Addition and subtraction of rational expressions


 Wendy Lambert
 1 years ago
 Views:
Transcription
1 Lecture 5. Addition nd subtrction of rtionl expressions Two rtionl expressions in generl hve different denomintors, therefore if you wnt to dd or subtrct them you need to equte the denomintors first. The common denomintor with the smllest possible degree is the Lest Common Multiple of the originl ones. On this lecture we will consider the procedure of finding the Lest Common Multiple nd solve some exmples on ddition nd subtrction of rtionl expressions Addition nd subtrction of rtionl expressions The rules for dding nd subtrcting rtionl expressions re the sme s rules for dding nd subtrcting frctions. b + c b + c, b b c b c. b In this exmple both the denomintors re equl nd re not zero. Exmple 5.1. (3.4 ex.13 ) (3.4 ex.16 ) x 2 x x2 + 1 x x2 x 2 1 x x x + 5 x 4 2x x 4x + 5 x 4 2x + 4 (x 4) 4x + 5 x 4 + 2x + 4 x 4 6x + 9 x 4. If denomintors of two rtionl expressions re not equl, we cn use following generl rule for dding nd subtrcting quotients Let us prove this formul b + c d d + bc, b 0, d 0. bd b + c d b d d + c d b b d bd + cb db d + cb. bd
2 Lecture 5. Addition nd subtrction of rtionl expressions 23 Similrly, b c d d bc. bd Exmple 5.2. (3.4 ex.22 ) x x x + 1 x2 (x + 1) + 3x 3(x + 1) x3 + x 2 + 3x 3(x + 1) x (x2 + x + 3 3(x + 1). (3.4 ex.36 ) 2x 1 x 1 2x + 1 (2x 1)(x + 1) (x 1)(2x + 1) x + 1 (x 1)(x + 1) 2x2 x + 2x 1 2x 2 + 2x x + 1 (x 1)(x + 1) 2x (x 1)(x + 1). (3.4 ex.41 ) x (x 2) + x + 1 x 1 + x + 1 x 2 x(x 1)(x 2) + (x 2)(x 2)(x 1) + (x + 1)(x + 1)(x 1) (x + 1)(x 1) x(x2 3x + 3) + (x + 1)(x 2 4x + 4) + (x 2 + 2x + 1)(x 1) (x + 1)(x 1)(x 2) x3 3x 2 + 3x + x 3 4x 2 + 4x + x 2 4x x 3 + 2x 2 + x x 2 2x 1 (x + 1)(x 1)(x 2) 3x3 5x 2 + 2x + 3 (x + 1)(x 1)(x 2) (x 1)x(3x 2) + 3 (x + 1)(x 1)(x 2). One cn check tht there is no extr cnceling by trying long division of the numertor by the fctors (x + 1), (x 1), nd (x 2) of the denomintor Lest common multiple The generl formul for summing up two rtionl expressions will work in every sitution. But not in every sitution the generl formul is the best. Let us consider following exmple Exmple 5.3. (3.4 ex.69 ) x + 4 x 2 x 2 2x + 3 x 2 + 2x 8 generl formul (x + 4)(x2 + 2x 8) (x 2 x 2)(2x + 3). (x 2 x 2)(x 2 + 2x 8)
3 Lecture 5. Addition nd subtrction of rtionl expressions 24 A little bit smrter wy to do it is to fctor the denomintors, first: (x + 4) (x 2)(x + 1) (2x + 3) (x + 4)(x 2). Then, obviously, we cn equte the denomintors by multiplying the first rtionl expression by nd the second one by (x + 4) (x + 1) (x + 4) (x + 1) : (x + 4)(x + 4) (2x + 3)(x + 1) (x 2)(x + 1)(x + 4) x2 + 8x x 2 3x 2x 3 (x 2)(x + 1)(x + 4) x 2 + 3x 13 (x 2)(x + 1)(x + 4). Let us summrize wht we did. We wnted to equte the denomintors nd we wnted to keep our work simple. The work will be kept simple if common denomintor hs the smllest possible degree. In this exmple polynomil (x 2)(x+1)(x+4) is the one nd it is clled the Lest Common Multiple of polynomils (9x 2)(x 1) nd (x 2)(x + 4). Definition 5.1. The polynomil P (x) is the lest common multiple of polynomils P 1 (x), P n (x) if ) ech of p i is fctor of P (x); b) there is no polynomil q(x) with degree less thn degree of P (x) such tht ) is stisfied. Remrk 5.1. Prt b) sys tht P (x) hs the lest possible degree for property ). Question 5.1. Let P (x) be the lest common multiple of polynomils P 1 (X), P 2 (X), P 3 (x). ) Consider P (x) P 1 (x), will P 1(x)  cncel? b) Consider P (x) P 2 (x), will P 2(x)  cncel? c) Consider P (x) P 1 (x)p 2 (x), will P 1(x)P 2 (x)  cncel? Let us discuss the process of finding the lest common multiple of two or more polynomils. The first step is to fctor ech polynomil completely. Then to construct the lest common multiple we successively combine prime fctors of the originl polynomils. For ech originl polynomil we dd only those of it s fctor which re missing in the
4 Lecture 5. Addition nd subtrction of rtionl expressions 25 lest common multiple. In prticulr, let polynomils M(x), N(x), P (x) hve fctoring, correspondingly M(x) (x)b(x)c(x), N(x) b(x)c(x)d(x), P (x) b(x)d(x)e(x), then the lest common multiple for M (x), N (x), nd P (x) will be LCM (x)b(x)c(x)d(x)b(x)e(x). Here initilly we dded to LCM ll fctors from M(x) since LCM should hve ll fctors of M(x) nd hd nothing initilly. On the second step we dded d(x) from N(x) since other two multipliers lredy were LCM. On the third step we dded e(x) nd b(x) from P (x). Although LCM lredy hd one b(x), still P (x) hd two of them so one ws missing, thus, we dded it. Exmple 5.4. (3.4 ex.51 ) Find the LCM to x 3 x, x 3 2x 2 + x, nd x 3 1. First, we will fctor three polynomils The LCM is then given by (3.4 ex.72 ) x 3 x x(x + 1)(x 1), x 3 2x 2 + x x(x 1)(x 1), x 3 1 (x 1)(x 2 + x + 1). LCM x(x + 1)(x 1)(x 1)(x 2 + x + 1). x (x 1) x x + 1 x 3 x 2 x (x 1) x x + 1 x 2 (x 1) x x2 + 2 x (x 1) 2 (x + 1)(x 1) (x 1) 2 x 2 x 3 + 2x 3 4x 2 + 2x x x3 5x 2 + 2x + 1 (x 1) 2 x 2. By doing long division of the numertor over the (x 1) one cn see tht (x 1) does not cncels.
5 Lecture 6. Mixed quotients When the numertor or the denomintor of quotient contins combintions of rtionl functions we cll it mixed quotient. On this lecture we will consider how to simplify mixed quotients Mixed quotients Definition 6.1. When sums nd/or differences of rtionl expressions pper s the numertor nd/or denomintor of quotient, the quotient is clled mixed quotient. To simplify mixed quotient mens to write it s rtionl expression reduced to lowest terms. Exmple of Mixed Quotient b + c d e f + k, where, b, c, d, e, f, k, l re polynomils. l There re two fvorite methods for simplifying mixed quotients. The first methods is to simplify mixed quotient step by step by considering rtionl expressions is the numertor, then in the denomintor nd, finlly, the whole thing. In the second pproch you will need to compute the lest common multiplied of the denomintors of ll rtionl expressions entering the mixed quotient. Then the mixed quotient is simplified in one step by multiplying both numertor nd denomintor with the lest common multiple. We will consider some exmples for ech of two methods Simplifying by prts Method Summry: Consider numertor, simplify; consider denomintor, simplify, etc. b + c d e f + k l d + bc bd el + kf fl (d + bc) fl (el + fk) bd
6 Lecture 6. Mixed quotients 27 Exmple 6.1. (3.5 ex.1 ) x x x + 1 x x + 4 x + 1 x (x + 4) (x + 1)(x + 4) 2 x Simplifying by one strike! Method Summry: Tke LCM of ll little denomintors (b d f e) nd multiply both the numertor nd the denomintor of the mixed quotient by LCM. b + c d e f + k l LCM LCM b LCM + c d LCM e f LCM + k. e LCM Exmple 6.2. (3.5 ex.16 ) 1 x x x 1 x LCM (x + 1)x, Multiplying both the numertor nd the denomintor by LCM 1 x x x 1 x (3.5 ex.22 ) (x + 1)x (x + 1)x (x + 1)x x 2 2(x + 1)x 1(x 1)(x + 1) x 2 + x x 2 2x 2 + 2x x (2x + 5) x x x 3 x 2 (x + 1)2 x 3 x + 3 (2x + 5)(x 3) x 2 x(x 3) x 2 (x + 3) (x + 1) 2 x 3 (x 3)(x + 3) ((2x + 5)(x 3) x2 )(x 3)(x + 3) x(x 3)(x 2 (x + 3) (x + 1) 2 (x 3)) x (x + 1) 2. (2x2 + 5x 6x 15 x 2 )(x + 3) x(x 3 + 3x 2 ) (x 2 + 2x + 1)(x 3) (x 2 x 15)(x + 3) x(x 3 + 3x 2 x 3 2x 2 x + 3x 2 + 6x + 3) (x2 x 15)(x + 3). x(4x 2 + 5x + 3)
7 Lecture 6. Mixed quotients 28 (3.5 ex.27 ) x 1 1 x 1 x 1 x x 1 x 1 x x 1 1 x x. Solving the sme exmple by the Method 2 looks little more strightforwrd x x x x 1 x x 1 x 1 x x x.
8 Lecture 7. Negtive exponents. Scientific nottion. Squre root On this lecture we will discuss the negtive exponents. Properties of negtive exponent turn out to be the sme s tht of usul positive exponent. Therefore, both negtive nd positive exponents represent the sme object. We will lso consider the scientific nottions nd how to convert numbers into scientific nottion nd will discuss properties of squre root. In the end of the lecture we will discuss rtionlizing of mthemticl expressions Negtive exponents Let us strt from some simple remrks. If n is counting number then rised to the power n is defined by n... }{{} n fctors. The number n is lso cn be referred s the exponents of. Let us derive simple properties for the exponents n+m n m {}}{ } {{... } } {{... } n+m ; n m (b) n (b)(b)... (b) }{{} n b n ; n times ( ) n ( ) ( ) ( )... n b b b b b ; n n m {}}{ ( n ) m (... ) (... )... (... ) }{{}}{{}}{{} n m ; n n n }{{} n m n {}}{ }{{} m m n m, m < n.
9 Lecture 7. Negtive exponents. Scientific nottion. Squre root 30 Let us tke better look t the lst formul: m < n is n obvious the limittion for it. The formul is not working for m n simply becuse it is undefined for tht cses. From the other side wht if we could extend existing definitions to new cses? If this is possible, we will improve the performnce of our old formul. The first bd cse is m n for which 0 is undefined. Obviously, the cse m n cn be cptured by defining 0 1. You cn check tht the whole set of five formuls will still fit together. Similrly, to mke the lst formul meningful for m > n we need to define 1 1, 2 1 2, 3 1 3, etc. Eventully, we rrive to the definition of the negtive exponent n 1 n. It is esy do check tht properties of exponents re the sme for negtive exponents nd we cn improve the lst formul: m n m n m+( n) m n for ny numbers m nd n. Exmple 7.1. (4.1 ex.11 ) (4.1 ex.13 ) As n exercise, let us derive one more property of the exponents (4.1 ex.44 ) ( b ) n 1 ( b ) n 1 n b n b n ( b ) n n 4x 2 (yz) 1 ( 5) 2 x 4 y 2 z 2 4 ( 5) 2 x 2 y 1 z 1 x 4 y 2 z 2 4( 5) 2 x 6 y 3 x 1 4z 25x 6 y 3. (4.1 ex.52 ) (3xy 1 ) 2 (2x 1 y) 3 32 x 2 y x 3 y 3 32 x 2 x y 3 y 2 32 x y 5.
10 Lecture 7. Negtive exponents. Scientific nottion. Squre root Scientific Nottion Numbers you meet in pplictions my very from very lrge to very smll. As n exmple let us tke or these numbers re tedious to write nd difficult to red. Therefore people use exponents to mke better representtion of such numbers. Recll if you hve some number in decimls. the multipliction by 10 will move the point to the right nd division by 10 will move comm to the left. Therefore,. 10 (. 1 ) (. 10) Definition 7.1. When number hs been written s product of it is sid to be written in scientific nottion. 10 S, 1 10 Exmple 7.2. (4.1 ex.69 ) Write in scientific nottion: (4.1 ex.81 ) Write s deciml Question 7.1. Try to put the following number in ny redble form: Squre roots A rel number is sid to be squred when it is rised to the power 2. Tking squre root is the opertion inverse to squring. Definition 7.2. The squre root of number is number b such tht b 2
11 Lecture 7. Negtive exponents. Scientific nottion. Squre root 32 Question 7.2. If b is squre root of, wht could be sid bout b? How mny roots hs if is negtive? Let us list some properties of squre root if > 0 there exist two squre roots; if 0 the squre root of 0 is 0; if < 0 there re no squre roots. The positive squre root is clled the Principl squre root nd the symbol clled rdicl sign is used to denote the principle (or nonnegtive) squre root. So, > 0; nother squre root is given by. Definition 7.3. In some cses n expression cn be presented s squre of something not ended with the rdicl sign. Such expression re clled perfect squres. Exmples: ( + x) 2, ( x + ) 2 re perfect squres, ( 3) 2 is not perfect squre. Remrk 7.1. Tking squre root or perfect squres is esy nd plesnt. Note nevertheless, tht some cre still should be tken. You should keep in mind tht 2. As n exercise, try to prove why bsolute vlue is need to be putted. Exmple 7.3. (4.2 ex.6 ) Evlute Some more properties of squre roots b b, b b to prove the bove identities we need to recll the definition of squre root. The definition sys tht b ( ) 2 b if nd only if b b. Let us check it ( b ) 2 ( ) 2 ( b ) 2 b. Exmple 7.4. (4.2 ex.16 )
12 Lecture 7. Negtive exponents. Scientific nottion. Squre root 33 Next we will consider some exmples on sums nd differences of squre roots. The ide is to combine the like terms. (4.2. ex.27 ) ( ) 3( ) 3(19 3 2). (4.2. ex.36 ) (3 2)(3 + 2) 3 2 ( 2) Rtionlizing There re severl conventions in mthemtics describing how to write mthemticl expressions. In prticulr, there is rule sying tht the denomintor of quotient should not contin rdicls. The process of removing the rdicls from the denomintor is clled rtionlizing. Exmple 7.5. Rtionlize the following expressions. (4.2. ex.43 ) ; (4.2. ex.56 ) (4.2. ex.62 ) 1 x + h 1 x h Solution 2: x + h x h ( 5 3) x x + h x + h x h h( x x + h) x + h x h(x x + h (x + h) x) (x + h)x ( x + h x ) h x + h ; x 1 x + h h 5 3; h x + h x( x x + h)h (x + h)x h (x x + h (x + h) x) (x + h)x 1 x h h h ( x + h h x + h x (x + h) Let us discuss the properties of the nth root. It will tke some dvnced nlysis to x x ).
13 Lecture 7. Negtive exponents. Scientific nottion. Squre root 34 First, let us summrize some fcts bout the exponents > 0, if b > 0 nd n is even b n > 0, if b < 0 nd n is even is > 0, if b > 0 nd n is odd < 0, if b < 0 nd n is odd
14 Lecture 8. Rdicls On this lecture we will define root of degree n nd discuss it s properties. In the end we will dd some review exmples on squre roots since techniques for working with nth roots nd with squre roots re very similr Rdicls Definition 8.1. An nth root of number is number b such tht b n. n b b n. Note, tht the squre root is specil cse of this definition. Let us discuss the properties of the nth root. By doing some not very dvnced nlysis one cn prove tht there exists t lest one nth root of if > 0. Here we will ccept this fct without proof nd will summrize other informtion which is vilble for rdicls. First, if the number n is even then with ny root b there will be root b since ( b) n (b) n, if n is even. Second, if the number n is odd then if is negtive then its nth root will be negtive nd if is positive then its nth root will be positive. We summrize this informtion in the next tble two, > 0, < 0 if > 0 nd n is even none, if < 0 nd n is even the number of nth roots of is exctly one, > 0 if > 0 nd n is odd one, < 0 if < 0 nd n is odd Definition 8.2. The principl nth root of number denoted by the rdicl sign is n {positive nth root if n is even, nth root if n is odd.
15 Lecture 8. Rdicls 36 Exmple 8.1. (4.3 ex.1 ) (4.3 ex.2 ) (4.3 ex.4 ) (4.3 ex.10 ) x (x 2 ) 3 3 (3x 2 ) 3 3x Properties of rdicls The first property follows from the definition of rdicls nd the definition of the principl root { n if n is odd n if n is even Exmple ( 3)3 3, 2 ( 2)2 2 2 Another properties follow from the definition of rdicls nd properties of exponents n b n n b, n n b n, b ( ) n m n m, m n mn. Remrk 8.1. Note tht these identities re only welldefined when both nd b re positive. For exmple, the first one is meningless for < 0, b < 0, nd n even. Question 8.1. How to prove these identities? Let us prove the lst one. When we rise both sides to degree nm. Using the properties of integer exponents nd the definition of rdicls we rewrite the left s ( ) m n m n (( ) m n m ) n ( ) n n. At the sme time, ccording to the definition of the nth root the right side is ( nm ) mn. The identity is proved.
16 Lecture 8. Rdicls 37 Exmple 8.3. (4.3 ex.27 ) 3x 2 12x 3 x x x 6 x x (4.3 ex.31 ) (4.3 ex.30 ) 4 9x 2 y 4 3 x2 y 3 125x 3 3 8x3 y x 2 y x y x 5 y 23 x 3 y 5x 3 x 2 3 y 4 3 2xy 3 y 5 3 x 2 2y. (4.3 ex.47 ) 8x3 3 50x + 2x 5 2 x 2x 3 5 2x x 2 2x (2 x 15 x 2 ) 2x. (4.3 ex.51 ) (4.3 ex.54 ) (3 3 6)(2 3 9) 3( 3 3 2)2( 3 3 3) ( 4 4 2)( ) ( 4 4) 2 (2) Review exmples Rtionlize the denomintor in ech expression (4.2 ex.55 ) ( 3 + 2) (4.2 ex.57 ) ( 3 2)( ) ( 3 + 2) Rtionlize the numertor: (4.2 ex.59 ) (2 5)(2 + 5) (3 2 5)(2 + 5) Solve eqution x + 3 2x 5 0, x + 3 2x 5, x + 3 2x 5, x 8.
17 Lecture 9. Rtionl exponents On this lecture we will give the definition of rtionl exponent nd will discuss its properties which re, in fct, similr to the properties of integer exponent Rtionl exponents Let us compre two identifies, one for exponents nd one for rdicls ( m ) n mn m n mn Since the similrity is obvious it give us hope tht both the nturl exponents nd rdicls represent the sme but more generl object. The reltion is even more obvious if we consider the following identity for rdicls ( n ) n ( n 1 n) (n 1n) ( ) n n Finlly, the gp between integer exponents nd rdicls will be eliminted if we define 1 n n, or, more generl, for ny rtionl number m n, n > 0 define m n ( n ) m Exmple 9.1. (4.4 ex.6 ) (64) 3 2 (4 16) 3 2 (2 6 ) (2 6 ) (4.4 ex.8 ) (25) 5 2 (5 2 ) (5 2 ) 5 1 (52 )
18 Lecture 9. Rtionl exponents 39 Let us list bsic identifies for the rtionl exponents. These identifies will combine together properties of integer exponents nd rdicls. r s r+s, ( r ) s rs, (b) r r b r, ( b ) r r b r, r 1 r, r s r s. As n exercise let us check the second one r s m n p ( ) q n m ( ) q p ( n ( q ) m ( ) q q ( n ) p ) n ( ) n q qm ( ) q n pn ( nq ) qm ( nq ) pn ( nq ) qm+pn qm+pn nq r+s. Agin, the identifies re welldefined only for positive nd b. When nd b re negtive some identifies will fil. Exmple 9.2. (4.4 ex.32 ) (4.4 ex.16 ) (4.4 ex.28 ) ( x 3 2 y 5 2 ) 4 3 ( x 1 3 y 1 3 ( 1 9 ) 3 ( 3 x y ) 1.5 ( 1 9)15 10 ( ) 4x 1 y )( 1 x 3 ( 3) ) x 2 x 1 y 1 3 ( 3) y 10 3 y 1 1 x 3 y 7 3 ( 1 9) (x 1 ) (1 9 ( ) y ) 3 ( 1 3 ) x 3 2 y y x 3 1 x 3 3 y 7. Question 9.1. Wht is equl to? Answer: (x 2 ) 3 2 (x 2 ) 3 2 x 3
19 Lecture 10. Geometry topics On this lecture we will consider some re formuls, prove the Pythgoren theorem, nd stte some extr formuls which will be used lter Are formuls For rectngle of length l nd width w re is defined to be Are lw. For the right tringle with the bse l nd ltitude the re ccording to the picture will be Are 1 2 lh. For n rbitrry tringle: Are 1 2 l 1h l 2h 1 2 lh. Let us use re formuls to prove the Pythgoren Theorem The Pythgoren theorem
20 Lecture 10. Geometry topics 41 Definition A right tringle is one tht contins right ngle  tht is, n ngle of 90. The side of the tringle opposite to 90 ngle is clled the hypotenuse; the remining two sides re clled legs. Theorem Pythgoren Theorem. In right tringle, the squre of the length of the hypotenuse is equl to the sum of the squres of the lengths of the legs. c b 2 Proof. The side of the big squre is + b. Note tht ll four tringles re right nd congruent. The inside figure is squre with the side c. Obviously, the re of the big squre equls to the sum of res of tringle nd the smll squre. the Pythgoren Theorem is proved. ( 1 ( + b) 2 4 b + c 2) b + b 2 2b + c b 2 c 2 Exmple (4.6 ex.34 ) The re of squre ABCD is S ABCD 100 squre feet, the re of squre BEF G is S BEF G 16 squre feet Wht is the re of the tringle CGF. Solution. Since S ABCD 100 then side BC 10. Since S BEF G 16 then side BG GF 4, therefore GC BC BG 6 nd S CGF (4.6 ex.8 ) Check if the tringle with the sides below is right one: Tke 6, b 8, nd c 10, then 6, 8, b c 2.
21 Lecture 10. Geometry topics 42 Answer: the tringle is right. (4.6 ex.14 ) Check if the tringle with the sides below is right one: Tke 5, b 4, nd c 7, then Answer: the tringle is not right. 5, 4, b c 2. Question Why the hypotenuse is lwys the longest side of the tringle. Prove using Pythgoren theorem Some more geometricl formuls Volume of rectngulr box of length l, width w, nd height h: Volume lwh For the circle with rdius r (dimeter d 2r) Are πr 2, Circumference 2πr πd Perimeter of rectngulr of length l nd width w is given by Perimeter 2l + 2w. Exmple (4.6 ex.38 ) How fr person cn see. R is the rdius of the Erth. h is the height of tower. d is the distnce the person cn see.
22 Lecture 10. Geometry topics 43 According to Pythgoren theorem d 2 + R 2 (R + h) 2 or d (R + h) 2 R 2.
Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
More informationPROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive
More informationAlgebra Review. How well do you remember your algebra?
Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then
More informationSection A4 Rational Expressions: Basic Operations
A Appendi A A BASIC ALGEBRA REVIEW 7. Construction. A rectngulr opentopped bo is to be constructed out of 9 by 6inch sheets of thin crdbord by cutting inch squres out of ech corner nd bending the
More informationSPECIAL PRODUCTS AND FACTORIZATION
MODULE  Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationMath 135 Circles and Completing the Square Examples
Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for
More informationFactoring Polynomials
Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles
More informationFUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation
FUNCTIONS AND EQUATIONS. SETS AND SUBSETS.. Definition of set. A set is ny collection of objects which re clled its elements. If x is n element of the set S, we sy tht x belongs to S nd write If y does
More informationBinary Representation of Numbers Autar Kaw
Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse rel number to its binry representtion,. convert binry number to n equivlent bse number. In everydy
More informationRIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS
RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is
More informationAppendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:
Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you
More information1 Numerical Solution to Quadratic Equations
cs42: introduction to numericl nlysis 09/4/0 Lecture 2: Introduction Prt II nd Solving Equtions Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mrk Cowlishw Numericl Solution to Qudrtic Equtions Recll
More informationLINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES
LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of
More informationMATH 150 HOMEWORK 4 SOLUTIONS
MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive
More information4.11 Inner Product Spaces
314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces
More informationTHE RATIONAL NUMBERS CHAPTER
CHAPTER THE RATIONAL NUMBERS When divided by b is not n integer, the quotient is frction.the Bbylonins, who used number system bsed on 60, epressed the quotients: 0 8 s 0 60 insted of 8 s 7 60,600 0 insted
More informationExample A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding
1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde
More informationMathematics Higher Level
Mthemtics Higher Level Higher Mthemtics Exmintion Section : The Exmintion Mthemtics Higher Level. Structure of the exmintion pper The Higher Mthemtics Exmintion is divided into two ppers s detiled below:
More informationBabylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity
Bbylonin Method of Computing the Squre Root: Justifictions Bsed on Fuzzy Techniques nd on Computtionl Complexity Olg Koshelev Deprtment of Mthemtics Eduction University of Texs t El Pso 500 W. University
More informationUnit 6: Exponents and Radicals
Eponents nd Rdicls : The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N):  counting numers. {,,,,, } Whole Numers (W):  counting numers with 0. {0,,,,,, } Integers (I): 
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments  they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationSquare Roots Teacher Notes
Henri Picciotto Squre Roots Techer Notes This unit is intended to help students develop n understnding of squre roots from visul / geometric point of view, nd lso to develop their numer sense round this
More informationThe remaining two sides of the right triangle are called the legs of the right triangle.
10 MODULE 6. RADICAL EXPRESSIONS 6 Pythgoren Theorem The Pythgoren Theorem An ngle tht mesures 90 degrees is lled right ngle. If one of the ngles of tringle is right ngle, then the tringle is lled right
More informationLecture 15  Curve Fitting Techniques
Lecture 15  Curve Fitting Techniques Topics curve fitting motivtion liner regression Curve fitting  motivtion For root finding, we used given function to identify where it crossed zero where does fx
More informationGraphs on Logarithmic and Semilogarithmic Paper
0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl
More information4: RIEMANN SUMS, RIEMANN INTEGRALS, FUNDAMENTAL THEOREM OF CALCULUS
4: RIEMA SUMS, RIEMA ITEGRALS, FUDAMETAL THEOREM OF CALCULUS STEVE HEILMA Contents 1. Review 1 2. Riemnn Sums 2 3. Riemnn Integrl 3 4. Fundmentl Theorem of Clculus 7 5. Appendix: ottion 10 1. Review Theorem
More information1.2 The Integers and Rational Numbers
.2. THE INTEGERS AND RATIONAL NUMBERS.2 The Integers n Rtionl Numers The elements of the set of integers: consist of three types of numers: Z {..., 5, 4, 3, 2,, 0,, 2, 3, 4, 5,...} I. The (positive) nturl
More informationGeometry 71 Geometric Mean and the Pythagorean Theorem
Geometry 71 Geometric Men nd the Pythgoren Theorem. Geometric Men 1. Def: The geometric men etween two positive numers nd is the positive numer x where: = x. x Ex 1: Find the geometric men etween the
More informationCurve Sketching. 96 Chapter 5 Curve Sketching
96 Chpter 5 Curve Sketching 5 Curve Sketching A B A B A Figure 51 Some locl mximum points (A) nd minimum points (B) If (x, f(x)) is point where f(x) reches locl mximum or minimum, nd if the derivtive of
More informationand thus, they are similar. If k = 3 then the Jordan form of both matrices is
Homework ssignment 11 Section 7. pp. 24925 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If
More informationRational Functions. Rational functions are the ratio of two polynomial functions. Qx bx b x bx b. x x x. ( x) ( ) ( ) ( ) and
Rtionl Functions Rtionl unctions re the rtio o two polynomil unctions. They cn be written in expnded orm s ( ( P x x + x + + x+ Qx bx b x bx b n n 1 n n 1 1 0 m m 1 m + m 1 + + m + 0 Exmples o rtionl unctions
More informationSection 54 Trigonometric Functions
5 Trigonometric Functions Section 5 Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form
More informationAssuming all values are initially zero, what are the values of A and B after executing this Verilog code inside an always block? C=1; A <= C; B = C;
B26 Appendix B The Bsics of Logic Design Check Yourself ALU n [Arthritic Logic Unit or (rre) Arithmetic Logic Unit] A rndomnumer genertor supplied s stndrd with ll computer systems Stn KellyBootle,
More informationReasoning to Solve Equations and Inequalities
Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing
More informationAREA OF A SURFACE OF REVOLUTION
AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.
More informationFinite Automata. Informatics 2A: Lecture 3. John Longley. 25 September School of Informatics University of Edinburgh
Lnguges nd Automt Finite Automt Informtics 2A: Lecture 3 John Longley School of Informtics University of Edinburgh jrl@inf.ed.c.uk 25 September 2015 1 / 30 Lnguges nd Automt 1 Lnguges nd Automt Wht is
More informationMathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100
hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by
More informationIntegration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More information6.2 Volumes of Revolution: The Disk Method
mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine soclled volumes of
More informationRegular Sets and Expressions
Regulr Sets nd Expressions Finite utomt re importnt in science, mthemtics, nd engineering. Engineers like them ecuse they re super models for circuits (And, since the dvent of VLSI systems sometimes finite
More informationLecture 5. Inner Product
Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right
More information19. The FermatEuler Prime Number Theorem
19. The FermtEuler Prime Number Theorem Every prime number of the form 4n 1 cn be written s sum of two squres in only one wy (side from the order of the summnds). This fmous theorem ws discovered bout
More information9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes
The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is soclled becuse when the sclr product of two vectors
More informationHomework 3 Solutions
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.
More informationCHAPTER 11 Numerical Differentiation and Integration
CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods
More informationA new algorithm for generating Pythagorean triples
A new lgorithm for generting Pythgoren triples RH Dye 1 nd RWD Nicklls 2 The Mthemticl Gzette (1998); 82 (Mrch, No. 493), p. 86 91 (JSTOR rchive) http://www.nicklls.org/dick/ppers/mths/pythgtriples1998.pdf
More information200506 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration
Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 256 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting
More informationSection 74 Translation of Axes
62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 74 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More informationSquare & Square Roots
Squre & Squre Roots Squre : If nuber is ultiplied by itself then the product is the squre of the nuber. Thus the squre of is x = eg. x x Squre root: The squre root of nuber is one of two equl fctors which
More informationIntegration by Substitution
Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is
More informationMultiplication and Division  Left to Right. Addition and Subtraction  Left to Right.
Order of Opertions r of Opertions Alger P lese Prenthesis  Do ll grouped opertions first. E cuse Eponents  Second M D er Multipliction nd Division  Left to Right. A unt S hniqu Addition nd Sutrction
More informationChapter 2 The Number System (Integers and Rational Numbers)
Chpter 2 The Number System (Integers nd Rtionl Numbers) In this second chpter, students extend nd formlize their understnding of the number system, including negtive rtionl numbers. Students first develop
More informationRadius of the Earth  Radii Used in Geodesy James R. Clynch February 2006
dius of the Erth  dii Used in Geodesy Jmes. Clynch Februry 006 I. Erth dii Uses There is only one rdius of sphere. The erth is pproximtely sphere nd therefore, for some cses, this pproximtion is dequte.
More informationP.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn
33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More informationLecture 2: Matrix Algebra. General
Lecture 2: Mtrix Algebr Generl Definitions Algebric Opertions Vector Spces, Liner Independence nd Rnk of Mtrix Inverse Mtrix Liner Eqution Systems, the Inverse Mtrix nd Crmer s Rule Chrcteristic Roots
More informationMODULE 3. 0, y = 0 for all y
Topics: Inner products MOULE 3 The inner product of two vectors: The inner product of two vectors x, y V, denoted by x, y is (in generl) complex vlued function which hs the following four properties: i)
More informationPhysics 43 Homework Set 9 Chapter 40 Key
Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nmwide region t x
More information10.6 Applications of Quadratic Equations
10.6 Applictions of Qudrtic Equtions In this section we wnt to look t the pplictions tht qudrtic equtions nd functions hve in the rel world. There re severl stndrd types: problems where the formul is given,
More informationEQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in pointdirection nd twopoint
More informationMA 15800 Lesson 16 Notes Summer 2016 Properties of Logarithms. Remember: A logarithm is an exponent! It behaves like an exponent!
MA 5800 Lesson 6 otes Summer 06 Rememer: A logrithm is n eponent! It ehves like n eponent! In the lst lesson, we discussed four properties of logrithms. ) log 0 ) log ) log log 4) This lesson covers more
More informationwww.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)
www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More informationRepeated multiplication is represented using exponential notation, for example:
Appedix A: The Lws of Expoets Expoets re shorthd ottio used to represet my fctors multiplied together All of the rules for mipultig expoets my be deduced from the lws of multiplictio d divisio tht you
More information1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator
AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.
More informationPHY 140A: Solid State Physics. Solution to Homework #2
PHY 140A: Solid Stte Physics Solution to Homework # TA: Xun Ji 1 October 14, 006 1 Emil: jixun@physics.ucl.edu Problem #1 Prove tht the reciprocl lttice for the reciprocl lttice is the originl lttice.
More informationALGEBRAIC FRACTIONS,AND EQUATIONS AND INEQUALITIES INVOLVING FRACTIONS
CHAPTER ALGEBRAIC FRACTIONS,AND EQUATIONS AND INEQUALITIES INVOLVING FRACTIONS Although people tody re mking greter use of deciml frctions s they work with clcultors, computers, nd the metric system, common
More information9 CONTINUOUS DISTRIBUTIONS
9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete
More information5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.
5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous relvlued
More informationVolumes of solids of revolution
Volumes of solids of revolution We sometimes need to clculte the volume of solid which cn be obtined by rotting curve bout the xxis. There is strightforwrd technique which enbles this to be done, using
More informationMath Review 1. , where α (alpha) is a constant between 0 and 1, is one specific functional form for the general production function.
Mth Review Vribles, Constnts nd Functions A vrible is mthemticl bbrevition for concept For emple in economics, the vrible Y usully represents the level of output of firm or the GDP of n economy, while
More informationLinear Equations in Two Variables
Liner Equtions in Two Vribles In this chpter, we ll use the geometry of lines to help us solve equtions. Liner equtions in two vribles If, b, ndr re rel numbers (nd if nd b re not both equl to 0) then
More informationBasic Analysis of Autarky and Free Trade Models
Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently
More informationReview guide for the final exam in Math 233
Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered
More informationSCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics. Basic Algebra
SCHOOL OF ENGINEERING & BUILT ENVIRONMENT Mthemtics Bsic Alger. Opertions nd Epressions. Common Mistkes. Division of Algeric Epressions. Eponentil Functions nd Logrithms. Opertions nd their Inverses. Mnipulting
More informationMATLAB Workshop 13  Linear Systems of Equations
MATLAB: Workshop  Liner Systems of Equtions pge MATLAB Workshop  Liner Systems of Equtions Objectives: Crete script to solve commonly occurring problem in engineering: liner systems of equtions. MATLAB
More informationDouble Integrals over General Regions
Double Integrls over Generl egions. Let be the region in the plne bounded b the lines, x, nd x. Evlute the double integrl x dx d. Solution. We cn either slice the region verticll or horizontll. ( x x Slicing
More information5.6 POSITIVE INTEGRAL EXPONENTS
54 (5 ) Chpter 5 Polynoils nd Eponents 5.6 POSITIVE INTEGRAL EXPONENTS In this section The product rule for positive integrl eponents ws presented in Section 5., nd the quotient rule ws presented in Section
More informationThe Math Learning Center PO Box 12929, Salem, Oregon 97309 0929 Math Learning Center
Resource Overview Quntile Mesure: Skill or Concept: 1010Q Determine perimeter using concrete models, nonstndrd units, nd stndrd units. (QT M 146) Use models to develop formuls for finding res of tringles,
More informationA Note on Complement of Trapezoidal Fuzzy Numbers Using the αcut Method
Interntionl Journl of Applictions of Fuzzy Sets nd Artificil Intelligence ISSN  Vol.  A Note on Complement of Trpezoidl Fuzzy Numers Using the αcut Method D. Stephen Dingr K. Jivgn PG nd Reserch Deprtment
More information4 Approximations. 4.1 Background. D. Levy
D. Levy 4 Approximtions 4.1 Bckground In this chpter we re interested in pproximtion problems. Generlly speking, strting from function f(x) we would like to find different function g(x) tht belongs to
More informationThe Chain Rule. rf dx. t t lim " (x) dt " (0) dx. df dt = df. dt dt. f (r) = rf v (1) df dx
The Chin Rule The Chin Rule In this section, we generlize the chin rule to functions of more thn one vrible. In prticulr, we will show tht the product in the singlevrible chin rule extends to n inner
More informationMath 314, Homework Assignment 1. 1. Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.
Mth 4, Homework Assignment. Prove tht two nonverticl lines re perpendiculr if nd only if the product of their slopes is. Proof. Let l nd l e nonverticl lines in R of slopes m nd m, respectively. Suppose
More informationTreatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3.
The nlysis of vrince (ANOVA) Although the ttest is one of the most commonly used sttisticl hypothesis tests, it hs limittions. The mjor limittion is tht the ttest cn be used to compre the mens of only
More information0.1 Basic Set Theory and Interval Notation
0.1 Bsic Set Theory nd Intervl Nottion 3 0.1 Bsic Set Theory nd Intervl Nottion 0.1.1 Some Bsic Set Theory Notions Like ll good Mth ooks, we egin with definition. Definition 0.1. A set is welldefined
More informationAA1H Calculus Notes Math1115, Honours 1 1998. John Hutchinson
AA1H Clculus Notes Mth1115, Honours 1 1998 John Hutchinson Author ddress: Deprtment of Mthemtics, School of Mthemticl Sciences, Austrlin Ntionl University Emil ddress: John.Hutchinson@nu.edu.u Contents
More information. At first sight a! b seems an unwieldy formula but use of the following mnemonic will possibly help. a 1 a 2 a 3 a 1 a 2
7 CHAPTER THREE. Cross Product Given two vectors = (,, nd = (,, in R, the cross product of nd written! is defined to e: " = (!,!,! Note! clled cross is VECTOR (unlike which is sclr. Exmple (,, " (4,5,6
More informationQUANTITATIVE REASONING
Guide For Exminees InterUniversity Psychometric Entrnce Test QUNTITTIVE RESONING The Quntittive Resoning domin tests your bility to use numbers nd mthemticl concepts to solve mthemticl problems, s well
More informationEcon 4721 Money and Banking Problem Set 2 Answer Key
Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in
More informationApplications to Physics and Engineering
Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics
More informationLecture 3 Gaussian Probability Distribution
Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike
More informationThinking out of the Box... Problem It s a richer problem than we ever imagined
From the Mthemtics Techer, Vol. 95, No. 8, pges 568574 Wlter Dodge (not pictured) nd Steve Viktor Thinking out of the Bo... Problem It s richer problem thn we ever imgined The bo problem hs been stndrd
More informationNQF Level: 2 US No: 7480
NQF Level: 2 US No: 7480 Assessment Guide Primry Agriculture Rtionl nd irrtionl numers nd numer systems Assessor:.......................................... Workplce / Compny:.................................
More informationCOMPONENTS: COMBINED LOADING
LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of
More informationPROBLEMS 13  APPLICATIONS OF DERIVATIVES Page 1
PROBLEMS  APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.
More informationMechanics Cycle 1 Chapter 5. Chapter 5
Chpter 5 Contct orces: ree Body Digrms nd Idel Ropes Pushes nd Pulls in 1D, nd Newton s Second Lw Neglecting riction ree Body Digrms Tension Along Idel Ropes (i.e., Mssless Ropes) Newton s Third Lw Bodies
More information