a.) prove that the sequence is geometric.\n\nb)  Find U1 and r.\n\nc)  Find the 16th term to 3 significant figures.
119570
This seems not a geometric sequence:\n\nU(1) = 6*(1/2)^1-1 = 2.0\nU(2) = 6*(1/2)^2-1 = 0.5\nU(3) = 6*(1/2)^3-1 = -0.25\n\nU(3) / U(2) is not the same as U(2) / U(1)\n\nDo you mean geometric sequence or geometric series?\n\n------- more note ---------\n\nThe answer by another user below indicated that the formula was: U(n) = 6*(1/2)^(n-1).\n\nSomeone needs to learn more about math operators and their precedence :-).
446734
1.\nn | n + 20\n------------\n0 | 20\n1 | 21\n2 | 22\n3 | 23\n4 | 24\n\nn | 3n - 35\n-------------\n1 | -32 \n2 | -29 \n3 | -26 \n4 | -23
559127
1. yes.\n\n2. one\n\n3. 2 equal segments (this one's a theorem)
625932
35\n\nThe sequence is every even-indexed Fibbonacci number plus 1.  So from the original sequence whose value is the sum of the two previous numbers:\n\n0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...\n\nYou take every second number and add 1:\n\n2, 3, 6, 14, 35, 90, 234, ...
418905
HOMEWORK\nTo find f, just collapse A. To find g, make sure the image is inside of A.\nSeparability is easy: in fact, there is a dense *point*. To answer about pseudometrizability, figure out what balls would have to look like.\n\nYou really should do your own work for math at this level. It is no help  to your understanding if you just ask others. If you really need help, ask your professor.
510608
Wow...OK, I'll take a shot at them.\n\n1)  Call the ages E, C and D.  We're told E = 4C, D = 2C and E + D + C = 140.  Substituting the first two equations into the third gives 4C + 2C + C = 140, or 7C = 140.  Therefore C = 20.  Plugging this back into the first two equations gives E = 80 and D = 40.  Eduardo is 80, Don is 40 and Clayton is 20.\n\n2) This uses the Pythagorean Theorem.  The height is 18 feet and the width is 7.5, so the diagonal is the square root of (18^2 + 7.5^2).  That works out to 19.5 feet.\n\n3) Slope-intercept form is y = mx + b.  We're given m = 5.  When x = 0, y = -1, so b is -1.\n\n4) -6a * (12a + 4) = -6a * 12a + -6a * 4 = -72a^2 - 24a.\n\n5) (3x + 4y) + (-7y + 16x) = 3x + 4y -7y + 16x = 19x - 3y.\n\n6) -2z - 4 > 20.  -2z > 24.  -z > 12.  z < 12.\n\n7. Zadie's graph would start out at (0, 10) and rise with a slope of 10.  Doug's starts at (0, 75) and rises with a slope of 5.  So his graph starts out higher and rises less steeply.\n\n8) 34x + 15y = 51.  15y = 51 - 34x.  y = 51/15 - 34x/15.  y = -(2 4/15)x + 3 2/5.\n\n9) The next step is to group like terms by adding 20 + x to each side, giving you 28 = -7x, or by adding 8x and subtracting 8, giving 7x = -28.  They're equivalent.
394112
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
383443
26. The pattern has to start 1, 2, 4, 8, 15. I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3. The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.\n\nAs for why this makes sense: Consider the question in two dimensions, i.e. into how many parts will many random lines diivide a plane?  That pattern starts 1, 2, 4, 7, 11 - the same numbers on the differences line above.  If you want to get really abstruse, I would submit that the number parts random spaces divide the fourth dimension is 1, 2, 4, 8, 16, 31, 57.  :)\n\n* * * * * *\n\nHey, you really do ask an interesting question, but it appears I'm the only one willing to answer it.  (Yes, I see you asked the same question a few days ago.)  OK, then, let's look at it more systematically:\n\n\nOne-dimensional question: Into how many parts will random points divide a line?\n\nEasy question, obviously:  Zero points leave one part (i.e. the whole line); one point breaks it into two parts, two points into three parts, etc.  The sequence is 1, 2, 3, 4, 5, etc.  Generally, n random points divide a line into n + 1 parts. \n\n\nTwo-dimensional question: Into how many parts will random lines divide a plane?\n\nHarder question, but you can work this one out on paper.  Zero lines leave one part (the whole plane); one line breaks it in two; two lines breaks it in four.  Then it gets tricky - the third line can't go thru all four parts, but it can go through three of them, so you end up with 7 parts.  The fourth line can go through 4 of the 7 parts.  Now, how you'd prove this rigorously I don't know, but the sequence is 1, 2, 4, 7, 11, 16, etc. and is clearly derived from the one-dimensional sequence: start with 1, then add 1 to get 2, then add 2 to get 4, then add 3 to get 7, then add 4 to get 11, etc.  To put this into a formula, n random lines divide a plane into (n^2 + n + 2)/2 parts.        \n\n\nNow for the three-dimensional question: Into how many parts will random planes divide a three-dimensional space?\n\nTo visualize this one, look at a corner of your room.  Obviously, zero planes leave one part (the whole space), one plane  breaks it in two (visualize the floor of your room as the plane, for example), two planes breaks it in four (visualize the floor plus one wall), and three planes breaks it in eight (visualize the floor plus two adjoining walls).  Now, try to imagine the fourth plane - it can't go through all eight parts, but it can go through seven of them, making a total of 15 parts.  The three-dimensional  sequence starts 1, 2, 4, 8, 15, and again we see it is derived from the sequence for the previous dimension: start with 1, then add 1 to get 2, then add 2 to get 4, then add 4 to get 8, then add 7 to get 15, etc.  To get the number for 5 planes, add 11 to get 26.  To put this into a formula, n random planes divide space into (n^3 + 5n + 6)/6 parts.   \n\nBased on what we are seeing, what the heck - why not carry it forward to the fourth dimension.  Using the numbers from the three-dimensional sequence, the sequence becomes 1, 2, 4, 8, 16, 31, 57, etc.  We can conclude, for example, that 6 random three-dimensional spaces divide the fourth dimension in 57 parts.  Now that should really impress your instructor!
271465
thats a geometric sequence with a common ratio of -1... you can get the common ratio by getting the first set of differences... thats 10-12... 12-11.. 11-15.. 15-12... then have the first set of difference: -2 1 -4 3.. then get the second set of difference... -2-1.. 1--4... -4-3... and u get -1 -1.. thats how u get to chek if its a geometric sequence by checking if they have the same second diff... chek the net and search for the formula for solving geometric sequences cuz i forgot bout those already.. goodluck!
502547
1.4u+ut-6t=-13\nu(4+t)-6t=-13\nu(4+t)=-13+6t\nu=(-13+6t)/(4+t)\n\n2.c(a+3)+(c+7)=5a\nc(a+3+1)=5a-7\nc(a+4)=5a-7\nc=(5a-7)/(a+4)\n\n3.you haven't completed the question.solve interms of what?\nif it is w\n4uw-3w=v(v-1)\nw(4u-3)=v(v-1)\nw=v(v-1)/4u-3)
612255
1113213211\n\nAfter starting the sequence with 1, each term in the sequence consists of groups of two numbers based on the previous term - the first being the quantity and the second specifying which digit.
6373
Geometric -- multiplying the previous term by 4
229292
a. (x+9)(x-9)\nb.(t-2)(t+2)\nc. (b-12)(b+12)
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1. B\n2. C\n5. C\n11. C\n17. A\n\nwe're studying this now too!!\nbut these were all I had time for!!\ngood luck on your exam!!
228322
U1=4-2*1=4-2=2\nU2=4-2*2=4-4=0\nU3=4-2*3=4-6=-2\nU4=4-2*4=4-8=-4\nU5=4-2*5=4-10=-6\n\nSo, the first five terms are 2,0,-2,-4,-6.\n\nb)There are a few ways of writing this one, but you can simply see that every number is in the form 4n+3 where n goes from 1 to infinity\nso you have 7,11,15,19,23,27,...\nso nth term would be an=4n+3 or an=an-1 + 4\n\nor you can write nth term as 7+4n where n goes from zero (from zero because you must have 7 included) to infinity.
505572
(1) \n6\nΣ (5+3i) = \ni=1\n\n8\nΣ (5+3(i - 2)) (i from 2 later → i in the term must decrease by 2) =\ni=3\n\n8\nΣ (3i - 1))\ni=3\n\n\n\n(2) \n6\nΣ (i^3-3) (i from 1 earlier → i in the term must increase by 1) =\ni=1\n\n5\nΣ ((i+1)^3-3)\ni=0
26424
1. A(N)<2 for any N:\n\na) A(0) = sqrt(2) < 2\nb) induction: A(N)<2 => A(N+1) = sqrt(2)^A(N) < sqrt(2)^2 = 2\n\n2. The sequence A(N) increases:\n\na) 1<sqrt(2) => sqrt(2)^1<sqrt(2)^sqrt(2) => A(0)<A(1)\nb) induction: A(N)<A(N+1) => sqrt(2)^A(N)<sqrt(2)^A(N+1) => A(N+1)<A(N+2)\n\nFrom 1 and 2 it follows that A(N) HAS a limit X, with sqrt(2)<X<=2\n\n3. sqrt(2)^X=X:\nX = lim(N->inf.)A(N) = lim(N->inf.)A(N+1) = lim(N->inf.)sqrt(2)^A(N) = sqrt(2)^X\n\n4. X=2 or X=4:\nWe already know that 2 and 4 are solutions for sqrt(2)^X=X. Let's show that they are the ONLY solutions:\nThe second derivative of f(X)=sqrt(2)^X-X is f"(X) = ln(sqrt(2))^2 * sqrt(2)^X > 0 for any X. Then f is a strictly convex function, so f(X)=0 can have AT MOST two solutions. We already found two, so they must be the only ones.\n\n5. X=2:\nWe showed that X<=2 and (X=2 or X=4), you draw the conclusion.