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what is 4/17 of 15/16 of 19.63 acres?

i have an interest in 19.63 acres. My interest is 14/17 of 15/16. What is the percentage of my interest?

Research, Knowledge and Information :


17.63 Acres - Nature Lovers Retreat : Land for Sale: Orosi ...


Land for Sale with 17.63 acres in Orosi, Costa Rica at LANDFLIP.com. Home. Home; Buyer Profile; Advertise Land; Land For Sale; Land Auctions; Land for Lease; Land ...
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Area Converter: How Big is an Acre? | World Land Trust


How Big is an Acre? Find out below. Area Converter. ... These support 17% of the world’s bird ... covers approximately 260,000 acres - 4% of the total area of the ...
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What is 0.19 acre in square feet - Answers.com


What is 0.19 acre in square feet? ... 0.19 acres = 8,276.4 square feet ... 479,160 / 12 | 522,720 / 13 | 566,280 / 14 | 609,840 / 15 | 653,400 / 16 | 696,960 / 17 ...
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USDA Crop Progress: Corn Condition/Progress-Jul 10


... acres, not the number of ... 16 17 63 64 17 16 Mo 2 2 5 5 25 26 56 56 12 11 Nebr 2 1 7 4 23 20 56 63 12 12 NC 1 0 2 4 15 16 63 57 19 23 ND 8 5 12 11 28 29 49 52 3 ...
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Jonah 1:1-4; The Lord said#tn Heb “The word of ... - Bible


Occupying about 1800 acres, ... (Lev 9:24; Num 17:11, 24; cf. Gen 4:16). ... (Josh 19:46; 2 Chr 2:15 [16]; Ezra 3:7).
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Convert 16 Square Kilometers to Acres - CalculateMe.com


Convert 16 Square Kilometers to Acres. ... 16.15: 3,990.8: 16.16: 3,993.2: 16.17: 3,995.7: 16.18: 3,998.2: ... 16.62: 4,106.9: 16.63: 4,109.4: 16.64: 4,111.8: 16.65:
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Acres to Square Feet - Metric Conversion charts and calculators


Acres to Square Feet. ... 15 ac: 653400.00 ft²: 16 ac ... Square Feet to Acres; Acres to Square Feet;
Read More At : www.metric-conversions.org...

2009 Total Certified Seed Accepted Acres - USA


2009 Total Certified Seed Accepted Acres ... 25 30.00 38.05 295.23 4.43 17.15 20.50 382.69 896.30 Chipeta 128.63 122.70 ... 27.00 16.00 17.70 60.70 Mondial 15.00 ...
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Convert Acres to Hectares - CalculateMe.com


Acres to Hectares Conversion Table inverse table ... 15: 6.0703: 16: 6.4750: 17: 6.8797: 18: 7.2843: acres ha; 19: 7.6890: 20: 8.0937: 21: 8.4984: 22: 8.9031: 23:
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Suggested Questions And Answer :


what are the possible combination of making 30 with 1, 3, 5, 7, 9

I started off by figuring how many ways to use nines: 30   3 nines, 27 total   2 nines, 18 total   1 nine, 9 total   0 nines, 0 total I then figured how many ways to use sevens: 30   3 nines, 27 total     0 sevens, 27 total   2 nines, 18 total     1 seven, 25 total     0 sevens, 18 total   1 nine, 9 total     3 sevens, 30 total     2 sevens, 23 total     1 seven, 16 total     0 sevens, 9 total   0 nines, 0 total     4 sevens, 28 total     3 sevens, 21 total     2 sevens, 14 total     1 seven, 7 total     0 sevens, 0 total I then figured how many ways to use fives, then threes, ending up with this: 30   3 nines, 27 total     0 sevens, 27 total       0 fives, 27 total         1 three, 30 total         0 threes, 27 total   2 nines, 18 total     1 seven, 25 total       1 five, 30 total         0 threes, 30 total       0 fives, 25 total         1 three, 28 total         0 threes, 25 total     0 sevens, 18 total       2 fives, 28 total         0 threes, 28 total       1 five, 23 total         2 threes, 29 total         1 three, 26 total         0 threes, 23 total       0 fives, 18 total         4 threes, 30 total         3 threes, 27 total         2 threes, 24 total         1 three, 21 total         0 three, 18 total   1 nine, 9 total ​    3 sevens, 30 total       0 fives, 30 total         0 threes, 30 total     2 sevens, 23 total       1 five, 28 total         0 threes, 28 total       0 fives, 23 total         2 threes, 29 total         1 three, 26 total         0 threes, 23 total     1 seven, 16 total       2 fives, 26 total         1 three, 29 total         0 threes, 26 total       1 five, 21 total         3 threes, 30 total         2 threes, 27 total         1 three, 24 total         0 threes, 21 total       0 fives, 16 total         4 threes, 28 total         3 threes, 25 total         2 threes, 22 total         1 three, 19 total     0 threes, 16 total     0 sevens, 9 total       4 fives, 29 total     0 threes, 29 total       3 fives, 24 total     2 threes, 30 total     1 three, 27 total     0 threes, 24 total       2 fives, 19 total     3 threes, 28 total     2 threes, 25 total     1 three, 22 total     0 threes, 19 total       1 five, 14 total     5 threes, 29 total     4 threes, 26 total         3 threes, 23 total     2 threes, 20 total     1 three, 17 total     0 threes, 14 total       0 fives, 9 total     7 threes, 30 total     6 threes, 27 total     5 threes, 24 total     4 threes, 21 total     3 threes, 18 total     2 threes, 15 total     1 three, 12 total     0 threes, 9 total   0 nines, 0 total     4 sevens, 28 total       0 fives, 28 total     0 threes, 28 total     3 sevens, 21 total       1 five, 26 total     1 three, 29 total     0 threes, 26 total       0 fives, 21 total     3 threes, 30 total     2 threes, 27 total     1 three, 24 total     0 threes, 21 total     2 sevens, 14 total       3 fives, 29 total     0 threes, 29 total       2 fives, 24 total     2 threes, 30 total     1 three, 27 total     0 threes, 24 total       1 five, 19 total     3 threes, 28 total     2 threes, 25 total     1 three, 22 total     0 threes, 19 total       0 fives, 14 total     5 threes, 29 total     4 threes, 26 total     3 threes, 23 total     2 threes, 20 total     1 three, 17 total     0 threes, 14 total     1 seven, 7 total       4 fives, 27 total     1 three, 30 total     0 threes, 27 total       3 fives, 22 total     2 threes, 28 total     1 three, 25 total     0 threes, 22 total       2 fives, 17 total     4 threes, 29 total     3 threes, 26 total     2 threes, 23 total     1 three, 20 total     0 threes, 17 total       1 five, 12 total     6 threes, 30 total     5 threes, 27 total     4 threes, 24 total     3 threes, 21 total     2 threes, 18 total     1 three, 15 total     0 threes, 12 total       0 fives, 7 total     7 threes, 28 total     6 threes, 25 total     5 threes, 22 total     4 threes, 19 total     3 threes, 16 total     2 threes, 13 total     1 three, 10 total     0 threes, 7 total     0 sevens, 0 total       6 fives, 30 total     0 threes, 30 total       5 fives, 25 total     1 three, 28 total     0 threes, 25 total       4 fives, 20 total     3 threes, 29 total     2 threes, 26 total     1 three, 23 total     0 threes, 20 total       3 fives, 15 total     5 threes, 30 total     4 threes, 27 total     3 threes, 24 total     2 threes, 21 total     1 three, 18 total     0 threes, 15 total       2 fives, 10 total     6 threes, 28 total     5 threes, 25 total     4 threes, 22 total     3 threes, 19 total     2 threes, 16 total     1 three, 13 total     0 threes, 10 total       1 five, 5 total     8 threes, 29 total     7 threes, 26 total     6 threes, 23 total     5 threes, 20 total     4 threes, 17 total     3 threes, 14 total     2 threes, 11 total     1 three, 8 total     0 threes, 5 total       0 fives, 0 total     10 threes, 30 total     9 threes, 27 total     8 threes, 24 total     7 threes, 21 total     6 threes, 18 total     5 threes, 15 total     4 threes, 12 total     3 threes, 9 total     2 threes, 6 total     1 three, 3 total     0 threes, 0 total (more to follow)
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how many ways can you have coins that total 18p using 1p , 2p, 5p, and 10p coins

in how many ways can you have coins that total exactly 18pence using 1p, 2p, 5p and 10p coins but you may wish to use as many of each sort as you wish. Start with the biggest coin, then the next biggest, etc. This is so that you always end up adding on just single coins of 1p. 10    we can only have 1*10 because 2*10 is greater than 18. 10 + 5   we can only add on 1*5 because adding on 2*5 will make the sum greater than 18 10 + 5 + 2   again we can only add on 1*2 10 + 5 + 2 + 1   our first arrangement  (1*10, 1*5, 1*2, 1*1) Now we modify the 2p intp 2*1p, the 5p into 2*2p + 1p and the 10 p into 2*5p (as well as 2p's and 1p). 10 + 5 + (1+1) + 1    our 2nd arrangement   (1*10, 1*5, 0*2, 3*1) Now modify the 5, in the 1st arrangement. 10 + (2+2+1) + 2 + 1    3rd arrangement    (1*10, 0*5, 3*2, 2*1) 10 + (2+2+1) + (1+1) + 1    etc.   (1*10, 0*5, 2*2, 4*1) 10 + (2+(1+1)+1) + (1+1) + 1    etc.  (1*10, 1*5, 1*2, 1*1) 10 + ((1+1)+(1+1)+1) + (1+1) + 1     (1*10, 0*5, 0*2, 8*1) Now modify the 10, in the 1st arrangement. (5+5) + 5 + 2 + 1   our 7th arrangement     (0*10, 3*5, 1*2, 1*1) (5+5) + 5 + (1+1) + 1               (0*10, 3*5, 0*2, 3*1) (5+5) + (2+2+1) + (1+1) + 1      (0*10, 2*5, 2*2, 4*1) (5+5) + (2+(1+1)+1) + (1+1) + 1    (0*10, 2*5, 1*2, 6*1) (5+5) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 2*5, 0*2, 8*1)  -- 11th arrangment (5+(2+2+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 1*5, 2*2, 9*1) (5+(2+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 1*5, 1*2, 11*1) (5+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 1*5, 0*2, 13*1)   --- 14th arrangememt ((2+2+1)+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 0*5, 2*2, 14*1) ((2+(1+1)+1)+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 0*5, 1*2, 16*1) (((1+1)+(1+1)+1)+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 0*5, 0*2, 18*1) --17th arrangement So, in total there are 17 arrangements of 1p, 2p, 5p and 10p coins to give 18p
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show that Z15 is isomorphic to Z5 external Z3

Z15 can be represented as a pair of quantities a and b so that a has a range of 0 to 4 (Z5) and b has a range of 0 to 2 (Z3). So we have a scheme of representation: (0,0)=0 (1,1)=11 (2,2)=7 (0,3)=3 (1,4)=14 (2,0)=10 (0,1)=6 (1,2)=2 (2,3)=13 (0,4)=9 (1,0)=5 (2,1)=1 (0,2)=12 (1,3)=8 (2,4)=4. In general (X,Y) maps to (5X+6Y) modulo 15, where X is confined to modulo 3 and Y to modulo 5.  EXAMPLES:  ADDITION: (2,3)+(1,2)=(13+2) mod 15 = 15 mod 15 = 0 (2,3)+(1,2)=((2+1) mod 3, (3+2) mod 5)=(0,0)=0 (1,4)+(0,2)=(14+12) mod 15 = 26 mod 15 = 11 (1,4)+(0,2)=((1+0) mod 3, (4+2) mod 5)=(1,1)=11 SUBTRACTION: (2,3)-(1,2)=(13-2) mod 15 = 11 (2,3)-(1,2)=((2-1) mod 3, (3-2) mod 5)=(1,1)=11 (1,4)-(0,2)=(14-12) mod 15 = 2 (1,4)-(0,2)=((1-0) mod 3, (4-2) mod 5)=(1,2)=2 MULTIPLICATION (2,3)*(1,2)=(13*2) mod 15 = 26 mod 15 = 11 (2,3)*(1,2)=(2,3)+(2,3)=(1,1)=11 (1,4)*(0,2)=(14*12) mod 15 = 168 mod 15 = 3 (1,4)*(0,2)=((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4))+((1,4)+(1,4)+(1,4)) =(0,2)+(0,2)+(0,2)+(0,2)=(0,3)=3 Or is this cheating?! The point is that the mapping shows for each of all the combinations of Z3 and Z5 there is one and only one Z15 element, and no elements of Z15 have been omitted, so demonstrating isomorphism.
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how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0 I think gaussian reduction is similar to back substitution, but with all the equation shaving different variables... just not sure how to do it. You create a matrix using the constants in the equations. Below, you will see approximately what the matrix would look like. Unfortunately, it is impossible to get it to display properly on this page. ┌                       ┐ │ 4   0   1  |   3  │ │ 2  -1  0   |  2   │ │ 0   3   2  |   0   │ └                        ┘ The idea is to perform the same math procedures on these matrix rows that you would perform on the full equations. We want the first row to be  1 0 0 | a, meaning that whatever value appears as the a entry is the value of x. The second row has to be  0 1 0 | b,  and the third row has to be  0 0 1 | c. Multiply row 2 by 2. ┌                      ┐ │ 4   0   1  |   3  │ │ 4  -2   0  |  4   │ │ 0   3   2  |   0   │ └                       ┘ Now subtract row 2 from row 1, and replace row 2 with the result. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  2   1   |  -1   │ │ 0   3   2  |   0   │ └                       ┘ Multiply row 2 by 2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  4    2  |  -2   │ │ 0   3   2  |   0   │ └                       ┘ Subtract row 3 from row 2, replacing row 2. ┌                        ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   3   2  |   0   │ └                        ┘ Multiply row 2 by 3 and subtract row 3, replacing row 3. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0  -2  |   -6  │ └                       ┘ Divide row 3 by -2. ┌                       ┐ │ 4   0   1  |   3   │ │ 0  1    0  |  -2   │ │ 0   0   1  |   3   │ └                       ┘ Subtract row 3 from row 1, replacing row 1. ┌                      ┐ │ 4   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0  1  |   3   │ └                      ┘ Divide row 1 by 4. ┌                      ┐ │ 1   0   0  |   0  │ │ 0  1    0  |  -2  │ │ 0   0   1  |   3  │ └                      ┘ Row 1 shows that x = 0 Row 2 shows that y = -2 Row 3 shows that z = 3 Plug those values into the original equations to check the answer. 4x + z = 3 4(0) + 3 = 3 0 + 3 = 3 3 = 3 2x – y = 2 2(0) – (-2) = 2 0 + 2 = 2 2 = 2 3y + 2z = 0 3(-2) + 2(3) = 0 -6 + 6 = 0 0 = 0
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Find the point(s) where the line through the origin with slope 6 intersects the unit circle.

The line passes through the origin so its equation is y=6x. The equation of the unit circle is x^2+y^2=1, which has centre (0,0) and radius 1. Substituting y=6x in the equation of the circle we have 37x^2=1 and x=+sqrt(1/37). Therefore the y values for the intersections are y=+6sqrt(1/37). The points of intersection are (-sqrt(1/37),-6sqrt(1/37)), (sqrt(1/37),6sqrt(1/37)).  If the unit circle has centre (h,k) the equation is (x-h)^2+(y-k)^2=1 and substituting y=6x we get (x-h)^2+(6x-k)^2=1, which gives the x value of the intersection. So x^2-2xh+h^2+36x^2-12xk+k^2=1; 37x^2-2x(h+6k)+h^2+k^2-1=0. There are only two factors of 37, which is prime, so to factorise rationally we must have (37x+a)(x+b)=0; 37x^2+x(37b+a)+ab=0. Using the quadratic formula: x=(h+6k+sqrt((h+6k)^2-37(h^2+k^2-1))/37. The square root can only be evaluated if the expression is positive, so (h+6k)^2>37(h^2+k^2-1). This requirement applies so that the line intersects the unit circle. When the expression is zero, the line is a tangent to the circle, so there is only one intersection point. h^2+36k^2+12hk>37h^2+37k^2-37; 36h^2-12hk+k^2<37; (6h-k)^2<37 and (6h-k)< +sqrt(37) (=+6.08). This connects the coordinates of the centre of the unit circle: k>6h+sqrt(37). On equality the line y=6x will be tangential to the circle. For example, if h=0 (centre of the circle is on the y axis), k=+sqrt(37) and the unit circle will lie above or below the axis with y=6x running tangentially on the right of the circle; or on the left touching the circle below the x axis. Ideally, we want the square root to be rational so 37-(6h-k)^2=a^2. If a=+1, 6h-k=6 and k=6(h-1); or 6h-k=-6, so k=6(h+1). That gives many possible values for h and k represented by pairs: (1,0), (2,6), (3,12), (0,-6), (-1,-12), (-2,-18), (0,6), (1,12), (-1,0), (-2,-6),  to mention but a few. Using (1,0) in the quadratic: 37x^2-2x(h+6k)+h^2+k^2-1=0 we have 37x^2-2x=0=x(37x-2) giving intersection points (0,0) and (2/37,12/37). The equation for the circle is (x-1)^2+y^2=1 or y^2=2x-x^2. Let's try (2,6). 37x^2-76x+39=0, (37x-39)(x-1)=0 giving intersection points (39/37,234/37) and (1,6). The equation of the circle is (x-2)^2+(y-6)^2=1 or x^2-4x+y^2-12x+39=0. If a=+6, 6h-k=1 or -1, so k=6h-1 or 6h+1. This generates more possible intersection points. There are clearly an infinite number of positions for the unit circle centre (h,k) and an infinite number of intersection points. However, the relationship between h and k so as to produce rational intersection points has been established. k=6(h+1), k=6h+1 are the equations linking the coordinates of the centre of the unit circle. With these equations in mind the quadratic determining the intersection points (x,6x) can be solved: 37x^2-2x(h+6k)+h^2+k^2-1=0. There are four variations of this quadratic because there are four equations linking h and k. Recap There are 2 values of a^2 where a^2=37-(6h-k)^2 and x=(h+6k+a)/37; a^2=1 or 36. When a^2=1, k=6(h+1). The equation of the circle is (x-h)^2+(y-6(h+1))^2=1 and x=(37h+36+1)/37. So the points of intersection are (h+1,6(h+1)), ((37h+35)/37,6(37h+35)/37), ((37h-35)/37,6(37h-35)/37), (h-1,6(h-1)). When a^2=36, k=6h+1. The equation of the circle is (x-h)^2+(y-6h+1)^2=1 and x=(37h+6+6)/37. The points of intersection are ((37h+12)/37,6(37h+12)/37), (h,6h), ((37h-12)/37,6(37h-12)/37). Note that (h,6h) is the result of (37h+6-6)/37 and (37h-6+6)/37. We can check the (h,k) values we used earlier. These were (1,0) and (2,6). We used the formula k=6(h-1) in each case (a=+1), so intersection points for h=1, k=6(h-1)=0, should be x=(h+6k+1)/37, giving (2/37,12/37) and x=(h+6k-1), giving (0,0). For (2,6) h=2 and k=6, giving intersection points x=(2+36+1)/37, giving (39/37,234/37) and x=(2+36-1)/37=1, giving (1,6). The values of h and k are not restricted to integers.
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what is the answer to 2x-y-2z=1, 4x+y-z=5, x+y+4z=-5 using matrices

what is the answer to 2x-y-2z=1, 4x+y-z=5, x+y+4z=-5 using matrices how to solve this math problem using matrices?   2  -1  -2  │  1   4   1  -1  │  5   1   1   4  │ -5 Multiply line 2 by 2.   2  -1  -2  │  1   8   2  -2  │ 10   1   1   4  │ -5 Subtract line 2 from line 1, replace line 1.  -6  -3   0  │ -9   8   2  -2  │ 10   1   1   4  │ -5 Multiply line 2 by 2.  -6  -3   0   │ -9  16   4  -4  │ 20    1   1   4  │ -5 Add line 3 to line 2, replace line 2.  -6   -3   0  │ -9  17   5   0  │ 15    1   1   4  │ -5 Multiply line 3 by 6.  -6  -3   0  │  -9  17   5   0  │  15   6   6  24  │ -30 Add line 1 to line 3, replace line 3.  -6  -3   0  │  -9  17   5   0  │  15   0   3  24  │ -39 Multiply line 1 by 5; multiply line 2 by 3. -30 -15   0  │ -45  51  15   0  │  45    0   3  24  │ -39 Add line 2 to line 1, replace line 1.  21    0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Divide line 1 by 21.   1     0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Multiply line 1 by 51.  51    0   0  │   0  51  15   0  │  45    0   3  24  │ -39 Subtract line 1 from line 2, replace line 2.  51   0   0  │   0   0  15   0  │  45   0   3  24  │ -39 Divide line 2 by 5.  51   0   0  │   0    0   3   0  │   9   0   3  24  │ -39 Subtract line 2 from line 3, replace line 3.  51   0   0  │   0    0   3   0  │   9   0   0  24  │ -48 Three-in-one: divide line 1 by 51, divide line 2 by 3, divide line 3 by 24.   1   0   0  │   0   0   1   0  │   3   0   0   1  │  -2 This shows that x = 0, y = 3, z = -2  
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how to integrate (ln(cosx)) using polylogarithms

Differential of ln(cos(x)) is -sin(x)/cos(x)=-tan(x)=-(a0+a1x+a2x^2+a3x^3+...+a[n]x^n), where a0, a1, etc., are the coefficients in an infinite series, general nth term a[n]x^n. The coefficients can be found through differentiation and substitution of x=0. When x=0, tan(0)=0 so a0=0. Differential of -tan(x)=-(sec(x))^2; the next differential is -2sec(x)sec(x)tan(x)=-2(sec(x))^2tan(x)=-2(tan(x)+(tan(x))^3) 4th differential: -2((sec(x))^2+3(tan(x))^2(sec(x))^2). If y=tan(x) then y'=1+y^2 and y"=2yy'=2y(1+y^2)=2y+2y^3. y'"=2y'+6y^2y'=2+2y^2+6y^2+6y^4=2+8y^2+6y^4=2(1+4y^2+3y^4)=2(1+3y^2)(1+y^2). Also when x=0, y=0. The 4th differential is 16yy'+24y^3y'=8yy'(2+3y^2)=8y(1+y^2)(2+3y^2). As a power series the consecutive differentials at x=0 are a1; 2a2; 6a3; 24a4; ... r!a[r] where r is the rth differential. Therefore, we deduce that a[r]=0 when r is even. Take a4, for example. Because of the factor 8y the 4th differential is zero. Take a3: y'"=2=6a3, so a3=1/3. The 5th differential is: 8y'(1+y^2)(2+3y^2)+8y(2yy')(2+3y^2)+8y(1+y^2)(6yy')=8(1+y^2)^2(2+3y^2)+16y^2(1+y^2)(2+3y^2)+48y^2(1+y^2)^2. This becomes: 8(1+y^2)((1+y^2)(2+3y^2)+2y^2(2+3y^2)+6y^2(1+y^2))= 8(1+y^2)(2+5y^2+3y^4+2y^2(5+6y^2))= 8(1+y^2)(2+15y^2+15y^4) At y=0, this becomes 16. a5=16/5!=16/120=2/15. The 6th differential is: 16yy'(2+15y^2+15y^4)+8(1+y^2)(30yy'+60y^3y')= 16y(1+y^2)(2+15y^2+15y^4)+240y(1+y^2)^2(1+2y^2)= 16y(1+y^2)(2+15y^2+15y^4+15(1+y^2)(1+2y^2))= 16y(1+y^2)(2+15y^2+15y^4+15(1+3y^2+2y^4))= 16y(1+y^2)(17+60y^2+45y^4). As expected, this zero when y=0. The 7th differential is: 16(1+y^2)^2(17+60y^2+45y^4)+16y(2y(1+y^2)(17+60y^2+45y^4)+16y(1+y^2)(120y(1+y^2)+180y^3(1+y^2)). When y=0 this becomes: 16*17=272. Therefore, a7=272/7!=272/5040=17/315. If we can express the rth differential in terms of y we can calculate a[r]= (rth differential when y=0)/r!. Once we have a series for -tan(x), we can integrate it to get a series for ln(cos(x)) and integrate it again to get the integral of ln(cos(x)). So far, the series for -tan(x) is -(x+x^3/3+2x^5/15+17x^7/315+...) Integrating this we get: c-(x^2/2+x^4/12+x^6/45+17x^8/2520+...), where c is a constant. Another approach is to use the expansion of ln(1+u)=u-u^2/2+u^3/3-... . If we write ln(cos(x))=1/2ln(cos^2(x))=1/2ln(1-sin^2(x))=-(1/2)(sin^2(x)-sin^4(x)/2+sin^6(x)/3-...). We can write sin^2(x)=(1-cos(2x))/2. cos(4x)=2cos^2(2x)-1, so cos^2(2x)=(cos(4x)+1)/2 sin^4(x)=((1-cos(2x))/2)^2=(1+cos^2(2x)-2cos(2x))/4=(1+(cos(4x)+1)/2-2cos(2x))/4. The powers of sin(x) can be converted to sines and cosines of multiple angles in a series. Each term is easily integrated, even if somewhat painstakingly, as indicated above. More...
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is there a solution to (a) x1 + 2x2 + 3x3 = 1 2x1 + 3x2 + 4x3 = 3 x1 + 2x2 + x3 = 3

Matrix format (Gauss method): ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 1 2 1 | 3 ) R3-R1: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 -2 | 2 ) R3/-2: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 1 | -1 ) R2-R1: ( 1 2 3 | 1 ) ( 1 1 1 | 2 ) ( 0 0 1 | -1 ) R1⇔R2: ( 1 1 1 | 2 ) ( 1 2 3 | 1 ) ( 0 0 1 | -1 ) R2-R1: ( 1 1 1 | 2 ) ( 0 1 2 | -1 ) ( 0 0 1 | -1 ) R2-R3: ( 1 1 1 | 2 ) ( 0 1 1 | 0 ) ( 0 0 1 | -1 ) R1-R2: ( 1 0 0 | 2 ) ( 0 1 1 | 0 ) ( 0 0 1 | -1 ) R2-R3: ( 1 0 0 | 2 ) ( 0 1 0 | 1 ) ( 0 0 1 | -1 ) x1=2; x2=1; x3=-1. So there is a unique solution.
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Solve by Gauss-Jordan elimination (a) x1 + 2x2 + 3x3 = 1 2x1 + 3x2 + 4x3 = 3 x1 + 2x2 + x3 = 3

Matrix format: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 1 2 1 | 3 ) R3-R1: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 -2 | 2 ) R3/-2: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 1 | -1 ) R2-R1: ( 1 2 3 | 1 ) ( 1 1 1 | 2 ) ( 0 0 1 | -1 ) R1⇔R2: ( 1 1 1 | 2 ) ( 1 2 3 | 1 ) ( 0 0 1 | -1 ) R2-R1: ( 1 1 1 | 2 ) ( 0 1 2 | -1 ) ( 0 0 1 | -1 ) R2-R3: ( 1 1 1 | 2 ) ( 0 1 1 | 0 ) ( 0 0 1 | -1 ) R1-R2: ( 1 0 0 | 2 ) ( 0 1 1 | 0 ) ( 0 0 1 | -1 ) R2-R3: ( 1 0 0 | 2 ) ( 0 1 0 | 1 ) ( 0 0 1 | -1 ) x1=2; x2=1; x3=-1.
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How to find equation of the circle passing through (9,1), (-3,-1) and (4,5)

The equation of a circle is more specific: (x-h)^2+(x-k)^2=a^2, where (h,k) is the centre and a the radius. Plug in the points: (9-h)^2+(1-k)^2=(-3-h)^2+(-1-k)^2=(4-h)^2+(5-k)^2=a^2 This can be written (9-h)^2+(1-k)^2=(3+h)^2+(1+k)^2=(4-h)^2+(5-k)^2=a^2 Leaving a out of it for the moment we can use pairs of equations, using difference of squares: 12(6-2h)+2(-2k)=0, 36-12h-2k=0, 18-6h-k=0, so k=18-6h. 7(-1-2h)+6(-4-2k)=0, -7-14h-24-12k=0, -31-14h-12k=0 or 31+14h+12k=0, 31+14h+12(18-6h)=0, 31+14h+216-72h=0. 247-58h=0 so h=247/58 This looks suspiciously ungainly. So rather than continuing, I'm going to call the three points: (Q,R), (S,T), (U,V) to provide a general answer to all questions of this sort. (Q-h)^2+(R-k)^2=(S-h)^2+(T-k)^2=(U-h)^2+(V-k)^2=a^2 [(equation 1)=(equation 2)=(equation 3)=a^2] Take the equations in pairs and temporarily ignore a^2. Equations 1 and 2: (Q-S)(Q+S-2h)+(R-T)(R+T-2k)=0 Q^2-S^2-2h(Q-S)+R^2-T^2-2k(R-T)=0 2k(R-T)=Q^2+R^2-(S^2+T^2)-2h(Q-S), so k=(Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T)) Equations 2 and 3: S^2-U^2-2h(S-U)+T^2-V^2-2k(T-V)=0, so k=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) At this point, we can substitute for k and end up with an equation involving the unknown h only: (Q^2+R^2-(S^2+T^2)-2h(Q-S))/(2(R-T))=(S^2+T^2-(U^2+V^2)-2h(S-U))/(2(T-V)) (T-V)(Q^2+R^2-(S^2+T^2)-2h(Q-S))=(R-T)(S^2+T^2-(U^2+V^2)-2h(S-U)) (T-V)(Q^2+R^2-(S^2+T^2))-2h(T-V)(Q-S)=(R-T)(S^2+T^2-(U^2+V^2))-2h(R-T)(S-U) 2h((R-T)(S-U)-(T-V)(Q-S))=(R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)) h=((R-T)(S^2+T^2-(U^2+V^2))-(T-V)(Q^2+R^2-(S^2+T^2)))/((R-T)(S-U)-(T-V)(Q-S)). Once h is found we can calculate k, and then we can substitute the values for h and k in any equation to find a^2 which is equal to each of the three equations. When we use Q=9, R=1, S=-3, T=-1, U=4, V=5 or -5, we appear to get very ungainly solutions. One way to find which values need to be changed may be to plot the values and work out where the circle should fit and where its centre has less complex values. If V=6 or -6, there is a simple solution: (x-3)^2+y^2=37 (centre at (3,0), a^2=37=6^2+1^2. (-3-3)^2=(9-3)^2=6^2; (-6)^2=6^2; (2-3)^2=(4-3)^2=1; (-1)^2=1^2 shows the combination of points that would lie on the circle: (9,1), (9,-1), (-3,1), (-3,-1), (4,6), (4,-6), (2,6), (2,-6) and this includes points A and B, but not C. Note that the equation k=18-6h is valid for h=3, k=0.
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