X .A. 4 .2

Nov 15, 2014

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Programfejlesztési Központ. X .A. 4 .2. hasonmások. Eminem. Mark Garment. Jennifer Lopez. Juliette Samuels. Mariah Carey. Michaela Morris. Melyikük az igazi?. Paul McCartney. Neil Tudor.

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X .A. 4 .2

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ProgramfejlesztésiKözpont X.A.4.2 hasonmások
Eminem Mark Garment
Jennifer Lopez Juliette Samuels
Mariah Carey Michaela Morris
Melyikük az igazi? Paul McCartney Neil Tudor

Warm Up 1. 2( w + 1) 2. 3 x ( x 2 – 4)

Warm Up 1. 2( w + 1) 2. 3 x ( x 2 – 4). Simplify. . 2 w + 2. 3 x 3 – 12 x. Find the GCF of each pair of monomials. 2 h. 3. 4 h 2 and 6 h. 13 p. 4. 13 p and 26 p 5. Find the value of each power of 10. b. 10 5. c. 10 10. a. 10 –2. 100,000. 0.01. 10,000,000,000.

379 views • 19 slides

f(x) = x 3 + 2x +3 g(x) = 2 x + x 2 + x + 4 h(x) = x 2 + 3x 3 + 2x + 1 t(x) = x -2 + x + 2

Do Now: Do we have a polynomial function? Is it in standard form? What is the degree? What is the leading coeficient ?. f(x) = x 3 + 2x +3 g(x) = 2 x + x 2 + x + 4 h(x) = x 2 + 3x 3 + 2x + 1 t(x) = x -2 + x + 2. Chapter 2.2/ 2. SWBAT Evaluate and Graph Polynomial Functions

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x 2 + 4 = 0

números. COMPLEXOS. x 2 + 5 = 0. x 2 + 4 = 0. x 2 + 5x +8 = 0. Fred Tavares NÚMEROS COMPLEXOS. NÚMEROS COMPLEXOS. Os números Complexos constituem o maior conjunto numérico existente.

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x 2 - 5 x + 4

I. x 2 - 5 x + 4. I. x 2 - 5 x + 4. x. x. I. x 2 - 5 x + 4. x. +4. +2. +1. - 4. - 2. - 1. x. +1. +2. +4. - 1. - 2. - 4. I. x 2 - 5 x + 4. x. +4. +2. +1. - 4. - 2. - 1. x. +1. +2. +4. - 1. - 2. - 4. I. x 2 - 5 x + 4. x. +4. +2. +1. - 4. - 2.

483 views • 41 slides

Warm Up Evaluate each expression for x = 12. 1. x + 2 2. 3. x – 8 4. 10 x – 4 5. 2 x + 12

Equations and Their Solutions. 1-10. Course 2. Warm Up Evaluate each expression for x = 12. 1. x + 2 2. 3. x – 8 4. 10 x – 4 5. 2 x + 12 6. 5 x + 7. 14. x 4. 3. 4. 116. 36. 67. Equations and Their Solutions. 1-10. Course 2.

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a 1 2 x 2 +

Matrizes. Conceitos Básicos. a 3 1. a 3 2. a 3 3. a 3 n. a m 1. a m 2. a m 3. a m n. Consideremos o sistema. ... +. a 1 1 x 1 +. a 1 2 x 2 +. a 1 3 x 3 +. a 1 n x n =. b 1. ... +. a 2 1 x 1 +. a 2 2 x 2 +. a 2 3 x 3 +. a 2 n x n =. b 2. ... +. a 3 1 x 1 +.

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Multiply: a) x ( x + 7) b) 6 x ( x 2  4 x + 5) Solution

Example. Multiply: a) x ( x + 7) b) 6 x ( x 2  4 x + 5) Solution a) x ( x + 7) = x  x + x  7 = x 2 + 7 x b) 6 x ( x 2  4 x + 5) = (6 x ) ( x 2 )  (6 x ) (4 x ) + (6 x ) (5) = 6 x 3  24 x 2 + 30 x. L. F. I. O. Example.

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Matching. |x + 2| = 4 a. x < 2 and x > -6 | x + 2| < 4 b. x = 2 or x = -6

Matching. |x + 2| = 4 a. x < 2 and x > -6 | x + 2| < 4 b. x = 2 or x = -6 | x + 2| < -4 c. x > 2 or x < - 6 | x + 2| > -4 d. there is no solution | x + 2| > 4 e. all real numbers | x + 2| = -4 f. none of the above.

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Solve. 16 + 4 x 2 – 7 x 3 2. 16 ÷ 4 x 2 – 8 + 2

Warm-up. Solve. 16 + 4 x 2 – 7 x 3 2. 16 ÷ 4 x 2 – 8 + 2. NOTES - Angles. Types of Angles:. Naming Angles – Four ways: 1. Use angle number alone <2 2. Use middle letter alone <G 3. Use letter in ABC order, vertex in the middle <FGH

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4( x 2 – 2 x + ? ) + y 2 = 8 + 4( ? )

Note that A = 4, B = 0, and C = 1 , so the value of the discriminant is:. B 2 – 4 AC = 0 2 – 4(4)(1) = –16. Because B 2 – 4 AC < 0 and A = C, the conic is an ellipse. To graph the ellipse, first complete the square in x. EXAMPLE 6. Classify a conic.

210 views • 12 slides

Permintaan = X 1 * X 2 * X 3 * X 4 * X 5

Permintaan = X 1 * X 2 * X 3 * X 4 * X 5. X 1 : Jumlah penduduk di Pulau Jawa. X 2 : Persentase SES A dan B di Pulau Jawa. X 3 : Persentase wanita berusia 25-50 tahun. X 4 : Persentase ibu rumah tangga di Pulau Jawa yang memiliki kepedulian terhadap kesehatan.

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=x 2 +5 x +6 =x 2 -3 x -4 =x 2 -11 x + 30

活动与探索： ① ( x +2)( x +3) ②( x -4)( x +1) ③ ( x -5)( x -6) ( x + p )( x + q ). =x 2 +5 x +6 =x 2 -3 x -4 =x 2 -11 x + 30. =x 2 +( p + q ) x + pq. 利用你发现的规律口答： ① ( y +4 )( y -2 ) ②( y -5 )( y -3 ) ③( a -2 )( a +5 ) ④( a +9 )( a -9 ). 口答 : (1)(P+1) 2 ＝ (P+1)(P+1 )= . (2)(m+2) 2 = .

260 views • 13 slides

Warm-up: 1.) 2 x 2 x 2 =8 2.) 5 x 5 =25 3.) 4 x 4 x 4 =64 4.) 10 x 10 x 10 =1000

Warm-up: 1.) 2 x 2 x 2 =8 2.) 5 x 5 =25 3.) 4 x 4 x 4 =64 4.) 10 x 10 x 10 =1000. Remember that any number can be written as a product of Prime Factors. Example: 32. 2. 16. 2. 2. 8. 2. 2. 2. 4. 2. 2. 2. 2. 2.

430 views • 20 slides

f ( x ) = 2 x 4 - 8 x 2 + 5 x - 7 when x = 3.

Therefore, we can use synthetic division to help us evaluate functions through a process called synthetic substitution. Evaluate. f ( x ) = 2 x 4 - 8 x 2 + 5 x - 7 when x = 3. REMAINDER THEOREM.

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Warm Up Solve. 1. 2 x + 8 = x – 7 2. –4( x + 3) = –5 x – 2

Warm Up Solve. 1. 2 x + 8 = x – 7 2. –4( x + 3) = –5 x – 2 3. 5 x + x + ( - 11) = 25 – 3 x 4. 6 n + 9 – 4 n = 3 n. x = –15. x = 10. x = 4. n = 9.

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Solve: 2 x + 4 = 8

Solve: 2 x + 4 = 8. =. Solve: 2 x + 4 = 8. x. x. =. Solve: 2 x + 4 = 8. x. x. =. Solve: 2 x + 4 = 8. x. x. =. Solve: 2 x + 4 = 8. 1.) Remove 4 counters from each side of the scale Is the scale still balanced?. x. x. =. Solve: 2 x + 4 = 8.

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X 4 -3X 2 - 4 = 0

X 4 -3X 2 - 4 = 0. 3X 3 + 8X 2 = 9X - 2. Attachments.

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tbp x y y -2/(xx) + 4/(xx*x) arrows 20 0 8 -2 2

tbp x y y -2/(x*x) + 4/(x*x*x) arrows 20 0 8 -2 2 .

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Warm Up Solve each equation. 1. 2 x + 3 = 2 x + 4 2. 2( x + 1) = 2 x + 2

Warm Up Solve each equation. 1. 2 x + 3 = 2 x + 4 2. 2( x + 1) = 2 x + 2 3. Solve 2 y – 6 x = 10 for y. no solution. infinitely many solutions. y = 3 x + 5. Solve by using any method. y = 3 x + 2. x – y = 8. 5. 4. (6, –2). (1, 5). 2 x + y = 7. x + y = 4.

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Warm Up Simplify. 1. 4 x – 10 x 2. –7( x – 3) 3. 4. 15 – ( x – 2) Solve. 5. 3 x + 2 = 8

Warm Up Simplify. 1. 4 x – 10 x 2. –7( x – 3) 3. 4. 15 – ( x – 2) Solve. 5. 3 x + 2 = 8 6. –6 x. –7 x + 21. 2 x + 3. 17 – x. 2. 28. Objective. Solve equations in one variable that contain variable terms on both sides. Vocabulary. identity contradiction.

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4 x 2 – 9 y 2 = 36

2. 1. x 2. y 2. x 2. y 2. 25. 4. 16. 9. – = 1. – = 1. Reminder: Multiplying both sides of each equation by the least common multiple eliminates the fractions. 4 x 2 – 9 y 2 = 36. 16 y 2 – 25 x 2 = 400. Objectives. Write the standard equation for a hyperbola.

267 views • 25 slides

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X .A. 4 .2

X .A. 4 .2

Presentation Transcript

Warm Up 1. 2( w + 1) 2. 3 x ( x 2 – 4)

f(x) = x 3 + 2x +3 g(x) = 2 x + x 2 + x + 4 h(x) = x 2 + 3x 3 + 2x + 1 t(x) = x -2 + x + 2

x 2 + 4 = 0

x 2 - 5 x + 4

Warm Up Evaluate each expression for x = 12. 1. x + 2 2. 3. x – 8 4. 10 x – 4 5. 2 x + 12

a 1 2 x 2 +

Multiply: a) x ( x + 7) b) 6 x ( x 2  4 x + 5) Solution

Matching. |x + 2| = 4 a. x &lt; 2 and x &gt; -6 | x + 2| &lt; 4 b. x = 2 or x = -6

Solve. 16 + 4 x 2 – 7 x 3 2. 16 ÷ 4 x 2 – 8 + 2

4( x 2 – 2 x + ? ) + y 2 = 8 + 4( ? )

Permintaan = X 1 * X 2 * X 3 * X 4 * X 5

=x 2 +5 x +6 =x 2 -3 x -4 =x 2 -11 x + 30

Warm-up: 1.) 2 x 2 x 2 =8 2.) 5 x 5 =25 3.) 4 x 4 x 4 =64 4.) 10 x 10 x 10 =1000

f ( x ) = 2 x 4 - 8 x 2 + 5 x - 7 when x = 3.

Warm Up Solve. 1. 2 x + 8 = x – 7 2. –4( x + 3) = –5 x – 2

Solve: 2 x + 4 = 8

X 4 -3X 2 - 4 = 0

tbp x y y -2/(x*x) + 4/(x*x*x) arrows 20 0 8 -2 2

Warm Up Solve each equation. 1. 2 x + 3 = 2 x + 4 2. 2( x + 1) = 2 x + 2

Warm Up Simplify. 1. 4 x – 10 x 2. –7( x – 3) 3. 4. 15 – ( x – 2) Solve. 5. 3 x + 2 = 8

4 x 2 – 9 y 2 = 36

Matching. |x + 2| = 4 a. x < 2 and x > -6 | x + 2| < 4 b. x = 2 or x = -6

tbp x y y -2/(xx) + 4/(xx*x) arrows 20 0 8 -2 2