## Åbo Sentence Examples

- Rhyn didn.t wait for him to settle himself but struck first with his long, oak bo, a blow that caught the Immortal by surprise.
- Tamer emerged from Kris's chamber into the hallway first, armed with a scythe and a bo, while Erik followed with a long sword.
- Instinctive, powerful, and light-footed, he twirled the bo as if it was an extension of him, adapting to his opponent and absorbing any blows that fell to him without flinching.
- The two exchanged some short communication before Jule tossed the bo to none other than Gerry.
- Bo.*)
- The ph~ me organizers of the movement were King Franciss uncle, the bo mt of Trapani, and Mons.
- Parker.) bo, Basi-occipital.
- Bo, Basi-occipital.
- Porphyry says of Origen, Kara Tds rrepi lrpay f caTWV Kai Belot) bo s as `EXX vt cav (Euseb.
- Burra is a contraction of Bo?gar-oy, meaning "Broch island."
- P s boea? ?.° bo ere I W G.
- Thus to obtain the resultant of aox 3 +a i x 2 +a 2 x+a 3, 4, =box2+bix+b2 we assume the identity (Box+Bi)(aox 3 +aix 2 +a2x+a3) = (Aox 2 +Aix+ A 2) (box2+bix+b2), and derive the linear equations Boa ° - Ac b o = 0, Boa t +B i ao - A 0 b 1 - A 1 bo =0, Boa t +B 1 a 1 - A0b2 - A1b1-A2b° = 0, Boa3+Bla2 - A l b 2 -A 2 b 1 =0, B 1 a 3 - A 2 b 2 =0, = = (y l, y2,...ynl `x1, x2,...xnl for brevity.
- And by elimination we obtain the resultant ao 0 bo 0 0 al ao b1 bo 0 a 2 a i b 2 b 1 bo a numerical factor being disregarded.
- Taking the same example as before the process leads to the system of equations acx 4 +alx 3 +a2x 2 +a3x =0, aox 3 +a1x 2 +a2x+a 3 = 0, box +bix -1-b2x =0, box' +b i x 2 -{-h 2 x = 0, box + b i x + b:: = 0, whence by elimination the resultant a 0 a 1 a 2 a 3 0 0 a 0 a 1 a 2 a3 bo b 1 b 2 0 0 0 bo b 1 b 2000 bo b 1 b2 which reads by columns as the former determinant reads by rows, and is therefore identical with the former.
- Put (aox 3 -}-a l x 2 +a 2 x +a 3) (box' +b1x'+b2) - (aox'3+aix'2+a2x'+a3) (box' + bix + b2) = 0; after division by x-x the three equations are formed aobcx 2 = aobix+aob2 =0, aobix 2 + (aob2+a1b1-a2bo) x +alb2 -a3bo = 0, aob2x 2 +(a02-a3bo)x+a2b2-a3b1 =0 and thence the resultant aobo ao aob2 aob 1 aob2+a1b1-a2bo alb2-a3b0 aob 2 a1b2 - a 3 bo a2b2 - a3b1 which is a symmetrical determinant.
- Being formed, the expansion being carried out, an operand ao or bo or co ...
- Solving the equation by the Ordinary Theory Of Linear Partial Differential Equations, We Obtain P Q 1 Independent Solutions, Of Which P Appertain To S2Au = 0, Q To 12 B U =0; The Remaining One Is Ab =Aobl A 1 Bo, The Leading Coefficient Of The Jacobian Of The Two Forms. This Constitutes An Algebraically Complete System, And, In Terms Of Its Members, All Seminvariants Can Be Rationally Expressed.
- 1 And The Actual Forms For The First Three Weights Are 1 Aobzo, (Ao B 1 A 1 B O) Bo, (A O B 2 A 1 2 0 Bo, Ao(B2, 3 A1B2 A2B1 A O (B L B 2 3B O B 3) A I (B 2 1 2B 0 B 2); Amongst These Forms Are Included All The Asyzygetic Forms Of Degrees 1, 1, Multiplied By Bo, And Also All The Perpetuants Of The Second Binary Form Multiplied By Ao; Hence We Have To Subtract From The 2 Generating Function 1Z And 1 Z Z2, And Obtain The Generating Function Of Perpetuants Of Degrees I, 2.
- (7) If B m denote the brightness of the mth lateral image, and Bo that the central image, we have amp 'cosx' dx= a d (1) (-) m7r B.: Bo= a+d am?r sin' a4 d (1).
- A+d If B denotes the brightness of the central image when the whole of the space occupied by the grating is transparent, we have Bo:B =a2:(a+d)2, and thus (2).