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Bilkent University - CS - 421
CS 421: Computer NetworksFALL 2007MIDTERM II December 4, 2007 120 minutesName: Student No: Show all your work very clearly. Partial credits will only be given if you carefully state your answer with a reasonable justification.Q1 Q2 Q3 TOT1) a) (12 pt
Bilkent University - CS - 421
CS 421: Computer Networks MIDTERM II April 28, 2005 120 minutes Name: Student No:SPRING 2005Show all your work very clearly. Partial credits will only be given if you carefully state your answer with a reasonable justification.1) a) (10 pts) Execute th
Bilkent University - CS - 421
CS 421: Computer Networks MIDTERM II April 27, 2006 120 minutes Name: Student No:SPRING 2006Show all your work very clearly. Partial credits will only be given if you carefully state your answer with a reasonable justification.1) a) (10 pts) Execute th
Bilkent University - CS - 421
CS 421: Computer NetworksSPRING 2007MIDTERM II April 26, 2007 120 minutesName: Student No: Show all your work very clearly. Partial credits will only be given if you carefully state your answer with a reasonable justification.Q1 Q2 Q3 TOT1) a) (12 pt
Riverside Community College - ACCOUNTING - 165A
Anderson Graduate School of ManagementUniversity of California, RiversideBSAD 165A - INTERMEDIATE ACCOUNTINGWinter 2012 #20725(updated 01-12-12)INSTRUCTORBruce SamuelsonOFFICEOLMH 2331OFFICE HOURS 12:40-2:00 p.m. TRCOURSEROOMHOURSe-mailBUS 1
Riverside Community College - ACCOUNTING - 165A
EXAM 2 TOPICSBSAD 165ASpring 20121.Define the field of financial accountingwhat does it prepare?(1)theidentification,measurement,andcommunicationoffinancialinformationabout(2)economic entitiesto(3)interestedparties.Financialaccountingistheprocesstha
Riverside Community College - ACCOUNTING - 165A
Solutions to exercises in lecture on Ch. 5, part aBRIEF EXERCISE 5-1Current assetsCash.Accounts receivable.Less: Allowance for doubtful accounts.Inventories.Prepaid insurance.Total current assets.$30,000$110,000(8,000)102,000290,0009,500$43
Riverside Community College - ACCOUNTING - 165A
Solutions to exercises in lecture: Ch. 6BRIEF EXERCISE 6-18% annual interesti = 8%PV = $15,000FV = ?0123n=3FV = $15,000 (FVF3, 8%)FV = $15,000 (1.25971)FV = $18,895.658% annual interest, compounded semiannuallyi = 4%PV = $15,0000FV = ?1
Riverside Community College - ACCOUNTING - 165A
Solutions to exercises in lecture: Ch. 8aBRIEF EXERCISE 8-2Inventory (150 X $34).Accounts Payable.5,100Accounts Payable (6 X $34).2045,100Inventory.Accounts Receivable (125 X $50).2046,250Sales.Cost of Goods Sold (125 X $34).Inventory.6,250
Riverside Community College - ACCOUNTING - 165A
Solutions to exercises in lecture: Ch. 8bBRIEF EXERCISE 8-92010 inventory at base amount ($22,140 1.08)2009 inventory at base amountIncrease in base inventory$20,500(19,750)$ 7502010 inventory under LIFOLayer oneLayer two$19,750 X 1.00$ 750 X
Riverside Community College - ACCOUNTING - 165A
BUS 165A 001Intermediate Financial AccountingInstructor:Office:Phone:Fax:E-mail:Bruce Samuelson, DBA & CPA2331 Olmstednone949-831-8112bus165a@cox.netQuarter:Spring 2012Lecture time: 2:10-3:30 p.m. TRClassroom:1212 OlmstedCourse Website: h
Riverside Community College - ACCOUNTING - 165A
BUS 165A 001Intermediate Financial Accountingcls268115Instructor:Office:Phone:Fax:E-mail:Bruce Samuelson, DBA & CPA2331 Olmstednone949-831-8112bus165a@cox.netQuarter:Spring 2012Lecture time: 2:10-3:30 p.m. TRClassroom:1212 OlmstedCourse
Riverside Community College - BUS - 147
Fixed investment = $3.5 millionFixed costs other than depreciation = $600,000Net working capital = 25%Variable cost = 35%7. Lillian Jordon is considering using some of the cash generated from her mail-order business to open aretail store. The fixed i
Golden Gate - ACCOUNTING - 360
1Marks:10Wagesaresubjecttothefederalincometaxandnotsubjecttoanyothertax.Chooseoneanswer.A. TrueB. FalseWagesaresubjecttoFICAandMedicaretaxwhichiscollectedbytheemployer.CorrectMarksforthissubmission:10/10.Question2Marks:10Whichofthefollowingisli
Golden Gate - ACCT - 360
Session 1 SolutionsChapter 1 - 14, 20, 45, 5214.Herman could have been overcharged, but at least part of the excessprobably is attributable to a hotel occupancy tax and a car rental tax. Inthe major cities, these types of excise taxes have become a p
Universidad del Norte - ING - 105
Diferentes tipos de conocimientosExtrado de http:/www.monografias.com/trabajos12/marcono/marcono.shtmlEl hombre, a lo largo de su existencia, ha sentido curiosidad por conocer el mundo que lerodea y ha pretendido dar explicaciones a una serie de interr
Technion - CS - 101
II1::: v1 w1 v2 vw2v:v = , w = M n1 ( F ) = n F - A = ( aij ) M mn ( F )MMv w n nvvA ( v ) = Av .1vvvvA ( v + w ) = Av + Aw .2kkA ci vi = ci Avi i =1 i =1..Auv- ei A .Ai:1. A M m n ( F ).vvAx = 0.3uvAei . A M
Technion - CS - 101
II2:.:1:2cfw_1, 2,., n". cfw_1, 2,., nSn ( k, l)1 k, l n:312.n (1) ( 2 ) . ( n ) ..-:5even 1sg = 1det A = A =:6 ( sg ) a ( ) a ( ) L a ( )11 Sn22nn:B M n ( F ) - A = ( aij ) M n ( F ):" A-B = A().1B=A .2B = Aa
Technion - CS - 101
II3::Ak 1 Ak 2 ,k =,A n.bAx = bA 0k-A:1:3 y + 2 x = z + 13 x + 2 z = 8 5 y3 z 1 = x 2 y:1 adj ( A ) = ( bij )adj ( A).i - j. A Mn (F )A-AM j ,i , bij = ( 1)i+ jM j ,i:24 5 3. A = 2 1 0 9 5 3 . adj ( A ):A1 = A ad
Technion - CS - 101
II4::1T. T :V W ,FW- V "W- Vv, v V T ( v + v ) = T ( v ) + T ( v ) ,T ( cv ) = cT ( v ) ,.1c F ,v V :.2:1T : [ x ] [ x ]T ( p ( x ) ) = p ( x ):2T : [ x ] [ x]T ( p ( x ) ) = p ( x )dx:3T : 2 [ x] , T ( p ( x ) ) = p ( ):4A M mn
Technion - CS - 101
II5::1T :V WT,cfw_. ker (T ) = v V T ( v ) = 0 V : ".V"ker (T ):1T :V W,:2"Im (T )T :V WT,. Im (T ) = cfw_Tv v V W : ".W:2T :V W,:13T : 3T ( x, y, z ) = ( x, y, 0 ).T :23:T : T (1, 0,1) = (1,1,1) .1( 0,1,1) Im T( 0,
Technion - CS - 101
II6::1T :V W.F" W- Vcfw_v , v ,., v cfw_T ( v ) , T ( v ) ,.,T ( v ),V.W112n2n:1T : 32T ( x, y, z ) = ( 3x + 2 y, x + z ).T -cfw_T ( v ) , T ( v ) ,.,T ( v )."12n". T ( p ( x ) ) = p ( ) "."TWcfw_v , v ,., v 12n":1
Technion - CS - 101
II7:v = a1 v1 + a2 v2 ++ an vn ,F.BV"cfw_B = v1 , v2 ,., vn a1 v = B a nv:1, cfw_ai i =1 Fn:1v , B.[ ]B : V .nF.V "2.R -cfw_u , u ,., u 12cfw_,B'- B -cfw_1:2B = v1 , v2 ,., vncfw_u ,vV Pcfw_A+ anj vnB, u2
Technion - CS - 101
II8:,Wcfw_C = w1 , w2 ,., wm - VC- B,( [Tvi ]CE1 [T ]E2T :V W(1):1T : 2 [ x ] 1 [ x ]:2T : ,. T ( x, y, z ) = x + y + z "E2) [T ]B = [Tv1 ]C , , [Tvn ]CCiC = cfw_1 + x,1 x , B = cfw_1, x, x. [T ]BCcfw_B = v1 , v2 ,., vnT
Technion - CS - 101
II9:1F .0 v V(") 0 v nF-A.FT :V V 0 v VT. Tv = v . "A Mn (F )F .(")"VT( "). Av = v. T ( x, y ) = ( 2 x + 2 y, x + 3 y ) ". "- "1. B = P AP -:2A, B M n ( F ), dim F V = n ,FV". [T ]D = B - [T ]C = A -"V:1A, B M n ( F )P M
Technion - CS - 101
II10:1.A:1mA ( x ):2. m( x)mA ( x ) / g ( x ) ""AmA ( x )g ( x ) F [ x]g ( A) = 0"A":.Mn (F ) -":11 0A=0 0":3.A":"A":2 5 6 6 A = 1 4 2 3 6 4 "":310A = 00021000002000000100310" 1 2A= 1 4
Technion - CS - 101
II11:1V())F .( VF - V V - :::(). < v1 + v2 , w > = < v1 , w > + < v2 , w > F , v1, v2 , w V . w, v V< v , w >=< w, v > :(. v = 0 < v,v > = 0 < v, v > 0 , v V:v = (v1 ,., vn ), w = ( w1 ,., wn ) ()():1.V = 2:2( n 1 ) V = nv = ( x
Technion - CS - 101
I N<Z<Q<R<C)2(a1 , b1) (a2 , b2)=(a1a2-b1b2 , b1a2+a1b . 1) i + j M ji( = [ adj( A) ] i j>(u1+u2,v)=<u1,v>+<u2,vA*adjA=(detA)I><u1,v>=<v1,u: A -0 0=> ><u,u .A0=> <u,u" 0=u: detAB=detA detB a-bi a+bi .): Zn=rn (cos(n) + i sin(n-
Technion - CS - 101
1:Web: http:/www.ash-college.ac.ilE-Mail: shapird@cs.brandeis.edu: G :1. 2. 3. 4. .: F +, ::1.2.3.4.5.:6. 7. 8. ) (9. 01.
Technion - CS - 101
2 :: = :1. )(a,b)+(c,d)=(a+c,b+d2. )(a,b)(c,d)=(ac-bd,ad+bc )1,0( - .i: ) z=(a,b . a z - Re z b z - Im z z=a+bi: z=a+biC z - z = a bi :1. z = z2. 2(
Technion - CS - 101
3 :: V )"( F:1. : u , v V - u + v V v V- cF - cv V2. : u , v , w V - )( u + v ) + w = u + ( v + w3. : 0 V v V - 0 + 0 + v = v4. : v V u V - 0 = u + v = v + u5. : u , v V - u + v = v + u6. u , v V- cF - c ( u + v ) =
Technion - CS - 101
4 , - :: V .Fk . u1 , u2 ,., uk V . U = u V u = ci ui , c1 , c2 ,., ck F 1= icfw_1( U " V . A = u1 , u2 ,., uk2( U" ,A .UU3( U " .A: U" " " A - ).Sp(A A U:
Technion - CS - 101
5 :: V" .F B = u1 , u2 ,., uk V V:cfw_1. )V=Sp(B2. B":1. 2. ][ xnn3. C" R C32234.: V V )dim(V:-n A -n ei ) A =
Technion - CS - 101
6 )(:: V" .F cfw_ A = v1 , v2 ,., vn:1. A2. A )( 3. A " : A", : V" .F A V " B V a A b B -: A = v1 , v2 ,., vk " - V-: " V
Technion - CS - 101
7 : F1= b2= b= bm+ . ++ . +1a11 x1a21 x+ . + amn xn1am1 xa1n xna2 n xn ( i = 1.m, j = 1.n ) aij , ( i = 1.m ) bi . bi = 0 i , .:: :1. 2.
Technion - CS - 101
8 :: A -0 A ) rank(A).r(A: A v1 ,., vn- B . w1 ,., wn B - A" A v1 ,., vn: A - Ac . ": A n.
Technion - CS - 101
9 :: V" U- W V )dim(U+W)=dim U +dim W dim(UW 1 : 0 = U = cfw_( x, y, z ) x + y + z- 2: ) 0 ,1,1 ( = 2 U = Sp cfw_u1 = (1, 0,1) , u-3.1. W = cfw_( 0, 0, z ) z 1 cfw_w1 , w2 , u -33. ?UW)1,0 ,2 ( = 2 W = Sp cfw_w1 = (1,1, 0 ) , w"
Technion - CS - 101
I 01:: :) A = ( aij ) , B = ( bij ) M mn ( F: ) A = ( a1 , a2 ,., an ) M 1n ( F n.) A + B = ( aij + bij ) M mn ( F 1 aa ) A = 2 M n1 ( F n. a n: : ) A = ( aij ) M mn ( F- F ) A = ( aij ) M mn ( F : ) A = ( aij ) , B = ( bij ) , C =
Technion - CS - 101
" 1 ' 1) : 90.11.11(1( ) , (:( a b 2= a + b- . a b = 3ab( ) 2 = cfw_2 z | z ( .2( ( 3 . ( 6,5,4,3,2 - 7 - 11. ( : 2 = 5 x 5 7 ? 3( : : Z1 = 3 4i , Z 2 = 5 + 12i , Z3 = 1 3i , Z 4 = 6
Technion - CS - 101
" 1 :1( ( , :a b ba= a + b . a b=b+a22( , , .2( (220111200012+012202110120000*012( 7 : 4 = 125 = 25 = 134 = 32 = 143 = 43 = 156 = 1611 : 9 = 28 = 37 = 46 = 55 = 62 = 51 = 64 = 125 = 132 = 143 =
Technion - CS - 101
" ' 2) 90.11.81(1( :. 3 + 3i. 5 5i21 4. )(2 + i )( i552( :. z 3 = i. z 6 = 1 + 3i013( : 1 + 3i 1 3i 4( z . z 3 = i ) (:.... . z = i0 < ) . Im( z0 = ) Re( z ) . Im( z1 =| | z . z
Technion - CS - 101
": 2 . 3 + 3i = 6cis.1( 47.. 5 5i = 50cis44 127. (2 + i )( i) = 4 4i = 32cis.554i = cis23i = cis 2z0 = cis61 + 3i = 2cis 363, k = 0,1, 2, z1 = cis.2( 53, z2 = cis62+ 2 k66, k = 0,1, 2,3, 4,5713, z2 = 6 2cis, .1818
Technion - CS - 101
" ' 3) 90.21.2(1( 4 V = R . R - U i ? V U i"- . - .4( = 0, ai Rai) 4, U1 = cfw_(a1 , a2 , a3 , a1= i( U 2 = cfw_(a1 , a2 , a3 , a4 ) ai > 0, ai R( . U 3 = cfw_(a1 , a2 , a3 , a4 ) a1 + 3a2 = a3 , a2 + a3 = a4 , ai R( 0 = U 4 = cfw_(a, b, c,
Technion - CS - 101
" 3 :1( . 1 U 4 . : 1 x, y U . 40 = x = ( x1 , x2 , x3 , x4 ), xi1= i40 = y = ( y1 , y2 , y3 , y4 ), yi1= i 1: x + y U) 4x + y = ( x1 , x2 , x3 , x4 ) + ( y1 , y2 , y3 , y4 ) = ( x1 + y1 , x2 + y2 , x3 + y3 , x4 + y4= 4= x1 + y1 + x2 + y2 + x
Technion - CS - 101
" ' 4) 90.21.9(1( S- . T :. ) Sp ( S ) + Sp (T ) = SP ( S T. ) Sp ( S ) + Sp (T ) = SP( S + T. 0 S ) Sp ( S ) Sp (T ) SP ( S + T2( S = cfw_s1 ,., sm , T = cfw_t1 ,., tn" 0cfw_ = ) Sp ( S ) Sp (T S T".3( ":. )0 ,1,2( ,)6 ,5,4( ,)3,2 ,1(, 3f1 (
Technion - CS - 101
" 4 :1( . . : . S = cfw_v1 ,., vk ; T = cfw_u1 ,., um . : . 1v1 + . + k vk + 1u1 + . + mum :) Sp ( S) Sp (T S T = cfw_v1 ,., vk , u1 ,., um : . 1v1 + . + k vk + k +1u1 + . + k + mum .".. . :)1,0(cfw_ = S = cfw_(1, 0); T)1,1(cfw_ = S + T) (
Technion - CS - 101
" ' 5) 90.21.61(1( : 4 v1 = (1, 0,1, 1) - 4 , v2 = (1,1, 1, 0) :4 . v = (a 2 , 2, a + 2, a) . a v ) 4 ( " 2? v1 , v. a ', v 2. v1 , v2( )0 ,3 ,1,1( ,)1,0 ,2 ,1( ,)1,3 ,1,2(cfw_ U = Sp 4 . . U3( " )!( ]: 3 [ x 2 cfw_1 + x 2 ,5 x x3 , 2 4 x 2 +
Technion - CS - 101
" 5 :1( . v 2 , spancfw_v1 , v :)1 ,1,0 ,1(a 2 , 2, a + 2, a ) = b(1,1, 1, 0) + c a2 = b + c2=ba +2 = cba = c2 = a. )1 ,1,0 ,1(2 + )0 ,1 ,1,1(2 = )2 ,0 ,2 ,4( .2( )0 ,3,1,1( ,)1,0 ,2 ,1( ,)1,3 ,1,2(cfw_. U = Sp ": 2 1 3 1 1 2 0 1 1 2 0 1 1
Technion - CS - 101
" ' 6) 90.21.03(1( ] R 4 [t - t -4. ? - .41 + . g1 (t ) = 2t 3 4t 2 + 9t + 5 , g 2 (t ) = t 3 t 2 8t + 2 , g 3 (t ) = 10t 2 + 2t2( ) cfw_(1 + i, 3 + 8i, 5 + 7i ), (1 i, 5, 2 + i ), (1 + i, 3 + 2i, 4 i - 3 C ? C 3 C ? R3( : 1 = x + 3z
Technion - CS - 101
" 6 :1( , , ":0 = )41 + a(2t 3 4t 2 + 9t + 5) + b(t 3 t 2 8t + 2) + c(10t 2 + 2tt 3 (2a + b) = 0 t 3 b = 2at 2 (4a b 10c) = 0 t 2 a = 5 c3t (9a 8b + 2c) = 0 t0 = 5a + 2b + 14c = 0 15c = 14c = 0 c = 0; a = 0; b ", 3.2( ) cfw_(1 + i, 3 + 8i, 5 + 7
Technion - CS - 101
" ' 7) 01.1.6(1( : 7 = x + 2 y 4z. 8 = 2 x y + 5 z3 = x + 12 y 30 z7 = 3 x + 3 y + z + 2 w3= x+ y+z.2= x+ y+w2( : 1 = (5 k ) x 2 y z2 = 2 x + (2 k ) y 2 z1 = x 2 y + (5 k ) z k ?3( 3 :
Technion - CS - 101
" 7 : 1 2 4 7 1 2 4 7 1 2 4 7 1( . 2 1 5 8 0 5 13 6 0 5 13 6 1 12 30 3 0 10 26 4 0 0 0 16 ..x4 = t1 + x3 = tx2 = s1x1 = (6 3t 3s ) = 2 t s33 3 1 2 7 3 3 1 2 7 3 3 1 2 7 1 1 1 0 3 0 0 2 2 2 0 0 2 2 2 1 1 0 1 2 0 0 1 1 1 0 0 0 0 0
Technion - CS - 101
" ' 8) 01.1.02(1( :1 2 1 0 3 2 1 2 2 1 2 5 1 3 1 3 2( . W :0 = 42 x1 + x2 + 3 x3 x0 = 43 x1 + 2 x2 2 x0 = 3 x + x + 9 x x342 1. : A ) . rankA + dim W = n (3( :. 01 V ,W 8 = . dim V = 9, dim W 7 = ) . dim(V W. 01 V ,W
Technion - CS - 101
" 8 :1( :10 1 2 1 0 2 11 2 1 0 1 2 1 0 2 0 4 2 2 2 4 0 3 2 1 2 0 4 2 2 2 1 2 5 0 5 0 5 0 0 10 10 0 0 10 10 1 3 1 3 0 1 2 3 0 0 10 10 0 0 0 0 3. )3 (.2(. ) - ( A: 2 1 3 1 2 1 3 1 2 1 3 1
Technion - CS - 101
" ' 9) 01.1.02(1( ) ( At , A + B , AB- , Bt:5. 03. 0 16. 9 2 43 1 1 A = 1 5 , B = 2 2 4 3 4 1 0 11 12 0A = 2 1 0 , B = 0 3 6 0 01 7 9 1 0A=, B = 5 0 13 0 42( 1
Technion - CS - 101
": 9 2 4 18 6 2 10 1 1 3 5 AB = 1 5 = 19 11 13 5 .1( 3 4 4 2 2 0 19 51 15 1 4 2 1 3 t 1 2 At = ,B = 3 2 4 5 45 0 1 0 1 1 1AB = 2 1 0 0 2 0 3 6 0 0 1 0 1 1 1A+ B = 2 1 0 + 0 20 3 6 0 03 1 1 20 = 2 4 6 .1 0 6 63 2 1 20 = 2 3 0
Technion - CS - 102
ASCIIdec32hex20chardechexchardechexchardechexchardechexchar[space]523447248H925C\11270p3321!533557349I935D]11371q3422"54366744AJ945E^11472r3523#55377754BK955F_11573s3624$5638
Technion - CS - 102
Recursive FunctionA function defined in terms of itself iscalled a recursive function.Introduction to Computer Sciencereturn_value rec_func(type arg1, type arg2, )cfw_rec_func();RecursionTo Iterate is Human, to Recurse, DivineL. Peter Deutsch1
Technion - CS - 102
Declaring variables - Remindertype variable_name;Introduction to Computer Science Before using a variable, one must declare it. The declaration introduces the variable type, then its name. When a variable is declared, its value is undefined.Pointers
Technion - CS - 102
Pointer Relational OperatorsComparing pointers:Two pointers can be compared if they point to the same type.Any pointer may be compared to the NULL (zero) pointer.Introduction to Computer ScienceExamples:Pointersint *p1 = NULL, *p2 = NULL;if (p1 =
Technion - CS - 102
What does this program do?Introduction to Computer ScienceMemory Allocation1Memory Allocation#include <stdio.h>char* suffix(char *str, int suffsize)cfw_#include <string.h>int length = strlen(str);#define MaxSize 50if(length<suffsize)char* suff