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Hmmm, never thought someone would have a difficulty visualizing the diagram. (Especially if you are a mathematician yourself!!) I decided to colour the cube to maybe help you guys out abit, though I'm no professional at art. You should notice that the cutted plane is being tilted 45 degrees diagonally in the cube. I'll give you the first magnitude: [math]\sqrt{32}[/math]. It's the length of the middle line of the hexagon. ([math]\sqrt{4^{2} + 4^{2}} = \sqrt{32}[/math]). Now, try to solve the surface area. By the way, The Thing: Good luck on your exams, as I just completed mine last week.

Not neccessarily. I worked on the problem in school today for fun, and it's not really much of triangles to figure out. You just have to find the surface area for the side where the cut is been done at 45 degrees diagonally in the cube. First hint: You figure out the length of the hexagon, by an imaginary triangle as its hypotenuse touch along the middle line of the hexagon. If you don't understand, then I can clarify more later.

[ATTACH]1205[/ATTACH] A solid cube of side length 4 cm is cut into two pieces by a plane that passed through the midpoints of six edges, as shown. To the nearest square centimetre, the surface area of each half cube created is ?

I'll have another math question for tomorrow.

Wow very clever. I never thought of getting the radius of the circle ([math]\sqrt{340}[/math]). So the area of the semi-circle is [math]A_{circle} = \frac{r^{2}\pi}{2}[/math]. The answer is: [math]534.0707511[/math]. The area of the triangle is: [math]336[/math]. Subract the area of the rectangle from the semi-circle is an error I just realized. (There are all those small spaces left). ________________________________________________________ Omit above. ________________________________________________________ All I say is that I want to applause The Thing for providing his method of thinking. I had to actual get a pen and paper and draw it out and do the math. Good job The Thing. The quadratic equation is: [math]2x^{2} + 28x - 144[/math]. After using the quadratic formula, x yielded [math]-18[/math] and [math]4[/math]. Obviously, the 4 is the dimension for one of the side for the square STUV, therefore the area of the square is 16!! Whew!

Those areas doesn't really matter, because it's a part of the semi-circle. Yeah I think it's a question from the last section that worths 10 marks for the correct answer. Also, how did you get the 340??

The course is actually entitled, "Advanced Math & Calculus 120". The topics that will be teach are: Sequences, Functions, and Complex Numbers. Do you think those topics apply to calculus? I don't think very much so. Otherwise, I guess I could fine my math skills to prepare for mathematics in university in my own spare times (and on SFN of course), which I'm rather exciting for.

Use Google! I searched for those Rubik's Cube techniques before, and there's plenty of informations on how to solve the cube.

That's wierd.

Yeah, I'm curious about Dak's question. "Does the reciever need the plug-in as well, or just the sender?" Because of both sender and reciever need both of the plug-ins, that will be too much work, as both need to get gaim, then the plug-ins, and so on. So..?

That's the problem: I can't figure out the length of line XP (or SY). Can anybody help me here? Note: Today was my first day of second semester. One of my class was Advanced Math & Calculus 120, and after the 20 minutes of briefing, I was not very impressed by the teacher. I looked on all the units that are offered, and I already learnt those before?! I asked the teacher, if there will be calculus, and she said, "just a bit". I was like, oh come on! So I headed straight to my guidance counselor and he said that I can stay there and get good mark, or change my whole schedule. He offered that I see how it's going for a week, then I will go back to him again. So I should buy a Calculus book for myself from Chapters for self-learning experience. I found a really good book called "Barron's Easy Way to Learn Calculus". You guys ever heard of that book before? The price is $21.99 CAN, but I'm broke right now. Anyway, I know you guys will help me if I have any questions relate to Calculus.

Thanks Klaynos. Can anyone answer my challenge question? (post #28)

Should I download gaim to make the plug-in work? Do it require a resignation (username and password)?

You are not alone. I live in New Brunswick, Canada, and it's have been a frequent storyline in the daily newspaper. It is said to be 4 degrees warmer than normal around here. (I can notice it myself, as there is oddly not very much of snow).

Thanks a bunch!! It's a big lifesaver!! Now I can explain complex things clearly to my friends!! The file appears to be a .bz2 file. What program can I open it with?

Here is the challenge question: In the diagram, a semi-circle has diameter XY. Rectangle PQRS is inscribed in the semi-circle with PQ = 12 and QR = 28. Square STUV has T on RS, U on the semi-circle and V on XY. The area of STUV is ?

Google?! You're not a mathematician lover, are you? (Just kidding). Anyhow the answer of [math]1. \overline{1}[/math] is correct, or rather [math]\frac{10}{9}[/math]. This accounts to be figured out by [math](-1)^{-4} + (-3)^{-2}[/math]. Also, don't you mind explaining me the above calculus formula? I have never taken Calculus yet, well actually will start take it tomorrow. Show me how the provided cubic formula worked out to be the answer, please.

What is the derivative of [math]3x^{3} - 5x^{2} + 2x + 13[/math]? As I know that [math]k \cdot x^{n}[/math] is equal to [math]n \cdot k \cdot x^{n-1}[/math]. Therefore: [math](3 \cdot 3 \cdot x^{3-1})-(2 \cdot 5 \cdot x^{2-1})+(1 \cdot 5 \cdot x^{1-1}) + 13...?[/math] [math]\rightarrow 9x^{2} - 10x + ....?[/math] I have no idea how to get the final value? Because the answer said to be [math]9x^{2} - 10x + 2[/math].

Stop wasting my time! Do you have any finite answer to my question? If anybody got the same answer as I do, then there's a challenge question next.

Nope. A hint: It's greater than 1.

Four different numbers a, b, c, and d are chosen from the list -1, -2, -3, -4, and -5. The largest possible value for the expression [math]a^{b} + c^{d}[/math] is ? I already figured out the answer, but wanted to see what you guys got.

No. It was a question extracted from an old Fermat Math Competition booklet. I was skeptical at the requestion of the question because English is my second language. "then P, as a percent of Q", is rather confusing to me. I think it means, what is the answer for P, while Q is a percent too. By the way, I am in grade 12, so apparently I have been forgetting grade 10 math, so I'm doing this to refresh my memory

I never thought I could solve for 2 simultaneous equations that have 3 different variables, but it comes to that x is only 1. So I used substution method to discover the answer, not the elimination that Klaynos recommended. I got an answer for t= 0.4. This is the answer for Q. I multipled 0.5 to the answer, and it yielded 0.2 for the value of P. Am I correct?

What is "simultaniouse equations"?