answersLogoWhite

0


Best Answer

The only factor is 2.

2*(t3 + 2t2 + 4x)

User Avatar

Wiki User

โˆ™ 2012-09-02 14:03:53
This answer is:
User Avatar
Study guides

Algebra

20 cards

A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

โžก๏ธ
See all cards
3.81
โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…
2278 Reviews

Add your answer:

Earn +20 pts
Q: What is the factor of the polynomial of 2t3 plus 4t2 plus 8x?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Continue Learning about Math & Arithmetic
Related questions

How do you factor 4t2 - 12?

4(t^2 - 3)


How do you factor 1-128t7?

-(2t - 1)(64t6 + 32t5 + 16t4 + 8t3 + 4t2 + 2t + 1)


What is the value of 4t2 plus 6r-tr when t equals -3 and r equals 5?

21 4t2+6r-tr 4[-3]2+6[5]-[-3][5] 8[-3]+30-[-15] -24+30+15=21


Where can you play Lego mars mission?

no site 4t2 profolio procure


Let L be the line in R3 that consists of all scalar multiples of the vector -2 -1 2T Find the reflection of the vector v equal to 7 2 7T in the line L?

I'm not going to lie, I'm not 100% sure if I'm going to do this correctly. It's been a while since I've done something like this, so you may want to double check my answer. Also, the letter T in your question is really throwing me off, so I'm just going to give you two answers. The first answer will treat T as a variable, the second answer will ignore it completely. Both answers, however, use the following equation for the reflection of a vector about a line:RefL(v) = 2L(v ● L)/(L ● L) - vFor my first answer, I'll use the following vectors for Land v:L = -2i - j + 2Tk, andv = 7i + 2j + 7Tk,where i, j, and k are the unit vectors in the direction of the x, y, and z axes in R3, respectively.Thus,v ● L = -14 - 2 + 14T2 = 14T2 - 16.L ● L = 4 + 1 + 4T2 = 4T2 + 5.Therefore, 2L(v ● L)/(L ● L) =2L(14T2 - 16)/(4T2 + 5) = L(28T2 - 32)/(4T2 + 5) =-8(7T2 - 8)/(4T2 + 5)i - 4(7T2 - 8)/(4T2 + 5)j + 8T(7T2 - 8)/(4T2 + 5)k.Let A = 2L(v ● L)/(L ● L).(A - v)i = [-8(7T2 - 8)/(4T2 + 5) - 7(4T2 + 5)/(4T2 + 5)]i =(-56T2 + 64 - 28T2 - 35)/(4T2 + 5)i = (-84T2 + 29)/(4T2 + 5)i.(A - v)j = [-4(7T2 - 8)/(4T2 + 5) - 2(4T2 + 5)/(4T2 + 5)]j =(-28T2 + 32 - 8T2 - 10)/(4T2 + 5)j = (-36T2 + 22)/(4T2 + 5)j.(A - v)k = [8T(7T2 - 8)/(4T2 + 5) - 7T(4T2 + 5)/(4T2 + 5)]k =(56T3 - 64T - 28T3 - 35T)/(4T2 + 5)k = (28T3 - 99T)/(4T2 + 5)k.Let b = 1/(4T2 + 5), thenRefL(v) = b[(-84T2 + 29)i + (-36T2 + 22)j + (28T3 - 99T)k]That expression for RefL(v) looks pretty ugly, so I'm going to do the problem again, this time without the variable T.L = -2i - j + 2k, andv = 7i + 2j + 7k.v ● L = -14 - 2 + 14 = -2L ● L = 4 + 1 + 4 = 9Therefore, 2L(v ● L)/(L ● L) =2L(-2/9) = L(-4/9) =(8/9)i + (4/9)j - (8/9)k.RefL(v) = 2L(v ● L)/(L ● L) - v = (8/9)i + (4/9)j - (8/9)k - 7i - 2j - 7k =-(55/9)i - (14/9)j - (71/9)k.While this expression does look much nicer, I'm not sure if it's right. So, like I recommended above, please double check my work!


What is balanced equation for tin nitric acid ammonium hydroxide and citric acid?

NaNo3 Cu-4T2


What are the factors of 4t squared minus 12?

4t2 - 12 = 4*(t2 - 3) which has no further rational factors


What is 4t2 minus 16 over 8 divided by t minus 2 over 6 simplified?

(4t2 - 16)/8 ÷ (t - 2)/6 = [4(t2 - 4)/8] x 6/(t - 2) = (t2 - 22)/2 x 6/(t - 2) = (6/2)[(t + 2)(t - 2)/(t - 2)] = 3(t + 2) = 3t + 6


Suppose that an object is traveling along a straight line Suppose that its velocity function is given by vt equals 4t2 -8t where t is in seconds Find the objects average acceleration along the time?

The object's instantaneous acceleration is (8t - 8) at any time.We can't calculate the average acceleration, because you haven't defined a periodof time over which to average it. We need the start and finish times in order tofind an average.


Liam carrs auto free rider 2?

-18 1i 18 1i,-3c 1p -6l 4f -8p 5g,-8p 5i -be 66 -ei 62,-bg 66 -en 5b -el 61 -b4 77 -8a 6h,-7c 6c -5b 56,-87 6h -79 6c,-5b 54 -40 3m,-41 3i -3d 1q,-p8 1hk -ov 1i4 -og 1i1,-p8 1hn -p4 1i8 -og 1i3,-m2 1n4 -lr 1mo -m0 1nm,-lv 1nm -ms 1nb -m2 1n4,-rq 1qi -rd 1qt -qu 1qu -rd 1rd -rr 1qi,-r3 21t -qj 229 -q9 225 -qi 22l -r5 21r,-ms 23g -mi 23i -ma 237 -MD 23n -mt 23h,-mv 2ap -ml 2b3,-ml 2b1 -m9 2b1 -ml 2bb -mv 2as,-qn 2il -q6 2j6 -pk 2is,-pk 2iu -q8 2jg,-qa 2jg -qn 2in,-bv 2gr -bi 2hu -c1 2is -cj 2ln -dk 2o1 -g3 2qi,-c3 2gu -av 2hb -ad 2im,-ai 2ip -bf 2ki -c6 2o9,-c6 2o6 -eb 2ps -eo 2qm -g9 2qh,-t8 3kd -sg 3k6 -rs 3k1 -s0 3kl,-ru 3kk -t8 3kd,-168 4lj -16f 4r3 -16i 5hr,-16i 5hp -142 61n -10s 68c,-108 681 -13c 61f -15o 5i1 -15p 565,-110 686 -15t 61j -18l 5gq,-18n 5gs -169 4lm,-3pt 8pl -3qc 8ou,-3qc 8p1 -3pg 8ok,-3pv 8pl -3qo 8om -3pj 8om,-533 ar0 -52o arb -527 ar0 -52p ari,-52r ari -530 ar1,-4nu be6 -4o1 bem -4oi bel -4nu bep -4nu be7,-4t4 bka -4t0 bkp -4sf bkp,-4sf bko -4t2 bku -4t3 bkc,-4j3 d2a -4j3 d2p -4jp d2m -4j1 d2v,-4j1 d2u -4j3 d29,-4t6 d3v -4t5 d4b -4sl d4f,-4sl d4g -4t7 d4f -4ta d4c -4t6 d41,-4nc df2 -4nf dfi -4o2 dfc -4nb dfp -4nd df4,-4qg dfl -4qo dg6 -4r3 dg3 -4ql dgc -4qg dfl,-4uh dlr -4uh dll,-4un dm7 -4v7 dlv,-4um dm7 -4ui dln,-4uh dln -4ui dmc -4v8 dlv,-516 dnv -51a doa -51p do9 -517 doe -517 do1,-55v due -561 dup,-55v due -55t du9,-561 duo -56h duj -55t duu -55v dub,-5b1 e2t -5b8 e34 -5bt e37 -5b7 e39 -5ar e2l -5b2 e2t,-5gk e6k -5gq e78 -5hi e7b -5gk e7e -5gm e6k,-5n4 ebd -5nb ec0 -5nr ebu -5n6 ec5 -5n3 ebe,-5ss efe -5t1 efq,-5t2 efp -5tk efv,-5tk eg0 -5st eg0 -5ss efe,-637 ejo -63b ek6 -63q eka -63a ekb -636 ejq,-9f4 g93 -9fn g9f -9g8 g9u -9go gae -9h6 gb0 -9hj gbi -9i0 gc3 -9ic gco -9im gdf,-9is gdk -9is gk6 -9is gkg -9h0 gpq -9d8 gug -9fe gti -9hu gqo -9j6 gmc -9jq gkg,-9i8 gf6 -9i8 gjs -9i8 gkg -9gc gpg -9cu gts,-9jq gk6 -9jq ge8 -9is gag -9fe g98#-99q gdk -988 gb4 -988 g8k -98s g80 -996 g7c -98s g6e -97u g6o -97k g7c -988 g8k,-988 gb4 -97a gdk,-99g g9i -96m g9i -96c g8u -95o g8u -95e g8u -954 g9i -95e g9i -95e g98 -962 g98 -96m g9i,-99g g9i -99q g98 -9ae g98 -9ae g9i -9a4 g9i -9ae g9s -9a4 ga6 -99g g9i#B -9ha ge8 9g,B -9ha gd0 99,G -9h3 gba 5j,G -6a4 f43 8c,G -47f 8sc 5h,G -jt 7ar 8o,B -15m 545 8e,B -15v 51i 82,B -15q 4vn 89,B -15t 4t7 8c,B -15r 4r0 8d,B -15o 4oo 8d,B -ds 2l8 4e,B -dd 2kd 2l,B -d1 2j5 3l,B -p9 2jb u,B d 1e a4


Liam carrs loopy free rider 2 track?

1b 1i 24 1h 2t 1m 3l 25 47 2n 4e 3i 4b 4d 3k 57 2s 5s 1t 6f t 77 1 8c -j 9h -17 aq -1q c6 -2c dl -2t f7,-2q f7 -30 g4 -30 h5 -2o i8 -2d jc -1t kd -1a lc -h mf 7 n9 16 o3 27 on 3c pb 4g pp 5m q7 6t qi 8d qu 9p r8 b4 rg cg ro dt ru f6 s4 gg s8 hu sc jd sd,o8 12k oc 13f ok 14k p1 15j pv 16g r5 17k sg 18e tu 198 vm 19j 11h 19p 13d 1a0,13b 19u 146 1a0 156 19v 165 19t 17d 19e 18h 18v 19k 18i 1ak 189 1bi 18d 1ch 18h,1cf 18g 1d7 18e 1du 189 1em 18b 1fq 18b 1gv 185 1i5 183 1j7 188 1k4 18h,1k4 18i 1kk 192 1l5 19j 1lo 1a1 1mt 1ab 1o3 1aj 1pi 1ak 1qt 1ag 1s8 1ad 1th 1ad 1uj 1ai 1vo 1ak 215 1ak 22g 1ao 23t 1at 25d 1b2 26v 1ba 28j 1bk 2a7 1c1 2bs 1ce 2dj 1co 2fc 1d1 2h7 1d9 2j2 1dg 2km 1d0 2mb 1c9 2o0 1bg,2o1 1bh 2os 1bg 2pv 1bg 2qt 1bp,2qe 1bk 2r4 1bb 2rq 1b9 2sj 1bf 2tb 1bt 2ts 1ce 2ud 1d6 2uu 1dr,31f 1gb 30u 1ft 30d 1fd 2vu 1eu 2vi 1ef 2v6 1e2 2uq 1dm,31d 1g8 32a 1gk 33g 1h6 34q 1hh 366 1hl 37j 1ho 392 1hm 3ad 1h9 3bq 1gr 3d8 1gd,3hn 1h3 3ic 1gl 3j0 1ft 3jg 1eq,3jf 1ep 3jq 1e0 3jv 1d6 3jn 1c9 3j9 1be 3ik 1am 3hs 1a6 3gu 19s 3g1 19t 3f8 1a6 3ee 1am 3dr 1ba 3da 1c2 3d1 1cq 3d3 1dh,3d6 1dj 3d9 1e3,3f5 2te 3fl 2u5 3ga 2us 3gt 2vk 3hn 309 3ih 30t 3jg 31g 3kh 31s 3lf 322 3mg 327 3nf 32a 3od 32b 3pb 329 3q8 322 3r3 31m 3rp 319,3ro 31c 3rn 31v 3rm 32g 3r0 32l 3q7 32f 3pg 32b 3or 328,3tt 311 3tu 31l 3tq 328,3tu 314 3ug 31e 3v4 31p 3vp 31v 40g 322,3tp 325 3ue 327 3v4 327 3vn 325 407 324,40d 322 416 322 41t 320 42l 31r 43c 31h 442 315 44o 30m 45f 300 463 2v4 46l 2u6,46m 2u3 479 2u0 47u 2tv 48g 2ta,47u 2u0 489 2tf 48c 2sq 48d 2s6 48e 2rj,4a3 2s6 4a2 2st 4a4 2tj,4a5 2ti 4ab 2u3,48b 2rk 490 2rk 49i 2rk,48i 2ov 48k 2lv,4a4 2p2 4a2 2m2,48i 2lu 481 2lj 47t 2kr 481 2k6 48g 2jl 494 2ja 49s 2jd 4ae 2k0 4ap 2kn 4ar 2lf 4ag 2lt 4a1 2m4,48q 2lo 48q 2l4 48q 2kj 49b 2kd 49o 2ko 49r 2lb 49n 2ls,4ab 2u2 4b1 2tu 4bn 2u2 4ch 2ue 4d6 2v4 4ds 2vu 4ek 30g 4fh 31b 4gn 325 4ht 32d 4j2 32f 4k3 32g 4l9 32i 4mm 32p 4o7 332 4pk 337 4r3 33d 4sk 33k 4u7 33q,4u7 33m 4uv 33i,52k 33t 53a 33m 53t 337 54d 32f 54l 31l 54l 30p 54d 2vr 53v 2v1 53e 2u9 52o 2tl 51s 2tb 50v 2t8 507 2td 4vh 2tt 4v0 2uj 4ul 2vc 4ue 305 4uf 30t 4up 31h,543 4v1 54j 4vq 554 50g 55q 513 56j 51e 57f 51k 58e 51m 59m 51l 5ap 51h 5br 516 5cp 50i 5df 4vq 5du 4uu 5ea 4tv 5ei 4t2 5eq 4rv 5fb 4qo 5fq 4p8 5g9 4nt 5gm 4me 5h6 4ks 5hh 4j9 5hq 4hk 5i3 4fu 5ib 4e5 5ij 4ca 5is 4ae 5j5 48i 5jf 46k 5jq 44n 5k5 42p 5kh 40q 5kn 3uu 5ks 3t1 5l1 3r3 5l7 3p5 5ld 3n8 5lj 3la 5lp 3jc 5lv 3he,1b 1i 16 1k,15 1k -8 1g -s 1q -1a 1h,66d 5g4 678 5gu 685 5hr 698 5im 6ag 5jd 6bl 5k1 6d0 5ko 6ef 5li 6g2 5mf 6hp 5nc 6jh 5oc 6la 5pf 6n4 5ql 6og 5ri 6ps 5si 6r7 5tj 6si 5uk 6ts 5vn 6v5 60s 70d 623 71l 63a 72u 64i 745 65q 75d 673 76l 68c 77t 69m 795 6b0 7ac 6ca 7bk 6dk 7cr 6eu 7e3 6g8 7fa 6hi 7gh 6is 7hp 6k7 7j0 6li 7k7 6mt 7lf 6o8 7mm 6pj 7nv 6qu 7p8 6sa 7qg 6tl 7rq 6v0 7t9 706 7uo 71b 807 72h 81m 73m 834 74s 84i 75i,4a3 2s7 49m 2rm,934 6cm 93m 6db 94k 6du 95m 6ee 96r 6f2 98c 6fm 99m 6g9 9b7 6gh 9cq 6gq 9eg 6h1 9g8 6h5 9i1 6h9 9jq 6hb 9lj 6h6 9nd 6gs 9p6 6gf 9qu 6fr 9sm 6f7 9uc 6ec,a1u 6fe a2i 6fu a3b 6gh a47 6h0,a43 6h1 afj 6nh,afk 6ng apv 6qg,aq0 6qg b63 6ra,b63 6rb bh6 6pt bre 6me,bh4 6os br2 6li,br1 6ll c0c 6hv,bre 6mg c0u 6im,c0s 6im c6d 6gc,c0a 6hu c5s 6fl,c6d 6gb cem 6gh,c5v 6fl ceo 6fp,cek 6gi d76 6gl#48d 2rj 48g 2p2,4a3 2rj 4a5 2p0,4a2 2p0 48f 2p0,472 2kp 3so 2i9,472 2lp 3sj 2nj 3so 2i8,4b1 2km 4mo 2j4,4al 2lp 4mo 2na 4mo 2j3,470 2lq 480 2lj,46t 2kn 47r 2kv,4b8 2km 4ao 2kp,81u 6rd 7vj 6rh 7vl 6si 80m 6sn 80o 6to 805 6tq 800 6to,833 6s5 833 6te,82g 6s6 83s 6s6,84c 6tb 84l 6rs,84q 6rp 857 6t3,84h 6sk 855 6sk,862 6s6 862 6t1,871 6su 871 6s5 87p 6s5,863 6rf 863 6rp,89c 6rk 891 6rk 891 6s0,891 6rs 89g 6ru 89g 6sg 88q 6sg,89c 6ri 89q 6rk,82v 6ve 82l 6ub,82d 6uc 83j 6ub,84h 6ul 83s 6um 83s 6v5 84l 6v5 84l 6um,84l 6ue 84c 6uj,81i 71l 81c 70r,81h 716 82a 718,82a 70r 826 71f,82q 70r 82q 71g 836 71g,82s 711 833 711,82p 70n 839 70m,83i 71d 83j 70n 83v 716,83i 713 83t 713,84j 70p 84l 716,84l 715 84r 70p,854 70f 854 713,854 70u 85j 70u,855 70k 85f 70k,855 70c 85p 70a,86d 70s 86c 700,86c 705 86n 70n,86n 70i 86s 703,cmk 6fg cl0 6fe,ckv 6fe ckq 6ds cmh 6du cnf 6d6 cqr 6d3,cqr 6d1 crf 6di css 6di css 6f1 cmd 6fc,cmd 6f9 clu 6fh clu 6ft cm3 6g7 cmi 6gc cmr 6g4 cmu 6ft cmu 6fg cmc 6f9,cqp 6d1 cqp 6di,cqp 6dk crh 6dk,cra 6f6 cr7 6fc cr5 6fj cr5 6fo,cr5 6fl cr7 6fo,cr7 6fr crc 6fv crd 6g2,crf 6g2 crn 6g5,crm 6g4 cru 6gc cri 6gc crc 6g4 cra 6fv,cru 6gc cs3 6g5,cs3 6g4 cs3 6fq cs0 6fc,cs0 6fe crp 6f4#B 8c9 6ul b5,B 8bq 6v4 3,B 8bg 6v7 b5,B 8b7 6ve 5,B 8b1 6vr b8,B 8an 705 b5,B 8ae 70c b8,B 8a3 70o b8,B 89s 717 b4,B 89f 717 b8,B 898 71c b8,B 893 71r au,B 88n 723 b8,B 88a 72d b8,B 885 72g b8,B 87p 72q b8,B 87e 731 b1,B 872 738 b8,B 86r 73l b8,B 86e 73t 7,B 866 746 b8,B 85n 74j b8,B 85e 74o b8,B 854 755 b8,B 84p 75a b8,B 5cr 4ve 21,B 5dm 4tr 1q,B 5dj 4sj 1k,B 5e3 4r4 1l,B 5ei 4ps 1o,B 5f1 4o8 1t,B 5fl 4mt 1t,B 5ge 4l6 24,B 5gv 4jd 1p,B 5hc 4h2 1j,B 5h9 4i9 1q,B 5hh 4fq 20,B 5hh 4e0 1t,B 5hq 4ch 29,B 5i7 4as 1q,B 5ic 498 18,B 5ij 47q 22,B 5io 46n 2g,B 5j7 45q 1e,B 5jf 445 20,B 5jf 42j 1q,B 5k1 40s 1j,B 5k1 3v7 1p,B 5k8 3u2 1h,B 5ka 3t7 1j,B 5kd 3s8 1g,B 5ka 3r8 23,B 5ki 3qd 20,B 5kl 3pe 1v,B 5kn 3ol 2q,B 5ki 3nt 25,B 5kl 3m6 24,B 5ka 3ms 1v,B 5kp 3ld 1t,B 5kv 3k5 1v,B 5l4 3j6 1n,B 5l8 3i4 27,B 4v4 2u9 6d,B 518 2t2 7p,B 54j 30g ag,B 539 33o 1l,B 3d7 1bs 5p,B 3ga 19m 7f,B 3jc 1bg 9n,B 3jo 1dr 8,B 3im 1ge t,B -15 gk 7f,B -1a fo 87,B -13 el 8s,B -m d8 8r,B 3 bu 7v,B f b0 88,B 1b 9j 8r,B 26 7u 8o,B 3o 6h ah,B 4m 5h 9k,B 55 46 8e,B 5f 31 7j,B 53 25 76,B 3c 9 6f


Draw the architecture of dbms and explain each component?

Relational Data Model (Course Library)1. H.Maurer and N.Scherbakov "From Databases to Hypermedia"Part 1: Relational Data Model12. Chapter 12 "Architecture of Database Systems"12.1 IntroductionSoftware systems generally have an architecture, ie. possessing of a structure (form) and organisation (function). The former describes identifiable components and how they relate to one another structurally; the latter describes how the functions of the various structural components interact to provide the overall functionality of the system as a whole. Since a database system is basically a software system (albeit complex), it too possesses an architecture. A typical architecture must define a particular configuration of and interaction between data, software modules, meta-data, interfaces and languages (see Figure 12-1).The architecture of a database system determines its capability, reliability, effectiveness and efficiency in meeting user requirements. But besides the visible functions seen through some data manipulation language, a good database architecture should provide:Figure 12-1: General database system architectureThe features listed above become especially important in large organisations where corporate data are held centrally. In such situations, no single user department has responsibility over, nor can they be expected to know about, all of the organisation's data. This becomes the job of a Database Administrator (DBA) who has a daunting range of responsibilities that include creating, expanding, protecting and maintaining the integrity of all data while adressing the interests of different present and future user communities. To create a database, a DBA has to analyse and assess the data requirements of all users and from these determine its logical structure (database schema). This, on the one hand, will need to be efficiently mapped onto a physical structure that optimises retrieval performance and the use of storage. On the other, it would also have to be mapped to multiple user views suited to the respective user applications. For large databases, DBA functions will in fact require the full time services of a team of many people. A good database architecture should have features that can significantly facilitate these activities.12.2 Data AbstractionTo meet the requirements above, a more sophisticated architecture is in fact used, providing a number of levels of data abstraction or data definition. The database schema, also known as Conceptual Schema, mentioned above represents an information model at the logical level of data definition. At this level, we abstract out details like computer storage structures, their restrictions, or their operational efficiencies. The view of a database as a collection of relations or tables, each with fixed attributes and primary keys ranging over given domains, is an example of a logical level of data definition.The details of efficiently organising and storing objects of the conceptual schema in computers with particular hardware configurations are dealt with at the internal (storage) level of data definition. This level is also referred to as the Internal Schema. It maps the contents of the conceptual schema onto structures representing tuples, associated key organisations and indexes, etc, taking into account application characteristics and restrictions of a given computer system. That is, the DBA describes at this level how objects of the conceptual schema are actually organised in a computer. Figure 12-2 illustrates these two levels of data definition.Figure 12-2: The logical and internal levels of data abstractionAt a higher level of abstraction, objects from the conceptual schema are mapped onto views seen by end-users of the database. Such views are also referred to as External Schemas. An external schema presents only those aspects of the conceptual schema that are relevant to the particular application at hand, abstracting out all other detaiils. Thus, depending on the requirements of the application, the view may be organised differently from that in the conceptual schema, eg. some tables may be merged, attributes may be suppressed, etc. There may thus be many views created - one for each type of application. In contrast, there is only one conceptual and one internal schema. All views are derived from the same conceptual schema. This is illustrated in Figure 12-3 which shows two different user views derived from the same conceptual schema.Thus, modern database systems support three levels of data abstraction: External Schemas (User Views), Conceptual Schema, and Internal (Storage) Schema.The DDL we discussed in earlier chapters is basically a tool only for conceptual schema definition. The DBA will therefore usually need special languages to handle the external and internal schema definitions. The internal schema definition, however, varies widely over different implementation platforms, ie. there are few common principles for such definition. We will therefore say little more about them in this book.Figure 12-3: User views (external schema)As to external schema definitions, note that in the relational model, the Data Sub- Languages can be used to both describe and manipulate data. This is because the expressions of a Data Sub-Language themselves denote relations. Thus, a collection of new (derived) relations can be defined as an external schema.For example, suppose the following relations are defined:Customer( C#, Cname, Ccity, Cphone )Product( P#, Pname, Price )Transaction( C#, P#, Date, Qnt )We can then define an external view with a construct like the following:Define View My_Transaction_1 As which defines the relation (view):My_Transaction_1( Cname, Ccity, Date, Total_Sum )This definition effectively maps the conceptual database structure into a form more convenient for a particular user or application. The extension of this derived table is itself derived from the extensions of the source relations. This is illustrated in Figure 12-4 below.Figure 12-4: External view definitionThis is a very important property of the relational data model: a unified approach to data definition and data manipulation.12.3 Data AdministrationFunctions of a DBA include:Creation of the database To create a database, a DBA has to analyse and assess the requirements of the users and from these determine its logical structure. In other words, the DBA has to design a conceptual schema and a first variant of an internal schema. When the internal schema is ready, the DBA must load the database with actual data.Acting as intermediary between users and the database A DBA is responsible for all user facilities determined by external schemas, ie. the DBA is responsible for defining all external schemas or user views.Ensuring data privacy, integrity and security In analysing user requirements, a DBA must determine who should have access to which data and subsequently arrange for appropriate privacy locks (passwords) for identified individuals and/or groups. The DBA must also determine integrity constraints and arrange for appropriate data validation to ensure that such constraints are never violated. Last, but not least, the DBA must make arrangements for data to be regularly backed up and stored in a safe place as a measure against unrecoverable data losses for one reason or another.At first glance, it may seem that a database can be developed using the conventional "waterfall" technique. That is, the development process is a sequence of stages, with work progressing from one stage to the next only when the preceding stage has been completed. For relational database development, this sequence will include stages like eliciting user requirements, analysing data relationships, designing the conceptual schema, designing the internal schema, loading the database, defining user views and interfaces, etc, through to the deployment of user facilities and database operations. In practice, however, when users start to work with the database, the initial requirements inevitably change for a number of reasons including experience gained, a growing amount of data to be processed, and, in this fast changing world, changes in the nature of the business it supports. Thus, a database need to evolve, learning from experience and allowing for changes in requirements. In particular, we may expect periodic changes to:improve database performance as data usage patterns changes or becomes cleareradd new applications to meet new processing requirementsmodify the conceptual schema as understanding of the enterprise's perception of data improvesChanging a database, once the conceptual and internal schemas have been defined and data actually loaded, can be a major undertaking even for seemingly small conceptual changes. This is because the data structures at the storage layer will need to be reorganised, perhaps involving complete regeneration of the database. A good DBMS should therefore provide facilities to modify a database with a minimum of inconvenience. The desired facilities can perhaps be broadly described to cover: performance monitoringdatabase reorganisationdatabase restructuringBy performance monitoring we mean the collection of usage statistics and their analysis. Statistics necessary for performance optimisation generally fall under two headings: static and dynamic statistics. The static statistics refer to the general state of the database and can be collected by special monitoring programs when the database is inactive. Examples of such data include the number of tuples per relation, the population of domains, the distribution of relations over available storage space, etc. The dynamic statistics refer to run-time characteristics and can be collected only when the database is running. Examples include frequency of access to and updating of each relation and domain, use of each type of data manipulation operator and associated response times, frequency of disk access for different usage types, etc. It is the DBA's responsibility to analyse such data, interpret them and where necessary take steps to reorganise the storage schema to optimise performance. Reorganising the storage schema also entails the subsequent physical reorganisation of the data themselves. This is what we mean by database reorganisation.The restructuring of the conceptual schema implies changing its contents, such as:adding/removing data items (ie. columns of a relation)adding/removing entire relationssplitting/recombining relationschanging a relation's primary keysetcFor example, assuming the relations as on page 3, suppose we now wish to record also for each purchase transaction the sales representative responsible for the sale. We will need therefore to add a column into the Transaction relation, say with column name R# :Transaction( C#, P#, R# , Date, Qnt ) : new attribute addedThe intention, of course, is to record a unique value under this column to denote a particular sales representative. Details of such sales representatives will then be given in a new relation:Representative( R#, Rname, Rcity, Rphone)A retructured conceptual schema will normally be followed by a database reorganisation in the sense explained above. 12.4 Data IndependenceData independence refers to the independence of one user view (external schema) with respect to others. A high degree of independence is desirable as it will allow a DBA to change one view, to meet new requirements and/or to optimise performance, without affecting other views. Relational databases with appropriate relational sub-languages have a high degree of data independence.For example, suppose that the viewMy_Transaction_1( Cname, Ccity, Date, Total_Sum )as defined before no longer meet the user's needs. Let's say that Ccity and Date are no longer important, and that it is more important to know the product name and quantity purchased. This change is easily accommodated by changing the select- clause in the definition thus: Define View My_Transaction_1 AsSelect Cname, Pname , Qnt, Total_Sum=Price*QntFrom Customer, Transaction, ProductWhere Customer.C# = Transaction.C#& Transaction.P# = Product.P#Pname replaces the original specification of Ccity and Date itemsIf each view is defined separately over the conceptual schema, then as long as the conceptual schema does not change, a view may be redefined without affecting other views. Thus the above change will have no effect on other views, unless they were built upon My_Transaction_1.Data independence is also used to refer to the independence of user views relative to the conceptual schema. For example, the reader can verify that the change in the conceptual schema in the last section (adding the attribute R# to Transaction and adding the new relation Representative), does not affect My_Transaction_1 - neither the original nor the changed view!. In general, if the relations and attributes referred to in a view definition are not removed in a restructuring, the view will not be affected. Thus we can accommodate new (additive) requirements without affecting existing applications.Lastly, data independence may also refer to the extent to which we may change the storage schema without affecting the conceptual or external schemas. We will not elaborate on this as we have pointed out earlier that the storage level is too diverse for meaningful treatment here.12.5 Data ProtectionThere are generally three types of data protection that any serious DBMS must provide. These were briefly described in Chapter 1 and we summarise them here:1. Authorisational Security2. Operational Security3. Physical SecurityIn the context of the relational data model, we can use relational calculus as a notation to define integrity constraints, ie. we define them as formulae of relational calculus. In this case, however, all variables must be bound variables as we are specifying properties over their ranges rather than looking for particular instantiations satisfying some predicate. For example, suppose that for the Product relation, the Price attribute should only have a value greater than 100 and less than 99999. This can be expressed (in DSL Alpha style) as:Range Product X ALL;(X.Price > 100 & X.Price < 99999 )This is interpreted as an assertion that must always be true. Any data manipulation that would make it false would be disallowed (typically generating messages informing the user of the violation). Thus, not only does the relational data model unify data definition and manipulation, but its control as well.In the area of physical security, database backups should of course be done periodically. For this purpose, it is perhaps best to view a database as a large set of physical pages, where each page is a block of fixed size serving as the basic unit of interaction between the DBMS and storage devices. A database backup is thus essentially a copy of the entire set of pages onto another storage medium that is kept in a secure and safe place. Aside from the obvious need for backups against damage of storage devices, theft, natural disasters and the like, backups are necessary to recover a consistent database in the event of a database "crash". Such crashes can occur in the course of a sequence of database transactions, particularly transactions that modify the database content.Suppose, for example, that the last backup was done at time t0, and subsequent to that, a number of update transactions were applied one after another. Suppose further that the first ntransactions were successfully completed, but during the (n+1)th transaction a system failure occurred (eg. disk malfunction, operating system crash, power failure, etc) leaving some pages in a corrupted state.In general, it is not possible to just reapply the failed transaction - the failure could have corrupted the updates performed by previous transactions as well, or worse, it could have damaged the integrity of the storage model as to make some pages of the database unreadable! We have no recourse at this point but to go back to the last known consistent state of the database at time t0, ie. the entire contents of the last backup is reinstated as the current database. Of course, in doing so, all the transactions applied after t0 are lost.At this point it may seem reasonable that, to guard against losing too much work, backups should perhaps be done after each transaction - then at most only the work of one transaction is lost in case of failure. However, many database applications today are transaction intensive typically involving many online users generating many transactions frequently (eg. online airline reservation system). Many databases, on the other hand, are very large and an entire backup could take hours to complete. While backup is being performed the database must be inactive. Thus, it should be clear that this proposition is impractical.As it is clearly desirable that transactions since the last backup are also somehow saved in the event of crashes, an additional mechanism is needed. Essentially, such mechanisms are based on journalling successful transactions applied to a database. This simply means that a copy of each transaction (or affected pages) is recorded in a sequential file as they are applied to the database.The simplest type of journalling is the Forward System Journal. In this, whenever a page is modified, a copy of the modified page is also simultaneously recorded into the forward journal.To illustrate this mechanism, let the set of pages in a database be P = {p1, p2, ... pn}. If the application of an update transaction T on the database changes PT, where PT &Iacute; P, then T(PT) will be recorded in the forward journal. We use the notation T(PT) to denote the set of pages PTafter the transaction T has changed each page in PT. Likewise, we write T(pi) to denote a page pi after it has been changed by transaction T. Furthermore, if T was applied successfully (ie. no crash during its processing), a separator mark, say ';', would be written to the journal. Thus, after a number of successful transactions, the journal would look as follows:< T(PT1) ; T(PT2) ; ... T(PTk) ; >As a more concrete example, suppose transactionT1 changed {p1, p2, p3},T2 changed {p2, p3, p4}, andT3 changed {p3, p4, p5},in that order and all successfully carried out. Then the journal would contain: < T1( {p1, p2, p3} ) ; T2({T1(p2), T1(p3), p4} ) ; T3( {T2(T1(p3)), T2(p4), p5} ) ; >Now suppose a crash occurred just after T3 has been applied. The recovery procedure consists of two steps:replace the database with the latest backupread the system journal in the forward direction (hence the term 'forward' journal) and, for each set of journal pages that precedes the separator ';', use it to replace the corresponding pages in the database.Effectively, this duplicates the effect of applying transactions in the order they were applied prior to the crash. The technique is applicable even if the crash occurred during the last transaction. In this case, the journal for the last transaction would be incomplete and, in particular, the separator ';' would not be written out. Say that transaction T3 was interrupted after modifying pages p3 and p4 but before it could complete modifying p5. Then the journal would look as follows:T1( 123) ; T2( 234) ; T3( 34) In this case, recovery is exactly as described above except that the last incomplete block of changes will be ignored (no separator ';'). Of course, the work of the last transaction is lost, but this is unavoidable. It is possible, however, to augment the scheme further by saving the transaction itself until its effects are completely written to the journal. Then T3 above can be reapplied, as a third step in the recovery procedure. While the forward journal can recover (almost) fully from a crash, its disadvantage is that it is still a relatively slow process - hundreds or even thousands of transactions may have been applied since the last full backup, and the corresponding journals of each of these transactions must be copied back in sequence to restore the state of the database. In some applications, very fast recovery is needed.In these cases, the Backward System Journal will be the more appropriate journalling and recovery technique. With this technique, whenever a transaction changes a page, the page contents before the update is saved. As before, if the transaction succesfully completes, a separator is written. Thus the backward journal for the same example as above would be:123T1( 2) T1(3) 4T2( 3)T2( 4) Since each block of journal pages represents the state immediately before a transaction is applied, recovery consists of only one step: read the journal in the backward direction until the first separator and replace the pages in the database with the corresponding pages read from the journal. Thus, the backward journal is like an 'undo' file - the last block cancels the last transaction, the second last cancels the second last transaction, etc. Features such as those discussed above can significantly facilitate the management of corporate data resources. Such features, together with the overall architecture and the Data Model examined in previous chapters, determine the quality of a DBMS and are thus often used as part of the principal criteria used in critical evaluation of competing DBMSs.12.6 Further ReadingThose of you wanting a more detailed introduction to the relational data model, are strongly recommended to read [Date 1995]. The book is written in a tutorial style with many of the issues, such as the database languages, presented through a progressive series of examples. Another book by the same author [Date 1993] is a readable account of the SQL standard. The book contains many examples and a reader can learn enough to be able to write simple SQL programs or to understand an existing program.The books [Kroenke 1995], [Benyon 1997] and [Ullmann & Widom1997] are written in a similar tutorial style and describe the Relational Data Model in greater detail providing useful recommendations on actual database design.Database theory is well presented in [Elmaseri and Navathe 1994] and [Ullmann 1988].The book [Smith & Barnes 1987] bridges the gap between physical data structures and logical database models. Assuming knowledge of elementary data structures, it describes file processing techniques and introduces fundamentals of database systems as you become aware of design and implementation issues.The book [Silverston, Inmon, and Graziano 1997] can be useful for those who are interested in the actual application of database systems. It provides a common set of database structures for specific functions common to most businesses, such as sales, marketing, order processing, budgeting, and accounting. The book presents and discusses in greater details different design and implementation techniques. Readers can apply one of such data structures to their own company to meet specific data needs.References:[Benyon 1997] Benyon D. Information and Data-Modelling(2nd. Edition), Mc Graw Hill (1997).[Date 1995] Date, C.J. Introduction to Database Systems(6th Edition), Addison- Wesley Publ. company (1995).[Date 1993] Date, C.J. A Guide to the SQL Standard. (3rd Edition), Addison-Wesley Publ. company (1993).[Elmaseri and Navathe 1994] Elmaseri, K. and Navathe, D. Fundamentals of Database Systems (2nd. Edition), Benjamin / Cumming (1994).[Kroenke 1995] Kroenke D.M. Database Processing. (5th. Edition), Prentice Hall (1995).[Silverston, Inmon, and Graziano 1997] Silverston L., Inmon W. H. and Graziano K. The Data Model Resource Book: A Library of Logical Data Models and Data Warehouse Designs, John Willey & Sons (1997).[Smith & Barnes 1987] Smith, P.D. and Barnes, M.G. Files and Databases: An Introduction, Addison-Wesley Publ. company (1987).[Ullmann 1988] Ullmann, J.D. Principles Of Database and Knowledge-Base Systems, Volume I: Classical Database Systems, Computer Science Press (1988).[Ullmann & Widom1997] Ullmann, J.D. and Widom, S. A First Course in Database Systems, Prentice Hall (1997).

People also asked