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  2. Let me see if i can ask my question better. Suppose you have twelve rods of two different sizes. One size is 4 in. The other is 8 in. There are 8 - 4in. Rods and 4 - 8 in rods. Use the 4 in rods to construct two squares. Stand your squares up perpendicular to your table and parallel to each other. Now take one 8 in rod and connect it to the top left of one square (first square) and bottom left of the other. Take your second rod, connect it to the top right of the first square and top left of the other. Third rod, bottom right of first and top right of second. Connect the last rod to the remaining vertices. All rods will remain straight through this process. Find the volume and compare it to the volume of 4x4x8
  3. Today
  4. I tried to write an essay and now I am taking some knowledge from , I can make it successful but I am in need of help to complete it. So there is anyone who know well how to write an good essay. please knock me, Thanks in advance
  5. It depends on what is constant and what can vary. Do the edges all keep their original lengths? Does the cross section remain everywhere square? By "connecting 8" edges" do you mean deforming the shape and "gluing" the square faces to each other, a to b, b to c, c to d and d to a so that the new shape has no vertices (corners)? Do the 8" edges end up as circular arcs? What keeps us from doing this and then squashing the final shape flat so that its volume is zero? I can't visualize a process that gives a final shape having a definitive volume.
  6. If we add 2 matrices from the answer of the puzzle, we will get interesting result, all cells have same result (number 10) Is There any logic explanation of this ? 3 6 4 7 4 6 10 10 10 8 9 2 + 2 1 8 = 10 10 10 1 5 7 9 5 3 10 10 10
  7. Ok, sorry, I 'm wrong, I miss 2 constraints to set the rule..
  8. Yesterday
  9. I concur: There are no solutions.
  10. In the one-elevator case, we can reasonably assume that the elevator is equally likely to be at any point between floor 1 and floor 15 at any point in time. We can also assume that the probability that the elevator is exactly on the 13th floor when Smith arrives is negligible. This gives the probability 2/14 = 1/7 0.1429 that it is above floor 13 (which is when it will go down when it goes by this floor) when Smith wants to go home. Let’s have n elevators now. Call the unbiased portion the part of the elevators route up from floor 9 to the top and then down to floor 13. Any elevator at a random spot of the unbiased portion is equally likely to go up or down when it goes by the 13th floor. Moreover, if there is at least one elevator in the unbiased portion, all elevators out of it do not matter. However, if no elevator is in the unbiased portion, then the first one to reach the 13th floor goes up. Therefore the probability that the first elevator to stop at 13th floor goes down equals 1 2 (1 − (10/14)n). (For n = 2 it equals approximately 0.2449.)
  11. The square faces are the front and back. If you think of the congruent squares each having vertices abcd in the same clockwise rotation each resting parallel to the other. Then connect 8 inch edges in the following manner, front A to back B, front B to back C, front C to back D, and front D to back A. How does this volume compare to a standard rectangular prism of 4x4x8?
  12. Hmm...trying to figure out where I've gone wrong...
  13. Ah, I missed a few of the diagonals (I accounted for the center square but not all):
  14. Some of your solutions is wrong, "adjacent" also means adjacent diagonally. Re filter your answers. Reflection works vertically
  15. Last week
  16. "the middle-middle square is adjacent to all other square" This implies that the restriction applies to diagonally adjacent squares. There are only two solutions, along with their vertical reflections, that qualify in this case.
  17. Hi Buddyboy, I've changed clue 106 to read like this: "Do-wop and grunge music are listened to in the fifth and sixth houses, but not necessarily in that respective order." Thus, I'm referring to #9 and 11 or #10 and 12 (E-W). Clue 157 specifically states "directly opposite house #5", not the "fifth" house! ["Opposite" meaning across the road or N-S] Regards, C
  18. One question did come up to me, because something does not add up. 106 says do-wop and grunge are listened to in the fifth or sixth house. 157 says that the house opposite of house 5 listens to metal, which would be the sixth house. These say that the sixth house listens to two types of music.
  19. Hey guys try this game! Quick Sort V1 Link GooglePlay: Link appStore: Do you get more than 3000 points in easy mode? I challenge you =)
  20. See attached suggestion. HINT: Determine all known "odds" and "evens" (either stated or linked, but all of them can certainly be determined). Once you have 8 odds and 8 evens in each category, you ought to know the remaining 8 odds and 8 evens for each. The Zodiac signs, incidentally, will sort themselves out! I'm sure this will help! Grid1.xlsx
  21. There are 2 solutions, if we rule out Reflection
  22. Are we looking at the back and front 4x4 faces from the same direction? Or, does left and right, as they refer to the back face, assume we have turned the shape around so we're looking at outside of the back 4x4 face? That is, if we label the vertices ABCD on the front face, clockwise starting from the upper left corner like this A B D C Then if the 8" edges connect these to the back face vertices EFGH as A-E, B-F, C-G, D-H would your connections be A to H, D to G, and so on? It sounds as if we are joining the front and back faces of a prism to form a torus that has a square cross section, only we're twisting the shape by 90 degrees before joining the faces. (Kind of like constructing a Mobius strip from a piece of paper. This new shape would have only three sides, just as a Mobius strip has only one side. Interesting.) I wonder, though, with the length (8) being only twice the side (4) whether this is even possible. One thought:
  23. I noted the number(s) of the clue(s) that refer to each of the items. This is making it easier to piece together little clusters of relationships. Next I'm trying to fit these clusters into a table where one column is filled in (e.g. house numbers). Or probably into several tables, each with a different column filled in, depending on the "shape" of the clusters. It's a huge puzzle any way you look at it.
  24. The way I learned to do grid puzzles was to make a grid with every variable by every variable. The two scroll bars were for the width and height. There seems to be another way to do this to make it smaller. If any of you could give some insight on how to do this correctly, it would be great.
  25. And that is the solution for the fewest number of sides, as described by Buddyboy3000
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