EvoN1020v Posted February 4, 2006 Author Posted February 4, 2006 I'm looking for more participation from other SFN members than just you. The surface on the top of the cube plus the surface on the bottom adds to another side of the cube' date=' so another [math']4^2[/math] You only want to figure out the surface area for a half piece of the cutted cube. I found the center line of the hexagon to be [math]\sqrt{32}[/math], and all the edges to be [math]\sqrt{8}[/math], from using the pythagroem theorem: [math]\sqrt{2^{2} + 2^{2}} = \sqrt{8}[/math]. (But I don't think this measure is really necessary to know). I figured out the surface area of the hexagon to be 14.627417. Plus the 3 surfaces with 14 units. (They are not 16 units anymore because the cut caused 3 small areas to be with the other half). 14.627417 + 14 + 14 + 14 = 56.627417. The offered multiple choices are: (A) 69 (B) 48 © 32 (D) 65 (E) 58 So obviously my answer is the closest to (E). Do you guys agree with my answer or not?
The Thing Posted February 4, 2006 Posted February 4, 2006 That's funny - I got 69... Lemme go through my train of thoughts again. The area of the hexagon is calculated first. How? Well, there are 6 equilaterial triangles in the hexagon, each with side length of 2 root 2 (root 8). Their areas can be found with the formula [math]\frac{1}{2}bcsinA[/math], so multiply that by 6, and we have [math]3*2\sqrt{2}*2\sqrt{2}*sin60[/math]. That's about 20.7846. We have 1 complete face for the cube, namely the face of the cube that is facing you. That's a complete face - the plane doesn't cut through it, so [math]4^2[/math] The right face of the cube is missing a corner (the small triangle cut by the plane). The left face IS a corner! So we can put those together to get another complete face. Again, the top face is missing a corner, while the bottom face is a corner. Add those two together. So overall, we have another [math]2*4^2[/math] Adding them all together yields 68.7846.
EvoN1020v Posted February 4, 2006 Author Posted February 4, 2006 I couldn't find the hexagon formula on the Internet before I saw it on your last post, and I used it and calculated the area of the hexagon. I got [math]20.78460969[/math] too. However, I don't agree with your addition of the faces. Let's look at a different perspective: A normal cube have 8 corners right? Well, when it's cutted, each piece should have 4 corners. It's sure is hard to imagine the cube being already cut. It would definetly be easier if I had a real cube model and cut it to see it more clearly.
The Thing Posted February 4, 2006 Posted February 4, 2006 Let's look at a different perspective: A normal cube have 8 corners right? Well' date=' when it's cutted, each piece should have 4 corners.[/quote'] I'm not sure if I understood you correctly. But you're saying that a normal cube has 8 vertices and a cut cube has 4 therefore? I'm sorry, but that makes no sense. By your logic, if you have a desk with 4 corners, slice off each corner, you have how many vertices now? 2? That's impossible. No, its 8. Try it. If that's not what you mean, sorry. Could you elaborate? Also, try shading in each face. That helps.
EvoN1020v Posted February 5, 2006 Author Posted February 5, 2006 What I meant was that when you cut the cube according to the lines, you won't cut the verticles. Therefore it leaves a small triangle, which should be 4 corners for each piece. That's my theory.
The Thing Posted February 5, 2006 Posted February 5, 2006 Sooo...can you find the answer to the question? As in, the answer given by whoever created the question. Next question please!
EvoN1020v Posted February 5, 2006 Author Posted February 5, 2006 My answer still remains at 58. I counted the right, left, and top to have 12 surface area units. Plus, the alone triangle at the bottom which is just 2. So the total surface areas of the faces is 38. 38 + 20.784060969 is 58.78460969. Those questions that I got was created by Waterloo University in Ontario so I don't have the answer for the questions. I have another question, but much easier to give you guys a break. You should answer it with no problem. The weight of a lioness is six times the weight of her female cub and four times the weight of her male cub. If the difference between the weights of the male and female cub is 14 kg, the weight of the lioness, in kg, is ?
The Thing Posted February 5, 2006 Posted February 5, 2006 168 kg. Come on, I'm sure you have harder questions . And I'm 99.999% sure that my solution for the cube is right (69 for the surface area, choice A).
EvoN1020v Posted February 5, 2006 Author Posted February 5, 2006 That's why we need more inputs from the other SFN members to see who favour in my answer or yours, but nobody seems to participate except The Thing. Anyway, the answer of 168 kg is incorrect. Did I scare there for a second? Ha, no, it is actually correct, and we both know that. Next question: The lines [math]y = -2x + 8[/math] and [math]y = \frac{1}{2}x - 2[/math] meet at (4,0). The area of the triangle formed by these two lines and the line [math]x = -2[/math] is ?
Connor Posted February 5, 2006 Posted February 5, 2006 45 units squared I'll work on the cube problem and get back to you
EvoN1020v Posted February 5, 2006 Author Posted February 5, 2006 Good job Connor. I'd appreciate your opinion on the cube question. I'll post a new question tomorrow.
EvoN1020v Posted February 5, 2006 Author Posted February 5, 2006 [ATTACH]1208[/ATTACH] Rectangle ABCD is made up of six squares. (Notice: my drawing is not good, as you can see ). The areas of two of the squares are shown. The perimeter of rectangle ABCD, in centimetres, is ? This problem shouldn't be a hassle for you guys.
cosine Posted February 5, 2006 Posted February 5, 2006 48cm first you can say the edge of the 36cm^2 square is 6, and the edge of the 25cm^2 square is 5. So that middle square is 1x1. So the top right square has edges of 7. So the width of the rectangle is 7+6 = 13, and from before the height is 6+5 = 11. So 13+13+11+11=48
The Thing Posted February 6, 2006 Posted February 6, 2006 I demand harder questions! No seriously, good job cosine.
EvoN1020v Posted February 6, 2006 Author Posted February 6, 2006 It's nice to see other person's solution. You did it differently than I did it, as I used algerba to get my answer. For the 2 medium squares, I put in "y" for the sides, and "x" for the single big square. You notice that the left side's total length is 11 cm. Thus [math]x + y = 11[/math] for the opposite side. I changed it to [math]y = 11 - x[/math]. Now for the equality of both top and bottom length I got [math]5 + 2y = 6 + x[/math]. Hence: [math]5 + 2(11-x) = 6 + x[/math] [math]5 + 22 - 2x = 6 + x[/math] [math]27 = 6 + 3x[/math] [math]x = 7[/math] Therefore for the length of y:[math] y= 11 - (7)[/math] is 4. [math]7 + 6 + 11 + 5 + 8 + 4 + 7 = 48 cm[/math].
EvoN1020v Posted February 6, 2006 Author Posted February 6, 2006 I am starting to running out of questions on the Fermat booklet that are challenged enough for you guys.. Next Question: The arithmetic sequence, a, a+2d, a+3d,...,a+(n-1)d has the following properties: -When the first, third, and fifth, and so on terms are added, up to and including the last term, the sum is 320. -When the first, fourth, seventh, and so on, terms are added, up to and including the last term, the sum is 224. What is the sum of the whole sequence?
The Thing Posted February 7, 2006 Posted February 7, 2006 I'm inclined to think that I need to come up with two equations for each of the statements you gave me and then cancel out some variables leaving only two. Either that or I'm totally on the wrong track. I have an exam in a few hours. Must get ready. I'll work on this problem when I get back.
EvoN1020v Posted February 7, 2006 Author Posted February 7, 2006 The answer for [math]\sum_{n=4}^{10}(3n-3) - \sum_{n=5}^{10}(3n-3)[/math] is [math]9[/math]. Is this correct?
EvoN1020v Posted February 7, 2006 Author Posted February 7, 2006 Anyone who can help me here? What is the difference between [math]\sum_{n=1}^{\infty}[/math] and [math]\sum_{n=1}^{n}[/math]? My teacher said that the "n" means any number, and the infinity means forever. What's the difference?
Klaynos Posted February 7, 2006 Posted February 7, 2006 n implies a finite number of sumations where as the infinity symbol implies just carrying on forever...
EvoN1020v Posted February 8, 2006 Author Posted February 8, 2006 Now, I'm sure nobody want a mathematician thread anymore?
Klaynos Posted February 8, 2006 Posted February 8, 2006 Can anybody answer post #69? Your answer was correct, n=5,6,...10 all cancel out, leaving just n=4.
EvoN1020v Posted February 10, 2006 Author Posted February 10, 2006 Today, I rather got in an arguement with my math teacher. We did geometric sequence review today, where she gave us this formula: [math]t_{n}=ar^{n-1}[/math] where a = first term, and n = any number. The question was: How long will it take for the population of the town to double? (Note: the first term is 26,235 and the rate is .09% which is 1.009 to signify growth.) My answer was: [math]52,470 = 26,235(1.009)^{n}[/math] [math]\log_{1.009}{2}=n[/math] [math]n=77.36240945[/math] or rather 77. My teacher answer was: [math]52,470 = 26,235(1.009)^{n-1}[/math] [math]n = 78[/math] (Note: Her answer was rounded up form 78.362...) I agrued that her answer was wrong, because I checked the graphic calcular, and I input my answer of 77.36240945, and it yielded 52,470. So who do you think is correct? Because [math]t_{n}=ar^{n-1}[/math] is rather a geometric sequence formula, not a real life formula like mine. Let me know.
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