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valina [46]
4 years ago
8

For questions 1 - 2, what are the excluded functions of the function?

Mathematics
1 answer:
PilotLPTM [1.2K]4 years ago
6 0

1.) Function of x.f (x) =  \frac{4x}{9} -45
     Function of y.f (y) = \frac{405}{4} +  \frac{9y}{4}
2.)Function of y.f (y)=  \frac{7}{2(y-48)}
    Function of x.f (x) = \frac{7}{2x} + 48
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Identifying and Graphing Sequences: lesson 4-1
goldfiish [28.3K]

Answer:

1) Domain: {   1,2,3,4,5,6}

Range: {12,24,36,48,60,72}

2) First four terms are: 2,5,8,11

3) First four terms are: 8,13,20,29

4) First four terms are: 0,0,2,6

5) First four terms are: 0,1,1.41,1.73

Step-by-step explanation:

1.

n       1      2     3     4      5      6

f(n)    12    24    36   48    60    72

Domain: {   1,2,3,4,5,6}

Range: {12,24,36,48,60,72}

Write first four terms of each sequence

2) f(n)=3n-1

Put n = 1, 3(1)-1 = 3-1 = 2

Put n = 2, 3(2)-1 = 6-1 = 5

Put n = 3, 3(3)-1 = 9-1 = 8

Put n = 4, 3(4)-1 = 12-1 = 11

So, First four terms are: 2,5,8,11

3) f(n)=n^2+2n+5

Put n = 1, (1)^2+2(1)+5= 1+2+5 = 8

Put n = 2, (2)^2+2(2)+5 = 4+4+5= 13

Put n = 3, (3)^2+2(3)+5 = 9+6+5 = 20

Put n = 4, (4)^2+2(4)+5 = 16+8+5 = 29

So, First four terms are: 8,13,20,29

4) f(n)=(n-1)(n-2)

Put n = 1, (1-1)(1-2) = 0(-1) = 0

Put n = 2, (2-1)(2-2) = 1(0) = 0

Put n = 3, (3-1)(3-2) = 2(1) = 2

Put n = 4, (4-1)(4-2) = 3(2) = 6

So, First four terms are: 0,0,2,6

5) f(n) = \sqrt{n-1}

Put n=1 \sqrt{1-1}=\sqrt{0}=0

Put n=2 \sqrt{2-1}=\sqrt{1}=1

Put n=3 \sqrt{3-1}=\sqrt{2}= 1.41

Put n=4 \sqrt{4-1}=\sqrt{3}=1.73

So, First four terms are: 0,1,1.41,1.73

8 0
3 years ago
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