Challenging decreasing Geometric Sequence?
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The common ratio is simply 1.5.\n\nf(t) = 6 * (1.5)^t\n\nDay 0: 6 pellets\nDay 1: 9 pellets\nDay 2: 13.5 pellets\nDay 3: 20.25pellets\nDay 4: 30.375 pellets\nDay 5: 45.5625 pellets\nDay 6: 68.34375 pellets\n\nIf you add it up (not counting day 0), I get that she would eat through a total of 187.03125 pellets by the end of day 6, so she runs out before the end of day 6.
844185
square-ish in shape\ncome in different sizes\nmany varieties
661969
Adjust until the diagonals are equal.
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Start here:\nhttp://www.math.pitt.edu/~sussmanm/2071Spring05/lab07/lab07.html#TheGramSchmidtMethod
286004
What is the relationship between the items in the sequence? If you are adding the same thing over and over, it is arithmetic (2,4,6,8,10). If you are multiplying by the same thing over and over, it is geometric (2,4,8,16,32).
79896
Like in pythagorean therom? Hypotenuse
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You can construct one simply by interleaving a bounded sequence with an unbounded sequence. For example, the sequence 1/n with the sequence n to form \n\nS(i) = if i is odd 1/i if i is even i. \n\nThis has 1/i as a convergent subsequence.
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1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
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Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.
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Yes, it is possible.\n\nDenote the number of black dots belonging to a red dot by N.\n\nThe present problem is formulated for N=10.\n\nLet us first solve a simpler problem with N=2.\n\nStep 1. \nTake a square, that is a four-sided regular polygon, with the side equal to R. Put the two red dots at the opposite [i.e., connected by a diagonal] vertices of the square. Put the two black dots at the remaining two vertices. If you draw a circle of radius R around each of the two red dots, you will find exactly 2 black dots lying precisely on the circle line. The problem is almost solved, but the number of red dots (2) is equal to the number of black dots (2).\n\nStep 2. \nTake another square and repeat the above procedure. \nNow we have the two square compounds. In total we have 4 red dots and 4 black dots.\n\nStep 3. \nPut the 5th red dot anywhere in the plane and draw a circle of radius R around it. \nDrag each of the above compounds as a whole closer to this circle. \nPlace one of the black dots belonging to the first compound and one of the black dots belonging to the second compound on the circle line surrounding the 5th red dot.\n\nWhile doing so, make sure that you move and rotate each compound as a whole, keeping all the sides of the squares equal to R. \n\nThe problem with N=2 has been solved: in total we have 5 red dots and 4 black dots satisfying the condition of the problem, 5>4. \n\nLet us solve the same problem for N=3.\n\nIn this case, one has to extend the Step 1.\n\nStep 1a. \nHaving constructed the square compound according to the Step 1 above, draw parallel vectors of length R starting at the vertices of the square. Make sure that the vectors do NOT coincide with the sides of the square. \nIf the vector starts at the red vertex, put the black dot at the end of the vector. If the vector starts at the black vertex, put the red dot at the end of the vector. \n\nNow we have one compound consisting of the two shifted squares containing 4 red dots and 4 black dots. If you draw a circle of R around each of the four red dots, you will find exactly 3 black dots lying precisely on the circle line.\n\nStep 2. Repeat the procedure 1+1a.\nStep 2a. Repeat the procedure 1+1a.\n\nNow that we have 3 compounds with equal numbers of red and black dots (in total 12 red dots and 12 black dots), put the 13th red dot anywhere in the plane and follow the instructions from Step 3 above.\n\nThe problem with N=3 has been solved: in total we have 13 red dots and 12 black dots satisfying the condition of the problem, 13>12. \n\nFor N=4, take the compound constructed for N=3 at Step 1a and translate ALL 8 vertices by a vector of length R.\nAgain, if the vector starts at the red vertex, put the black dot at the end of the vector and vice versa. You will have 8 red dots and 8 black dots, etc.\n\nThe above procedure applies to an arbitrary N, in particular to N=10.\n\n\nAn interesting question is: what is the minimal number of dots needed to solve the problem for a given N.
17334
Geometric -- multiplying the previous term by 4
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The solution set is (-infinity, -7) U (-3/2, infinity)
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the formula for the sum of a geometric series depends whether sequence is finite or infinite.\nFor finite sequence formula is:\n   \n    Sn = (1st term(1-r^n))/(1-r)\n\nFor example, for the sequence 1,2,4,8,16...\n   If we wanted to add the first 10 terms of sequence, we would know that a1 = 1, r = 2 and n = 10. \n     Sn = [1st term(1-r^n)]/(1-r) \n    S10 = [1(1-2^10)]/(1-2)\n    S10 = (1-1024)/-1\n    S10 = -1023/-1\n  =>S10 = 1023\n\nFor infinite sequence,formula is:\n      Sum(of an infinite serise) = 1st term/(1-r)\nSum of an infinite serise only exist if the ratio of that sequence lies between -1 and 1. Since the above sequence has a ratio=2 so its sum for infinite sequence does not exist.\n now lets see a geo sequence 16,-8,4,-2,1... \nhere r = -1/2 which is lies between -1 and 1 so its infinite sequence sum exists. Its sum will be:\n    Sum = 1st term/(1-r)\n    Sum = 16/(1-(-1/2))\n    Sum = 16/(1+(1/2))\n    Sum = 16/(3/2)\n    Sum = 32/3\n or Sum = 10.666666667
683735
The path of least resistance?
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I recommend Unit 5 of this course:\nhttp://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/\n\n(Construction of Bézier Curves, De Casteljau's Algorithm, Subdividing a Bézier Curve)\n\nand the software "DesignMentor V2: Curve":\nhttp://www.cs.mtu.edu/~shene/NSF-2/DM2-BETA/index.html
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I can remember upto 4 to 5 ways of doing it and not more than that.  btw you are talking about halving a square (diagram)! Right ?
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Arithmetic sequence:  1, 2, 3, 4, 5, 6, 7\nGeometric sequence:  1, 2, 4, 8, 16, 32, 64