integration can also find you the "original" function. For example: [math]f(x)=x^2[/math] is the orginial function. First derivative is: [math]2x[/math]. Suppose you have the function of [math]2x[/math] and you want the integration of it, you use the Integration Rule which is [math]\frac{x^{n+1}}{n+1}[/math]. So you get: [math]\frac{2x^{2}}{2}[/math]. The "2" will cancel each other so you have the original function: [math]x^2[/math]. Integration is just a kind of math that gives you the "orginial" function of the function that you have in firsthand.