Klaynos Posted January 31, 2006 Share Posted January 31, 2006 I got the answer of 1.11 Wasting your time? Yet ask a question where the answer can be found on google... okie doke... Not that I mind o coursef... For differntials with multiple terms treat each term seperately [math]\frac {d}{dx}(ax+bx) = \frac {d}{dx}ax+\frac {d}{dx}bx[/math] Link to comment Share on other sites More sharing options...
EvoN1020v Posted January 31, 2006 Author Share Posted January 31, 2006 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. Link to comment Share on other sites More sharing options...
EvoN1020v Posted January 31, 2006 Author Share Posted January 31, 2006 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 ? Link to comment Share on other sites More sharing options...
Klaynos Posted January 31, 2006 Share Posted January 31, 2006 Google?! You're not a mathematician lover' date=' 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. Happy to explain. A differntiation is a mathmatical operator. Applying it to differnt functions works in differnt ways, [math] n \cdot k \cdot x^{n-1} [/math] Is the correct form for that type of equation. if you've got some function: [math] d+ax+bx^2+cx^3 [/math] Apply your above rule to each of the individule bits and them add them together so you would get: [math] a+2bx+3cx^2 [/math] If you notice the d disapears because the differntial of a number is 0. Also: x0 = 1 This is always true nomatter what x is. Link to comment Share on other sites More sharing options...
EvoN1020v Posted February 1, 2006 Author Share Posted February 1, 2006 Thanks Klaynos. Can anyone answer my challenge question? (post #28) Link to comment Share on other sites More sharing options...
cosine Posted February 1, 2006 Share Posted February 1, 2006 Four different numbers a' date=' 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. [math](-1)^{-4} + (-3)^{-2} = \tfrac{10}{9}[/math] Link to comment Share on other sites More sharing options...
cosine Posted February 1, 2006 Share Posted February 1, 2006 Here is the challenge question: In the diagram' date=' 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 ? [ATTACH']1201[/ATTACH] well the radius is the hypoteneuse of a 12, 14 right triangle, so... radius is the square root or 340. ... ah too tired to analyze anymore... hopefully future mathematicians can take off where I left off one day...(or I'll continue when I wake up) Link to comment Share on other sites More sharing options...
The Thing Posted February 1, 2006 Share Posted February 1, 2006 Er...a wild guess: a=-1, b=-4, c=-3 and d=-2? EDIT: OMFG There's a second page...I'm an idiot. Link to comment Share on other sites More sharing options...
EvoN1020v Posted February 1, 2006 Author Share Posted February 1, 2006 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. Link to comment Share on other sites More sharing options...
cosine Posted February 1, 2006 Share Posted February 1, 2006 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' date=' 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. [/quote'] Be Careful! Many students think they have learned Advanced Calculus's topics before, but they have not developed them in the rigorous way Advanced calculus does. Advanced Calculus teaches very little "new material" from regular calculus sequences, but it is very important to understand analytical theory. Link to comment Share on other sites More sharing options...
EvoN1020v Posted February 1, 2006 Author Share Posted February 1, 2006 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. Link to comment Share on other sites More sharing options...
Klaynos Posted February 1, 2006 Share Posted February 1, 2006 The course I took last year described series expansions, complex numbers, linear equations and matricies as basic algebra and calculus in the sylabus Link to comment Share on other sites More sharing options...
The Thing Posted February 2, 2006 Share Posted February 2, 2006 That's the problem: I can't figure out the length of line XP (or SY). Can anybody help me here? SY or XP will not get you the side of the square. XP is the radius of the circle minus 14, so root 340 - 14. But what's more important than that is the length of VY. If you somehow got VY, the problem's solved. By the way, what grade level is this question? Link to comment Share on other sites More sharing options...
EvoN1020v Posted February 2, 2006 Author Share Posted February 2, 2006 SY or XP will not get you the side of the square. XP is the radius of the circle minus 14' date=' so root 340 - 14. But what's more important than that is the length of VY. If you somehow got VY, the problem's solved. By the way, what grade level is this question?[/quote'] This question hailed from an old Fermat Math Competition booklet (Grade 11). The question is rather tricky. You DO have to find the length of SY or XP, because then with it, you can find the area of the semi-circle. So you subract the area of the rectangle from the area of the semi-circle. Therefore, you will find the area of the square STUV. Link to comment Share on other sites More sharing options...
The Thing Posted February 2, 2006 Share Posted February 2, 2006 XP and SY are easily found. The area of the semicircle too. However, to get STUV you need to subtract not only the rectangle but the shape RTU and UVY, and how do you calculate their areas? Is it the last question in the booklet or one from the last section? Link to comment Share on other sites More sharing options...
EvoN1020v Posted February 2, 2006 Author Share Posted February 2, 2006 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?? Link to comment Share on other sites More sharing options...
The Thing Posted February 2, 2006 Share Posted February 2, 2006 Haha I figured out the answer right after I thought of how to explain that the radius equals 340 to you. So let's start with 340 then get to my solution. 340 is derived from the fact that if you draw a line from the center of the semicircle to the point R, you will get a right angle triangle. The center of the semicircle is also the center of side PS for the rectangle, thus you will get a right angle triangle with sides 14 and 12, and by Pythagoras the hypotenuse, or the radius of the circle, is root 340. Mind that the radius of the circle is the same from the center to any point, so from center to point X or Y is root 340, which is why I said it was easy to figure out XP and SY, because you they're then root 340 - 14. But let's ignore that because, like I said, it is not very important. So let's draw a line from the center of the semicircle to the circumference at point U. Now you have yourself another right-angled triangle, namely triangle O(center of the semicircle)UV. You know that the radius, or the hypotenuse of this triangle is root 340. Let's make the side of the square STUV x, and thus the remaining side of that triangle is 14+x. Now we have an equation: sqrt(340)=sqrt(x^2+(14+x)^2) Square both sides to get the quadratic equation 340=x^2+(14+x)^2 Solve. Link to comment Share on other sites More sharing options...
EvoN1020v Posted February 2, 2006 Author Share Posted February 2, 2006 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! Link to comment Share on other sites More sharing options...
EvoN1020v Posted February 2, 2006 Author Share Posted February 2, 2006 I'll have another math question for tomorrow. Link to comment Share on other sites More sharing options...
EvoN1020v Posted February 3, 2006 Author Share Posted February 3, 2006 [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 ? Link to comment Share on other sites More sharing options...
The Thing Posted February 3, 2006 Share Posted February 3, 2006 Aaaarg, lots of tedious work. The hardest part is wrapping my mind around this model. I simply can't visualize this in 3D space. Maybe it has to do with my speakers on full volume and very discordant music playing. Link to comment Share on other sites More sharing options...
EvoN1020v Posted February 3, 2006 Author Share Posted February 3, 2006 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. Link to comment Share on other sites More sharing options...
The Thing Posted February 4, 2006 Share Posted February 4, 2006 Well, while you were having fun doing the problem , I contented myself with about 6 different final exams from 4 different teachers. It's the end of the semester for me, and in about 2 days I have 2 provincial exams to write. So I didn't really think about the problem. But now, are you sure you drew the cube correctly? I can't visualize what this thing looks like. Is the hexagonal plane like a knife cut, as in, one flat plane (well duh a plane's flat)? Link to comment Share on other sites More sharing options...
EvoN1020v Posted February 4, 2006 Author Share Posted February 4, 2006 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. Link to comment Share on other sites More sharing options...
The Thing Posted February 4, 2006 Share Posted February 4, 2006 Ahahahahahaha. Mathematician, me? No I'm just a grade 10 that is not particulary bad at math. Okay so the big hexagon has the area of [math]3bcsinA,[/math], so [math]3*2\sqrt{2}*2\sqrt{2}*sin60[/math]. What is that calculated? Iunno. Okay, we have two large sides of the cube, so that's [math]2*4^2[/math]. The surface on the top of the cube plus the surface on the bottom adds to another side of the cube, so another [math]4^2[/math] Basically all sides missing a corner has a corresponding part on the opposite side that is a CORNER missing the REST of the side. So yeah, more of those. So are there a few triangles? Each has area of 2. Just add them all together? Tell me if I did something wrong - I can't think anymore. Too tired. Link to comment Share on other sites More sharing options...
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