Spart

The ambitious project, Mars One (as I expect most of you know about) lead by the founder Bas Lansdorp plans to send humans to colonize Mars in the year 2023. It's a one-way ticket for them as the technology for a return vehicle isn't available yet. Even though I always dreamed of visiting our neighboring planet, I would have to take in consideration that I will never see my family and friends ever again. It's a daunting decision for me but a recent survey collected about 80'000 applicants that would be interested to permanently stay and explore on the unforgiving surface of Mars.

I can imagine that an intelligent specie would be less biological and more mechanical, otherwise fully robotic. If an alien specie would need to travel to or live on other worlds, it would be more efficient for them adapting to the extreme/hazardous environments rather than altering them (eg. outer space, mars).

Desperate people do surgery just to get 3 inches taller. What they do is they chainsaw their bones in their legs and extend them with a metal rod. Many people who did this couldn't walk for the rest of their lives.

Maybe I should have used the concept of a zip-line than a rail track where the moon and the cable don't have to be perpendicular or directly above the train. It's stupid nonetheless but at least it's great for a sci-fi novel.

If that's the case then the Earth's rotation would also slow down and the tilt would slowly be straightened up. However, you should think about the energy which is associated in the Earth's tidal force. I'd imagine it being quite large and I'm surprised that the Moon is drifting away from Earth. I would be more worried about how the cable would be brought up to the moon since it's orbiting around Earth at an average of 1,022km/s.

To help you find the area of a triangle given three vertices you will need to use this formula: [latex]area = \left | \frac{x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{1} \right )+x_{3}\left ( y_{1}-y_{2} \right )}{2} \right |[/latex] Since I'm nice, I wrote this line that you will need to use and remember to return the area if you are using a method. double area = Math.abs((x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))*0.5);

How about we just cover the Earth's equator with a friggen long railway track, plant a train on it, attached on an indestructible cable which is hooked up on the moon? Do I need to go any further than this?

I haven't heard of it as far as I know. Am I obliged to use it or is there another way?

I have an issue solving this limit: [latex]\lim_{x\to 0}\frac{\; ln\left (1+3x \right )\; }{x}[/latex] The answer is 3, however I'm not sure how it's obtained. I know that [latex]\lim_{x\to 0}\frac{\; ln\left (1+x \right )\; }{x} = 1[/latex] but I can't quite figure out how to implement this property on the problem above. Could someone please give me a step-by-step explanation on how it should be solved? That would be great!

I'd suggest GameMaker by Mark Overmars. It has a very friendly and interactive drag and drop interface. It is based on object orientated programming but you could also use its scripting language if you want to be more advanced. It also has an inbuilt sprite editor and a level editor which are all essential in making great 2D games.

I'd say that objects can accelerate towards the speed of light without ever reaching it.

Just stunning. What language did you write this in?

[latex]\lim_{x\to+\infty }\frac{1}{\; \sqrt{x+1}-\sqrt{x+7}\; } \times \frac{\sqrt{x+1}+\sqrt{x+7}}{\sqrt{x+1}+\sqrt{x+7}}[/latex] [latex]\lim_{x\to+\infty }\frac{\sqrt{x+1}+\sqrt{x+7}}{\left ( \sqrt{x+1}-\sqrt{x+7} \right )\left ( \sqrt{x+1}+\sqrt{x+7} \right )}[/latex] [latex]\lim_{x\to+\infty }\frac{\sqrt{x+1}+\sqrt{x+7}}{\sqrt{x+1}^{\: 2}-\sqrt{x+7}^{\: 2}}[/latex] [latex]\lim_{x\to+\infty }\frac{\sqrt{x+1}+\sqrt{x+7}}{\left ( x+1 \right )-\left ( x+7 \right )}[/latex] [latex]\lim_{x\to+\infty }\frac{\sqrt{x+1}+\sqrt{x+7}}{-6} =\frac{+\infty +\infty }{-6}=-\infty[/latex] Simple! Thank you!

Hello, I'm having a bit of difficulty calculating this limit. [latex]\lim_{x\to+\infty }\frac{1}{\; \sqrt{x+1}-\sqrt{x+7}\; }[/latex] [latex]\lim_{x\to+\infty }\frac{1}{\; \sqrt{+\infty +1}-\sqrt{+\infty +7}\; }[/latex] [latex]\lim_{x\to+\infty }\frac{1}{\; \sqrt{+\infty}-\sqrt{+\infty}\; }[/latex] [latex]\lim_{x\to+\infty }\frac{1}{\;+\infty-\infty \; }[/latex] Well this is where I'm stuck. Naturally the square root of infinity will always remain infinity however it results in an indeterminate form since positive infinity cannot be summed with negative infinity. According to my text book the answer should be [latex]-\infty[/latex] but I have no idea how to work it out.