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IN 49=3.8918

Convert to an exponential equation IN 49=3.8918 the equivalent equation is?     =49

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8918 PRA SANDS CAMBRIA 3 cm 64.98 in 49.79 in - Remnant


8918 PRA SANDS CAMBRIA 3 cm 64.98 in 49.79 in. Last Updated: Mon 2017-01-23: Seller Name: KB Surfaces View all remnants from this seller : Contact Info:
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Instructions for Form 8918 (Rev. January 2017)


Note: Instructions for Form 8918 (Rev. January 2017) begins on the next page. The zip code for where to file the completed Form 8918 has changed
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Form 8918 (Rev. December 2011) - IRS tax forms


Note: Form 8918 (Rev. December 2011) begins on the next page. The zip code for where to file the completed Form 8918 has changed from 84404 to 84201.
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fi8918w | eBay


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Erowid.org: Erowid Reference 8918 : The Potential Religious ...


The Index page for the reference article: Richards WA The Potential Religious Relevance of Entheogens Zygon 2014 49(3):652-665
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How to File Form 8918 - Irs information - Irs Tax Help


Requirements for filing IRS Form 8918, Material Advisor Disclosure Statement and answers to frequently asked questions about Form 8918 and the material advisor role.
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8918 Series (in) - Belden


8918 Hook-up/Lead - UL AWM Style 1015. Put Ups and Colors: Item # Putup Ship Weight Color Notes Item Desc 8918 001100 100 FT 1.300 LB BROWN 18 STR PVC BRN
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Human Cytomegalovirus Resistance to Antiviral Drugs


Two of those are located at the 3′ extremity of the gene (Fig. 2), and the ... 10.1128/AAC.49.3.873-883.2005 Antimicrob. Agents Chemother. March 2005 ...
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Pictures of CN 8918 - rrpicturearchives.net


Pictures of CN 8918, Model: SD70M-2 : 7/11/2017 ... 3/22/2013 Upload Date: 4/25/2013 4:08:49 AM : Location: Paducah, KY : Author: Rodney Pidcock: Categories:
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Suggested Questions And Answer :


Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0.

Question: Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0. The tangent plane to the surface z=f(x,y)=xy^3+8/y is given by 2x+7y+2z=D Where D is some constant. The equation of the tangent plane to the surface f(x,y), at the point (x_0 y_0 ) can be written as z=f(x_0,y_0 ) + f_x (x_0 y_0 )∙(x-x_0 ) + f_(y ) (x_0,y_0 )∙(y-y_0 ) Where (x_0,y_0) is some point on that plane andf_x (x_0 y_0 ) is the slope of the tangent line to the surface at that point in the x-direction and similarly for f_(y ) (x_0,y_0 ). We have f(x,y)=xy^3+8/y. f_x=y^3,  f_y=3xy^2-8/y^2 . So, f_x (x_0 y_0 )=y_0^3, f_(y ) (x_0,y_0 )=3x_0 y_0^2-8/(y_0^2 ), f(x_0,y_0 )=x_0 y_0^3+8/y_0 . Substituting for the above into the equation of the tangent plane, z = x_0 y_0^3+8/y_0 + y_0^3∙(x-x_0 ) + (3x_0 y_0^2 - 8/(y_0^2 ))∙(y-y_0 ) z = x_0 y_0^3+8/y_0 + y_0^3 x - x_0 y_0^3 + (3x_0 y_0^2 - 8/(y_0^2 ))y - 3x_0 y_0^3 + 8/y_0 z = y_0^3 x + (3x_0 y_0^2 - 8/(y_0^2 ))y + {x_0 y_0^3 + 8/y_0 - x_0 y_0^3 - 3x_0 y_0^3 + 8/y_0 } z = y_0^3 x + (3x_0 y_0^2 - 8/(y_0^2 ))y + {16/y_0 - 3x_0 y_0^3 } -2y_0^3 x - 2(3x_0 y_0^2 - 8/(y_0^2 ))y + 2z = 32/y_0 - 6x_0 y_0^3 We have just written down the equation of the tangent plane, which has also been written down as 2x+7y+2z=D Comparing the two forms of these equations, -2y_0^3 = 2     ----------------------------------- (1) 2(3x_0 y_0^2 - 8/(y_0^2 )) = 7     ---------- (2) 32/y_0 - 6x_0 y_0^3 = D   ------------------- (3) From (1), y_0 = -1 Substituting for y_0=-1 into (2), we get x_0 = 3/2. Substituting for x_0 and y_0 into (3), we get D = -23, and z = f(x_0,y_0 ) = (3/2) (-1)^3 + 8/(-1) = (-9.5) The equation of the tangent plane is thus 2x+7y+2z=-23 And this plane is tangential to the surface f(x,y) at one point only, which is: (1.5,-1,-9.5)  
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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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What is inverse

If B= ( b 0 0 ) ( 0 b 0 ) = bI, where I is the identity matrix, ( 0 0 b ) then B^7= ( b^7 0 0 ) ( 0 b^7 0 ) = ( 0 0 b^7 ) So  ( b^7 0 0 )   ( 3b 0 0 )    ( 1 0 0 )    ( 0 0 0 ) ( 0 b^7 0 ) - ( 0 3b 0 ) + ( 0 1 0 ) = ( 0 0 0 ). ( 0 0 b^7 )   ( 0 0 3b )     ( 0 0 1 )    ( 0 0 0 ) b^7-3b+1=0 applies right across the matrices and this equation has 3 real roots which we can call b1, b2, b3, where b1, b2, b3 =-1.249, 0.333, 1.133 approx. B=bI so the inverse of B=B'=I÷b, where b can be any one of b1, b2, b3. B'= ( 1/b 0 0 )    ( -0.801 0 0 )   ( 3.003 0 0 )  ( 0.8826 0 0 ) ( 0 1/b 0 ) = ( 0 -0.801 0 ),  ( 0 3.003 0 ), ( 0 0.8826 0 ) ( 0 0 1/b)     ( 0 0 -0.801)    ( 0 0 3.003 )  ( 0 0 0.8826 ).  
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adx/(b-c)yz=bdy/(c-a)zx=cdz/(a-b)xy

(1) I am assuming this is not a calculus question where dx, dy and dz have a different meaning. Variable d can be removed as a common factor: ax/((b-c)yz)=by/((c-a)zx)=cz/((a-b)xy) Take the first pair and remove common factor z: ax/((b-c)y)=by/((c-a)x); ax^2(c-a)=by^2(b-c); ax^2=by^2(b-c)/(c-a); by^2=ax^2(c-a)/(b-c) The second pair have common factor x: by^2(a-b)=cz^2(c-a); by^2=cz^2(c-a)/(a-b); cz^2=by^2(a-b)/(c-a) First and last have common factor y: ax^2(a-b)=cz^2(b-c); ax^2=cz^2(b-c)/(a-b); cz^2=ax^2(a-b)/(b-c) We have a pair of equations for each of the three quantities ax^2, by^2, cz^2. So, ax^2=by^2(b-c)/(c-a)=cz^2(b-c)/(a-b), by^2/(c-a)=cz^2/(a-b) or (A) by^2/(cz^2)=(c-a)/(a-b) by^2=ax^2(c-a)/(b-c)=cz^2(c-a)/(a-b), ax^2/(b-c)=cz^2/(a-b) or (B) ax^2/(cz^2)=(b-c)/(a-b) cz^2=by^2(a-b)/(c-a)=ax^2(a-b)/(b-c), by^2/(c-a)=ax^2/(b-c) or (C) by^2/(ax^2)=(c-a)/(b-c) The relative sizes of a, b and c can be: (1) a0 because b0 implies ac>0 and a>0 because c>0 (1C)<0 implies ab<0 and b<0 so ba: INCONSISTENT (2A)<0 implies bc<0 and b>0, c<0 because b>c (2B)<0 implies ac<0 and a>0 because c<0 (2C)>0 implies ab>0 and b>0 but c<0 and a>0 making ca: INCONSISTENT (3A)>0 implies bc>0 (3B)<0 implies ac<0 and a<0, c>0 and b>0 because a0 and so b>a, but b0 because b0 (4C)>0 implies ab>0, true, because a and b are both negative, but c0: INCONSISTENT (5A)>0 implies bc>0 (5B)<0 implies ac<0 and c<0, b<0, a>0 because a>c (5C)<0 implies ab<0, but b>a which cannot be if b<0: INCONSISTENT (6A)<0 implies bc<0 and c<0, b>0 because b>c (6B)>0 implies ac>0 and a<0, but b0 which cannot be: INCONSISTENT So there are no solutions because there is inconsistency throughout. (2) I am assuming this is a calculus problem. Assume a, b and c are constants and x, y and z are variables. adx/((b-c)y)=bdy/((c-a)x); b(b-c)ydy=a(c-a)xdx Integrating: b(b-c)y^2/2=a(c-a)x^2/2 (constant to be added later) a(c-a)x^2-b(b-c)y^2=k a constant. A similar result follows for the other two pairs of variables (x,z and y,z) by separation of variables and using the other two pairs of equations: a(a-b)x^2-c(b-c)z^2=p and b(a-b)y^2-c(c-a)z^2=q  
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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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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 many different distributions can the manager make if every employee receives at least one voucher?

There must be 100 vouchers because each is worth RM5 and the total value is 500 ringgits or RM500. (i) Each employee receives at least 1 voucher, so that means there are 95 vouchers left to distribute. We can write each distribution as {A,B,C,D,E}: starting with {0,0,0,0,95}, then {0,0,0,1,94}, {0,0,0,2,93}, ..., {0,0,0,95,0}, {0,0,1,0,94}, ..., {0,0,95,0,0}, ..., ..., {95,0,0,0,0}. So when A=B=C=0, {D,E} range from {0,95}, {1,94}, ..., to {95,0}, 96 ways. When A=B=0 and C=1, {D,E} range from {0,94} to {94,0}, 95 ways. Finally when C=95 so {C,D,E}={95,0,0} we will have covered 96+95+...+1=96*97/2=4656 ways. (The sum of the whole numbers from 1 to n is given by S=n(n+1)/2.) That was for B=0; when B=1, {C,D,E} ranges from {0,0,94} to {94,0,0} to cover 95*96/2=4560 ways. When B=2 it's 94*95/2=4465 ways. So for A=0 we have 4656+4560+4465+...+3+1 = 96^2+94^2+...+2^2 = 4(48^2+47^2+46^2+...+2^2+1^2)=4*48*49*97/6=152096 (the sum of the squares of the whole numbers from 1 to n is given by S=n(n+1)(2n+1)/6. Also, the sum of whole numbers between 1 and n taken in pairs gives us: n(n+1)/2+(n-1)n/2 for each pair. This is n^2/2+n/2+n^2/2-n/2=n^2. For the next pair we get (n-2)^2 and so on.) That was just for A=0! For A=1 we have {1,0,0,0,94} to {1,94,0,0,0}. This will give us 4560+4465+...+10+6+3+1 = 95^2+93^2+...+5^2+3^2+1. There is a formula for this sum. It is S=(n+1)(2n+1)(2n+3)/3, where 2n+1=95, so n=47. So for A=1, the number of ways is 48*95*97/3=147440. For A=2 we have {2,0,0,0,93} to {2,93,0,0,0} which produces 94^2+92^2+...+4^2+2^2 = 4(47^2+46^2+...+1) = 4*47*48*95/6 = 142880. So we alternate between two formulae as A continues to go from 3 to 95.  More to follow...
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d square y divided by dt square+2 dy divided by dx_3y=sint, given that y=dy divided by dx=0 when t=0

d(dy/dt)/dt+2dy/dx-3y=sin(t); y=dy/dx=0 when t=0. y=f(t); x=g(t); dy/dt=f'; dx/dt=g'; d(dy/dt)/dt=f"; dy/dx * dx/dt=dy/dt, g'dy/dx=f'; dy/dx=f'/g'. f"+2f'/g'-3f=sin(t) or g'f"+2f'-3fg'=g'sin(t). f(0)=0; f'(0)/g'(0)=0, so f'(0)=0 (because dy/dx=0=f'(0)/g'(0)). Also, g'(0) cannot be zero, otherwise dy/dx would be undefinable at t=0. But for f'/g', we would have an expression on the left that could be solved as an auxiliary, or characteristic, equation and we could apply the homogeneous solution fH=Ae^-3t + Be^t where fH denotes the homogeneous component of the solution, and A and B are constants and the exponents are simply the solutions of m^2+2m-3=(m+3)(m-1)=0 giving us roots -3 and 1. If g'=1 then clearly g' is non-zero, which is a requirement. But if g'=1 x=g(t)=t+C where C is a constant and fH=Ae^-3t + Be^t. xH=gH=t, where H denotes the homogeneous component. Could this be a solution? This is the homogeneous component of the solution fH, where complete f=fH+fP, where fP denotes the particular solution. Now we need to resolve the right-hand side. This is the particular solution fP and we can assume that it is of the form fP=Csin(t)+Dcost(t). Applying the derivatives we have: Ccos(t)-Dsin(t) as 1st, -Csin(t)-Dcos(t) as 2nd. These must satisfy the original DE. So, -Csin(t)-Dcos(t)+(2/g')(Ccos(t)-Dsin(t))-3(Csin(t)+Dcos(t))=sin(t). Comparing coefficients: sin(t): -C-2D/g'-3C=1; cos(t): -D+2C/g'-3D=0. 4C+2D/g'+1=0 and 2C/g'=4D, so C=2Dg'. 8Dg'+2D/g'+1=0, 8Dg'^2+2D+g'=0. 2D(4g'^2+1)=-g'. This gives us fP=-sin(t)/5-g'cos(t)/(8g'^2+2). Combining the auxiliary and particular components, y=f(t)=fH+fP=Ae^-3t + Be^t-sin(t)/5-g'cos(t)/(8g'^2+2). If we can replace g' with 1 then D=-1/10, C=-1/5, y=Ae^-3t + Be^t-sin(t)/5-cos(t)/10 and x=t. The initial conditions have f(0)=0=f'(0), so 0=A+B-1/10 and -3A+B-1/5=0. From these, -3A+1/10-A-1/5=0, -4A-1/10=0, A=-1/40, 0=-1/40+B-1/10, making B=1/8. Thus, y=(e^t)/8-(e^-3t)/40-sin(t)/5-cos(t)/10, x=t or y=(e^x)/8-(e^-3x)/40-sin(x)/5-cos(x)/10. CHECK: y'=(e^t)/8+3e^(-3t)/40-cos(t)/5+sin(t)/10; x'=g'=1, y'=f', dy/dx=dy/dt=f'; y"=(e^t)/8-9e^(-3t)/40+sin(t)/5+cos(t)/10.  y"+2dy/dx-3y becomes: (e^t)/8-9e^(-3t)/40+sin(t)/5+cos(t)/10 [this is y"] +(e^t)/4+3e^(-3t)/20-2cos(t)/5+sin(t)/5 [this is 2dy/dx=2y'] -3(e^t)/8+3(e^-3t)/40+3sin(t)/5+3cos(t)/10= [this is -3y] (e^t)(1/8+1/4-3/8) [=0] +e^(-3t)(-9/40+3/20+3/40) [=0] +sin(t)(1/5+1/5+3/5) [=sin(t)*1]  +cos(t)(1/10-2/5+3/10)= [=0] sin(t) [ADDENDUM: Why the auxiliary, or characteristic, equation works. Consider the equation y=Ae^ax+Be^bx, where a and b are constants. y'=aAe^ax+bBe^bx; y"=a^2Ae^ax+b^2Be^bx. Now consider a 2nd degree DE: y"+Py'+Qy=0, where P and Q are constants. We can replace the derivatives: a^2Ae^ax+b^2Be^bx+PaAe^ax+PbBe^bx+QAe^ax+QBe^bx=0, which can be written: Ae^ax(a^2+Pa+Q)+Be^bx(b^2+Pb+Q)=0. Since the exponential components cannot be <0, the quadratics in a and b must each be zero and they must be the roots of the same equation: e.g., z^2+Pz+Q=0 where the roots are z=a and b, with P and Q as the coefficients. In the question, P=2 and Q=-3 so z^2+2z-3=0 is the quadratic, so (z+3)(z-1)=0 and a=-3 and b=1, so fH=Ae^-3t+Be^t.]
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How do I solve z= 50 x+ 70 y?

  z= 50 x+ 70 y?     Constraints are : x >= 0, y >= 0, x + 2y <= 1500,5x + 2y <= 3500 1) x> = 0; y >= 0 represents first quadrant i.e Q1. 2)Draw a line x + 2y =1500 x 0 1500 y 750 0 (x,y) (0,750) (1500,0) 3) plot above points on graph sheet and draw  a line. 4)substitute (0,0) in inequation x+ 2y <= 1500  you get 0 <= 1500 ( it is true) therefore    (0.0) lies x+ 2y <= 1500 region.shade the region. 5)Draw a line 5x + 2y = 3500. x 0 700 y 1750 0 (x,y) (0,1750) (700,0) plot above points on the graph and draw a line passing through them 6)substitute (0,0) in inequation 5x + 2y <= 3500 you get 0 <= 3500 ( it is true ) therefore (0,0) lies 5x + 2y < = 3500 region. shade that region. 7) x + 2y = 1500 and 5x + 2y = 3500 lines intersect at (500,500) from graph. 8) required region is bounded by the points (0,0),(700,0) (500,500),(0,750). point z=50x+70y value (0,0) z=50*0+70*0 0 (700,0) z=50*700+70*0 35000 (500,500) z=50*500+70*500 60000 (0,750) z=50*0+70*750 52500   therfore z is minimum at (0,0) and maximum at (500,500)  
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Given that the ODE is: x"+16x=0 Comment on the nature of solution if the ODE has the following boundary conditions:?

x"+16x=0 can be written x"=-16x which is classic simple harmonic motion, like a pendulum or oscillating spring. Its solution is a wave, a sine wave or cosine wave. If x=Asin(nt+c)+Bcos(nt+c), x'=nAcos(nt+c)-nBsin(nt+c) and x"=-n^2Asin(nt+c)-n^2Bcos(nt+c)=-n^2x. So if we put n^2x=16x, n=4 and x(t)=Asin(4t+c)+Bcos(4t+c). (i) x(0)=0, x(π/2)=0: Asin(c)+Bcos(c)=0, tan(c)=-B/A; (ii) x(0)=0, x(π/8)=0: Asin(c+π/2)+Bcos(c+π/2)=Acos(c)-Bsin(c)=0, tan(c)=A/B; (iii) x(0)=0, x(π/2)=1: Asin(c)+Bcos(c)=1. Another approach is to use the characteristic equation which gives us x=Ae^4it+Be^-4it where i=√-1. e^iy=cos(y)+isin(y) and e^-iy=cos(y)-isin(y) and e^iny=cos(ny)+isin(ny)=(cos(y)+isin(y))^n. Therefore, x=A(cos(4t)+isin(4t))+B(cos(4t)-isin(4t)). (i) x(0)=0, x(π/2)=0: 0=A+B, B=-A, x=2Aisin(4t) (ii) x(0)=0, x(π/8)=0; 0=A+B, 0=iA-iB, A=B=0, x=0 (iii) x(0)=0, x(π/2)=1; 0=A+B, 1=A+B, which is inconsistent. Should the question have x' instead of x in the second of the pair of boundary conditions?
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