Assignment. Name. Factor each completely. 1) a w a yk a k a yw 2) m z mnc m c mnz. 3) uv p u pv 4) w a w x k w a k w x

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Assignment. Name. Factor each completely. 1) a w a yk a k a yw 2) m z mnc m c mnz. 3) uv p u pv 4) w a w x k w a k w x"

Transcription

1 Assignment ID: 1 Name Date Period Factor each completely. 1) a w a yk a k a yw 2) m z mnc m c mnz 3) uv p u pv 4) w a w x k w a k w x 5) au xv av xu 6) xbz x c xbc x z 7) a w axk a k axw 8) mn xm m xn 9) x u xyv x v xyu 10) x y x k x x ky 11) p z p qc p c p qz 12) xc yk xk yc 13) x y x x xy 14) pu q v pv q u 15) mn m n 16) p qw pq k p qk pq w 17) x y x x xy 18) x y x x xy 19) xy x y 20) xy x x y 21) mn m n 22) xy x y 23) x y xy x xy 24) ab b a b

2 Answers to Assignment (ID: 1) 1) a ( a y)( w k) 2) m(m n)( z c) 3) ( u p)( v ) 4) w ( a x )( w k) 5) ( a x)( u v) 6) x(b x)( z c) 7) a( a x)( w k ) 8) ( m x)( n m) 9) x( x y)(u v) 10) x ( x k)( y ) 11) p (p q)( z c) 12) (x y)( c k) 13) x( x )( y x) 14) ( p q )( u v ) 15) (m )(n ) 16) pq( p q)( w k) 17) x( x )( y ) 18) x( x )( y ) 19) (x )( y ) 20) ( x )( y x) 21) ( m )(n ) 22) ( x )(y ) 23) x( x y )( y ) 24) ( a b )(b )

3 Assignment ID: 2 Name Date Period Factor each completely. 1) m n m m mn 2) bc xk bk xc 3) m n m b m m bn 4) xz yh xh yz 5) p qc p q d p qd p q c 6) p kh p k q p k p kqh 7) p qz p q h p qh p q z 8) ac b d ad b c 9) x y xn x xny 10) ah bk ak bh 11) az bh ah bz 12) nxy n n x n y 13) xy r x ry 14) a z ayc a c ayz 15) a h y k a k y h 16) a h ayk a k ayh 17) mz nc mc nz 18) ab n an nb 19) m n m p m mpn 20) a h a x k a k a x h 21) b h bxk b k bxh 22) u v u u uv 23) m n mn m mn 24) x y x y x x y

4 Answers to Assignment (ID: 2) 1) m( m )( n ) 2) ( b x)( c k) 3) m ( m b)( n m) 4) ( x y)( z h) 5) p q(p q)( c d ) 6) p k( p q)( h k) 7) p q( p q)( z h) 8) ( a b )(c d ) 9) x(x n)( y ) 10) ( a b)( h k) 11) ( a b)( z h) 12) n( x n)( y n) 13) ( x r)( y ) 14) a( a y)( z c) 15) ( a y )( h k ) 16) a(a y)(h k) 17) (m n)( z c) 18) ( a n)( b n) 19) m(m p)( n m ) 20) a (a x )( h k) 21) b(b x)( h k) 22) u( u )( v ) 23) m(m n)( n ) 24) x ( x y)( y )

5 Assignment ID: 3 Name Date Period Factor each completely. 1) xy x x y 2) m n m n m n m n 3) x y x x xy 4) x y x y x y x y 5) uv u v 6) x y x a x xay 7) xy y x y 8) a c ayk a k ayc 9) ab a b 10) a w a w y k a w k a w y 11) a u yv a v yu 12) u a u yv u av u y 13) ab v av vb 14) a u axv a v axu 15) bc xk bk xc 16) bc xk bk xc 17) b xy b x b x b y 18) x y xya x y xy a 19) xy b xb by 20) xc yk xk yc 21) xy rx x ry 22) xy n x ny 23) p c pqd p d pqc 24) a h b k a k b h

6 Answers to Assignment (ID: 3) 1) ( x )( y x) 2) m n( m )( n m) 3) x(x )( y ) 4) x y ( x )( y x ) 5) ( u )( v ) 6) x( x a)( y x) 7) ( x y)( y ) 8) a( a y)(c k) 9) ( a )( b ) 10) a w ( a y )(w k) 11) ( a y)( u v) 12) u ( a y)( u v) 13) ( a v)(b v) 14) a(a x)( u v) 15) ( b x)( c k) 16) (b x)( c k) 17) b ( x b)(y x) 18) xy( x a)( y ) 19) ( x b)( y b) 20) ( x y)( c k) 21) (x r)( y x ) 22) ( x n)( y ) 23) p( p q)( c d) 24) ( a b )( h k )

7 Assignment ID: 4 Name Date Period Factor each completely. 1) a z ayh a h ayz 2) mc nd md nc 3) x y x m x x my 4) a z abh a h abz 5) uv u v 6) x y x x xy 7) ab b a b 8) m n m m m n 9) x y xy x y xy 10) x y xy x xy 11) x y x y x y xy 12) x y x x xy 13) u v u u u v 14) y x y yx y 15) p q z pq c p q c pq z 16) x y x n x x ny 17) p z qc p c qz 18) b ah b k b ak b h 19) x y x p x x py 20) au bv av bu 21) mw n k mk n w 22) mh nk mk nh 23) uv x u xv 24) x y xy a x y xy a

8 Answers to Assignment (ID: 4) 1) a( a y)( z h) 2) ( m n)( c d) 3) x (x m)( y ) 4) a(a b)(z h) 5) ( u )( v ) 6) x( x )( y ) 7) ( a b )( b ) 8) m ( m )( n m) 9) xy ( x )( y ) 10) x( x y )( y ) 11) xy( x )( y x) 12) x(x )( y ) 13) u ( u )(v ) 14) y( x )( y ) 15) pq ( p q)( z c) 16) x ( x n)( y x) 17) ( p q)( z c ) 18) b ( a b)( h k) 19) x (x p)(y x) 20) ( a b)( u v) 21) ( m n )( w k) 22) ( m n)( h k) 23) ( u x)( v ) 24) xy ( x a)( y )

9 Assignment ID: 5 Name Date Period Factor each completely. 1) a b a k a akb 2) az xc ac xz 3) bu xv bv xu 4) a b a bk a bk a b k 5) bu xv bv xu 6) m vn m v m v m v n 7) xy rx x ry 8) xy ax x ay 9) xc yk xk yc 10) x c xyk x k xyc 11) x y x y x y x y 12) x y x x x y 13) mn m n 14) x y x x xy 15) xy x y 16) a b a a a b 17) y x y yx y 18) x y x y x x y 19) mn m n 20) ab b a b 21) uv m um mv 22) a z ax h a h ax z 23) bc xd bd xc 24) bz xh bh xz

10 Answers to Assignment (ID: 5) 1) a( a k)(b a) 2) ( a x)( z c) 3) ( b x)( u v ) 4) a b( a k)(b k) 5) (b x)( u v) 6) m v( m v)( n m ) 7) ( x r)( y x ) 8) (x a)( y x) 9) ( x y)( c k) 10) x( x y)( c k) 11) x y( x )(y ) 12) x (x )( y ) 13) ( m )(n ) 14) x(x )( y x) 15) ( x )( y ) 16) a (a )( b a) 17) y(x )( y ) 18) x ( x y )(y ) 19) ( m )( n ) 20) ( a b)( b ) 21) ( u m)( v m) 22) a( a x )( z h) 23) ( b x)( c d ) 24) ( b x)( z h)

11 Assignment ID: 6 Name Date Period Factor each completely. 1) xh yk xk yh 2) x kh xk y x k xkyh 3) x y x n x xny 4) p qh pq k p qk pq h 5) xz yh xh yz 6) x y xn x xny 7) qpz q c qpc q z 8) x y x p x x py 9) a z ab h a h ab z 10) m kh mk n m k mknh 11) xy n xn ny 12) y a w y k y a k y w 13) a z azyc a zc az y 14) a h ayk a k ayh 15) au xv av xu 16) ab p ap pb 17) a b ax a axb 18) xy y x y 19) xy x y 20) m n m m m n 21) m n m n m n m n 22) xy x x y 23) u v u u u v 24) xy x x y

12 Answers to Assignment (ID: 6) 1) ( x y)( h k) 2) xk( x y)(h k) 3) x( x n)(y x ) 4) pq( p q)( h k) 5) (x y)( z h) 6) x(x n)( y ) 7) q( p q)( z c ) 8) x ( x p)(y x) 9) a( a b )( z h) 10) mk( m n)( h k) 11) ( x n)( y n) 12) y ( a y)( w k) 13) az( a y)( z c ) 14) a( a y)(h k) 15) ( a x)( u v ) 16) ( a p)( b p ) 17) a(a x)( b ) 18) (x y)(y ) 19) ( x )( y ) 20) m (m )( n ) 21) m n( m n )( n ) 22) ( x )( y x) 23) u ( u )( v ) 24) ( x )( y x)

13 Assignment ID: 7 Name Date Period Factor each completely. 1) ab a b 2) x y xy x xy 3) x y x x xy 4) b au b v b av b u 5) xy v x vy 6) ac y k ak y c 7) m u m nv m v m nu 8) a c ayd a d ayc 9) ac yk ak yc 10) mn b m bn 11) a b a x a a xb 12) az xh ah xz 13) b c x k b k x c 14) b z b xh b h b xz 15) xc yd xd yc 16) b xc bx d b xd bx c 17) x n y x n x n x n y 18) x y xn x n xny 19) p c qd p d qc 20) az bh ah bz 21) p c pqd p d pqc 22) mz n h mh n z 23) b a h b k b a k b h 24) xy ax x ay

14 Answers to Assignment (ID: 7) 1) ( a b)( b a) 2) x( x y)( y ) 3) x(x )( y ) 4) b ( a b)( u v) 5) ( x v)( y ) 6) (a y )( c k) 7) m ( m n)( u v) 8) a( a y)( c d ) 9) ( a y)( c k) 10) (m b)( n ) 11) a ( a x)( b ) 12) (a x)( z h) 13) ( b x )( c k) 14) b (b x)( z h ) 15) ( x y)( c d) 16) bx( b x)( c d) 17) x n ( x n)(y ) 18) x( x n)( y n) 19) (p q)( c d) 20) (a b)( z h) 21) p( p q)( c d) 22) ( m n )( z h) 23) b ( a b)( h k) 24) ( x a)( y x)

15 Assignment ID: 8 Name Date Period Factor each completely. 1) ab a b 2) xy y x y 3) xy x y 4) m n m n m mn 5) xy x y 6) mn m n 7) uv u v 8) x y x y x x y 9) x y x x x y 10) x u x yv x v x yu 11) xz yc xc yz 12) ph qk pk qh 13) uv u v 14) b aw b k b ak b w 15) mu nv mv nu 16) pw qk pk qw 17) y x yp yxp y p 18) m w m nk m k m nw 19) u a u yv u av u y 20) m u mn v m v mn u 21) xy kx x ky 22) uv x ux xv 23) aw xk ak xw 24) a xu ax v a xv ax u

16 Answers to Assignment (ID: 8) 1) ( a )(b ) 2) ( x y )( y ) 3) ( x )( y ) 4) m(m n )( n m) 5) ( x )( y ) 6) ( m )( n ) 7) ( u )( v ) 8) x ( x y)( y x ) 9) x ( x )( y ) 10) x ( x y)( u v) 11) ( x y)( z c ) 12) ( p q)( h k) 13) ( u )( v ) 14) b ( a b)( w k) 15) ( m n)( u v ) 16) ( p q)( w k) 17) y( x p)( y p) 18) m ( m n)( w k) 19) u ( a y)( u v) 20) m( m n )( u v) 21) (x k)(y x) 22) ( u x)( v x) 23) (a x)( w k) 24) ax( a x)( u v)

17 Assignment ID: 9 Name Date Period Factor each completely. 1) ac xd ad xc 2) k xc k y k x k yc 3) mn a ma an 4) a c axk a k axc 5) x c x yd x d x yc 6) xy x y 7) xy x y 8) x y x x xy 9) p c pqk p k pqc 10) xy x y 11) ab a a b 12) a b ab a ab 13) x y x x x y 14) mn m n 15) x ah x k x ak x h 16) x y x y x xy 17) xy x y 18) nuv n nu n v 19) c a cxd cad c x 20) bh xk bk xh 21) a z axh a h axz 22) b z b x c b c b x z 23) mn r mr rn 24) xy m xm my

18 Answers to Assignment (ID: 9) 1) (a x)( c d) 2) k ( x y)(c k) 3) ( m a)( n a) 4) a( a x)(c k) 5) x ( x y)( c d) 6) ( x )( y ) 7) (x )(y ) 8) x(x )(y ) 9) p( p q)( c k) 10) (x )(y ) 11) (a )( b a) 12) a( a b)( b ) 13) x ( x )( y ) 14) ( m n)( n m) 15) x ( a x)( h k) 16) x( x y)( y x ) 17) (x )( y ) 18) n( u n)( v ) 19) c( a x)(c d) 20) ( b x)( h k) 21) a( a x)( z h ) 22) b ( b x )( z c) 23) ( m r)( n r) 24) ( x m)(y m)

19 Assignment ID: 10 Name Date Period Factor each completely. 1) y x yn yxn y n 2) h p z h q h p h qz 3) a z a b c a c a b z 4) qpw q k qpk q w 5) ah bk ak bh 6) m w mnk m k mnw 7) x y xy p x y p xy p 8) x y x p x x py 9) a w bk a k bw 10) az y c ac y z 11) y aw y k y ak y w 12) x y x x x y 13) a b ak a akb 14) au yv av yu 15) ab a b 16) y x y yx y 17) mn m n 18) uv u v 19) xy x y 20) xy x y 21) ab a b 22) xy y x y 23) ab ba a b 24) a u abv a v abu

20 Answers to Assignment (ID: 10) 1) y( x n)(y n) 2) h ( p q)( z h ) 3) a ( a b )( z c) 4) q( p q)( w k) 5) ( a b)( h k) 6) m( m n)( w k) 7) xy ( x p)(y p) 8) x ( x p)( y ) 9) ( a b)(w k) 10) ( a y )( z c) 11) y ( a y)( w k) 12) x ( x )( y ) 13) a( a k)( b ) 14) ( a y)( u v ) 15) ( a )( b ) 16) y( x y)( y ) 17) ( m )( n ) 18) (u )( v ) 19) ( x )( y ) 20) ( x )(y ) 21) ( a )( b ) 22) ( x y)(y ) 23) ( a b)( b a ) 24) a(a b)( u v)

CHAPTER X. SIMULTANEOUS EQUATIONS.

CHAPTER X. SIMULTANEOUS EQUATIONS. CHAPTER X. SIMULTANEOUS EQUATIONS. 140. A SINGLE equation which contains two or more unknown quantities can be satisfied by an indefinite number of values of the unknown quantities. For we can give any

More information

SYMBOL NAME DESCRIPTION EXAMPLES. called positive integers) negatives, and 0. represented as a b, where

SYMBOL NAME DESCRIPTION EXAMPLES. called positive integers) negatives, and 0. represented as a b, where EXERCISE A-1 Things to remember: 1. THE SET OF REAL NUMBERS SYMBOL NAME DESCRIPTION EXAMPLES N Natural numbers Counting numbers (also 1, 2, 3,... called positive integers) Z Integers Natural numbers, their

More information

D ON MY HONOR, I WILL TRY.. AISI S E R S O DASS A B BR AM OWNI S E R S I GIRL SCOUTS! JUN SENIO IORS CAD E TTES

D ON MY HONOR, I WILL TRY.. AISI S E R S O DASS A B BR AM OWNI S E R S I GIRL SCOUTS! JUN SENIO IORS CAD E TTES ON MY HONOR, I WILL TRY.. DAISI ES AMBASSADORS BROWNI ES I JUNIORS 2017-2018 GIRL SCOUTS! CAD E TTES SENIORS Wm! W' I? W'v Pm G, v, m m G S x. I, y' m w x m, v, v v G S G W M. T v wm.. I y v y q q m, 888.474.9686

More information

Gaylord, Michigan, Thursday, May 17, 1945

Gaylord, Michigan, Thursday, May 17, 1945 » BCMV J &KJ V 70 N 0 B Q f $6 B U CCN K V F M M M J MCK %&«~ f - M K M B C M M B B f q z f q b f 7 b z f j f f C Nb f b f f M M J M f f B C f b J V f B C b b M M - q B M M M M ff b ; b b b b j f b b M

More information

Advanced Radiology Reporting and Analytics with rscriptor vrad results after 10 million radiology reports

Advanced Radiology Reporting and Analytics with rscriptor vrad results after 10 million radiology reports Av Ry R Ay wh vr 10 y I, w, v, - y y h, v, z yz y. I wh vy y v y hh-qy y z. A h h h N L P (NLP) h w y y. h w w vr Jy 2014 h h h 10 y. F h - w vr hv wk h v h qy wkw. h wh h h. I I /v h h wk wh vy y v w

More information

PRE-CALCULUS By: Salah Abed, Sonia Farag, Stephen Lane, Tyler Wallace, and Barbara Whitney

PRE-CALCULUS By: Salah Abed, Sonia Farag, Stephen Lane, Tyler Wallace, and Barbara Whitney PRE-CALCULUS By: Salah Abed, Sonia Farag, Stephen Lane, Tyler Wallace, and Barbara Whitney MATH 141/14 1 Pre-Calculus by Abed, Farag, Lane, Wallace, and Whitney is licensed under the creative commons attribution,

More information

elegant Pavilions Rediscover The Perfect Destination

elegant Pavilions Rediscover The Perfect Destination Pv Rv T Pf D W... T O Lv! Rv p f f v. T f p, f f, j f f, bw f p f f w-v. T f bk pv. Pv w b f v, pv f. W, w w, w f, pp w w pv. W pp v w v. Tk f w v k w w j x v f. W v p f b f v j. S f f... Tk Y! 2 3 p Pv

More information

So, eqn. to the bisector containing (-1, 4) is = x + 27y = 0

So, eqn. to the bisector containing (-1, 4) is = x + 27y = 0 Q.No. The bisector of the acute angle between the lines x - 4y + 7 = 0 and x + 5y - = 0, is: Option x + y - 9 = 0 Option x + 77y - 0 = 0 Option x - y + 9 = 0 Correct Answer L : x - 4y + 7 = 0 L :-x- 5y

More information

1 First Order Ordinary Differential Equation

1 First Order Ordinary Differential Equation 1 Ordinary Differential Equation and Partial Differential Equations S. D. MANJAREKAR Department of Mathematics, Loknete Vyankatrao Hiray Mahavidyalaya Panchavati, Nashik (M.S.), India. shrimathematics@gmail.com

More information

Section 5.8. (i) ( 3 + i)(14 2i) = ( 3)(14 2i) + i(14 2i) = {( 3)14 ( 3)(2i)} + i(14) i(2i) = ( i) + (14i + 2) = i.

Section 5.8. (i) ( 3 + i)(14 2i) = ( 3)(14 2i) + i(14 2i) = {( 3)14 ( 3)(2i)} + i(14) i(2i) = ( i) + (14i + 2) = i. 1. Section 5.8 (i) ( 3 + i)(14 i) ( 3)(14 i) + i(14 i) {( 3)14 ( 3)(i)} + i(14) i(i) ( 4 + 6i) + (14i + ) 40 + 0i. (ii) + 3i 1 4i ( + 3i)(1 + 4i) (1 4i)(1 + 4i) (( + 3i) + ( + 3i)(4i) 1 + 4 10 + 11i 10

More information

INTELLIGENT ROBOT USED IN THE FIELD OF PRACTICAL APPLICATION OF ARTIFICIAL NEURAL NETWORK & MACHINE VISION

INTELLIGENT ROBOT USED IN THE FIELD OF PRACTICAL APPLICATION OF ARTIFICIAL NEURAL NETWORK & MACHINE VISION T m3 D T g jmpg: www. g.m/ j NTEGENT OBOT UED N THE FED OF CTC CTON OF TFC NEU NETWO & MCHNE VON M. M d* Dpm M Egg..Cg Egg d Tgy C D 64 Tm Nd d E-M dd: pmmpd@gm.m T. V m Dpm M Egg..Cg Egg d Tgy C D 64

More information

Probleme propuse spre rezolvare. Solution by Mathematical Reflections and Mathematical Excalibu. Nicusor Zlota

Probleme propuse spre rezolvare. Solution by Mathematical Reflections and Mathematical Excalibu. Nicusor Zlota Revista Virtuala Info MateTehnic ISSN 69-7988 ISSN-L 69-7988 Probleme propuse spre rezolvare Solution by Mathematical Reflections and Mathematical Excalibu Nicusor Zlota O Let such that. Prove that O Solve

More information

10 KHALID AMIN 1. Introduction A famous theorem of Hajos [4] is the following: If G = A 1 :::A n is a factorization of a group G, where each of the su

10 KHALID AMIN 1. Introduction A famous theorem of Hajos [4] is the following: If G = A 1 :::A n is a factorization of a group G, where each of the su Acta Mathematica Academiae Paedagogicae Nyregyhaziensis 15 (1999), 9{18 www.bgytf.hu/~amapn THE FACTORIZATION OF ABELIAN GROUPS KHALID AMIN Abstract. If G is a nite abelian group and n > 1 is an integer,

More information

I can use properties of similar triangles to find segment lengths. I can apply proportionality and triangle angle bisector theorems.

I can use properties of similar triangles to find segment lengths. I can apply proportionality and triangle angle bisector theorems. Page! 1 of! 8 Attendance Problems. Solve each proportion. 12 1. 2. 3. 15 = AB 9.5 20 QR = 3.8 4.2 x 5 20 = x + 3 30 4.! y + 7 2y 4 = 3.5 2.8 I can use properties of similar triangles to find segment lengths.

More information

You don t need a better car, you need to learn how to drive

You don t need a better car, you need to learn how to drive O h Imp f Cyb-Df L Am Y d d b, y d hw dv E Lv, F Hm, Phpp Lwk m AG 2017 m.m Wh w? m AG 2017 Pg 3 Y d d b, y d hw dv Wh h k b Wh w dd Wh w h p Wh h k NOT b C p-by-p hw fx hg Cd Vd bhg A mk

More information

A THEOREM IN FINITE PROTECTIVE GEOMETRY AND SOME APPLICATIONS TO NUMBER THEORY*

A THEOREM IN FINITE PROTECTIVE GEOMETRY AND SOME APPLICATIONS TO NUMBER THEORY* A THEOREM IN FINITE PROTECTIVE GEOMETRY AND SOME APPLICATIONS TO NUMBER THEORY* BY JAMES SINGER A point in a finite projective plane PG(2, pn), may be denoted by the symbol (xi, x2, x3), where the coordinates

More information

53 NRODUCION I y m mzz y mzz ymm zz 155mm my US mzz vy m zz m mm ymm m my mzz ymm y j k j x vy m x m mzz k m 1800 W II W 199 my 0 CKGROUND B y C 9 C E

53 NRODUCION I y m mzz y mzz ymm zz 155mm my US mzz vy m zz m mm ymm m my mzz ymm y j k j x vy m x m mzz k m 1800 W II W 199 my 0 CKGROUND B y C 9 C E 52 6H INERNIONL SYPOSIU ON BLLISICS 2 SEPEBER 12 FL II 1 201 6 1 SYERICL UZZLE WER ISORICL PERSPECIVE H Bk Hm J E 1 B S E C Dvm R mm L Bé NY v W 9 218 1 m mzz ymm m m vv 155mm v m x z q v k mzz ymm mzz

More information

Math512 PDE Homework 2

Math512 PDE Homework 2 Math51 PDE Homework October 11, 009 Exercise 1.3. Solve u = xu x +yu y +(u x+y y/ = 0 with initial conditon u(x, 0 = 1 x. Proof. In this case, we have F = xp + yq + (p + q / z = 0 and Γ parameterized as

More information

Strongly chordal and chordal bipartite graphs are sandwich monotone

Strongly chordal and chordal bipartite graphs are sandwich monotone Strongly chordal and chordal bipartite graphs are sandwich monotone Pinar Heggernes Federico Mancini Charis Papadopoulos R. Sritharan Abstract A graph class is sandwich monotone if, for every pair of its

More information

THE ROOST. Thanks, Brad. Sad faced hurdy-gurdy girl, City of cobbles, Where the muddy Meuse Marks cathedral floors With fingers of flood.

THE ROOST. Thanks, Brad. Sad faced hurdy-gurdy girl, City of cobbles, Where the muddy Meuse Marks cathedral floors With fingers of flood. THE ROOST H f Bg, vy, b kw w--y f b cc. Hwv b cgz, g g, f y f p, w kw cy. W p cfy w pc bg pb cp w Uvy f Lg. H y w f c c f f Lg' b Fcp p (W pb pc fg F p 2007 fw cp vb f $6.50) Sc bg pb Lg I g I g c p I

More information

Musical Instruments. Answers

Musical Instruments. Answers m w Wkh 1 m my Wdwd my g my my m m V m 3. W h m f h m h Y V,, D 4. W h m f h m h Y. h h, 5. W h m f h m h U Y Dm, g, 6. W h m f h m h WDWD Y,, 7. whh fmy d h XY g? Wkh 2 g my h fwg m g m. m m hk Whh g

More information

PATH SUBCOALGEBRAS, FINITENESS PROPERTIES AND QUANTUM GROUPS

PATH SUBCOALGEBRAS, FINITENESS PROPERTIES AND QUANTUM GROUPS PATH SUBCOALGEBRAS, FINITENESS PROPERTIES AND QUANTUM GROUPS S.DĂSCĂLESCU1,, M.C. IOVANOV 1,2, C. NĂSTĂSESCU1 Abstract. We study subcoalgebras of path coalgebras that are spanned by paths (called path

More information

1. Z-transform: Initial value theorem for causal signal. = u(0) + u(1)z 1 + u(2)z 2 +

1. Z-transform: Initial value theorem for causal signal. = u(0) + u(1)z 1 + u(2)z 2 + 1. Z-transform: Initial value theorem for causal signal u(0) lim U(z) if the limit exists z U(z) u(k)z k u(k)z k k lim U(z) u(0) z k0 u(0) + u(1)z 1 + u(2)z 2 + CL 692 Digital Control, IIT Bombay 1 c Kannan

More information

Creating New Objects

Creating New Objects Wk Obj 1 C bj 2 A vb bj 3 bj C M & I I B & M 4 Cv bj v - 5 M b bj 6 A vv Jv b C N Obj A Jv C bj: bj k H bj : R R Rk V Rk Jv k bj 1 I v bj 2 I bj b C Vb bj Jv vb v 1 A vb v v 2 A vb Obj bj I R bj bj H x

More information

Australian Intermediate Mathematics Olympiad 2016

Australian Intermediate Mathematics Olympiad 2016 A u s t r a l i a n M at h e m at i c a l O ly m p i a d C o m m i t t e e a d e p a r t m e n t o f t h e a u s t r a l i a n m at h e m at i c s t r u s t Australian Intermediate Mathematics Olympiad

More information

defined on A satisfying axioms L 1), L 2), and L 3). Then (A, 1, A, In 1 we introduce an alternative definition o Lukasiewicz L 3) V(xA y) VxA Vy.

defined on A satisfying axioms L 1), L 2), and L 3). Then (A, 1, A, In 1 we introduce an alternative definition o Lukasiewicz L 3) V(xA y) VxA Vy. 676 [Vol. 41, 147. Boolean Elements in Lukasiewicz Algebras. II By Roberto CIGNOLI and Antonio MONTEIR0 Instituto de Matem,tica Universidad Nacional del Sur, Bahia Blanca, Argentina (Comm. by Kinjir6 Kuo,.J.A.,

More information

Web Solutions for How to Read and Do Proofs

Web Solutions for How to Read and Do Proofs Web Solutions for How to Read and Do Proofs An Introduction to Mathematical Thought Processes Sixth Edition Daniel Solow Department of Operations Weatherhead School of Management Case Western Reserve University

More information

Game Engineering: 2D

Game Engineering: 2D Game Engineering: 2D CS420-2010F-06 2D Math David Galles Department of Computer Science University of San Francisco 06-0: Back to Basics A Vector is a displacement Vector has both direction and length

More information

To Find the Product of Monomials. ax m bx n abx m n. Let s look at an example in which we multiply two monomials. (3x 2 y)(2x 3 y 5 )

To Find the Product of Monomials. ax m bx n abx m n. Let s look at an example in which we multiply two monomials. (3x 2 y)(2x 3 y 5 ) 5.4 E x a m p l e 1 362SECTION 5.4 OBJECTIVES 1. Find the product of a monomial and a polynomial 2. Find the product of two polynomials 3. Square a polynomial 4. Find the product of two binomials that

More information

Hilbert s Metric and Gromov Hyperbolicity

Hilbert s Metric and Gromov Hyperbolicity Hilbert s Metric and Gromov Hyperbolicity Andrew Altman May 13, 2014 1 1 HILBERT METRIC 2 1 Hilbert Metric The Hilbert metric is a distance function defined on a convex bounded subset of the n-dimensional

More information

Raman Amplifier Simulations with Bursty Traffic

Raman Amplifier Simulations with Bursty Traffic R S h y Tc h N., k F, Jy K. c D Ecc E C Scc Uy ch b 3, Oc b, ch 489 X Cc, c. 5 W. hy D, S, Tx 753 Th h h bh R h bjc by c. y c c h h z c c c. Sch by c h -k c cy b. cc c, h c -- h h ych c k SONET), hch c

More information

No-Bend Orthogonal Drawings of Subdivisions of Planar Triconnected Cubic Graphs

No-Bend Orthogonal Drawings of Subdivisions of Planar Triconnected Cubic Graphs N-B Oh Dw f Sv f P Tcc Cc Gh (Ex Ac) M. S Rh, N E, T Nhz G Sch f If Scc, Th Uvy, A-y 05, S 980-8579, J. {,}@hz.c.h.c. h@c.h.c. Ac. A h h wh fx. I - h w f h, ch vx w ch w hz vc. A h hv - h w f f h - h w.

More information

Partial Differential Equations

Partial Differential Equations Partial Differential Equations Lecture Notes Dr. Q. M. Zaigham Zia Assistant Professor Department of Mathematics COMSATS Institute of Information Technology Islamabad, Pakistan ii Contents 1 Lecture 01

More information

An LQ R weight selection approach to the discrete generalized H 2 control problem

An LQ R weight selection approach to the discrete generalized H 2 control problem INT. J. CONTROL, 1998, VOL. 71, NO. 1, 93± 11 An LQ R weight selection approach to the discrete generalized H 2 control problem D. A. WILSON², M. A. NEKOUI² and G. D. HALIKIAS² It is known that a generalized

More information

3.0 INTRODUCTION 3.1 OBJECTIVES 3.2 SOLUTION OF QUADRATIC EQUATIONS. Structure

3.0 INTRODUCTION 3.1 OBJECTIVES 3.2 SOLUTION OF QUADRATIC EQUATIONS. Structure UNIT 3 EQUATIONS Equations Structure 3.0 Introduction 3.1 Objectives 3.2 Solution of Quadratic Equations 3.3 Quadratic Formula 3.4 Cubic and Bioquadratic Equations 3.5 Answers to Check Your Progress 3.6

More information

Tips for doing well on the final exam

Tips for doing well on the final exam Algebra I Final Exam 01 Study Guide Name Date Block The final exam for Algebra 1 will take place on May 1 and June 1. The following study guide will help you prepare for the exam. Tips for doing well on

More information

LOWELL WEEKLY JOURNAL

LOWELL WEEKLY JOURNAL Y G Bk b $ 6 G Y 7 B B B B - BB -BY- B Bk B Qk Q k Q k B g (- -- k Bk G Bk k q B - - - - - $ gb q g bg g g b b q )( 6 B 7 B B k 6 g k 6 B b Y k b - b b k b b b g ( \ bg Y b b k b /% /% b k b b g Y Y k

More information

Frege s Proofs of the Axioms of Arithmetic

Frege s Proofs of the Axioms of Arithmetic Frege s Proofs of the Axioms of Arithmetic Richard G. Heck, Jr. 1 The Dedekind-Peano Axioms for Arithmetic 1. N0 2. x(nx y.p xy) 3(a). x y z(nx P xy P xy y = z) 3(b). x y z(nx Ny P xz P yz x = y) 4. z(nz

More information

INVERSE TERNARY CONTINUED FRACTIONS

INVERSE TERNARY CONTINUED FRACTIONS I93I-1 TERNARY CONTINUED FRACTIONS 565 INVERSE TERNARY CONTINUED FRACTIONS BY D. N. LEHMER In Jacobi's extension of the continued fraction algorithm* we are concerned with three series of numbers given

More information

SPECTRAL ORDER PRESERVING MATRICES AND MUIRHEAD'S THEOREM

SPECTRAL ORDER PRESERVING MATRICES AND MUIRHEAD'S THEOREM TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 200, 1974 SPECTRAL ORDER PRESERVING MATRICES AND MUIRHEAD'S THEOREM BY KONG-MING chongo.1) ABSTRACT. In this paper, a characterization is given

More information

Lecture Note #6 (Chap.10)

Lecture Note #6 (Chap.10) System Modeling and Identification Lecture Note #6 (Chap.) CBE 7 Korea University Prof. Dae Ryook Yang Chap. Model Approximation Model approximation Simplification, approximation and order reduction of

More information

Sound Correspondences in the World's Languages: Online Supplementary Materials

Sound Correspondences in the World's Languages: Online Supplementary Materials Sound Correspondences in the World's Languages: Online Supplementary Materials Cecil H. Brown, Eric W. Holman, Søren Wichmann Language, Volume 89, Number 1, March 2013, pp. s1-s76 (Article) Published by

More information

ii = jj = kk = 1 i.e. i2 = 1, j2 = 1, k2 = 1 and = - ij where =.IC1)' Lemma 2: They have the following multiplication table Definition 4

ii = jj = kk = 1 i.e. i2 = 1, j2 = 1, k2 = 1 and = - ij where =.IC1)' Lemma 2: They have the following multiplication table Definition 4 J. Natn.Sci.Foundation Sri Lanka 2005 33(1): 43-49 MATHEMATICAL TOOLS OF MIXED NUMBER ALGEBRA MD. SHAH ALAM1*, M. H. AHSAN1 and MUSHFIQ AHMAD2 Department of Physics, Shahjalal University of Science and

More information

Geometry 3 SIMILARITY & CONGRUENCY Congruency: When two figures have same shape and size, then they are said to be congruent figure. The phenomena between these two figures is said to be congruency. CONDITIONS

More information

Geometry Advanced Fall Semester Exam Review Packet -- CHAPTER 1

Geometry Advanced Fall Semester Exam Review Packet -- CHAPTER 1 Name: Class: Date: Geometry Advanced Fall Semester Exam Review Packet -- CHAPTER 1 Multiple Choice. Identify the choice that best completes the statement or answers the question. 1. Which statement(s)

More information

We review some high school arithmetic and algebra.

We review some high school arithmetic and algebra. Chapter Algebra Review We review some high school arithmetic and algebra. 6.1 Arithmetic Operations Result 1 Let a, b, c R. Then a + b b + a (Commutative Law of Addition) ab ba (Commutative Law of Multiplication)

More information

Undecidability of C(T 0,T 1 )

Undecidability of C(T 0,T 1 ) Undecidability of C(T 0,T 1 ) David A. Pierce 1997; recompiled, April 4, 2017 Mathematics Dept Mimar Sinan Fine Arts University, Istanbul david.pierce@msgsu.edu.tr http://mat.msgsu.edu.tr/~dpierce/ We

More information

(2) ^max^lpm g M/?"[l - ^-(l - /T1)2}

(2) ^max^lpm g M/?[l - ^-(l - /T1)2} PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 56, April 1976 SOME INEQUALITIES FOR POLYNOMIALS Q. I. RAHMAN Abstract. Let pn(z) be a polynomial of degree n. Given that pn(z) has a zero on the

More information

Inverse of a Square Matrix. For an N N square matrix A, the inverse of A, 1

Inverse of a Square Matrix. For an N N square matrix A, the inverse of A, 1 Inverse of a Square Matrix For an N N square matrix A, the inverse of A, 1 A, exists if and only if A is of full rank, i.e., if and only if no column of A is a linear combination 1 of the others. A is

More information

Expanding brackets and factorising

Expanding brackets and factorising Chapter 7 Expanding brackets and factorising This chapter will show you how to expand and simplify expressions with brackets solve equations and inequalities involving brackets factorise by removing a

More information

On Symmetric Property for q-genocchi Polynomials and Zeta Function

On Symmetric Property for q-genocchi Polynomials and Zeta Function Int Journal of Math Analysis, Vol 8, 2014, no 1, 9-16 HIKARI Ltd, wwwm-hiaricom http://dxdoiorg/1012988/ijma2014311275 On Symmetric Property for -Genocchi Polynomials and Zeta Function J Y Kang Department

More information

Jordan Γ * -Derivation on Semiprime Γ-Ring M with Involution

Jordan Γ * -Derivation on Semiprime Γ-Ring M with Involution Advances in Linear Algebra & Matrix Theory, 206, 6, 40-50 Published Online June 206 in SciRes. http://www.scirp.org/journal/alamt http://dx.doi.org/0.4236/alamt.206.62006 Jordan Γ -Derivation on Semiprime

More information

Calculus II - Basic Matrix Operations

Calculus II - Basic Matrix Operations Calculus II - Basic Matrix Operations Ryan C Daileda Terminology A matrix is a rectangular array of numbers, for example 7,, 7 7 9, or / / /4 / / /4 / / /4 / /6 The numbers in any matrix are called its

More information

A mathematical statement that asserts that two quantities are equal is called an equation. Examples: x Mathematics Division, IMSP, UPLB

A mathematical statement that asserts that two quantities are equal is called an equation. Examples: x Mathematics Division, IMSP, UPLB EQUATIONS A mathematical statement that asserts that two quantities are equal is called an equation. Examples: 1.. 3. 1 9 1 x 3 11 x 4xy y 0 Consider the following equations: 1. 1 + 9 = 1. x + 9 = 1 Equation

More information

NILPOTENCY IN AUTOMORPHIC LOOPS OF PRIME POWER ORDER

NILPOTENCY IN AUTOMORPHIC LOOPS OF PRIME POWER ORDER NILPOTENCY IN AUTOMORPHIC LOOPS OF PRIME POWER ORDER PŘEMYSL JEDLIČKA, MICHAEL KINYON, AND PETR VOJTĚCHOVSKÝ Abstract. A loop is automorphic if its inner mappings are automorphisms. Using socalled associated

More information

9-1 Skills Practice Factors and Greatest Common Factors Find the factors of each number. Then classify each number as prime or composite

9-1 Skills Practice Factors and Greatest Common Factors Find the factors of each number. Then classify each number as prime or composite 9-1 Skills Practice Factors and Greatest Common Factors Find the factors of each number. Then classify each number as prime or composite. 1. 10 2. 31 3. 16 4. 52 5. 38 6. 105 Find the prime factorization

More information

FK>E??=II 1DAHEJ=?A ; U ; U ; YQ BP P ; W W FHEL=JA IJHEC =A FHEL=JA IJHEC FK>E? IJHEC =A CAJ HAJKH =A IAJ =A L=KA FK>E

FK>E??=II 1DAHEJ=?A ; U ; U ; YQ BP P ; W W FHEL=JA IJHEC =A FHEL=JA IJHEC FK>E? IJHEC =A CAJ HAJKH =A IAJ =A L=KA FK>E Q;P K; W; ;Q ; PA> P ;R; P W BW= W 1DAHEJ=?A 2OHFDEI = W W BW; ; PA = WU BW= W ;R; P W FK>E??=II 5JK@AJ 1DAHEJ=?A ; U ; 5JK@AJ U ; YQ BXW @W; BP P ; W W FHEL=JA IJHEC =A FHEL=JA IJHEC IJK@AJ1@ FK>E? IJHEC

More information

The uvw method - Tejs. The uvw method. by Mathias Bæk Tejs Knudsen

The uvw method - Tejs. The uvw method. by Mathias Bæk Tejs Knudsen The uvw method - Tejs The uvw method by Mathias Bæk Tejs Knudsen The uvw method - Tejs BASIC CONCEPTS Basic Concepts The basic concept of the method is this: With an inequality in the numbers a, b, c R,

More information

Chapter 6. Worked-Out Solutions. Chapter 6 Maintaining Mathematical Proficiency (p. 299)

Chapter 6. Worked-Out Solutions. Chapter 6 Maintaining Mathematical Proficiency (p. 299) hapter 6 hapter 6 Maintaining Mathematical Proficiency (p. 99) 1. Slope perpendicular to y = 1 x 5 is. y = x + b 1 = + b 1 = 9 + b 10 = b n equation of the line is y = x + 10.. Slope perpendicular to y

More information

Cubic systems with invariant lines of total multiplicity eight and with four distinct infinite singularities

Cubic systems with invariant lines of total multiplicity eight and with four distinct infinite singularities Cubic systems with invariant lines of total multiplicity eight and with four distinct infinite singularities Cristina BUJAC and Nicolae VULPE Institute of Mathematics and Computer Science Academy of Science

More information

Solutions to Exercises Chapter 10: Ramsey s Theorem

Solutions to Exercises Chapter 10: Ramsey s Theorem Solutions to Exercises Chapter 10: Ramsey s Theorem 1 A platoon of soldiers (all of different heights) is in rectangular formation on a parade ground. The sergeant rearranges the soldiers in each row of

More information

1. Let r, s, t, v be the homogeneous relations defined on the set M = {2, 3, 4, 5, 6} by

1. Let r, s, t, v be the homogeneous relations defined on the set M = {2, 3, 4, 5, 6} by Seminar 1 1. Which ones of the usual symbols of addition, subtraction, multiplication and division define an operation (composition law) on the numerical sets N, Z, Q, R, C? 2. Let A = {a 1, a 2, a 3 }.

More information

SMU Equilibrium Heat Flow Data Contribution

SMU Equilibrium Heat Flow Data Contribution SMU Equilibrium Heat Flow Data Contribution Description Updated July 21, 2014 This document describes the column order and description for data provided by Southern Methodist University (SMU). This data

More information

2 BSRAC A v u m bm gu g S my G Cmmuy F my G Gv m FR k m vg uy m m my gvm u?, mmuy my gvm g m vb m bm u uy? mmuy z m qu m ug m 7 m 17 my A my g m gv 1

2 BSRAC A v u m bm gu g S my G Cmmuy F my G Gv m FR k m vg uy m m my gvm u?, mmuy my gvm g m vb m bm u uy? mmuy z m qu m ug m 7 m 17 my A my g m gv 1 O Y M G Su Suv: J Ryöm M y, 211 ROBLEMS FACED BY FEMALE RMARY SCHOOL EACHERS N VLLA GE KOH HUSSAN KHEL ESHAWAR F RONER REGON (FR) COMARA VE SUDY OF FEDERAL CO MMUNY RMARY SCHOOL AND G OVERNMEN G RLS RMARY

More information

Pre-AP Geometry 4-9 Study Guide: Isosceles and Equilateral triangles (pp ) Page! 1 of! 10

Pre-AP Geometry 4-9 Study Guide: Isosceles and Equilateral triangles (pp ) Page! 1 of! 10 Page! 1 of! 10 Attendance Problems. 1. Find each angle measure. True or False. Explain your choice. 2. Every equilateral triangle is isosceles. 3. Every isosceles triangle is equilateral. I can prove theorems

More information

( ) and then eliminate t to get y(x) or x(y). Not that the

( ) and then eliminate t to get y(x) or x(y). Not that the 2. First-order linear equations We want the solution (x,y) of () ax, ( y) x + bx, ( y) y = 0 ( ) and b( x,y) are given functions. If a and b are constants, then =F(bx-ay) where ax, y where F is an arbitrary

More information

Proofs. Chapter 2 P P Q Q

Proofs. Chapter 2 P P Q Q Chapter Proofs In this chapter we develop three methods for proving a statement. To start let s suppose the statement is of the form P Q or if P, then Q. Direct: This method typically starts with P. Then,

More information

Discrete-time first-order systems

Discrete-time first-order systems Discrete-time first-order systems 1 Start with the continuous-time system ẏ(t) =ay(t)+bu(t), y(0) Zero-order hold input u(t) =u(nt ), nt apple t

More information

CHAPTER 7 DIV, GRAD, AND CURL

CHAPTER 7 DIV, GRAD, AND CURL CHAPTER 7 DIV, GRAD, AND CURL 1 The operator and the gradient: Recall that the gradient of a differentiable scalar field ϕ on an open set D in R n is given by the formula: (1 ϕ = ( ϕ, ϕ,, ϕ x 1 x 2 x n

More information

Q.1. Which one of the following is scalar quantity? Displacement Option Electric field Acceleration Work Correct Answer 4 w = F.ds; it does not have any direction, it s a scalar quantity. Q.. Which one

More information

Roots and Coefficients of a Quadratic Equation Summary

Roots and Coefficients of a Quadratic Equation Summary Roots and Coefficients of a Quadratic Equation Summary For a quadratic equation with roots α and β: Sum of roots = α + β = and Product of roots = αβ = Symmetrical functions of α and β include: x = and

More information

For example, to get ~, type Option-Shift-0 for Mac and Alt- Shift-0 for Windows. H J K. m M. n N. (Alt) thinspace 255

For example, to get ~, type Option-Shift-0 for Mac and Alt- Shift-0 for Windows. H J K. m M. n N. (Alt) thinspace 255 Ä ê + $ & Å ö Ç ò Ü ; ^ ^ + c X x 22 12 2 H 2 2 2 ( n) 21 2 (n) cc cc n pn n n p cc cc n pn n n n p (c) cc cc n pn pn n pn n p cc cc n p pn n pn n n p xmp p pn f c n f n Unmf K I Kb ) 1 1 12 pn (c) (n)

More information

Group inverse for the block matrix with two identical subblocks over skew fields

Group inverse for the block matrix with two identical subblocks over skew fields Electronic Journal of Linear Algebra Volume 21 Volume 21 2010 Article 7 2010 Group inverse for the block matrix with two identical subblocks over skew fields Jiemei Zhao Changjiang Bu Follow this and additional

More information

Matrices and Determinants

Matrices and Determinants Chapter1 Matrices and Determinants 11 INTRODUCTION Matrix means an arrangement or array Matrices (plural of matrix) were introduced by Cayley in 1860 A matrix A is rectangular array of m n numbers (or

More information

MINIMAL REPRESENTING MEASURES ARISING FROM RANK-INCREASING MOMENT MATRIX EXTENSIONS

MINIMAL REPRESENTING MEASURES ARISING FROM RANK-INCREASING MOMENT MATRIX EXTENSIONS J. OPERATOR THEORY 42(1999), 425 436 c Copyright by Theta, 1999 MINIMAL REPRESENTING MEASURES ARISING FROM RANK-INCREASING MOMENT MATRIX EXTENSIONS LAWRENCE A. FIALKOW Communicated by Florian-Horia Vasilescu

More information

Study Guide/Practice Exam 3

Study Guide/Practice Exam 3 Study Guide/Practice Exam 3 This study guide/practice exam covers only the material since exam. The final exam, however, is cumulative so you should be sure to thoroughly study earlier material. The distribution

More information

CSE 468, Fall 2006 Homework solutions 1

CSE 468, Fall 2006 Homework solutions 1 CSE 468, Fall 2006 Homework solutions 1 Homework 1 Problem 1. (a) To accept digit strings that contain 481: Q ={λ,4,48, 481}, Σ ={0,1,...,9}, q 0 = λ, A ={481}. To define δ, weuse a for all letters (well,

More information

A L A BA M A L A W R E V IE W

A L A BA M A L A W R E V IE W A L A BA M A L A W R E V IE W Volume 52 Fall 2000 Number 1 B E F O R E D I S A B I L I T Y C I V I L R I G HT S : C I V I L W A R P E N S I O N S A N D TH E P O L I T I C S O F D I S A B I L I T Y I N

More information

8.333: Statistical Mechanics I Problem Set # 3 Due: 10/18/13. Kinetic Theory

8.333: Statistical Mechanics I Problem Set # 3 Due: 10/18/13. Kinetic Theory 8.333: Statistical Mechanics I Problem Set # 3 Due: 10/18/13 1. Poisson Brackets: Kinetic Theory (a) Show that for observable O(p(µ),q(µ)), do/dt = {O,H}, along the time trajectory of any micro state µ,

More information

Lecture Notes on PDEs

Lecture Notes on PDEs Lecture Notes on PDEs Alberto Bressan February 26, 2012 1 Elliptic equations Let IR n be a bounded open set Given measurable functions a ij, b i, c : IR, consider the linear, second order differential

More information

4 TANGENTS AND NORMALS OBJECTIVE PROBLEMS. For a curve = f(x) if = x then the angle made b the tangent at (,) with OX is ) π / 4 ) π / tan tan / ) 4). The angle made b the tangent line at (,) on the curve

More information

EXAMPLE CFG. L = {a 2n : n 1 } L = {a 2n : n 0 } S asa aa. L = {a n b : n 0 } L = {a n b : n 1 } S asb ab S 1S00 S 1S00 100

EXAMPLE CFG. L = {a 2n : n 1 } L = {a 2n : n 0 } S asa aa. L = {a n b : n 0 } L = {a n b : n 1 } S asb ab S 1S00 S 1S00 100 EXAMPLE CFG L = {a 2n : n 1 } L = {a 2n : n 0 } S asa aa S asa L = {a n b : n 0 } L = {a n b : n 1 } S as b S as ab L { a b : n 0} L { a b : n 1} S asb S asb ab n 2n n 2n L {1 0 : n 0} L {1 0 : n 1} S

More information

Characteristic Numbers of Matrix Lie Algebras

Characteristic Numbers of Matrix Lie Algebras Commun. Theor. Phys. (Beijing China) 49 (8) pp. 845 85 c Chinese Physical Society Vol. 49 No. 4 April 15 8 Characteristic Numbers of Matrix Lie Algebras ZHANG Yu-Feng 1 and FAN En-Gui 1 Mathematical School

More information

On reaching head-to-tail ratios for balanced and unbalanced coins

On reaching head-to-tail ratios for balanced and unbalanced coins Journal of Statistical Planning and Inference 0 (00) 0 0 www.elsevier.com/locate/jspi On reaching head-to-tail ratios for balanced and unbalanced coins Tamas Lengyel Department of Mathematics, Occidental

More information

A FIRST COURSE IN NUMBER THEORY

A FIRST COURSE IN NUMBER THEORY A FIRST COURSE IN NUMBER THEORY ALEXANDRU BUIUM Contents 1. Introduction 2 2. The integers 4 3. Induction 6 4. Finite sets, finite sums, finite products 7 5. The rationals 8 6. Divisibility and Euclid

More information

PAIR OF LINES-SECOND DEGREE GENERAL EQUATION THEOREM If the equation then i) S ax + hxy + by + gx + fy + c represents a pair of straight lines abc + fgh af bg ch and (ii) h ab, g ac, f bc Proof: Let the

More information

y mx 25m 25 4 circle. Then the perpendicular distance of tangent from the centre (0, 0) is the radius. Since tangent

y mx 25m 25 4 circle. Then the perpendicular distance of tangent from the centre (0, 0) is the radius. Since tangent Mathematics. The sides AB, BC and CA of ABC have, 4 and 5 interior points respectively on them as shown in the figure. The number of triangles that can be formed using these interior points is () 80 ()

More information

Optimization and Calculus

Optimization and Calculus Optimization and Calculus To begin, there is a close relationship between finding the roots to a function and optimizing a function. In the former case, we solve for x. In the latter, we solve: g(x) =

More information

2.4.8 Heisenberg Algebra, Fock Space and Harmonic Oscillator

2.4.8 Heisenberg Algebra, Fock Space and Harmonic Oscillator .4. SPECTRAL DECOMPOSITION 63 Let P +, P, P 1,..., P p be the corresponding orthogonal complimentary system of projections, that is, P + + P + p P i = I. i=1 Then there exists a corresponding system of

More information

GNSS-Based Orbit Determination for Highly Elliptical Orbit Satellites

GNSS-Based Orbit Determination for Highly Elliptical Orbit Satellites -Bd D f Hghy p Q,*, ug, Ch Rz d Jy u Cg f u gg, g Uvy f u d u, Ch :6--987, -:.q@ud.uw.du. h f uvyg d p If y, Uvy f w uh W, u : h Hghy p H ufu f y/yhu f h dgd hv w ud pg h d hgh ud pg h f f h f. Du h g

More information

2.4 Annihilators, Complemented Subspaces

2.4 Annihilators, Complemented Subspaces 40 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY 2.4 Annihilators, Complemented Subspaces Definition 2.4.1. (Annihilators, Pre-Annihilators) Assume X is a Banach space. Let M X and N X. We call the annihilator

More information

On directed versions of the Corrádi-Hajnal Corollary

On directed versions of the Corrádi-Hajnal Corollary On directed versions of the Corrádi-Hajnal Corollary Andrzej Czygrinow H. A. Kierstead heodore Molla October 4, 01 Abstract For k N, Corrádi and Hajnal proved that every graph G on k vertices with minimum

More information

2.25 Advanced Fluid Mechanics

2.25 Advanced Fluid Mechanics MIT Department of Mechanical Engineering.5 Advanced Fluid Mechanics Problem 6.0a This problem is from Advanced Fluid Mechanics Problems by A.H. Shapiro and A.A. Sonin Consider a steady, fully developed

More information

REMARKS ON THE CLASSIFICATION OF REVERSIBLE CUBIC SYSTEMS WITH CENTER. Henryk Żołądek

REMARKS ON THE CLASSIFICATION OF REVERSIBLE CUBIC SYSTEMS WITH CENTER. Henryk Żołądek Topological Methods in Nonlinear Analysis Journal of the Juliusz Schauder Center Volume 8, 1996, 5 4 REMARKS ON THE CLASSIFICATION OF REVERSIBLE CUBIC SYSTEMS WITH CENTER Henryk Żołądek The paper [8] from

More information

u n x m i K. > *. Th$ Bomt of V72TOL.

u n x m i K. > *. Th$ Bomt of V72TOL. =»»HHF _?/ + + D LU 38 H )L D ) R C F L R R q x 6 D q C! q H G F CP N P D C C P P [ D H_R D H_L C H Cq K F F N L F F D C C GC C / z Y D P D F C L C R C H / D G " HL & LR PRNG LLNRY 4D P R H G F HH N ~

More information

Chemistry 112 Name Exam I Form A Section January 29,

Chemistry 112 Name Exam I Form A Section January 29, Chemistry 112 Name Exam I Form A Section January 29, 2013 email IMPORTANT: On the scantron (answer sheet), you MUST clearly fill your name, your student number, section number, and test form (white cover

More information

MANY BILLS OF CONCERN TO PUBLIC

MANY BILLS OF CONCERN TO PUBLIC - 6 8 9-6 8 9 6 9 XXX 4 > -? - 8 9 x 4 z ) - -! x - x - - X - - - - - x 00 - - - - - x z - - - x x - x - - - - - ) x - - - - - - 0 > - 000-90 - - 4 0 x 00 - -? z 8 & x - - 8? > 9 - - - - 64 49 9 x - -

More information

1 Partial differentiation and the chain rule

1 Partial differentiation and the chain rule 1 Partial differentiation and the chain rule In this section we review and discuss certain notations and relations involving partial derivatives. The more general case can be illustrated by considering

More information

On the Feuerbach Triangle

On the Feuerbach Triangle Forum Geometricorum Volume 17 2017 289 300. FORUM GEOM ISSN 1534-1178 On the Feuerbach Triangle Dasari Naga Vijay Krishna bstract. We study the relations among the Feuerbach points of a triangle and the

More information

Examples: The (left or right) cosets of the subgroup H = 11 in U(30) = {1, 7, 11, 13, 17, 19, 23, 29} are

Examples: The (left or right) cosets of the subgroup H = 11 in U(30) = {1, 7, 11, 13, 17, 19, 23, 29} are Cosets Let H be a subset of the group G. (Usually, H is chosen to be a subgroup of G.) If a G, then we denote by ah the subset {ah h H}, the left coset of H containing a. Similarly, Ha = {ha h H} is the

More information