Unit 6: Modeling Mathematics 3 Ms. C. Taylor. Warm-Up.

Unit 6: ModelingMathematics 3

Ms. C. Taylor

Warm-Up

Solve Solve

Simplifying Radicals

Solving RadicalsStep 1: Get rid of anything that might

not be under the radical.Step 2: Square both sides to get rid

of the radical.Step 3: Isolate the variable.Step 4: SolveStep 5: Plug answer back into

equation and make sure that it works.

What is a Radical Expression?

A Radical Expression is an equation that has a variable in a radicand or has a variable with a rational exponent.

103 x

25)2( 3

2

x

103 x

yes

yes

no

EXAMPLE – Solving a Radical Equation0615 x

615 x2

( )2

615 x

3615 x

355 x

7x

Square both sides to get rid of the square root

EXAMPLE

11

431

1631516 x315x

x315x 2

( )2

)x3)(x3(15x

xx6915x

x6915

x624

x4 x16

NO SOLUTION Since 16 doesn’t plug inas a solution.

Let’s Double Check that this works

Note: You will get Extraneous

Solutions from time to time – always do a quick check

Let’s Try Some6232 x 50)2(2 3

2

x

Warm-Up

Can graphing calculators help?

SURE!

1. Input for Y1

2. Input x-2 for Y2

3. Graph

4. Find the points of intersection

x

2xx

One Solution at (4, 2)

To see if this is extraneous or not, plug the x value back into the equation. Does it work?

Graphing Radicals

5

Graphing Systems

Same type of deal as graphing radicals as in you are looking for the intersection point which will be your solution.

An Arithmetic Sequence is defined as a sequence in which there is a common

difference between consecutive terms.

Which of the following sequences are arithmetic? Identify the common

difference.

3, 1, 1, 3, 5, 7, 9, . . .

15.5, 14, 12.5, 11, 9.5, 8, . . .

84, 80, 74, 66, 56, 44, . . .

8, 6, 4, 2, 0, . . .

50, 44, 38, 32, 26, . . .

YES 2d

YES

YES

NO

NO

1.5d

6d

The common

difference is

always the

difference between

any term and the

term that proceeds

that term.26, 21, 16, 11, 6, . . .

Common Difference = 5

The general form of an ARITHMETIC sequence.

1aFirst Term:

Second Term: 2 1a a d

Third Term:

Fourth Term:

Fifth Term:

3 1 2a a d

4 1 3a a d

5 1 4a a d

nth Term: 1 1na a n d

Formula for the nth term of an ARITHMETIC sequence.

1 1na a n d

The nth termna

The term numbern

The common differenced

1 The 1st terma

If we know any three of these

we ought to be able to find the

fourth.

Given: 79, 75, 71, 67, 63, . . .Find: 32a

1 79

4

32

a

d

n

1

32

32

1

79 32 1 4

45

na a n d

a

a

IDENTIFY SOLVE

Given: 79, 75, 71, 67, 63, . . .

Find: What term number is -169?

1 79

4

169n

a

d

a

1 1

169 79 1 4

63

na a n d

n

n

IDENTIFY SOLVE

Given:10

12

3.25

4.25

a

a

1

3

3.25

4.25

3

a

a

n

1 1

4.25 3.25 3 1

0.5

na a n d

d

d

IDENTIFY SOLVE

Find: 1a

What’s the real question? The Difference

Given:10

12

3.25

4.25

a

a

10 3.25

0.5

10

a

d

n

1

1

1

1

3.25 10 1 0.5

1.25

na a n d

a

a

IDENTIFY SOLVE

Find: 1a

Arithmetic Series

Write the first three terms and the last two terms of the following arithmetic series.

50

1

73 2p

p

71 69 67 . . . 25 27

What is the sum

of this series?

71 69 67 . . . 25 27

27 25 . . . 67 69 71

44 44 44 . . . 44 44 44

50 71 27

2

110071 + (-27) Each sum is the same.

50 Terms

1 1 1 12 . . . 1a a d a d a n d

1 1 1 11 . . . 2a n d a d a d a

1

2nn a as

1

Sum

Number of Terms

First Term

Last Termn

S

n

a

a

1 1 1 1 1 11 1 . . . 1a a n d a a n d a a n d

Find the sum of the terms of this arithmetic series.

35

1

29 3k

k

1

2nn a a

S

1

35

35

26

76

n

a

a

35 26 76

2875

S

S

Find the sum of the terms of this arithmetic series. 151 147 143 139 . . . 5

1

2nn a a

S

1

40

40

151

5

n

a

a

40 151 5

22920

S

S

1 1

5 151 1 4

40

na a n d

n

n

What term is -5?

Alternate formula for the

sum of an Arithmetic

Series.

1

2nn a

Sa

1 1Substitute na a n d

1 1

1

1

2

2 1

2

n a a n dS

n a n dS

1

# of Terms

1st Term

Difference

n

a

d

Find the sum of this series 36

0

2.25 0.75j

j

2.25 3 3.73 4.5 . . .

12 1

2

n a n dS

It is not convenient to find the last term.

1

37

2.25

0.75

n

a

d

37 2 2.25 37 1 0.75

2582.75

S

S

35

1

45 5i

i

1

2nn a a

S

12 1

2

n a n dS

135 40 130nn a a 135 40 5n a d

35 40 130

21575

S

S

35 2 40 35 1 3

21575

S

S

Warm-Up

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 33

An infinite sequence is a function whose domain is the set of positive integers. a1, a2, a3, a4, . . . ,

an, . . .

The first three terms of the sequence an = 2n2 area1 = 2(1)2 = 2

a2 = 2(2)2 = 8

a3 = 2(3)2 = 18.

finite sequence

terms


A sequence is geometric if the ratios of consecutive terms are the same.

2, 8, 32, 128, 512, . . .

geometric sequence

The common ratio, r, is 4.

82

4

328

4

12832

4

512128

4


The nth term of a geometric sequence has the form

an = a1rn - 1

where r is the common ratio of consecutive terms of the sequence.

15, 75, 375, 1875, . . . a1 = 15

The nth term is 15(5n-1).

75 515

r

a2 = 15(5)

a3 = 15(52)

a4 = 15(53)


Example: Find the 9th term of the geometric sequence

7, 21, 63, . . .a1 = 7

The 9th term is 45,927.

21 37

r

an = a1rn – 1 = 7(3)n – 1

a9 = 7(3)9 – 1 = 7(3)8

= 7(6561) = 45,927


The sum of the first n terms of a sequence is represented by summation notation.

1 2 3 41

n

i ni

a a a a a a

index of summation

upper limit of summation

lower limit of summation

5

1

4n

n

1 2 3 4 54 4 4 4 4 4 16 64 256 1024 1364


The sum of a finite geometric sequence is given by 1

1 11

1 .1

n nin

i

rS a r ar

5 + 10 + 20 + 40 + 80 + 160 + 320 + 640 = ?

n = 8

a1 = 5

1

81 11

221

5n

nrS ar

5210r

1 25651 2 2555

1 1275


The sum of the terms of an infinite geometric sequence is called a geometric series.

a1 + a1r + a1r2 + a1r3 + . . . + a1rn-1 + . . .

If |r| < 1, then the infinite geometric series

11

0

.1

i

i

aS a r

r

has the sum

If 1 , then the series does not have a sum.r


Example: Find the sum of

1

1a

Sr

1 13 13 9

13

r

3

1 13

3 31 413 3

The sum of the series is 9 .4

3 934 4


Graphing Utility: Find the first 5 terms of the geometric sequence an = 2(1.3)n.

List Menu:

variable

Graphing Utility: Find the sum 10

1

22 .3

n

n

List Menu:

beginning value

end value

variable

upper limit

lower limit

IntroductionGeometric sequences are exponential functions that have a domain of consecutive positive integers. Geometric sequences can be represented by formulas, either explicit or recursive, and those formulas can be used to find a certain term of the sequence or the number of a certain value in the sequence.

42

3.8.2: Geometric Sequences

Guided Practice

Example 1

Find the constant ratio, write the explicit formula, and find the seventh term for the following geometric sequence.

3, 1.5, 0.75, 0.375, …

43


Guided Practice: Example 1, continued

1. Find the constant ratio by dividing two successive terms.

1.5 ÷ 3 = 0.5

44



2. Confirm that the ratio is the same between all of the terms.

0.75 ÷ 1.5 = 0.5 and 0.375 ÷ 0.75 = 0.5

45



3. Identify the first term (a1).a1 = 3

46



4. Write the explicit formula.an = a1 • r n – 1 Explicit formula for any given

geometric sequence

an = (3)(0.5)n – 1 Substitute values for a1 and n.

47



5. To find the seventh term, substitute 7 for n.a7 = (3)(0.5)7 – 1

a7 = (3)(0.5)6 Simplify.

a7 = 0.046875 Multiply.

The seventh term in the sequence is 0.046875.

48


✔

Guided Practice

Example 3

A geometric sequence is defined recursively by

, with a1 = 729. Find the first five terms of

the sequence, write an explicit formula to represent the

sequence, and find the eighth term.

49



1. Using the recursive formula:

50



The first five terms of the sequence are 729, –243, 81, –27, and 9.

51



2. The first term is a1 = 729 and the constant

ratio is , so the explicit formula is

.

52



3. Substitute 8 in for n and evaluate.

The eighth term in the sequence is .

53


✔

Warm-Up

Unit 6: Modeling Mathematics 3 Ms. C. Taylor. Warm-Up.

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Transcript of Unit 6: Modeling Mathematics 3 Ms. C. Taylor. Warm-Up.