Required: Solution: 1.4)
|
|
|
- Phillip Dickerson
- 5 years ago
- Views:
Transcription
1 CHAER.) ) k b) 000, 000W 000,000 W 000,000 W 00, 000 W W ) d) k s 4... s.) 8 r nn n s 9 s p p p.) Gen:.4) n 9 r re r re / dy / r re n n / s dy ) 8 b) nn dy 4 0 0se nds dy 8400s r re r re r re r re s ns ns 7 4 s, p p nn nn n p p p r r ) 7 d) k
2 .) ) 400k k b) 80,000 k 8 0, k ) s s 8 d) sends n dy dy 4 0 0s dy 8400 s dy dy dy s.)gen: n sle redn 0.0 erner dsn ndn 4 dsn n sle erner sle redn Zer errr 0.04 Zer rren )Gen: dsn n rulr sle=0dsn 0. es un es un dsns n rulr sle es un es un es un ) ).00 e snn qures n s quny re. b) k e snn qures n s quny re. ).0 7 e snn qures n s quny re. d) 0.0s e snn qures n s quny re 4..9) ).009 e snn qures n s quny re 4. b) k e snn qures n s quny re. ). 7 k e snn qures n s quny re. d) 00s e snn qures n s quny re 4.) Gen: en e rpper.7 Wd e rpper.4 Are e rpper Are en Wd A.7.4 A.8
3 CHAER.) Gen: k s s s s 0.) Gen: 0s k 00 0s (0)(0) ( )(0) 00 0 ( )(000) 00 0.s 000.) Gen: ( 0) (0.)(0) 0s s l nue 0s 0.s ()(0) (0.)(0) 00 (0.)(900) () (0.)(0) s.4) AR-() Gen: Gen: up 0s s s dn (0)() ( )() 90 ( )(9) s 4 s AR-()
4 4 (0)( ) ()( ) 4 0 dn 4 9 s up dn s s s.) AR-() Gen: Gen: 40s 0s s 0 (40) ( )() 40 4s ( 4)( ) (0) (40) s 40s AR-().) Gen: (40) () s 0.s 0 (0.)(0) (0) 0 s k.7) AR-() Gen: 0s 48k s 00 n 0s
5 Gen: Gen:. 0.s (.)(0) 800 r().s n 00s (.)(00) 4000 r().s 0s n 80s 0..s (.)(80) ) Gen: 0s s.9) Gen: s ( uprd) up up s 0 ( )() 0s ( ) (0) (0) " k k s.s ( )(800) (.) (.7) s
6 ( 0.4) (0) (.).) Gen: " " k k ".7s s ( )(800) (.) (.7) s (0) (.7) ( s 0.4)( ).) Gen: 0N s.) Gen: 0 k 47N s k CHAER.) Gen: k s ()() 0N
7 .4) Gen: 0N 0k.) Gen: up up up 0 0 s 0N s s.) Gen: 0 k ()() 4N 4 0 4N s k up 48k ()(48) () () N 00N.7) Gen: 48 () 48 4 () 0 0.4s 4k k s (4)() () 4 40 () 0 4.8N N 4 () 4 4.8s
8 .8) Gen: Ns 0N 0.9) Gen:.s k 0. s.) Gen: (0.)()() 0N 0.k r 0 0. s r (0.)() (0.) 9N CHAER 4 4.) Gen: N ln x-xs N ln y-xs 4N ln -e x-xs (As x N 4N x x y N N x y 7N 8.N n n n () 4 4.) Gen: 0N 0 x y x s x 0s(0 ) x 0(0.8) 4.4N x y x s n -e dren)
9 y sn x s 4.) Gen: 0sn(0 ) 0(0.) N N y N y y x y x y () 9N N n n () n (0.47) y x 4.) Gen: x s 0 s(0) N r 0. 0N rque uple r (0.) 0. 0(0.) N. x xs 4.4) Gen: 0N 0. 0(0.) N 4.) Gen: 0N 0 x 4.7) Gen:.8N 4.4N y N 0
10 4.8) Gen: k k ensn n upper srn ensn n ler srn y 0 ( ) 0 ( ) 4.9) Gen: ( )() 80N ()() 0N 00N 0. 0N (00)(0.) 0N ) Gen: k ()() 0N 0 0 0(0.) 0. 40N
11 CHAER.) Gen: 00k d 0. As e kn.) Gen: G d.7.7 d 0.007N As e kn 4 (00)(00) (0.) N G d G d. d G () k.) Gen:.4 R 70k.70 As e kn G R.7.77s.4) Gen:.s k.4 (.7 ) R 740k.740 As e kn.) Gen: G R R G (9.8)(.74 ) k k R 400k.4 00k.
12 4 As e kn G ( R ).7 4s.) Gen: (.4. ) k R 48700k 48.7 As e kn G R.7 0.7s.7) Gen: 4 (48.7 ) R 000k 4s G ( R ) As e kn.( R ) G 4 ( ) ) Gen: 4 As e kn.9) Gen: 4 G ( R ) G ( R ) G R 4 4G - R G - R As G R R= G pu n eq() R R R k -----() 80 k 8. As e kn ( R )
13 4 G ( R ).7 (.4 8. ) 7.4 s ( R ).) Gen: 7.4(.4 8. ) 740 s As e kn ( R ) G ( R ) s ( R ) 4000 k 4 4 (.4 4 ) 0.7(.4 4 ) 87 s CHAER.) Gen: 00N W As e kn W. W ( )(00 N) W 00J.) Gen: W 0N. E As e kn. E. E (0 N)( ). E 0J.) Gen: W kn N 0 s K. E As e kn K. E N
14 .4) Gen: s 00k K. E (00 k)(0 s ) K. E (00 k)(400 s ) K. E 40000J K. E 40kJ 00 0.k ) ) s. E K. E ) As e kn ). E N By rd equn n 0, ( ) (0) (). 0. E (0.)()(.). E.J As e kn K. E K. E (0.)() K. E (0.)() K. E.J.) Gen: ) ).s 40k K. E. E ) As e kn ) K. E K. E (40)(.) K. E (40)(.) K. E 4J As e kn. E.) Gen: Requured:. E 40()(). E 400J 4s 4000N As e kn W
15 ... (4000)(4) 000W kw.7) Gen: 0 00N 0s Requured: As e kn W. (00)(0) 0 0W.8) Gen: 0k 0s seps e ne sep 0..9) Gen: Requured: 0. 4 As e kn W (0)()(4) 0 0W 00k s As e kn W. W. (00)()() 00W
16 .) Gen: p 74W n 00s 800k W % Eeny As e kn W W. W (74)(00) W 44700J % Eeny upu npu upu upu (800)()() upu 0, 000W npu 44700W % Eeny % Eeny 0, , 00.8% ) Gen: CHAER k k densy 7.) Gen: k ss lue lre 00k 90k 00k k 90k k 90k k.09.09d.09lre
17 7.) ) Gen: b) Gen: k 800k k 800k k 800k ) Gen: k 00k 0.k 00k 0.k 00k.77 0.k 900k 0.k 900k 0.k 900k ) Gen: k 7.) Gen:.k 0 (.)(0) k 08k 7N. A A. 7.) Gen: A 7N. 0N 4 N A ( ) A A 4 0N 4 N
18 CA 9 HYIC NUERICA 7.7) Gen: A = =.. 4 A = =. 0 0 = = =. = 00 = k = 7. = 0.07 = ρ = = A = = = ( ) ks = N N =. = 778N ρ = 7.8) Gen: 4 k ρ =. ρ = 889k = ( ) = ss y ρ =. = ρ = " = = 0 (.) = ss u y = = 8.7 ss y = = =.7 7.9) Gen: ρ = = (.7) (.) = = 8N =.4N ρ = 00k Densy bje = ρ = erl = ρ As s lue s ery lse 700., erl s Alunu. 7.) Gen: = Densy blk = ρ = 0. ρ 8 ρ = ρ = (.77 00) k ρ = 77k 0. ρ = k = 0. k 00 Densy lqud = ρ = ρl = k = =.0N = = l =.0 = () = 0.0k l k ps://.urbrend.rdpress. e 8
19 (0.9 ) l () 7.) Gen: l l l l.0 (0.9 )() 40 D , 000N d 0.0 A A A r r D 0. (0.09) 4 d 0.0 (0.0009) 4 A 0, N (0.0009) 4 (0.09) 4 7.) Gen: A Y Y Y Y 4000N sress srn A A (4000)() Y ( )( Y Y N N )
20 8.) Gen: C 0 8.) Gen:.8C.8(0) C ( C) ( K) 8.) Gen: ( ) 98..8C 98..8C.8C 98. C..8 ( C) 7 C K K ( K) K 0 C 7K C CHAER 8 0 C 0 7 9K. ( ) K (. )()(9 7) ( )(0) ) Gen:. C 7 88K 40 C 40 7 K.7 K ( ) ( ) ( 88) K.{ (.7 )}.{ 9.7 }. 8.) Gen: 0.k C 7 8K C 7 8K 400Jk K ( ) (0.)(400)(8 8) (0)(), 00J 8.) Gen: 00Js 00 0.k 0 C 0 7 9K
21 90 C 90 7 K 400Jk K 8.7) Gen: (0.)(400)( 9) (70) 00 H 8.8s 8.8) Gen: 0000J, 000 jk H H 0000, k k C 7 K C 7 8K 0Jk K 400Jk K H, 000Jk He Requred ne eprure e r - C 0 C (0.)(0)(7 ) ()() 0J He Requred ne e 0 C er 0 C H (0.)(,000),00 J He Requred ne e eprure er r 0 C C (0.)(400)(8 7) (40)( ) 400J 0, J 8.9) Gen: 0 0.k H 0 8.) Gen:. C Jk (0.)(. ). 0.00k 00 0.k s H H C 400Jk K J. Jk
22 CA 9 HYIC NUERICA = = H (0.00)(. ) = 00J = (0.00)(400)(90) = 890J s = = = + s = 90J = = ( + )( )( ) s 90 = (0.0)(400)( ) 90 ( ) = =. + =. C 9.) Gen: = 0 = 0. A = 00 CHAER 9 = C = + 7 = 08K = C = + 7 = 88K k = 0.W K = ka( ) = (0.)(00)(08 88) = = 0. = 000Js 9.) Gen: = ur = 0 0se = 00s = 0.8 = A =. = = C = + 7 = 98K = C = + 7 = 78K k = 0.8W K = ka( ) = ka( ) = (00)(0.8)()(98 78) = = 0. 7 =. J ps://.urbrend.rdpress. e
lim Prob. < «_ - arc sin «1 / 2, 0 <_ «< 1.
ON THE NUMBER OF POSITIVE SUMS OF INDEPENDENT RANDOM VARIABLES P. ERDÖS AND M. KAC 1 1. Introduction. In a recent paper' the authors have introduced a method for proving certain limit theorems of the theory
LA PRISE DE CALAIS. çoys, çoys, har - dis. çoys, dis. tons, mantz, tons, Gas. c est. à ce. C est à ce. coup, c est à ce
> ƒ? @ Z [ \ _ ' µ `. l 1 2 3 z Æ Ñ 6 = Ð l sl (~131 1606) rn % & +, l r s s, r 7 nr ss r r s s s, r s, r! " # $ s s ( ) r * s, / 0 s, r 4 r r 9;: < 10 r mnz, rz, r ns, 1 s ; j;k ns, q r s { } ~ l r mnz,
w x a f f s t p q 4 r u v 5 i l h m o k j d g DT Competition, 1.8/1.6 Stainless, Black S, M, L, XL Matte Raw/Yellow
HELION CARBON TEAM S, M, L, XL M R/Y COR XC Py, FOC U Cn F, 110 T Innn Dn 27. AOS Snn Sy /F Ln, P, 1 1/8"-1 1/2" In H T, n 12 12 M D F 32 FLOAT 27. CTD FIT /A K, 110 T, 1QR, / FIT D, L & Rn A, T Ay S DEVICE
176 5 t h Fl oo r. 337 P o ly me r Ma te ri al s
A g la di ou s F. L. 462 E l ec tr on ic D ev el op me nt A i ng er A.W.S. 371 C. A. M. A l ex an de r 236 A d mi ni st ra ti on R. H. (M rs ) A n dr ew s P. V. 326 O p ti ca l Tr an sm is si on A p ps
P a g e 5 1 of R e p o r t P B 4 / 0 9
P a g e 5 1 of R e p o r t P B 4 / 0 9 J A R T a l s o c o n c l u d e d t h a t a l t h o u g h t h e i n t e n t o f N e l s o n s r e h a b i l i t a t i o n p l a n i s t o e n h a n c e c o n n e
= T. (kj/k) (kj/k) 0 (kj/k) int rev. Chapter 6 SUMMARY
Capter 6 SUMMARY e second la of termodynamics leads to te definition of a ne property called entropy ic is a quantitative measure of microscopic disorder for a system. e definition of entropy is based
Ranking accounting, banking and finance journals: A note
MPRA Munich Personal RePEc Archive Ranking accounting, banking and finance ournals: A note George Halkos and Nickolaos Tzeremes University of Thessaly, Department of Economics January 2012 Online at https://mpra.ub.uni-muenchen.de/36166/
AN UNDERGRADUATE LECTURE ON THE CENTRAL LIMIT THEOREM
AN UNDERGRADUATE LECTURE ON THE CENTRAL LIMIT THEOREM N.V. KRYLOV In the first section we explain why the central limit theorem for the binomial 1/2 distributions is natural. The second section contains
R th is the Thevenin equivalent at the capacitor terminals.
Chaper 7, Slun. Applyng KV Fg. 7.. d 0 C - Takng he derae f each erm, d 0 C d d d r C Inegrang, () ln I 0 - () I 0 e - C C () () r - I 0 e - () V 0 e C C Chaper 7, Slun. h C where h s he Theenn equalen
T h e C S E T I P r o j e c t
T h e P r o j e c t T H E P R O J E C T T A B L E O F C O N T E N T S A r t i c l e P a g e C o m p r e h e n s i v e A s s es s m e n t o f t h e U F O / E T I P h e n o m e n o n M a y 1 9 9 1 1 E T
Contents ... FREE! 4 6 BONUS Activity Pages! Additional worksheets for your students EASY MARKING ANSWER KEY GRAPHIC ORGANIZERS...
W A I? W? W S A? L C U C. W? W? W L C? R A I. W? W? A R U C. W A R? W A? W? W? W? W A I? U C W A? U S S. C, C. W? W? W? W? B S U C. W B S? W E U S? H A? W? W? W? W L C? W W z O R?, A C? E. H C U S W A
arxiv: v1 [math.pr] 12 Jan 2017
![arxiv: v1 [math.pr] 12 Jan 2017](/thumbs/94/119397937.jpg "arxiv: v1 [math.pr] 12 Jan 2017")
Central limit theorem for the Horton-Strahler bifurcation ratio of general branch order Ken Yamamoto Department of Physics and arth Sciences, Faculty of Science, University of the Ryukyus, 1 Sembaru, Nishihara,
trawhmmry ffimmf,f;wnt
r nsr rwry fff,f;wn My 26, $51 Swe, k "Te Srwberry Cp f e Vr,, c) [ re ers 6 (, r " * f rn ff e # s S,r,* )er*,3n*,.\ ) x 8 2 n v c e 6 r D r, } e ;s 1 :n..< Z r : 66 3 X f; 1r_ X r { j r Z r 1r 3r B s
OH BOY! Story. N a r r a t iv e a n d o bj e c t s th ea t e r Fo r a l l a g e s, fr o m th e a ge of 9
OH BOY! O h Boy!, was or igin a lly cr eat ed in F r en ch an d was a m a jor s u cc ess on t h e Fr en ch st a ge f or young au di enc es. It h a s b een s een by ap pr ox i ma t ely 175,000 sp ect at
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
mml LOWELL. MICHIGAN, THURSDAY, JULY 7,1932 Turkey Dinner Sets Lowell Board of Trade Going Strong For Show Boat on the Flat
D D U DN N U N 8 N UDY UY 79 U XXXX X D N [ ND D N D N D x N N ( U N : ( N D : z q N N ( D N ; N z q 5 x N N q x q x q q (: Y x z x N ( x ) N D N x \ x U x ; U x U ; x x : Y U q D N : ND NY ; NY ; z 5
Banach Spaces II: Elementary Banach Space Theory
BS II c Gabriel Nagy Banach Spaces II: Elementary Banach Space Theory Notes from the Functional Analysis Course (Fall 07 - Spring 08) In this section we introduce Banach spaces and examine some of their
h : sh +i F J a n W i m +i F D eh, 1 ; 5 i A cl m i n i sh» si N «q a : 1? ek ser P t r \. e a & im a n alaa p ( M Scanned by CamScanner
m m i s t r * j i ega>x I Bi 5 n ì r s w «s m I L nk r n A F o n n l 5 o 5 i n l D eh 1 ; 5 i A cl m i n i sh» si N «q a : 1? { D v i H R o s c q \ l o o m ( t 9 8 6) im a n alaa p ( M n h k Em l A ma
Why CEHCH? Completion of this program provides participants direct access to sit for the NAB HCBS exam.
Wy? T fus f s pns n us s p ssn pnns f w-bn & uny bs ss pg. Ts us s p n n ff s s bs p wn & uny bs ss pfssn. W uny n s f O s n n ns qu f sp u D, sussfu pn f s n us w nb yu ns n knwg bs. T O un f & sp n O
33 3 Vol.33,No JournaloftheMeteorologicalSciences Jun.,2013 M N,N,:, ES `.,2013,33(3): SUXiaoyong,GAOTaichang,LIUXichuan,etal.N
33 3 Vol.33,No.3 2013 6 JournaloftheMeteorologicalSciences Jun.,2013 M N,N,:,.3 4 3 ES `.,2013,33(3):282 288. SUXiaoyong,GAOTaichang,LIUXichuan,etal.Numericalsimulationoftheraindrophorizontalvelocityafectedbythewind.Jour
Abstract Algebra II Groups ( )
Abstract Algebra II Groups ( ) Melchior Grützmann / melchiorgfreehostingcom/algebra October 15, 2012 Outline Group homomorphisms Free groups, free products, and presentations Free products ( ) Definition
P a g e 3 6 of R e p o r t P B 4 / 0 9
P a g e 3 6 of R e p o r t P B 4 / 0 9 p r o t e c t h um a n h e a l t h a n d p r o p e r t y fr om t h e d a n g e rs i n h e r e n t i n m i n i n g o p e r a t i o n s s u c h a s a q u a r r y. J
Ground state half life. Ground state half life 34 Cl 32.2 minutes 1.53 seconds. 169 Re 16 seconds 8.1 seconds. 177 Lu days 6.
RDCH 70 Name: Quiz ssigned 5 Sep, Due 7 Sep Chart of the nuclides (up to and including page - of the lecture notes) Use the chart of the nuclides, the readings on the chart of the nuclides, table of the
Perturbative QCD. Chul Kim. Seoultech. Part I : Introduction to QCD Structure functions for DIS
Perturbative QCD Part I : Introduction to QCD Structure functions for DIS Chul Kim Seoultech Open KIAS, Pyeong-Chang Summer Institute 2013 Pyeong-Chang, Alpensia Resort, July 8, 2013 QCD QCD Lagrangian
Circle the letters only. NO ANSWERS in the Columns!
Chemistry 1304.001 Name (please print) Exam 5 (100 points) April 18, 2018 On my honor, I have neither given nor received unauthorized aid on this exam. Signed Date Circle the letters only. NO ANSWERS in
2015 Sectional Physics Exam Solution Set
. Crrec answer: D Ne: [quan] denes: uns quan WYSE cadec Challenge 05 Secnal Phscs Ea SOLUTION SET / / / / rce lengh lengh rce enu ass lengh e a) / ass ass b) energ c) wrk lengh e pwer energ e d) (crrec
Chemistry 431 Practice Final Exam Fall Hours
Chemistry 431 Practice Final Exam Fall 2018 3 Hours R =8.3144 J mol 1 K 1 R=.0821 L atm mol 1 K 1 R=.08314 L bar mol 1 K 1 k=1.381 10 23 J molecule 1 K 1 h=6.626 10 34 Js N A = 6.022 10 23 molecules mol
Exhibit 2-9/30/15 Invoice Filing Page 1841 of Page 3660 Docket No
xhibit 2-9/3/15 Invie Filing Pge 1841 f Pge 366 Dket. 44498 F u v 7? u ' 1 L ffi s xs L. s 91 S'.e q ; t w W yn S. s t = p '1 F? 5! 4 ` p V -', {} f6 3 j v > ; gl. li -. " F LL tfi = g us J 3 y 4 @" V)
/ " $ : Circle the letters only. NO ANSWERS in the Columns! (3 points each)
Chemistry 1304.001 Name (please print) Exam 4 (100 points) April 12, 2017 On my honor, I have neither given nor received unauthorized aid on this exam. Signed Date Circle the letters only. NO ANSWERS in
SAVE THESE INSTRUCTIONS
SAVE ESE NSUNS FFEE AE ASSEMY NSUNS SYE #: 53SN2301AS ASSEME N A FA, PEED SUFAE PPS EAD SEWDVE NEEDED F ASSEMY; N NUDED PA S FGUE UANY DESPN AA 1 P P 1 P EF SDE FAME 1 P G SDE FAME D 1 P A PANE E 2 PS
Magic Letterland. Welcome agic. the. Letterland!
Mi Leernd Weme i Leernd! e 5 Mi Leernd / 2. ire nd wrie e eers. s e z e u m mpuer 3. Jin e ds nd ur e piure. 18 17 19 20 21 22 23 24 16 11 25 26 14 15 13 10 12 6 9 1 8 2 3 30 7 4 29 5 28 27 2. ire nd wrie
Uniform Convergence and Series of Functions
Uniform Convergence and Series of Functions James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University October 7, 017 Outline Uniform Convergence Tests
221B Lecture Notes Steepest Descent Method
Gamma Function B Lecture Notes Steepest Descent Method The best way to introduce the steepest descent method is to see an example. The Stirling s formula for the behavior of the factorial n! for large
Erlkönig. t t.! t t. t t t tj "tt. tj t tj ttt!t t. e t Jt e t t t e t Jt
Gsng Po 1 Agio " " lkö (Compl by Rhol Bckr, s Moifi by Mrk S. Zimmr)!! J "! J # " c c " Luwig vn Bhovn WoO 131 (177) I Wr Who!! " J J! 5 ri ris hro' h spä h, I urch J J Nch rk un W Es n wil A J J is f
Contents ... FREE! 4 6 BONUS Activity Pages! Additional worksheets for your students EASY MARKING ANSWER KEY GRAPHIC ORGANIZERS...
I, Z. H,. R. B,. C,. x. I S Z,, P W.? P. G. G, S,. CONFLICT RESOLUTION SKITS.. L S G L. H T : P.. T H Tx. C - COPYING, ILLUSTRATING I - G L. EMBELLISHING. C ANTI-BULLYING STRATEGIES B I: G F,,. R H U S.
Chapter 27 Solutions. )( m 3 = = s = 3.64 h
Chapter 27 Solutions 27.1 I Q t N Q e Q I t (30.0 10 6 A)(40.0 s) 1.20 10 3 C 1.20 10 3 C 1.60 10 19 C/electron 7.50 1015 electrons *27.2 The atomic weight of silver 107.9, and the volume V is V (area)(thickness)
CHEM 172 EXAMINATION 1. January 15, 2009
CHEM 17 EXAMINATION 1 January 15, 009 Dr. Kimberly M. Broekemeier NAME: Circle lecture time: 9:00 11:00 Constants: c = 3.00 X 10 8 m/s h = 6.63 X 10-34 J x s J = kg x m /s Rydberg Constant = 1.096776 x
Lecture (5) Power Factor,threephase circuits, and Per Unit Calculations
Lecture (5) Power Factor,threephase circuits, and Per Unit Calculations 5-1 Repeating the Example on Power Factor Correction (Given last Class) P? Q? S? Light Motor From source 1000 volts @ 60 Htz 10kW
2010 Sectional Physics Solution Set
. Crrec nwer: D WYSE CDEMIC CHLLENGE Secnl hyc E 00 Slun Se y 0 y 4.0 / 9.8 /.45 y. Crrec nwer: y 8 0 / 8 /. Crrec nwer: E y y 0 ( 4 / ) ( 4.9 / ) 5.6 y y 4. Crrec nwer: E 5. Crrec nwer: The e rce c n
Series. Definition. a 1 + a 2 + a 3 + is called an infinite series or just series. Denoted by. n=1
Definition a 1 + a 2 + a 3 + is called an infinite series or just series. Denoted by a n, or a n. Chapter 11: Sequences and, Section 11.2 24 / 40 Given a series a n. The partial sum is the sum of the first
A Shrinking Core Model for Steam Hydration of CaO-Based Sorbents Cycled for CO2 Capture: Supplementary Information
A Shrinking Core Model for Steam Hydration of CaO-Based Sorbents Cycled for CO2 Capture: Supplementary Information John Blamey 1, Ming Zhao 2, Vasilije Manovic 3, Edward J. Anthony 3, Denis R. Dugwell
(C) Pavel Sedach and Prep101 1
(C) Pavel Sedach and Prep101 1 (C) Pavel Sedach and Prep101 1 (C) Pavel Sedach and Prep101 2 (C) Pavel Sedach and Prep101 2 (C) Pavel Sedach and Prep101 3 (C) Pavel Sedach and Prep101 3 (C) Pavel Sedach
TRASH ENCLOSURE WITH SOLID GATE 4 STORY BUSINESS / RESIDENTIAL BUILDING CONTAINING 2 BUSINESS SPACES AND 6 DWELLING UNITS 6' - 0"
NSN N. PUN WY R. P 0. SG S 4 SRY USNSS / RSN UNG NNNG USNSS SPS N 6 WNG UNS RS NSUR W S G.. RSRV PRKNG $50 N SGN RV S (7) UR PRKNG SPS ' - PRPRY N M N, YP PU Y SG RNGS S GNR NS 6" G UR rchitecture nteriors
NAME: SECOND EXAMINATION
1 Chemistry 64 Winter 1994 NAME: SECOND EXAMINATION THIS EXAMINATION IS WORTH 100 POINTS AND CONTAINS 4 (FOUR) QUESTIONS THEY ARE NOT EQUALLY WEIGHTED! YOU SHOULD ATTEMPT ALL QUESTIONS AND ALLOCATE YOUR
Numerical Sequences and Series
Numerical Sequences and Series Written by Men-Gen Tsai email: b89902089@ntu.edu.tw. Prove that the convergence of {s n } implies convergence of { s n }. Is the converse true? Solution: Since {s n } is
April 26, Applied mathematics PhD candidate, physics MA UC Berkeley. Lecture 4/26/2013. Jed Duersch. Spd matrices. Cholesky decomposition
Applied mathematics PhD candidate, physics MA UC Berkeley April 26, 2013 UCB 1/19 Symmetric positive-definite I Definition A symmetric matrix A R n n is positive definite iff x T Ax > 0 holds x 0 R n.
Geometric Predicates P r og r a m s need t o t es t r ela t ive p os it ions of p oint s b a s ed on t heir coor d ina t es. S im p le exa m p les ( i
Automatic Generation of SS tag ed Geometric PP red icates Aleksandar Nanevski, G u y B lello c h and R o b ert H arp er PSCICO project h ttp: / / w w w. cs. cm u. ed u / ~ ps ci co Geometric Predicates
Chapter 3. Second Order Linear PDEs
Chapter 3. Second Order Linear PDEs 3.1 Introduction The general class of second order linear PDEs are of the form: ax, y)u xx + bx, y)u xy + cx, y)u yy + dx, y)u x + ex, y)u y + f x, y)u = gx, y). 3.1)
: a. bahwa untuk mendukung kelancaran pelaksanaan
MTR PDAAUAA APARATUR ARA DA RORMAS BROKRAS RPUBK DOSA KPUTUSA MTR PDAAUAA APARATUR ARA DA RORMAS BROKRAS OMOR 29 TAHU 2018 TTA KBUTUHA PAA APARATUR SP ARA D KUA (ABUPAT ATUA TA.TTU AARA 2018 MTR PDAAUAA
Digital Integrated CircuitDesign
Dgal Inegraed CrcuDesgn Lecure 6 BJT Inverer Swchng Tmes µ s 01. 1 3 4 6 2 Adb Abrshamfar EE Deparmen IUST Cnens BJT Inverer Cuff Regn ( 1 ) Acve Regn ( 1 2 ) Sauran Regn ( 3 4 ) Acve Regn ( 4 ) Recvery
i;\-'i frz q > R>? >tr E*+ [S I z> N g> F 'x sa :r> >,9 T F >= = = I Y E H H>tr iir- g-i I * s I!,i --' - = a trx - H tnz rqx o >.F g< s Ire tr () -s
5 C /? >9 T > ; '. ; J ' ' J. \ ;\' \.> ). L; c\ u ( (J ) \ 1 ) : C ) (... >\ > 9 e!) T C). '1!\ /_ \ '\ ' > 9 C > 9.' \( T Z > 9 > 5 P + 9 9 ) :> : + (. \ z : ) z cf C : u 9 ( :!z! Z c (! $ f 1 :.1 f.
Solving Linear Systems Using Gaussian Elimination. How can we solve
Solving Linear Systems Using Gaussian Elimination How can we solve? 1 Gaussian elimination Consider the general augmented system: Gaussian elimination Step 1: Eliminate first column below the main diagonal.
221A Lecture Notes Steepest Descent Method
Gamma Function A Lecture Notes Steepest Descent Method The best way to introduce the steepest descent method is to see an example. The Stirling s formula for the behavior of the factorial n! for large
Autoregressive (AR) Modelling
Autoregressive (AR) Modelling A. Uses of AR Modelling () Alications (a) Seech recognition and coding (storage) (b) System identification (c) Modelling and recognition of sonar, radar, geohysical signals
THIS PAGE DECLASSIFIED IAW EO IRIS u blic Record. Key I fo mation. Ma n: AIR MATERIEL COMM ND. Adm ni trative Mar ings.
T H S PA G E D E CLA SSFED AW E O 2958 RS u blc Recod Key fo maon Ma n AR MATEREL COMM ND D cumen Type Call N u b e 03 V 7 Rcvd Rel 98 / 0 ndexe D 38 Eneed Dae RS l umbe 0 0 4 2 3 5 6 C D QC d Dac A cesson
Equation of State of Dense Matter
Department of Physics & Astronomy 449 ESS Bldg. Stony Brook University April 25, 2017 Nuclear Astrophysics James.Lattimer@Stonybrook.edu Zero Temperature Nuclear Matter Expansions Cold, bulk, symmetric
Direct Current Circuits
Eler urren (hrges n Moon) Eler urren () The ne moun of hrge h psses hrough onduor per un me ny pon. urren s defned s: Dre urren rus = dq d Eler urren s mesured n oulom s per seond or mperes. ( = /s) n
ieski. a n d H. A. Lange.
G 34 D 0 D 90 : 5S D Vz S D NEWS W Vz z F D < - ;»( S S C S W C - z z! L D F F V Q4 R U O G P O N G-34 q O G
CHEM 108 (Spring-2008) Exam. 3 (105 pts)
CHEM 08 (Spring-008) Exam. (05 pts) Name: --------------------------------------------------------------------------, CLID # -------------------------------- LAST NAME, First (Circle the alphabet segment
2.6 The optimum filtering solution is defined by the Wiener-Hopf equation
.6 The optimum filtering solution is defined by the Wiener-opf equation w o p for which the minimum mean-square error equals J min σ d p w o () Combine Eqs. and () into a single relation: σ d p p 1 w o
0# E % D 0 D - C AB
5-70,- 393 %& 44 03& / / %0& / / 405 4 90//7-90/8/3 ) /7 0% 0 - @AB 5? 07 5 >0< 98 % =< < ; 98 07 &? % B % - G %0A 0@ % F0 % 08 403 08 M3 @ K0 J? F0 4< - G @ I 0 QR 4 @ 8 >5 5 % 08 OF0 80P 0O 0N 0@ 80SP
Steady State Errors. Recall the closed-loop transfer function of the system, is
Steady State Errors Outline What is steady-state error? Steady-state error in unity feedback systems Type Number Steady-state error in non-unity feedback systems Steady-state error due to disturbance inputs
Cartan MASAs and Exact Sequences of Inverse Semigroups
Cartan MASAs and Exact Sequences of Inverse Semigroups Adam H. Fuller (University of Nebraska - Lincoln) joint work with Allan P. Donsig and David R. Pitts NIFAS Nov. 2014, Des Moines, Iowa Cartan MASAs
CHEM 10113, Quiz 5 October 26, 2011
CHEM 10113, Quiz 5 October 26, 2011 Name (please print) All equations must be balanced and show phases for full credit. Significant figures count, show charges as appropriate, and please box your answers!
Chapter 2 Speech Production Model
Chapter 2 Speech Production Model Abstract The continuous speech signal (air) that comes out of the mouth and the nose is converted into the electrical signal using the microphone. The electrical speech
FORMATION EVALUATION PETE 321
FORMATION EVALUATION PETE 321 CROSSPLOTS (Porosity and Lithology) Summer 2010 TWO-MEASUREMENT POROSITY CROSSPLOTS Two measurements determine two unknowns Formations with one lithology Lithology Porosity
Contents ... FREE! 4 6 BONUS Activity Pages! Additional worksheets for your students EASY MARKING ANSWER KEY GRAPHIC ORGANIZERS...
I, Z. H,. R. B,. C,. x. I S Z,, P W.? P. G. G, S,. CONLIC RESOLUION SKIS.. L S G L. H : P.. H x. C - COPYING, ILLUSRAING I - G L. EMBELLISHING. C ANI-BULLYING SRAEGIES B I: G,,. R H U S. N, N Mx A. P,
CAPWAP Introduction. 2016, Pile Dynamics, Inc. CAPWAP is a registered trademark of Pile Dynamics, Inc. Load (kn) Displacement (mm) Pile Top Bottom
CAPWAP Introduction Load (kn) 0.0 1000.0 2000.0 3000.0 4000.0 0.00 Pile Top Bottom Displacement (mm) 5.00 10.00 Ru = Rs = Rb = Dy = Dx = 3425.2 kn 1458.6 kn 1966.6 kn 18.8 mm 18.8 mm 15.00 20.00 2016,
Advanced Placement. Chemistry. Integrated Rates
Advanced Placement Chemistry Integrated Rates 204 47.90 9.22 78.49 (26) 50.94 92.9 80.95 (262) 52.00 93.94 83.85 (263) 54.938 (98) 86.2 (262) 55.85 0. 90.2 (265) 58.93 02.9 92.2 (266) H Li Na K Rb Cs Fr
Numerical Methods for Problems with Moving Fronts Orthogonal Collocation on Finite Elements
Electronic Text Provided with the Book Numerical Methods for Problems with Moving Fronts by Bruce A. Finlayson Ravenna Park Publishing, Inc., 635 22nd Ave. N. E., Seattle, WA 985-699 26-524-3375; ravenna@halcyon.com;www.halcyon.com/ravenna
F l a s h-b a s e d S S D s i n E n t e r p r i s e F l a s h-b a s e d S S D s ( S o-s ltiad t e D r i v e s ) a r e b e c o m i n g a n a t t r a c
L i f e t i m e M a n a g e m e n t o f F l a-b s ah s e d S S D s U s i n g R e c o v e r-a y w a r e D y n a m i c T h r o t t l i n g S u n g j i n L e, e T a e j i n K i m, K y u n g h o, Kainmd J
Elastic Fields of Dislocations in Anisotropic Media
Elastic Fields of Dislocations in Anisotropic Media a talk given at the group meeting Jie Yin, David M. Barnett and Wei Cai November 13, 2008 1 Why I want to give this talk Show interesting features on
o C *$ go ! b», S AT? g (i * ^ fc fa fa U - S 8 += C fl o.2h 2 fl 'fl O ' 0> fl l-h cvo *, &! 5 a o3 a; O g 02 QJ 01 fls g! r«'-fl O fl s- ccco
> p >>>> ft^. 2 Tble f Generl rdnes. t^-t - +«0 -P k*ph? -- i t t i S i-h l -H i-h -d. *- e Stf H2 t s - ^ d - 'Ct? "fi p= + V t r & ^ C d Si d n. M. s - W ^ m» H ft ^.2. S'Sll-pl e Cl h /~v S s, -P s'l
Answer Key THERMODYNAMICS TEST (a) 33. (d) 17. (c) 1. (a) 25. (a) 2. (b) 10. (d) 34. (b) 26. (c) 18. (d) 11. (c) 3. (d) 35. (c) 4. (d) 19.
HERMODYNAMICS ES Answer Key. (a) 9. (a) 7. (c) 5. (a). (d). (b) 0. (d) 8. (d) 6. (c) 4. (b). (d). (c) 9. (b) 7. (c) 5. (c) 4. (d). (a) 0. (b) 8. (b) 6. (b) 5. (b). (d). (a) 9. (a) 7. (b) 6. (a) 4. (d).
Chapter 5 Molecular Vibrations and Time-Independent Perturbation Theory Homework Solutions
Chapter 5 Molecular Vibrations and ime-independent Perturbation heory Homework Solutions. 4 y Form of solution: n 3 4 y anx a axax a3x a4x... n Specific Recursion Relations dy 3 aax3a3x 4 a4x... a 6a3x
Use 10 m/s 2 for the acceleration due to gravity.
ANSWERS Prjecle mn s he ecrl sum w ndependen elces, hrznl cmpnen nd ercl cmpnen. The hrznl cmpnen elcy s cnsn hrughu he mn whle he ercl cmpnen elcy s dencl ree ll. The cul r nsnneus elcy ny pn lng he prblc
Lecture 3: Resistive forces, and Energy
Lecure 3: Resisive frces, and Energy Las ie we fund he velciy f a prjecile ving wih air resisance: g g vx ( ) = vx, e vy ( ) = + v + e One re inegrain gives us he psiin as a funcin f ie: dx dy g g = vx,
8. Relax and do well.
CHEM 1515 Exam II John II. Gelder October 14, 1993 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 8 different pages. The last two pages include a periodic table, a
Faculty of Engineering
Faculty f Engneerng DEPARTMENT f ELECTRICAL AND ELECTRONIC ENGINEERING EEE 223 Crcut Thery I Instructrs: M. K. Uygurğlu E. Erdl Fnal EXAMINATION June 20, 2003 Duratn : 120 mnutes Number f Prblems: 6 Gd
This ODE arises in many physical systems that we shall investigate. + ( + 1)u = 0. (λ + s)x λ + s + ( + 1) a λ. (s + 1)(s + 2) a 0
Legendre equation This ODE arises in many physical systems that we shall investigate We choose We then have Substitution gives ( x 2 ) d 2 u du 2x 2 dx dx + ( + )u u x s a λ x λ a du dx λ a λ (λ + s)x
COMPOSITION OF LINEAR FRACTIONAL TRANSFORMATIONS IN TERMS OF TAIL SEQUENCES
PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume u7. Number 1, Mav 19K6 COMPOSITION OF LINEAR FRACTIONAL TRANSFORMATIONS IN TERMS OF TAIL SEQUENCES LISA JACOBSEN Abstract. We consider sequences
LECTURE 15, WEDNESDAY
LECTURE 15, WEDNESDAY 31.03.04 FRANZ LEMMERMEYER 1. The Filtration of E (1) Let us now see why the kernel of reduction E (1) is torsion free. Recall that E (1) is defined by the exact sequence 0 E (1)
SHARP BOUNDS FOR PROBABILITIES WITH GIVEN SHAPE INFORMATION
R u t c o r Research R e p o r t SHARP BOUNDS FOR PROBABILITIES WITH GIVEN SHAPE INFORMATION Ersoy Subasi a Mine Subasi b András Prékopa c RRR 4-006, MARCH, 006 RUTCOR Rutgers Center for Operations Research
INSTALLED BY C.F. C.I. TS-1 NO WORK LAY IN STACK PATTERN (STACK BOND) LAY IN BRICK PATTERN (RUNNING BOND) (CUSTOM ORDER -6 WEEK LEAD TIME)
H UNN - N K N NH HDU V.. VND UNHD.. N UNHD.. N ND PDU NUU DPN Z U N UPPD N K - W NKK. NU.... Q- K--HU QU D QU V 0Q40 D Z 5" x x /2" -3 U V H H N ( NDD) (U: H N) ( H N)..H. N K N Q- K--HU QU D QU 0Q40 D
General Mathematics Vol. 16, No. 1 (2008), A. P. Madrid, C. C. Peña
General Mathematics Vol. 16, No. 1 (2008), 41-50 On X - Hadamard and B- derivations 1 A. P. Madrid, C. C. Peña Abstract Let F be an infinite dimensional complex Banach space endowed with a bounded shrinking
Chemistry 2 Exam Roane State Academic Festival. Name (print neatly) School
Name (print neatly) School There are fifteen question on this exam. Each question is weighted equally. n the answer sheet, write your name in the space provided and your answers in the blanks provided.
Multicomponent diffusion in gases and plasma mixtures
High Temperatures ^ High Pressures, 2002, volume 34, pages 109 ^ 116 15 ECTP Proceedings pages 1337 ^ 1344 DOI:10.1068/htwu73 Multicomponent diffusion in gases and plasma mixtures Irina A Sokolova Institute
O M N I W I L L I A M P E N N H O T E L H O L I D A Y
N P N N D Y 2 0 1 8 P V N D Y P K G $70 P R P RN & NUD R y, y q x é - y y y NNN & $ 7 $300 2 x $ 1 1 $1 5 y y $ 1 0 U $ 7 * j 23% x x. Y D Y P K G $85 P R P RN & NUD R & y, y, q x é - y y y NNN & $ 7 $300
filing for office of CO AN Co L POlS R 4 home phone epqropriate elections officials ORS Filing for State Voters Pamphlet Fi rstday
Fn nddy npn nn L v 6 R 49 h nn pb d nd y b pbhd pdd p yp pn by n b k nk nb d ndd n d M LG hw n hd pp n b H MB L n A L P R 4 d dp pn nb ddd y 8 zp d R D R A b A ka DL d ny d h phn d AK L x dd W b W n ddwh
Physical Chemistry I CHEM 4641 Final Exam 13 questions, 30 points
Physical Chemistry I CHEM 4641 Final Exam 13 questions, 30 points Name: KEY Gas constant: R = 8.314 J mol -1 K -1 = 0.008314 kj mol -1 K -1. Boltzmann constant k = 1.381 10-23 J/K = 0.6950 cm -1 /K h =
If anything confuses you or is not clear, raise your hand and ask!
CHM 1045 Dr. Light s Section December 10, 2002 FINAL EXAM Name (please print) Recitation Section Meeting Time This exam consists of six pages. Make sure you have one of each. Print your name at the top
Chapter 3: Stoichiometry
Chapter 3: Stoichiometry Chem 6A Michael J. Sailor, UC San Diego 1 Announcements: Thursday (Sep 29) quiz: Bring student ID or we cannot accept your quiz! No notes, no calculators Covers chapters 1 and
Chapter 6 : Gibbs Free Energy
Wnter 01 Chem 54: ntrductry hermdynamcs Chapter 6 : Gbbs Free Energy... 64 Defntn f G, A... 64 Mawell Relatns... 65 Gbbs Free Energy G(,) (ure substances)... 67 Gbbs Free Energy fr Mtures... 68 ΔG f deal
I N A C O M P L E X W O R L D
IS L A M I C E C O N O M I C S I N A C O M P L E X W O R L D E x p l o r a t i o n s i n A g-b eanste d S i m u l a t i o n S a m i A l-s u w a i l e m 1 4 2 9 H 2 0 0 8 I s l a m i c D e v e l o p m e
Chemistry 2000 Fall 2017 Final Examination
Time: 3 hours hemistry 2000 Fall 2017 Final Examination Total marks: 105 Aids permitted: calculator (wireless communication capabilities OFF), molecular model kit. Significant figures: I will specifically
Contents ... FREE! 4 6 BONUS Activity Pages! Additional worksheets for your students EASY MARKING ANSWER KEY GRAPHIC ORGANIZERS...
R SLOGAN. H, I I! F, R. R G T. S S STAY IN OUR MINDS. W? Y CURSIVE LETTERS.,. W? M. K R R R: W,?. B. W? A? H? P- R. H, R. W? W? W? W? F I, R Q, A 8, R. Y. H R: F ( ). W? I L.?,. H? B C R Q, A 8 W? A. W.
Gary Callicoat Available All Day! Blue Plate Specials Monday: Gary s 3-Way Tuesday: Taco Salad Beef Stroganoff Wednesday: Spaghetti n Meatballs
Wc f fs fy c c js. W f k xcs f vs f s sff s G Hsy Bck fs y f cf s s s y k. Ec vy y v y Rsy Bck I s y s ks s! Gy Cc Dy! A Av s c P B s c Txs s fs c 3-Wy s s y c s G s My: c c.89 s 10 c c v s s s f f s ss
ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27
1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system
Use precise language and domain-specific vocabulary to inform about or explain the topic. CCSS.ELA-LITERACY.WHST D
Lesson eight What are characteristics of chemical reactions? Science Constructing Explanations, Engaging in Argument and Obtaining, Evaluating, and Communicating Information ENGLISH LANGUAGE ARTS Reading