2-3 The Remainder and Factor Theorems

Save this PDF as:

Size: px
Start display at page:

Download "2-3 The Remainder and Factor Theorems"

Transcription

1 - The Remaider ad Factor Theorems Factor each polyomial completely usig the give factor ad log divisio 1 x + x x 60; x + So, x + x x 60 = (x + )(x x 15) Factorig the quadratic expressio yields x + x x 60 = (x + )(x 5)(x + ) x + x 1x + 18; x So, x + x 1x + 18 = (x )(x + 5x 6) Factorig the quadratic expressio yields x + x 1x + 18 = (x )(x + 6)(x 1) x + x 18x 0; x So, x + x 18x 0 = (x )(x + 7x + 10) Factorig the quadratic expressio yields x + x 18x 0 = (x )(x + )(x + 5) esolutios Maual - Powered by Cogero x + 0x 8x 96; x + Page 1

2 - So, x + x 1x + 18 = (x )(x + 5x 6) The Remaider ad Factor Theorems Factorig the quadratic expressio yields x + x 1x + 18 = (x )(x + 6)(x 1) x + x 18x 0; x So, x + x 18x 0 = (x )(x + 7x + 10) Factorig the quadratic expressio yields x + x 18x 0 = (x )(x + )(x + 5) x + 0x 8x 96; x + So, x + 0x 8x 96 = (x + )(x + 8x ) Factorig the quadratic expressio yields x + 0x 8x 96 = (x + )(x + )(x ) 5 x + 15x + 108x 50; x 6 So, x + 15x + 108x 50 = (x 6)( x x + 90) Factorig the quadratic expressio yields x + 15x + 108x 50 = (x 6)(x + 6)(x 5) 6 6x 7x 9x 1; x + esolutios Maual - Powered by Cogero Page

3 So, x + 15x + 108x 50 = (x 6)( x x + 90) - The Remaider ad Factor Theorems Factorig the quadratic expressio yields x + 15x + 108x 50 = (x 6)(x + 6)(x 5) 6 6x 7x 9x 1; x + So, 6x 7x 9x 1 = (x + )(x 5x ) Factorig the quadratic expressio yields 6x 7x 9x 1 = (x + )(x + 1)(x ) 7 x + 1x + 8x + 1x 6; x + 6x + 9 So, x + 1x + 8x + 1x 6 = (x + 6x + 9)(x + 6x 7) Factorig both quadratic expressios yield x + 1x + 8x + 1x 6 = (x + ) (x + 7)(x 1) 8 x x 6x + 68x + 0; x x 1 So, x x 6x + 68x + 0 = (x x 1)(x + x 0) Factorig both quadratic expressios yield x + 1x + 8x + 1x 6 = (x 6)(x + )(x + 5)(x ) Divide usig log divisio esolutiosmaual- Powered by Cogero 9 (5x x + 6x x + 1) (x ) Page

4 - So, x x 6x + 68x + 0 = (x x 1)(x + x 0) The Remaider ad Factor Theorems Factorig both quadratic expressios yield x + 1x + 8x + 1x 6 = (x 6)(x + )(x + 5)(x ) Divide usig log divisio 9 (5x x + 6x x + 1) (x ) So, = 5x + 17x + 7x (x6 x5 + x x + x x + ) (x + ) So, 5 = x x + 9x 19x + 1x (x 8x + 1x 6x + 1) (x + ) esolutios Maual - Powered by Cogero Page

5 5 - So, The Remaider ad Factor = x Theorems x + 9x 19x + 1x (x 8x + 1x 6x + 1) (x + ) The remaider ca be writte as So, = x 8x + x (x 7x 8x + 10x + 60) (x ) So, = x x 1x 0 1 (6x6 x5 + 6x 15x + x + 10x 6) (x 1) esolutios Maual - Powered by Cogero Page 5

6 - So, The Remaider ad Factor = x Theorems x 1x 0 1 (6x6 x5 + 6x 15x + x + 10x 6) (x 1) = So, 1 (108x5 6x + 75x + 6x + ) (x + ) So, = 6x 6x + x + 9x (x + x + 6x + 18x 16) (x x + 18x 5) esolutios Maual - Powered by Cogero Page 6

7 - So, The Remaider ad Factor = 6x Theorems 6x + x + 9x (x + x + 6x + 18x 16) (x x + 18x 5) So, = x + 16 (x 1x 1x + 110x 8) (x + x 1) So, = x 8x So, = x x esolutios Maual - Powered by Cogero Page 7

8 - So, The Remaider ad Factor Theorems = x x ca be writte as The remaider So, = Divide usig sythetic divisio 19 (x x + x 6x 6) (x ) Because x, c = Set up the sythetic divisio as follows The follow the sythetic divisio procedure The quotiet is x + x + 5x (x + x x + 8x ) (x + ) Because x +, c = Set up the sythetic divisio as follows The follow the sythetic divisio procedure The quotiet is x x + x + 1 (x 9x x 8) (x ) Because x, c = Set up the sythetic divisio as follows, usig a zero placeholder for the missig x -term i the divided The follow the sythetic divisio procedure esolutios Maual - Powered by Cogero The quotiet is x + x + 1x + + Page 8

9 - The Thequotiet Remaider is x xad + xfactor + Theorems 1 (x 9x x 8) (x ) Because x, c = Set up the sythetic divisio as follows, usig a zero placeholder for the missig x -term i the divided The follow the sythetic divisio procedure The quotiet is x + x + 1x + + (x5 x + 6x + 9x + 6) (x + ) Because x +, c = Set up the sythetic divisio as follows, usig a zero placeholder for the missig x -term i the divided The follow the sythetic divisio procedure The quotiet is x x + x + x (1x5 + 10x 18x 1x 8) (x ) Rewrite the divisio expressio so that the divisor is of the form x c Because c = Set up the sythetic divisio as follows, usig a zero placeholder for the missig x-term i the divided The follow the sythetic divisio procedure The remaider ca be writte as So, the quotiet is 6x + 1x + 1x + 1x (6x 6x + 1x 0x 1) (x + 1) Rewrite the divisio expressio so that the divisor is of the form x c esolutios Maual - Powered by Cogero Page 9

10 ca be writte as The remaider So, the quotiet is 6x + 1x + 1x + 1x The Remaider ad Factor Theorems (6x 6x + 1x 0x 1) (x + 1) Rewrite the divisio expressio so that the divisor is of the form x c Set up the sythetic divisio as follows The follow the sythetic divisio procedure Because The quotiet is 1x 6x + 6x 1 5 (5x5 + 6x + x + 8x + 1) (x ) Rewrite the divisio expressio so that the divisor is of the form x c Because c = Set up the sythetic divisio as follows, usig a zero placeholder for the missig x -term i the divided The follow the sythetic divisio procedure The remaider ca be writte as So, the quotiet is 15x + 1x + 9x + 6x (8x5 + 8x + 68x + 11x + 6) (x + 1) Rewrite the divisio expressio so that the divisor is of the form x c esolutios Maual - Powered by Cogero Page 10

11 ca be writte as The remaider So, the quotiet is 15x + 1x + 9x + 6x + - The Remaider ad Factor Theorems + 6 (8x5 + 8x + 68x + 11x + 6) (x + 1) Rewrite the divisio expressio so that the divisor is of the form x c Set up the sythetic divisio as follows, usig a zero placeholder for the missig x -term i Because the divided The follow the sythetic divisio procedure ca be writte as The remaider So, the quotiet is 1x + x + 16x x (60x6 + 78x5 + 9x 1x 5x 0) (5x + ) Rewrite the divisio expressio so that the divisor is of the form x c Set up the sythetic divisio as follows, usig a zero placeholder for the missig x -term i Because the divided The follow the sythetic divisio procedure 5 The quotiet is 1x + 6x x 5 8 esolutios Maual - Powered by Cogero Page 11

12 - The Remaider Theorems 5 ad Factor The quotiet is 1x + 6x x 5 8 Rewrite the divisio expressio so that the divisor is of the form x c Set up the sythetic divisio as follows, usig a zero placeholder for the missig x -term i the Because divided The follow the sythetic divisio procedure 5 The quotiet is 8x 1x + 6x 15 9 EDUCATION The umber of US studets, i thousads, that graduated with a bachelor s degree from 1970 to ca be modeled by g(x) = 0000x 0016x + 051x 715x + 75x , where x is the umber of years sice 1970 Use sythetic substitutio to fid the umber of studets that graduated i 005 Roud to the earest thousad To fid the umber of studets that graduated i 005, use sythetic substitutio to evaluate g(x) for x = 5 The remaider is , so g(5) = Therefore, rouded to the earest thousad,,151,000 studets graduated i SKIING The distace i meters that a perso travels o skis ca be modeled by d(t) = 0t + t, where t is the time i secods Use the Remaider Theorem to fid the distace traveled after 5 secods To fid the distace traveled after 5 secods, use sythetic substitutio to evaluate d(t) for t = 5 The remaider is 50, so d(5) = 50 Therefore, 50 meters were traveled i 5 secods Fid each f (c) usig sythetic substitutio 1 f (x) = x5 x + x 6x + 8x 15; c = esolutios Maual - Powered by Cogero Page 1

13 - The Remaider ad Factor Theorems The remaider is 50, so d(5) = 50 Therefore, 50 meters were traveled i 5 secods Fid each f (c) usig sythetic substitutio 1 f (x) = x5 x + x 6x + 8x 15; c = The remaider is 711 Therefore, f () = 711 f (x) = x6 x5 + x x + 8x ; c = The remaider is 11,165 Therefore, f () = 11,165 f (x) = x6 + 5x5 x + 6x 9x + x ; c = 5 The remaider is 5,56 Therefore, f (5) = 5,56 f (x) = x6 + 8x5 6x 5x + 6x ; c = 6 The remaider is 7,88 Therefore, f (6) = 7,88 5 f (x) = 10x5 + 6x 8x + 7x x + 8; c = 6 The remaider is 67,978 Therefore, f ( 6) = 67,978 6 f (x) = 6x7 + x5 8x + 1x 15x 9x + 6; c = esolutios Maual - Powered by Cogero The remaider is 686 Therefore, f () = f (x) = x8 + 6x5 x + 1x 6x + ; c = Page 1

14 - The Remaider ad Factor Theorems The remaider is 67,978 Therefore, f ( 6) = 67,978 6 f (x) = 6x7 + x5 8x + 1x 15x 9x + 6; c = The remaider is 686 Therefore, f () = f (x) = x8 + 6x5 x + 1x 6x + ; c = The remaider is 15,18 Therefore, f () = 15,18 Use the Factor Theorem to determie if the biomials give are factors of f (x) Use the biomials that are factors to write a factored form of f (x) 8 f (x) = x x 9x + x + 6; (x + ), (x 1) Use sythetic divisio to test each factor, (x + ) ad (x 1) Because the remaider whe f (x) is divided by (x + ) is 0, (x + ) is a factor Test the secod factor, (x 1), with the depressed polyomial x x x + Because the remaider whe the depressed polyomial is divided by (x 1) is 1, (x 1) is ot a factor of f (x) Because (x + ) is a factor of f (x), we ca use the quotiet of f (x) (x + ) to write a factored form of f (x) as f (x) = (x + )(x x x + ) 9 f (x) = x + x 5x + 8x + 1; (x 1), (x + ) Use sythetic divisio to test each factor, (x 1) ad (x + ) Because the remaider whe f (x) is divided by (x 1) is 18, (x 1) is ot a factor Because the remaider whe f (x) is divided by (x + ) is 0, (x + ) is ot a factor 0 f (x) = x x + x + 18x + 15; (x 5), (x + 5) esolutios Maual - Powered by Cogero Use sythetic divisio to test each factor, (x 5) ad (x + 5) Page 1

15 - Because the remaider whe f (x) Because the remaider whe f (x) is is divided by (x 1) isad 18, (xfactor 1) divided by (x + ) is 0, (x + ) is The Remaider Theorems is ot a factor ot a factor 0 f (x) = x x + x + 18x + 15; (x 5), (x + 5) Use sythetic divisio to test each factor, (x 5) ad (x + 5) Because the remaider whe f (x) is divided by (x 5) is 100, (x 5) is ot a factor Because the remaider whe f (x) is divided by (x + 5) is 150, (x + 5) is ot a factor 1 f (x) = x x + 1x + 118x 0; (x 1), (x 5) Use sythetic divisio to test each factor, (x 1) ad (x 5) For (x 1), rewrite the divisio expressio so that the divisor is of the form x c Because Set up the sythetic divisio as follows The follow the sythetic divisio procedure Because the remaider whe f (x) is divided by (x 1) is 0, (x 1) is a factor Test the secod factor, (x 5), with the depressed polyomial x 7x + x + 0 Because the remaider whe the depressed polyomial is divided by (x 5) is 0, (x 5) is a factor of f (x) Because (x 1) ad (x 5) are factors of f (x), we ca use the fial quotiet to write a factored form of f (x) as f (x) = (x 1)(x 5)(x x 8) Factorig the quadratic expressio yields f (x) = (x 1)(x 5)(x )(x + ) f (x) = x x 6x 111x + 0; (x 1), (x 6) Use sythetic divisio to test each factor, (x 1) ad (x 6) For (x 1), rewrite the divisio expressio so that the divisor is of the form x c esolutios Maual - Powered by Cogero Page 15

16 Because the remaider whe the depressed polyomial is divided by (x 5) is 0, (x 5) is a factor of f (x) Because (x 1) ad (x 5) are factors of f (x), we ca use the fial quotiet to write a factored form of f (x) as f - The Remaider ad Factor Theorems (x) = (x 1)(x 5)(x x 8) Factorig the quadratic expressio yields f (x) = (x 1)(x 5)(x )(x + ) f (x) = x x 6x 111x + 0; (x 1), (x 6) Use sythetic divisio to test each factor, (x 1) ad (x 6) For (x 1), rewrite the divisio expressio so that the divisor is of the form x c Because Set up the sythetic divisio as follows The follow the sythetic divisio procedure Because the remaider whe f (x) is divided by (x 1) is 0, (x 1) is a factor Test the secod factor, (x 6), with the depressed polyomial x 9x 0 Because the remaider whe the depressed polyomial is divided by (x 6) is 1, (x 6) is ot a factor of f (x) Because (x 1) is a factor of f (x), we ca use the quotiet of f (x) (x 1) to write a factored form of f (x) as f (x) = (x 1)(x 9x 0) f (x) = x 5x + 8x + 56x + 6; (x ), (x + ) Use sythetic divisio to test each factor, (x ) ad (x + ) For (x ), rewrite the divisio expressio so that the divisor is of the form x c Because Set up the sythetic divisio as follows The follow the sythetic divisio procedure esolutios Maual - Powered by Cogero Because the remaider whe f (x) is divided by (x ) is Page 16, (x ) is ot a factor

17 - Because the remaider whe the depressed polyomial is divided by (x 6) is 1, (x 6) is ot a factor of f (x) Because (x 1) is a factor of f (x), we ca use the quotiet of f (x) (x 1) to write a factored form of f (x) as f The Remaider ad Factor Theorems (x) = (x 1)(x 9x 0) f (x) = x 5x + 8x + 56x + 6; (x ), (x + ) Use sythetic divisio to test each factor, (x ) ad (x + ) For (x ), rewrite the divisio expressio so that the divisor is of the form x c Because Set up the sythetic divisio as follows The follow the sythetic divisio procedure Because the remaider whe f (x) is divided by (x ) is, (x ) is ot a factor Test (x + ) Because the remaider whe f (x) is divided by (x + ) is, (x + ) is ot a factor f (x) = 5x5 + 8x 68x + 59x + 0; (5x ), (x + 8) Use sythetic divisio to test each factor, (5x ) ad (x + 8) For (5x ), rewrite the divisio expressio so that the divisor is of the form x c Because Set up the sythetic divisio as follows The follow the sythetic divisio procedure esolutios Maual - Powered by Cogero Because the remaider whe f (x) is divided by (5x ) is Page 17, (5x ) is ot a factor

18 - The Remaider ad Factor Theorems Because the remaider whe f (x) is divided by (x + ) is, (x + ) is ot a factor f (x) = 5x5 + 8x 68x + 59x + 0; (5x ), (x + 8) Use sythetic divisio to test each factor, (5x ) ad (x + 8) For (5x ), rewrite the divisio expressio so that the divisor is of the form x c Because Set up the sythetic divisio as follows The follow the sythetic divisio procedure Because the remaider whe f (x) is divided by (5x ) is, (5x ) is ot a factor Test (x + 8) Because the remaider whe f (x) is divided by (x + 8) is 1986, (x + 8) is ot a factor 5 f (x) = x5 9x + 9x + x + 75x + 6; (x + ), (x 1) Use sythetic divisio to test each factor, (x + ) ad (x 1) For (x + ), rewrite the divisio expressio so that the divisor is of the form x c Because Set up the sythetic divisio as follows The follow the sythetic divisio procedure + ) is 0, (x + ) is a factor Test the secod factor, (x 1),Page 18 with the depressed polyomial x x + 1x x + 1 esolutios Maualthe - Powered by Cogero Because remaider whe f (x) is divided by (x

19 - The Remaider ad Factor Theorems Because the remaider whe f (x) is divided by (x + 8) is 1986, (x + 8) is ot a factor 5 f (x) = x5 9x + 9x + x + 75x + 6; (x + ), (x 1) Use sythetic divisio to test each factor, (x + ) ad (x 1) For (x + ), rewrite the divisio expressio so that the divisor is of the form x c Set up the sythetic divisio as follows The follow the sythetic divisio procedure Because Because the remaider whe f (x) is divided by (x + ) is 0, (x + ) is a factor Test the secod factor, (x 1), with the depressed polyomial x x + 1x x + 1 Because the remaider whe the depressed polyomial is divided by (x 1) is 8, (x 1) is ot a factor of f (x) Because (x + ) is a factor of f (x), we ca use the quotiet of f (x) (x + ) to write a factored form of f (x) as f (x) = (x + ) (x x + 1x x + 1) 6 TREES The height of a tree i feet at various ages i years is give i the table a Use a graphig calculator to write a quadratic equatio to model the growth of the tree b Use sythetic divisio to evaluate the height of the tree at 15 years a Use the quadratic regressio fuctio o the graphig calculator f(x) = 0001x + x 69 esolutios Maual Powered by of Cogero b To fid -the height the tree at 15 years, use sythetic substitutio to evaluate f (x) for x = 15 Page 19

20 Because the remaider whe the depressed polyomial is divided by (x 1) is 8, (x 1) is ot a factor of f (x) Because (x + ) is a factor of f (x), we ca use the quotiet of f (x) (x + ) to write a factored form of f (x) as f - The Remaider Theorems ad Factor (x) = (x + ) (x x + 1x x + 1) 6 TREES The height of a tree i feet at various ages i years is give i the table a Use a graphig calculator to write a quadratic equatio to model the growth of the tree b Use sythetic divisio to evaluate the height of the tree at 15 years a Use the quadratic regressio fuctio o the graphig calculator f(x) = 0001x + x 69 b To fid the height of the tree at 15 years, use sythetic substitutio to evaluate f (x) for x = 15 The remaider is 985, so f (15) = 985 Therefore, the height of the tree at 15 years is about 985 feet 7 BICYCLING Patrick is cyclig at a iitial speed v0 of meters per secod Whe he rides dowhill, the bike accelerates at a rate a of 0 meter per secod squared The vertical distace from the top of the hill to the bottom of the hill is 5 meters Use d(t) = v0t + at to fid how log it will take Patrick to ride dow the hill, where d(t) is distace traveled ad t is give i secods Substitute v0 =, a = 0, ad d(t) = 5 ito d(t) = v0t + at Use the quadratic equatio to solve for t It will take Patrick 5 secods to travel the 5 meters esolutios Maual - Powered by Cogero Factor each polyomial usig the give factor ad log divisio Assume > 0 8 x + x 1x ; x + Page 0

21 - The Remaider ad Factor Theorems The remaider is 985, so f (15) = 985 Therefore, the height of the tree at 15 years is about 985 feet 7 BICYCLING Patrick is cyclig at a iitial speed v0 of meters per secod Whe he rides dowhill, the bike accelerates at a rate a of 0 meter per secod squared The vertical distace from the top of the hill to the bottom of the hill is 5 meters Use d(t) = v0t + at to fid how log it will take Patrick to ride dow the hill, where d(t) is distace traveled ad t is give i secods Substitute v0 =, a = 0, ad d(t) = 5 ito d(t) = v0t + at Use the quadratic equatio to solve for t It will take Patrick 5 secods to travel the 5 meters Factor each polyomial usig the give factor ad log divisio Assume > 0 8 x + x 1x ; x + So, x + x 1x = (x + )(x x 1) Factorig the quadratic expressio yields x + x 1x = (x + )(x )(x + ) 9 x + x 1x + 10; x 1 esolutios Maual - Powered by Cogero Page 1

22 - So, x + x 1x = (x + )(x x 1) The Remaider Factor Theorems Factorig the quadraticad expressio yields x + x 1x = (x + )(x )(x + ) 9 x + x 1x + 10; x 1 So, x +x 1x + 10 = (x 1)(x + x 10) 50 x + x 10x + ; x + So, x + x 10x + = (x + )(x Factorig the quadratic expressio yields x x + 1) + x 10x + = (x + )(x 1)(x 1) 51 9x + x 171x + 5; x 1 So, 9x + x 171x + 5 = (x 1)(x + 9x 5) Factorig the quadratic expressio yields 9x + x 171x + 5 = (x 1)(x + 6)(x ) 5 MANUFACTURING A 18-ich by 0-ich sheet of cardboard is cut ad folded ito a bakery box esolutios Maual - Powered by Cogero Page

23 - So, 9x + x 171x + 5 = (x 1)(x + 9x 5) Factorig the quadraticad expressio yields 9x + x 171x + 5 = (x 1)(x + 6)(x ) The Remaider Factor Theorems 5 MANUFACTURING A 18-ich by 0-ich sheet of cardboard is cut ad folded ito a bakery box a Write a polyomial fuctio to model the volume of the box b Graph the fuctio c The compay wats the box to have a volume of 196 cubic iches Write a equatio to model this situatio d Fid a positive iteger for x that satisfies the equatio foud i part c a The legth of the box is 18 x The height is x The width of the box is To fid the volume, calculate the product b Evaluate the fuctio for several x-values i its domai The height, width, ad legth of the box must all be positive values For the height, x > 0 For the legth, 18 x > 0 or x < 9 For the width, Thus, the domai of x is 0 < x < > 0 or x < x 0 1 v(x) Use these poits to costruct a graph c Substitute v(x) = 196 ito the origial equatio to arrive at 196 = x 7x +180x d Usig the trace fuctio o a graphig calculator, it appears that v(x) may be 196 i whe x = esolutios Maual - Powered by Cogero Page

24 - Remaider Factor cthe Substitute v(x) = 196ad ito the origialtheorems equatio to arrive at 196 = x 7x +180x d Usig the trace fuctio o a graphig calculator, it appears that v(x) may be 196 i whe x = If x = is a solutio for the equatio, it will also be a solutio to 0 = x 7x + 180x 196 Use sythetic substitutio to verify that x = is a solutio Because the remaider is 0, (x ) is a factor of x 7x + 180x 196 Thus, x = is a solutio to 196 = x 7x +180x Fid the value of k so that each remaider is zero 5 Because x, c = Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is 8 k Solve 8 k = 0 for k Whe k =, will have a remaider of 0 5 Because x +, c = Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is k + 68 Solve k + 68 = 0 for k Whe k =, esolutios Maual - Powered by Cogero 55 will have a remaider of 0 Page

25 k =, - Whe The Remaider ad will have a remaider of 0 Factor Theorems 5 Because x +, c = Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is k + 68 Solve k + 68 = 0 for k Whe k =, will have a remaider of 0 55 Because x + 1, c = 1 Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is k + Solve k + = 0 for k Whe k =, will have a remaider of 0 56 Because x 1, c = 1 Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is k + Solve k + = 0 for k Whe k =, will have a remaider of 0 57 SCULPTING Esteba will use a block of clay that is feet by feet by 5 feet to make a sculpture He wats to reduce the volume of the clay by removig the same amout from the legth, the width, ad the height a Write esolutios Maual Powered by Cogero a -polyomial fuctio to model the situatio Page 5 b Graph the fuctio c He wats to reduce the volume of the clay to of the origial volume Write a equatio to model the situatio

26 Whe k =, will have a remaider of 0 - The Remaider ad Factor Theorems 57 SCULPTING Esteba will use a block of clay that is feet by feet by 5 feet to make a sculpture He wats to reduce the volume of the clay by removig the same amout from the legth, the width, ad the height a Write a polyomial fuctio to model the situatio b Graph the fuctio c He wats to reduce the volume of the clay to of the origial volume Write a equatio to model the situatio d How much should he take from each dimesio? a Let the legth of the block equal feet, the width of the block equal feet, ad the height of the block equal 5 feet Let x equal the amout removed Thus, the legth is ( x), the width is ( x), ad the height is (5 x) The volume of the block is the product of the legth, width, ad height A polyomial fuctio to model the situatio is v(x) = x + 1x 7x + 60 b Evaluate the fuctio for several x-values i its domai The legth, width, ad height of the box must all be positive values For the legth, x > 0 or x < For the width, x > 0 or x < For the width, 5 x > 0 or x < 5 Thus, the domai of x is 0 < x < x v(x) Use these poits to costruct a graph c The volume of the origial block of clay is 5 or 60 cubic feet If Esteba reduces the volume by, the volume of the ew block will be 60 or 6 cubic feet A equatio to model the situatio is 6 = x + 1x 7x + 60 d To solve the equatio 6 = x + 1x 7x + 60 for x, use a graphig calculator to graph y = 6 ad y = x + 1x 7x + 60 o the same scree Use the itersect fuctio from the CALC meu to fid x esolutios Maual - Powered by Cogero Whe x is about 060, v(x) is 6 So, Esteba should take about 060 feet from each dimesio Page 6

27 volume of the ew block will be 60 - or 6 cubic feet A equatio to model the situatio is 6 = x + 1x 7x + 60 The Remaider ad d To solve the equatio 6 =Factor x + 1xTheorems 7x + 60 for x, use a graphig calculator to graph y = 6 ad y = x + 1x 7x + 60 o the same scree Use the itersect fuctio from the CALC meu to fid x Whe x is about 060, v(x) is 6 So, Esteba should take about 060 feet from each dimesio Use the graphs ad sythetic divisio to completely factor each polyomial 58 f (x) = 8x + 6x 10x 156x + 5 The graph suggests that (x + 5) ad (x ) may be factors of the polyomial Use sythetic divisio to test each factor, (x + 5) ad (x ) Because the remaider whe f (x) is divided by (x + 5) is 0, (x + 5) is a factor Test the secod factor, (x ), with the depressed polyomial 8x 1x x + 9 Because the remaider whe the depressed polyomial is divided by (x ) is 0, (x ) is a factor of f (x) Because (x + 5) ad (x ) are factors of f (x), we ca use the fial quotiet to write a factored form of f (x) as f (x) = (x + 5)(x )(8x + 10x ) Factorig the quadratic expressio yields f (x) = (x + 5)(x )(x 1)(x + ) 59 f (x) = 6x5 + 1x 15x + 5x + 7x 80 The graph suggests that (x + 6) ad (x ) may be factors of the polyomial Use sythetic divisio to test each factor, (x + 6) ad (x ) esolutios Maual - Powered by Cogero Page 7 Because the remaider whe f (x) is divided by (x + 6) is 0, (x + 6) is a factor Test the secod factor, (x ), with

28 Because the remaider whe the depressed polyomial is divided by (x ) is 0, (x ) is a factor of f (x) Because (x + 5) ad (x ) are factors of f (x), we ca use the fial quotiet to write a factored form of f (x) as f (x) ad Factor Theorems - =The (x + Remaider 5)(x )(8x + 10x ) Factorig the quadratic expressio yields f (x) = (x + 5)(x )(x 1)(x + ) 59 f (x) = 6x5 + 1x 15x + 5x + 7x 80 The graph suggests that (x + 6) ad (x ) may be factors of the polyomial Use sythetic divisio to test each factor, (x + 6) ad (x ) Because the remaider whe f (x) is divided by (x + 6) is 0, (x + 6) is a factor Test the secod factor, (x ), with the depressed polyomial 6x x 15x + 1x 10 Because the remaider whe the depressed polyomial is divided by (x ) is 0, (x ) is a factor of f (x) The depressed polyomial is 6x 11x 7x + 70 To fid factors of this polyomial, use a graphig calculator to observe the graph The graph suggests that (x ) may be a factor of the depressed polyomial Use sythetic divisio to test the factor (x ) Because the remaider whe the depressed polyomial is divided by (x ) is 0, (x ) is a repeated fact or of f (x) Because (x + 6) ad (x ) are factors of f (x), we ca use the fial quotiet to write a factored form of f (x) as f (x) = (x + 6)(x ) (6x + x 5) Factorig the quadratic expressio yields f (x) = (x + 6)(x ) (x 7)(x + 5) 60 MULTIPLE REPRESENTATIONS I this problem, you will explore the upper ad lower bouds of a fuctio a GRAPHICAL Graph each related polyomial fuctio ad determie the greatest ad least zeros The copy ad complete the table esolutios Maual - Powered by Cogero Page 8 b NUMERICAL Use sythetic divisio to evaluate each fuctio i part a for three iteger values greater tha the greatest zero

29 Because (x + 6) ad (x ) are factors of f (x), we ca use the fial quotiet to write a factored form of f (x) as f (x) = (x + 6)(x ) (6x + x 5) Factorig the quadratic expressio yields f (x) = (x + 6)(x ) (x 7)(x + 5) 60 MULTIPLE REPRESENTATIONS I this problem, you will explore the upper ad lower bouds of a fuctio - The Remaider ad Factor Theorems a GRAPHICAL Graph each related polyomial fuctio ad determie the greatest ad least zeros The copy ad complete the table b NUMERICAL Use sythetic divisio to evaluate each fuctio i part a for three iteger values greater tha the greatest zero c VERBAL Make a cojecture about the characteristics of the last row whe sythetic divisio is used to evaluate a fuctio for a iteger greater tha its greatest zero d NUMERICAL Use sythetic divisio to evaluate each fuctio i part a for three iteger values less tha the least zero e VERBAL Make a cojecture about the characteristics of the last row whe sythetic divisio is used to evaluate a fuctio for a umber less tha its least zero a Use a graphig calculator to graph y = x x 11x + 1 The least zero appears to The greatest zero appears to be Use a graphig calculator to graph y = x + 6x + x 10x The least zero appears to 5 The greatest zero appears to be 1 5 Use a graphig calculator to graph y = x x x The least zero appears to 1 The greatest zero appears to be b Sample aswer: For x x 11x + 1, use sythetic divisio for c = 5, 7, ad 8 esolutios Maual - Powered by Cogero Page 9

30 - The Remaider ad Factor Theorems b Sample aswer: For x x 11x + 1, use sythetic divisio for c = 5, 7, ad 8 f(5) =, f (7) = 180, f (8) = 08 For x + 6x + x 10, use sythetic divisio for c =,, ad f() = 56, f () = 0, f () = 68 5 For x x x, use sythetic divisio for c =, 5, ad 7 f() = 108, f (5) = 50, f (7) = 1,70 c Sample aswer: All of the elemets i the last row of the sythetic divisio are positive d Sample aswer: For x x 11x + 1, use sythetic divisio for c = 5, 7, ad 8 esolutios Maual - Powered by Cogero f( ) = 0, f ( 5) = 108, f ( 6) = 10 Page 0

31 - The Remaider ad Factor Theorems f( ) = 0, f ( 5) = 108, f ( 6) = 10 For x + 6x + x 10, use sythetic divisio for c =,, ad f( 6) = 168, f ( 7) = 560, f ( 9) = 50 5 For x x x, use sythetic divisio for c =, 5, ad 7 f( ) =, f ( ) = 70, f ( ) = 115 e Sample aswer: The elemets i the last row alterate betwee oegative ad opositive 61 CHALLENGE Is (x 1) a factor of 18x165 15x15 + 8x105 15x55 +? Explai your reasoig Yes; sample aswer: Usig the Factor Theorem, (x 1) is a factor if f (1) = 0 Sice f (1) = 0, (x 1) is a factor of the polyomial 6 Writig i Math Explai how you ca use a graphig calculator, sythetic divisio, ad factorig to completely factor a fifth-degree polyomial with ratioal coefficiets, three itegral zeros, ad two o-itegral, ratioal zeros Sample aswer: I would use my graphig calculator to graph the polyomial ad to determie the three itegral zeros, a, b, ad c I would the use sythetic divisio to divide the polyomial by a I would the divide the resultig esolutios Maual - Powered by Cogero Page 1 depressed polyomial by b, ad the the ew depressed polyomial by c The third depressed polyomial will have degree Fially, I would factor the secod-degree polyomial to fid the two o-itegral, ratioal zeros, d ad e So, the polyomial is either the product (x a)(x b)(x c)(x d)(x e) or this product multiplied by some ratioal

32 - The Remaider ad Factor Theorems Sice f (1) = 0, (x 1) is a factor of the polyomial 6 Writig i Math Explai how you ca use a graphig calculator, sythetic divisio, ad factorig to completely factor a fifth-degree polyomial with ratioal coefficiets, three itegral zeros, ad two o-itegral, ratioal zeros Sample aswer: I would use my graphig calculator to graph the polyomial ad to determie the three itegral zeros, a, b, ad c I would the use sythetic divisio to divide the polyomial by a I would the divide the resultig depressed polyomial by b, ad the the ew depressed polyomial by c The third depressed polyomial will have degree Fially, I would factor the secod-degree polyomial to fid the two o-itegral, ratioal zeros, d ad e So, the polyomial is either the product (x a)(x b)(x c)(x d)(x e) or this product multiplied by some ratioal umber 6 REASONING Determie whether the statemet below is true or false Explai If h(y) = (y + )(y + 11y ) 1, the the remaider of is 1 True; sample aswer: The Remaider Theorem states that if h(y) is divided by y ( ), the the remaider is r = h ( ), which is 1 CHALLENGE Fid k so that the quotiet has a 0 remaider 6 Because x + 7, c = 7 Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is 9k 9 Solve 9k 9 = 0 for k Whe k = 9, will have a remaider of 0 65 Because x 1, c = 1 Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is k + 0 Solve k + 0 = 0 for k esolutios Maual Powered by Cogero Whe k = - 0, will have a remaider of 0 Page

33 k = 9, will have a remaider of 0 - Whe The Remaider ad Factor Theorems 65 Because x 1, c = 1 Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is k + 0 Solve k + 0 = 0 for k Whe k = 0, will have a remaider of 0 66 Because x, c = Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is 8k 0 Solve 8k 0 = 0 for k will have a remaider of 0 Whe k = 5, 67 CHALLENGE If x dx + (1 d ) x + 5 has a factor x d, what is the value of d if d is a iteger? x dx + (1 d ) x + 5 ca be writte as x + ( d d + 1)x + 5 Because (x d) is a factor, c = d Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is d + d + 1d + 5 Let d + d + 1d + 5 = 0 ad solve for d Use a graphig calculator to graph y = d + d + 1d + 5 esolutios Maual - Powered by Cogero Page The graph suggests that possible values for d are 5, 0, ad 6 Use sythetic divisio to test each possible factor

34 - Whe The Remaider ad Factor Theorems k = 5, will have a remaider of 0 67 CHALLENGE If x dx + (1 d ) x + 5 has a factor x d, what is the value of d if d is a iteger? x dx + (1 d ) x + 5 ca be writte as x + ( d d + 1)x + 5 Because (x d) is a factor, c = d Set up the sythetic divisio as follows The follow the sythetic divisio procedure The remaider is d + d + 1d + 5 Let d + d + 1d + 5 = 0 ad solve for d Use a graphig calculator to graph y = d + d + 1d + 5 The graph suggests that possible values for d are 5, 0, ad 6 Use sythetic divisio to test each possible factor Sice the remaider is 0, (x ( 5)) or (x + 5) is a factor Test the two remaiig values usig the depressed polyomial Sice the remaider for both is 1, either oe is a factor Thus, d = 5 68 Writig i Math Compare ad cotrast polyomial divisio usig log divisio ad usig sythetic divisio Sample aswer: Both log divisio ad sythetic divisio ca be used to divide a polyomial by a liear factor Log divisio ca also be used to divide a polyomial by a oliear factor I sythetic divisio, oly the coefficiets are used I both log divisio ad sythetic divisio, placeholders are eeded if a power of a variable is missig Determie whether the degree of the polyomial for each graph is eve or odd ad whether its leadig coefficiet a is positive or egative 69 esolutios Maual - Powered by Cogero Page Sice is eve ad a is egative

35 - Sample aswer: Both log divisio ad sythetic divisio ca be used to divide a polyomial by a liear factor Log divisio ca also be used to divide a polyomial by a oliear factor I sythetic divisio, oly the coefficiets are The Remaider ad Factor Theorems used I both log divisio ad sythetic divisio, placeholders are eeded if a power of a variable is missig Determie whether the degree of the polyomial for each graph is eve or odd ad whether its leadig coefficiet a is positive or egative 69 is eve ad a is egative Sice 70 is odd ad a is positive Sice 71 is odd ad a is egative Sice 7 SKYDIVING The approximate time t i secods that it takes a object to fall a distace of d feet is give by Suppose a skydiver falls 11 secods before the parachute opes How far does the skydiver fall durig this time period? Substitute t = 11 ito esolutios Maual - Powered by Cogero ad solve for d Page 5

36 71 - Sice The Remaider ad Factor Theorems is odd ad a is egative 7 SKYDIVING The approximate time t i secods that it takes a object to fall a distace of d feet is give by Suppose a skydiver falls 11 secods before the parachute opes How far does the skydiver fall durig this time period? Substitute t = 11 ito ad solve for d The skydiver falls 196 feet 7 FIRE FIGHTING The velocity v ad maximum height h of water beig pumped ito the air are related by v =, where g is the acceleratio due to gravity ( feet/secod ) a Determie a equatio that will give the maximum height of the water as a fuctio of its velocity b The Mayfield Fire Departmet must purchase a pump that is powerful eough to propel water 80 feet ito the air Will a pump that is advertised to project water with a velocity of 75 feet/secod meet the fire departmet s eeds? Explai a Substitute g = ito v = ad solve for h A equatio that will give the maximum height of the water as a fuctio of its velocity is h = b Substitute v = 75 ito the equatio foud i part a The pump ca propel water to a height of about 88 feet So, the pump will meet the fire departmet s eeds esolutios Maual - Powered by Cogero Solve each system of equatios algebraically 7 5x y = 16 x + y = Page 6

37 - The Remaider ad Factor Theorems The skydiver falls 196 feet 7 FIRE FIGHTING The velocity v ad maximum height h of water beig pumped ito the air are related by v =, where g is the acceleratio due to gravity ( feet/secod ) a Determie a equatio that will give the maximum height of the water as a fuctio of its velocity b The Mayfield Fire Departmet must purchase a pump that is powerful eough to propel water 80 feet ito the air Will a pump that is advertised to project water with a velocity of 75 feet/secod meet the fire departmet s eeds? Explai a Substitute g = ito v = ad solve for h A equatio that will give the maximum height of the water as a fuctio of its velocity is h = b Substitute v = 75 ito the equatio foud i part a The pump ca propel water to a height of about 88 feet So, the pump will meet the fire departmet s eeds Solve each system of equatios algebraically 7 5x y = 16 x + y = 5x y = 16 ca be writte as y = 5x 16 Substitute 5x 16 for y ito the secod equatio ad solve for x Substitute x = ito y = 5x 16 ad solve for y esolutios Maual - Powered by Cogero The solutio to the system of equatio is (, 1) Page 7

38 - The Remaider ad Factor Theorems The pump ca propel water to a height of about 88 feet So, the pump will meet the fire departmet s eeds Solve each system of equatios algebraically 7 5x y = 16 x + y = 5x y = 16 ca be writte as y = 5x 16 Substitute 5x 16 for y ito the secod equatio ad solve for x Substitute x = ito y = 5x 16 ad solve for y The solutio to the system of equatio is (, 1) 75 x 5y = 8 x + y = 1 x + y = 1 ca be writte as x = 1 y Substitute 1 y for x ito the first equatio ad solve for y Substitute y = 1 ito x = 1 y ad solve for x The solutio to the system of equatio is ( 1, 1) 76 y = 6 x x = 5 + y Substitute 6 x for y ito the secod equatio ad solve for x esolutios Maual - Powered by Cogero Page 8

39 - The Remaider ad Factor Theorems The solutio to the system of equatio is ( 1, 1) 76 y = 6 x x = 5 + y Substitute 6 x for y ito the secod equatio ad solve for x Substitute x= 55 ito y = 6 x ad solve for y The solutio to the system of equatio is (55, 075) 77 x + 5y = x + 6y = 5 Elimiate x Solve for y Substitute y = ito the secod equatio ad solve for x The solutio to the system of equatio is 78 7x + 1y = 16 5y x = 1 esolutios Maual - Powered by Cogero 5y x = 1 ca be writte as x + 5y = 1 Elimiate x Page 9

40 - The Thesolutio Remaider adoffactor to the system equatio istheorems 78 7x + 1y = 16 5y x = 1 5y x = 1 ca be writte as x + 5y = 1 Elimiate x Solve for y Substitute y = 1 ito the first equatio ad solve for x The solutio to the system of equatio is (, 1) 79 x + 5y = 8 x 7y = 10 Elimiate x Solve for y Substitute ito the first equatio ad solve for x esolutios Maual - Powered by Cogero The solutio to the system of equatio is Page 0

41 - The Remaider ad Factor Theorems The solutio to the system of equatio is (, 1) 79 x + 5y = 8 x 7y = 10 Elimiate x Solve for y Substitute ito the first equatio ad solve for x The solutio to the system of equatio is 80 SAT/ACT I the figure, a equilateral triagle is draw with a altitude that is also the diameter of the circle If the perimeter of the triagle is 6, what is the circumferece of the circle? A6 B6 C 1 D 1 E 6 If the perimeter of the equilateral triagle is 6, the each side of the triagle measures 1 Also, each agle measures 60 We ca aalyze half of the equilateral triagle, usig the diameter of the circle as oe of the legs esolutios Maual - Powered by Cogero Page 1

42 - The Thesolutio Remaider adoffactor to the system equatio istheorems 80 SAT/ACT I the figure, a equilateral triagle is draw with a altitude that is also the diameter of the circle If the perimeter of the triagle is 6, what is the circumferece of the circle? A6 B6 C 1 D 1 E 6 If the perimeter of the equilateral triagle is 6, the each side of the triagle measures 1 Also, each agle measures 60 We ca aalyze half of the equilateral triagle, usig the diameter of the circle as oe of the legs Sice, this triagle is a right triagle, x = 6 x is also the diameter of the circle The circumferece of a circle is C = d So, the circumferece of the circle is C = The correct aswer is B (6 ) or 6 81 REVIEW If (, 7) is the ceter of a circle ad (8, 5) is o the circle, what is the circumferece of the circle? F 1 G 15 H 18 J 5 K 6 The distace from the ceter of a circle to a poit o the circle is equal to the radius of the circle Use the distace formula ad the two poits to fid the radius of the circle The circumferece of a circle is C = The correct aswer is K d or πr So, the circumferece of the circle is C = (1) or 6 8 REVIEW The first term i a sequece is x Each subsequet term is three less tha twice the precedig term esolutios Maual - Powered by Cogero What is the 5th term i the sequece? A 8x 1 B 8x 15 Page

43 of aad circlefactor is C = Theorems d or πr So, the circumferece of the circle is C = - The Thecircumferece Remaider (1) or 6 The correct aswer is K 8 REVIEW The first term i a sequece is x Each subsequet term is three less tha twice the precedig term What is the 5th term i the sequece? A 8x 1 B 8x 15 C 16x 9 D 16x 5 E x 9 The correct aswer is D 8 Use the graph of the polyomial fuctio Which is ot a factor of x5 + x x x x? F (x ) G (x + ) H (x 1) J (x + 1) The graph suggests that, 1, ad are zeros of the fuctio Thus, (x + ), (x + 1), ad (x ) are factors of f (x) The correct aswer is H esolutios Maual - Powered by Cogero Page

Soving Recurrence Relations

Soving Recurrence Relations Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree

More information

7.1 Finding Rational Solutions of Polynomial Equations

7.1 Finding Rational Solutions of Polynomial Equations 4 Locker LESSON 7. Fidig Ratioal Solutios of Polyomial Equatios Name Class Date 7. Fidig Ratioal Solutios of Polyomial Equatios Essetial Questio: How do you fid the ratioal roots of a polyomial equatio?

More information

Trigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is

Trigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is 0_0605.qxd /5/05 0:45 AM Page 470 470 Chapter 6 Additioal Topics i Trigoometry 6.5 Trigoometric Form of a Complex Number What you should lear Plot complex umbers i the complex plae ad fid absolute values

More information

3. If x and y are real numbers, what is the simplified radical form

3. If x and y are real numbers, what is the simplified radical form lgebra II Practice Test Objective:.a. Which is equivalet to 98 94 4 49?. Which epressio is aother way to write 5 4? 5 5 4 4 4 5 4 5. If ad y are real umbers, what is the simplified radical form of 5 y

More information

AP Calculus AB 2006 Scoring Guidelines Form B

AP Calculus AB 2006 Scoring Guidelines Form B AP Calculus AB 6 Scorig Guidelies Form B The College Board: Coectig Studets to College Success The College Board is a ot-for-profit membership associatio whose missio is to coect studets to college success

More information

FOUNDATIONS OF MATHEMATICS AND PRE-CALCULUS GRADE 10

FOUNDATIONS OF MATHEMATICS AND PRE-CALCULUS GRADE 10 FOUNDATIONS OF MATHEMATICS AND PRE-CALCULUS GRADE 10 [C] Commuicatio Measuremet A1. Solve problems that ivolve liear measuremet, usig: SI ad imperial uits of measure estimatio strategies measuremet strategies.

More information

Basic Elements of Arithmetic Sequences and Series

Basic Elements of Arithmetic Sequences and Series MA40S PRE-CALCULUS UNIT G GEOMETRIC SEQUENCES CLASS NOTES (COMPLETED NO NEED TO COPY NOTES FROM OVERHEAD) Basic Elemets of Arithmetic Sequeces ad Series Objective: To establish basic elemets of arithmetic

More information

Math 113 HW #11 Solutions

Math 113 HW #11 Solutions Math 3 HW # Solutios 5. 4. (a) Estimate the area uder the graph of f(x) = x from x = to x = 4 usig four approximatig rectagles ad right edpoits. Sketch the graph ad the rectagles. Is your estimate a uderestimate

More information

Sequences and Series

Sequences and Series CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their

More information

S. Tanny MAT 344 Spring 1999. be the minimum number of moves required.

S. Tanny MAT 344 Spring 1999. be the minimum number of moves required. S. Tay MAT 344 Sprig 999 Recurrece Relatios Tower of Haoi Let T be the miimum umber of moves required. T 0 = 0, T = 7 Iitial Coditios * T = T + $ T is a sequece (f. o itegers). Solve for T? * is a recurrece,

More information

Math 114- Intermediate Algebra Integral Exponents & Fractional Exponents (10 )

Math 114- Intermediate Algebra Integral Exponents & Fractional Exponents (10 ) Math 4 Math 4- Itermediate Algebra Itegral Epoets & Fractioal Epoets (0 ) Epoetial Fuctios Epoetial Fuctios ad Graphs I. Epoetial Fuctios The fuctio f ( ) a, where is a real umber, a 0, ad a, is called

More information

SAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx

SAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval

More information

Example 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here).

Example 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here). BEGINNING ALGEBRA Roots ad Radicals (revised summer, 00 Olso) Packet to Supplemet the Curret Textbook - Part Review of Square Roots & Irratioals (This portio ca be ay time before Part ad should mostly

More information

Section 11.3: The Integral Test

Section 11.3: The Integral Test Sectio.3: The Itegral Test Most of the series we have looked at have either diverged or have coverged ad we have bee able to fid what they coverge to. I geeral however, the problem is much more difficult

More information

3. Greatest Common Divisor - Least Common Multiple

3. Greatest Common Divisor - Least Common Multiple 3 Greatest Commo Divisor - Least Commo Multiple Defiitio 31: The greatest commo divisor of two atural umbers a ad b is the largest atural umber c which divides both a ad b We deote the greatest commo gcd

More information

http://www.webassign.net/v4cgijeff.downs@wnc/control.pl

http://www.webassign.net/v4cgijeff.downs@wnc/control.pl Assigmet Previewer http://www.webassig.et/vcgijeff.dows@wc/cotrol.pl of // : PM Practice Eam () Questio Descriptio Eam over chapter.. Questio DetailsLarCalc... [] Fid the geeral solutio of the differetial

More information

Infinite Sequences and Series

Infinite Sequences and Series CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...

More information

MATH 083 Final Exam Review

MATH 083 Final Exam Review MATH 08 Fial Eam Review Completig the problems i this review will greatly prepare you for the fial eam Calculator use is ot required, but you are permitted to use a calculator durig the fial eam period

More information

1. MATHEMATICAL INDUCTION

1. MATHEMATICAL INDUCTION 1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1

More information

5.4 Amortization. Question 1: How do you find the present value of an annuity? Question 2: How is a loan amortized?

5.4 Amortization. Question 1: How do you find the present value of an annuity? Question 2: How is a loan amortized? 5.4 Amortizatio Questio 1: How do you fid the preset value of a auity? Questio 2: How is a loa amortized? Questio 3: How do you make a amortizatio table? Oe of the most commo fiacial istrumets a perso

More information

Confidence Intervals. CI for a population mean (σ is known and n > 30 or the variable is normally distributed in the.

Confidence Intervals. CI for a population mean (σ is known and n > 30 or the variable is normally distributed in the. Cofidece Itervals A cofidece iterval is a iterval whose purpose is to estimate a parameter (a umber that could, i theory, be calculated from the populatio, if measuremets were available for the whole populatio).

More information

AP Calculus BC 2003 Scoring Guidelines Form B

AP Calculus BC 2003 Scoring Guidelines Form B AP Calculus BC Scorig Guidelies Form B The materials icluded i these files are iteded for use by AP teachers for course ad exam preparatio; permissio for ay other use must be sought from the Advaced Placemet

More information

CS103X: Discrete Structures Homework 4 Solutions

CS103X: Discrete Structures Homework 4 Solutions CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible six-figure salaries i whole dollar amouts are there that cotai at least

More information

Repeating Decimals are decimal numbers that have number(s) after the decimal point that repeat in a pattern.

Repeating Decimals are decimal numbers that have number(s) after the decimal point that repeat in a pattern. 5.5 Fractios ad Decimals Steps for Chagig a Fractio to a Decimal. Simplify the fractio, if possible. 2. Divide the umerator by the deomiator. d d Repeatig Decimals Repeatig Decimals are decimal umbers

More information

NATIONAL SENIOR CERTIFICATE GRADE 11

NATIONAL SENIOR CERTIFICATE GRADE 11 NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P NOVEMBER 007 MARKS: 50 TIME: 3 hours This questio paper cosists of 9 pages, diagram sheet ad a -page formula sheet. Please tur over Mathematics/P DoE/November

More information

Factoring x n 1: cyclotomic and Aurifeuillian polynomials Paul Garrett <garrett@math.umn.edu>

Factoring x n 1: cyclotomic and Aurifeuillian polynomials Paul Garrett <garrett@math.umn.edu> (March 16, 004) Factorig x 1: cyclotomic ad Aurifeuillia polyomials Paul Garrett Polyomials of the form x 1, x 3 1, x 4 1 have at least oe systematic factorizatio x 1 = (x 1)(x 1

More information

.04. This means $1000 is multiplied by 1.02 five times, once for each of the remaining sixmonth

.04. This means $1000 is multiplied by 1.02 five times, once for each of the remaining sixmonth Questio 1: What is a ordiary auity? Let s look at a ordiary auity that is certai ad simple. By this, we mea a auity over a fixed term whose paymet period matches the iterest coversio period. Additioally,

More information

Approximating Area under a curve with rectangles. To find the area under a curve we approximate the area using rectangles and then use limits to find

Approximating Area under a curve with rectangles. To find the area under a curve we approximate the area using rectangles and then use limits to find 1.8 Approximatig Area uder a curve with rectagles 1.6 To fid the area uder a curve we approximate the area usig rectagles ad the use limits to fid 1.4 the area. Example 1 Suppose we wat to estimate 1.

More information

5.3. Generalized Permutations and Combinations

5.3. Generalized Permutations and Combinations 53 GENERALIZED PERMUTATIONS AND COMBINATIONS 73 53 Geeralized Permutatios ad Combiatios 53 Permutatios with Repeated Elemets Assume that we have a alphabet with letters ad we wat to write all possible

More information

SEQUENCES AND SERIES CHAPTER

SEQUENCES AND SERIES CHAPTER CHAPTER SEQUENCES AND SERIES Whe the Grat family purchased a computer for $,200 o a istallmet pla, they agreed to pay $00 each moth util the cost of the computer plus iterest had bee paid The iterest each

More information

CS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations

CS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad

More information

In nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008

In nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008 I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces

More information

CHAPTER 11 Financial mathematics

CHAPTER 11 Financial mathematics CHAPTER 11 Fiacial mathematics I this chapter you will: Calculate iterest usig the simple iterest formula ( ) Use the simple iterest formula to calculate the pricipal (P) Use the simple iterest formula

More information

SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES

SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,

More information

WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER?

WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? JÖRG JAHNEL 1. My Motivatio Some Sort of a Itroductio Last term I tought Topological Groups at the Göttige Georg August Uiversity. This

More information

4.3. The Integral and Comparison Tests

4.3. The Integral and Comparison Tests 4.3. THE INTEGRAL AND COMPARISON TESTS 9 4.3. The Itegral ad Compariso Tests 4.3.. The Itegral Test. Suppose f is a cotiuous, positive, decreasig fuctio o [, ), ad let a = f(). The the covergece or divergece

More information

Solving equations. Pre-test. Warm-up

Solving equations. Pre-test. Warm-up Solvig equatios 8 Pre-test Warm-up We ca thik of a algebraic equatio as beig like a set of scales. The two sides of the equatio are equal, so the scales are balaced. If we add somethig to oe side of the

More information

COMPUTER LABORATORY IMPLEMENTATION ISSUES AT A SMALL LIBERAL ARTS COLLEGE. Richard A. Weida Lycoming College Williamsport, PA 17701 weida@lycoming.

COMPUTER LABORATORY IMPLEMENTATION ISSUES AT A SMALL LIBERAL ARTS COLLEGE. Richard A. Weida Lycoming College Williamsport, PA 17701 weida@lycoming. COMPUTER LABORATORY IMPLEMENTATION ISSUES AT A SMALL LIBERAL ARTS COLLEGE Richard A. Weida Lycomig College Williamsport, PA 17701 weida@lycomig.edu Abstract: Lycomig College is a small, private, liberal

More information

Listing terms of a finite sequence List all of the terms of each finite sequence. a) a n n 2 for 1 n 5 1 b) a n for 1 n 4 n 2

Listing terms of a finite sequence List all of the terms of each finite sequence. a) a n n 2 for 1 n 5 1 b) a n for 1 n 4 n 2 74 (4 ) Chapter 4 Sequeces ad Series 4. SEQUENCES I this sectio Defiitio Fidig a Formula for the th Term The word sequece is a familiar word. We may speak of a sequece of evets or say that somethig is

More information

BINOMIAL EXPANSIONS 12.5. In this section. Some Examples. Obtaining the Coefficients

BINOMIAL EXPANSIONS 12.5. In this section. Some Examples. Obtaining the Coefficients 652 (12-26) Chapter 12 Sequeces ad Series 12.5 BINOMIAL EXPANSIONS I this sectio Some Examples Otaiig the Coefficiets The Biomial Theorem I Chapter 5 you leared how to square a iomial. I this sectio you

More information

University of California, Los Angeles Department of Statistics. Distributions related to the normal distribution

University of California, Los Angeles Department of Statistics. Distributions related to the normal distribution Uiversity of Califoria, Los Ageles Departmet of Statistics Statistics 100B Istructor: Nicolas Christou Three importat distributios: Distributios related to the ormal distributio Chi-square (χ ) distributio.

More information

Chapter 5: Inner Product Spaces

Chapter 5: Inner Product Spaces Chapter 5: Ier Product Spaces Chapter 5: Ier Product Spaces SECION A Itroductio to Ier Product Spaces By the ed of this sectio you will be able to uderstad what is meat by a ier product space give examples

More information

Chapter 7: Confidence Interval and Sample Size

Chapter 7: Confidence Interval and Sample Size Chapter 7: Cofidece Iterval ad Sample Size Learig Objectives Upo successful completio of Chapter 7, you will be able to: Fid the cofidece iterval for the mea, proportio, ad variace. Determie the miimum

More information

Elementary Theory of Russian Roulette

Elementary Theory of Russian Roulette Elemetary Theory of Russia Roulette -iterestig patters of fractios- Satoshi Hashiba Daisuke Miematsu Ryohei Miyadera Itroductio. Today we are goig to study mathematical theory of Russia roulette. If some

More information

Review: Classification Outline

Review: Classification Outline Data Miig CS 341, Sprig 2007 Decisio Trees Neural etworks Review: Lecture 6: Classificatio issues, regressio, bayesia classificatio Pretice Hall 2 Data Miig Core Techiques Classificatio Clusterig Associatio

More information

INFINITE SERIES KEITH CONRAD

INFINITE SERIES KEITH CONRAD INFINITE SERIES KEITH CONRAD. Itroductio The two basic cocepts of calculus, differetiatio ad itegratio, are defied i terms of limits (Newto quotiets ad Riema sums). I additio to these is a third fudametal

More information

Department of Computer Science, University of Otago

Department of Computer Science, University of Otago Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS-2006-09 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly

More information

Multiple Representations for Pattern Exploration with the Graphing Calculator and Manipulatives

Multiple Representations for Pattern Exploration with the Graphing Calculator and Manipulatives Douglas A. Lapp Multiple Represetatios for Patter Exploratio with the Graphig Calculator ad Maipulatives To teach mathematics as a coected system of cocepts, we must have a shift i emphasis from a curriculum

More information

A probabilistic proof of a binomial identity

A probabilistic proof of a binomial identity A probabilistic proof of a biomial idetity Joatho Peterso Abstract We give a elemetary probabilistic proof of a biomial idetity. The proof is obtaied by computig the probability of a certai evet i two

More information

Hypergeometric Distributions

Hypergeometric Distributions 7.4 Hypergeometric Distributios Whe choosig the startig lie-up for a game, a coach obviously has to choose a differet player for each positio. Similarly, whe a uio elects delegates for a covetio or you

More information

FM4 CREDIT AND BORROWING

FM4 CREDIT AND BORROWING FM4 CREDIT AND BORROWING Whe you purchase big ticket items such as cars, boats, televisios ad the like, retailers ad fiacial istitutios have various terms ad coditios that are implemeted for the cosumer

More information

Chapter 6: Variance, the law of large numbers and the Monte-Carlo method

Chapter 6: Variance, the law of large numbers and the Monte-Carlo method Chapter 6: Variace, the law of large umbers ad the Mote-Carlo method Expected value, variace, ad Chebyshev iequality. If X is a radom variable recall that the expected value of X, E[X] is the average value

More information

Sequences and Series Using the TI-89 Calculator

Sequences and Series Using the TI-89 Calculator RIT Calculator Site Sequeces ad Series Usig the TI-89 Calculator Norecursively Defied Sequeces A orecursively defied sequece is oe i which the formula for the terms of the sequece is give explicitly. For

More information

Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed. This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at http://tutorial.math.lamar.edu/terms.asp. The olie versio

More information

Convexity, Inequalities, and Norms

Convexity, Inequalities, and Norms Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for

More information

Chapter 5 O A Cojecture Of Erdíos Proceedigs NCUR VIII è1994è, Vol II, pp 794í798 Jeærey F Gold Departmet of Mathematics, Departmet of Physics Uiversity of Utah Do H Tucker Departmet of Mathematics Uiversity

More information

1 Correlation and Regression Analysis

1 Correlation and Regression Analysis 1 Correlatio ad Regressio Aalysis I this sectio we will be ivestigatig the relatioship betwee two cotiuous variable, such as height ad weight, the cocetratio of a ijected drug ad heart rate, or the cosumptio

More information

1. C. The formula for the confidence interval for a population mean is: x t, which was

1. C. The formula for the confidence interval for a population mean is: x t, which was s 1. C. The formula for the cofidece iterval for a populatio mea is: x t, which was based o the sample Mea. So, x is guarateed to be i the iterval you form.. D. Use the rule : p-value

More information

Case Study. Normal and t Distributions. Density Plot. Normal Distributions

Case Study. Normal and t Distributions. Density Plot. Normal Distributions Case Study Normal ad t Distributios Bret Halo ad Bret Larget Departmet of Statistics Uiversity of Wiscosi Madiso October 11 13, 2011 Case Study Body temperature varies withi idividuals over time (it ca

More information

Question 2: How is a loan amortized?

Question 2: How is a loan amortized? Questio 2: How is a loa amortized? Decreasig auities may be used i auto or home loas. I these types of loas, some amout of moey is borrowed. Fixed paymets are made to pay off the loa as well as ay accrued

More information

Chapter 5 Unit 1. IET 350 Engineering Economics. Learning Objectives Chapter 5. Learning Objectives Unit 1. Annual Amount and Gradient Functions

Chapter 5 Unit 1. IET 350 Engineering Economics. Learning Objectives Chapter 5. Learning Objectives Unit 1. Annual Amount and Gradient Functions Chapter 5 Uit Aual Amout ad Gradiet Fuctios IET 350 Egieerig Ecoomics Learig Objectives Chapter 5 Upo completio of this chapter you should uderstad: Calculatig future values from aual amouts. Calculatig

More information

UC Berkeley Department of Electrical Engineering and Computer Science. EE 126: Probablity and Random Processes. Solutions 9 Spring 2006

UC Berkeley Department of Electrical Engineering and Computer Science. EE 126: Probablity and Random Processes. Solutions 9 Spring 2006 Exam format UC Bereley Departmet of Electrical Egieerig ad Computer Sciece EE 6: Probablity ad Radom Processes Solutios 9 Sprig 006 The secod midterm will be held o Wedesday May 7; CHECK the fial exam

More information

Overview on S-Box Design Principles

Overview on S-Box Design Principles Overview o S-Box Desig Priciples Debdeep Mukhopadhyay Assistat Professor Departmet of Computer Sciece ad Egieerig Idia Istitute of Techology Kharagpur INDIA -721302 What is a S-Box? S-Boxes are Boolea

More information

Building Blocks Problem Related to Harmonic Series

Building Blocks Problem Related to Harmonic Series TMME, vol3, o, p.76 Buildig Blocks Problem Related to Harmoic Series Yutaka Nishiyama Osaka Uiversity of Ecoomics, Japa Abstract: I this discussio I give a eplaatio of the divergece ad covergece of ifiite

More information

CHAPTER 7: Central Limit Theorem: CLT for Averages (Means)

CHAPTER 7: Central Limit Theorem: CLT for Averages (Means) CHAPTER 7: Cetral Limit Theorem: CLT for Averages (Meas) X = the umber obtaied whe rollig oe six sided die oce. If we roll a six sided die oce, the mea of the probability distributio is X P(X = x) Simulatio:

More information

Solutions to Exercises Chapter 4: Recurrence relations and generating functions

Solutions to Exercises Chapter 4: Recurrence relations and generating functions Solutios to Exercises Chapter 4: Recurrece relatios ad geeratig fuctios 1 (a) There are seatig positios arraged i a lie. Prove that the umber of ways of choosig a subset of these positios, with o two chose

More information

Solving Logarithms and Exponential Equations

Solving Logarithms and Exponential Equations Solvig Logarithms ad Epoetial Equatios Logarithmic Equatios There are two major ideas required whe solvig Logarithmic Equatios. The first is the Defiitio of a Logarithm. You may recall from a earlier topic:

More information

SEQUENCES AND SERIES

SEQUENCES AND SERIES Chapter 9 SEQUENCES AND SERIES Natural umbers are the product of huma spirit. DEDEKIND 9.1 Itroductio I mathematics, the word, sequece is used i much the same way as it is i ordiary Eglish. Whe we say

More information

Engineering 323 Beautiful Homework Set 3 1 of 7 Kuszmar Problem 2.51

Engineering 323 Beautiful Homework Set 3 1 of 7 Kuszmar Problem 2.51 Egieerig 33 eautiful Homewor et 3 of 7 Kuszmar roblem.5.5 large departmet store sells sport shirts i three sizes small, medium, ad large, three patters plaid, prit, ad stripe, ad two sleeve legths log

More information

Factors of sums of powers of binomial coefficients

Factors of sums of powers of binomial coefficients ACTA ARITHMETICA LXXXVI.1 (1998) Factors of sums of powers of biomial coefficiets by Neil J. Cali (Clemso, S.C.) Dedicated to the memory of Paul Erdős 1. Itroductio. It is well ow that if ( ) a f,a = the

More information

NATIONAL SENIOR CERTIFICATE GRADE 11

NATIONAL SENIOR CERTIFICATE GRADE 11 NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 007 MARKS: 50 TIME: 3 hours This questio paper cosists of pages, 4 diagram sheets ad a -page formula sheet. Please tur over Mathematics/P DoE/Exemplar

More information

Our aim is to show that under reasonable assumptions a given 2π-periodic function f can be represented as convergent series

Our aim is to show that under reasonable assumptions a given 2π-periodic function f can be represented as convergent series 8 Fourier Series Our aim is to show that uder reasoable assumptios a give -periodic fuctio f ca be represeted as coverget series f(x) = a + (a cos x + b si x). (8.) By defiitio, the covergece of the series

More information

Class Meeting # 16: The Fourier Transform on R n

Class Meeting # 16: The Fourier Transform on R n MATH 18.152 COUSE NOTES - CLASS MEETING # 16 18.152 Itroductio to PDEs, Fall 2011 Professor: Jared Speck Class Meetig # 16: The Fourier Trasform o 1. Itroductio to the Fourier Trasform Earlier i the course,

More information

GCE Further Mathematics (6360) Further Pure Unit 2 (MFP2) Textbook. Version: 1.4

GCE Further Mathematics (6360) Further Pure Unit 2 (MFP2) Textbook. Version: 1.4 GCE Further Mathematics (660) Further Pure Uit (MFP) Tetbook Versio: 4 MFP Tetbook A-level Further Mathematics 660 Further Pure : Cotets Chapter : Comple umbers 4 Itroductio 5 The geeral comple umber 5

More information

Chapter 7 Methods of Finding Estimators

Chapter 7 Methods of Finding Estimators Chapter 7 for BST 695: Special Topics i Statistical Theory. Kui Zhag, 011 Chapter 7 Methods of Fidig Estimators Sectio 7.1 Itroductio Defiitio 7.1.1 A poit estimator is ay fuctio W( X) W( X1, X,, X ) of

More information

THE ARITHMETIC OF INTEGERS. - multiplication, exponentiation, division, addition, and subtraction

THE ARITHMETIC OF INTEGERS. - multiplication, exponentiation, division, addition, and subtraction THE ARITHMETIC OF INTEGERS - multiplicatio, expoetiatio, divisio, additio, ad subtractio What to do ad what ot to do. THE INTEGERS Recall that a iteger is oe of the whole umbers, which may be either positive,

More information

Asymptotic Growth of Functions

Asymptotic Growth of Functions CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll

More information

where: T = number of years of cash flow in investment's life n = the year in which the cash flow X n i = IRR = the internal rate of return

where: T = number of years of cash flow in investment's life n = the year in which the cash flow X n i = IRR = the internal rate of return EVALUATING ALTERNATIVE CAPITAL INVESTMENT PROGRAMS By Ke D. Duft, Extesio Ecoomist I the March 98 issue of this publicatio we reviewed the procedure by which a capital ivestmet project was assessed. The

More information

Properties of MLE: consistency, asymptotic normality. Fisher information.

Properties of MLE: consistency, asymptotic normality. Fisher information. Lecture 3 Properties of MLE: cosistecy, asymptotic ormality. Fisher iformatio. I this sectio we will try to uderstad why MLEs are good. Let us recall two facts from probability that we be used ofte throughout

More information

Confidence Intervals for One Mean

Confidence Intervals for One Mean Chapter 420 Cofidece Itervals for Oe Mea Itroductio This routie calculates the sample size ecessary to achieve a specified distace from the mea to the cofidece limit(s) at a stated cofidece level for a

More information

CURIOUS MATHEMATICS FOR FUN AND JOY

CURIOUS MATHEMATICS FOR FUN AND JOY WHOPPING COOL MATH! CURIOUS MATHEMATICS FOR FUN AND JOY APRIL 1 PROMOTIONAL CORNER: Have you a evet, a workshop, a website, some materials you would like to share with the world? Let me kow! If the work

More information

The Stable Marriage Problem

The Stable Marriage Problem The Stable Marriage Problem William Hut Lae Departmet of Computer Sciece ad Electrical Egieerig, West Virgiia Uiversity, Morgatow, WV William.Hut@mail.wvu.edu 1 Itroductio Imagie you are a matchmaker,

More information

5 Boolean Decision Trees (February 11)

5 Boolean Decision Trees (February 11) 5 Boolea Decisio Trees (February 11) 5.1 Graph Coectivity Suppose we are give a udirected graph G, represeted as a boolea adjacecy matrix = (a ij ), where a ij = 1 if ad oly if vertices i ad j are coected

More information

CME 302: NUMERICAL LINEAR ALGEBRA FALL 2005/06 LECTURE 8

CME 302: NUMERICAL LINEAR ALGEBRA FALL 2005/06 LECTURE 8 CME 30: NUMERICAL LINEAR ALGEBRA FALL 005/06 LECTURE 8 GENE H GOLUB 1 Positive Defiite Matrices A matrix A is positive defiite if x Ax > 0 for all ozero x A positive defiite matrix has real ad positive

More information

Simple Annuities Present Value.

Simple Annuities Present Value. Simple Auities Preset Value. OBJECTIVES (i) To uderstad the uderlyig priciple of a preset value auity. (ii) To use a CASIO CFX-9850GB PLUS to efficietly compute values associated with preset value auities.

More information

Lecture 4: Cauchy sequences, Bolzano-Weierstrass, and the Squeeze theorem

Lecture 4: Cauchy sequences, Bolzano-Weierstrass, and the Squeeze theorem Lecture 4: Cauchy sequeces, Bolzao-Weierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits

More information

a 4 = 4 2 4 = 12. 2. Which of the following sequences converge to zero? n 2 (a) n 2 (b) 2 n x 2 x 2 + 1 = lim x n 2 + 1 = lim x

a 4 = 4 2 4 = 12. 2. Which of the following sequences converge to zero? n 2 (a) n 2 (b) 2 n x 2 x 2 + 1 = lim x n 2 + 1 = lim x 0 INFINITE SERIES 0. Sequeces Preiary Questios. What is a 4 for the sequece a? solutio Substitutig 4 i the expressio for a gives a 4 4 4.. Which of the followig sequeces coverge to zero? a b + solutio

More information

Determining the sample size

Determining the sample size Determiig the sample size Oe of the most commo questios ay statisticia gets asked is How large a sample size do I eed? Researchers are ofte surprised to fid out that the aswer depeds o a umber of factors

More information

Cooley-Tukey. Tukey FFT Algorithms. FFT Algorithms. Cooley

Cooley-Tukey. Tukey FFT Algorithms. FFT Algorithms. Cooley Cooley Cooley-Tuey Tuey FFT Algorithms FFT Algorithms Cosider a legth- sequece x[ with a -poit DFT X[ where Represet the idices ad as +, +, Cooley Cooley-Tuey Tuey FFT Algorithms FFT Algorithms Usig these

More information

Lesson 17 Pearson s Correlation Coefficient

Lesson 17 Pearson s Correlation Coefficient Outlie Measures of Relatioships Pearso s Correlatio Coefficiet (r) -types of data -scatter plots -measure of directio -measure of stregth Computatio -covariatio of X ad Y -uique variatio i X ad Y -measurig

More information

Tangent circles in the ratio 2 : 1. Hiroshi Okumura and Masayuki Watanabe. In this article we consider the following old Japanese geometry problem

Tangent circles in the ratio 2 : 1. Hiroshi Okumura and Masayuki Watanabe. In this article we consider the following old Japanese geometry problem 116 Taget circles i the ratio 2 : 1 Hiroshi Okumura ad Masayuki Wataabe I this article we cosider the followig old Japaese geometry problem (see Figure 1), whose statemet i [1, p. 39] is missig the coditio

More information

Systems Design Project: Indoor Location of Wireless Devices

Systems Design Project: Indoor Location of Wireless Devices Systems Desig Project: Idoor Locatio of Wireless Devices Prepared By: Bria Murphy Seior Systems Sciece ad Egieerig Washigto Uiversity i St. Louis Phoe: (805) 698-5295 Email: bcm1@cec.wustl.edu Supervised

More information

NATIONAL SENIOR CERTIFICATE GRADE 12

NATIONAL SENIOR CERTIFICATE GRADE 12 NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 04 MARKS: 50 TIME: 3 hours This questio paper cosists of 8 pages ad iformatio sheet. Please tur over Mathematics/P DBE/04 NSC Grade Eemplar INSTRUCTIONS

More information

Week 3 Conditional probabilities, Bayes formula, WEEK 3 page 1 Expected value of a random variable

Week 3 Conditional probabilities, Bayes formula, WEEK 3 page 1 Expected value of a random variable Week 3 Coditioal probabilities, Bayes formula, WEEK 3 page 1 Expected value of a radom variable We recall our discussio of 5 card poker hads. Example 13 : a) What is the probability of evet A that a 5

More information

FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. 1. Powers of a matrix

FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. 1. Powers of a matrix FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. Powers of a matrix We begi with a propositio which illustrates the usefuless of the diagoalizatio. Recall that a square matrix A is diogaalizable if

More information

Fast Fourier Transform

Fast Fourier Transform 18.310 lecture otes November 18, 2013 Fast Fourier Trasform Lecturer: Michel Goemas I these otes we defie the Discrete Fourier Trasform, ad give a method for computig it fast: the Fast Fourier Trasform.

More information

MEI Structured Mathematics. Module Summary Sheets. Statistics 2 (Version B: reference to new book)

MEI Structured Mathematics. Module Summary Sheets. Statistics 2 (Version B: reference to new book) MEI Mathematics i Educatio ad Idustry MEI Structured Mathematics Module Summary Sheets Statistics (Versio B: referece to ew book) Topic : The Poisso Distributio Topic : The Normal Distributio Topic 3:

More information

Solutions to Selected Problems In: Pattern Classification by Duda, Hart, Stork

Solutions to Selected Problems In: Pattern Classification by Duda, Hart, Stork Solutios to Selected Problems I: Patter Classificatio by Duda, Hart, Stork Joh L. Weatherwax February 4, 008 Problem Solutios Chapter Bayesia Decisio Theory Problem radomized rules Part a: Let Rx be the

More information

GCSE STATISTICS. 4) How to calculate the range: The difference between the biggest number and the smallest number.

GCSE STATISTICS. 4) How to calculate the range: The difference between the biggest number and the smallest number. GCSE STATISTICS You should kow: 1) How to draw a frequecy diagram: e.g. NUMBER TALLY FREQUENCY 1 3 5 ) How to draw a bar chart, a pictogram, ad a pie chart. 3) How to use averages: a) Mea - add up all

More information