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1 year ago

Construct a table of values for the following functions using the integers from -4 to 4.

Mathematics

1 answer:

jeka57 [31]1 year ago

3 0

Step-by-step explanation:

Find the table attached

a) Given

F(x) = 6/x-2

When x = -4

F(-4) = 6/-4-2

F(-4) = 6/-6

F(-4) = -1

F(x) = 6/x-2

When x = -3

F(-3) = 6/-3-2

F(-3) = 6/-5

F(-3) = -1.2

F(x) = 6/x-2

When x = -2

F(-2) = 6/-2-2

F(-2) = 6/-4

F(-2) = -1.5

F(x) = 6/x-2

When x = -1

F(-1) = 6/-1-2

F(-1) = 6/-3

F(-1) = -2.0

F(x) = 6/x-2

When x = 0

F(0) = 6/0-2

F(0) = 6/-2

F(0) = -3

F(x) = 6/x-2

When x = 1

F(1) = 6/1-2

F(1) = 6/-1

F(1) = -6

F(x) = 6/x-2

When x = 2

F(2) = 6/2-2

F(2) = 6/0

F(2) = infty

F(x) = 6/x-2

When x = 3

F(3) = 6/3-2

F(3) = 6/1

F(3) = 6

F(x) = 6/x-2

When x = 4

F(4) = 6/4-2

F(4) = 6/2

F(4) = 3

b) Given

r(x)=6x+12/x^-4

When x = -4

r(-4) = 6(-4)+12/(-4)^-4

r(-4) = -24+12/(1/256)

r(-4) = -12(256)

r(-4) = -3072

When x = -3

r(-3) = 6(-3)+12/(-3)^-4

r(-3) = -18+12/(1/81)

r(-3) = -6(81)

r(-3) = -486

When x = -2

r(-2) = 6(-2)+12/(-2)^-4

r(-2) = -12+12/(1/16)

r(-2) = -0(16)

r(-2) = 0

When x = -1

r(-1) = 6(-1)+12/(-1)^-4

r(-1) = -6+12/(1)

r(-1) = -6+12

r(-1) = 6

When x = 0

r(0) = 6(0)+12/(0)^-4

r(0) = 0+12/0

r(0) = 12/0

r(0) = infty

When x = 1

r(1) = 6(1)+12/(1)^-4

r(1) = 6+12/1

r(1) = 18(1)

r(1) = 18

When x = 2

r(2) = 6(2)+12/(2)^-4

r(2) = 12+12/1/16

r(2) = 24(16)

r(2) = 384

When x = 3

r(3) = 6(3)+12/(3)^-4

r(3) = 18+12/1/81

r(3) = 30(81)

r(3) = 2430

When x = 4

r(4) = 6(4)+12/(4)^-4

r(4) = 24+12/1/256

r(4) = 36(256)

r(4) = 9216

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Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, $L$ = 120 cm

Its wavelength $\lambda_1$ = 480 cm

$\lambda_2$ = 160 cm and $\lambda_3$ = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

The fundamental wavelength of a pipe which is open at both ends is 2L.
The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ $4L=4(120)=480\ cm$

It seems that the pipe is open at one end and closed at one end.

Now lets check with the subsequent wavelengths.

For one side open and one side closed pipe:

An odd-integer number of quarter wavelength have to fit into the tube of length L.

⇒ $\lambda_2=\frac{4L}{3}$ ⇒ $\lambda_3=\frac{4L}{5}$

⇒ $\lambda_2=\frac{4(120)}{3}$ ⇒ $\lambda_3=\frac{4(120)}{5}$

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jeka94

Answer:

a) Y 0 1 2

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b) mean= 0.45, S.D= 0.6718

c) mean= 1.285, S.D= 8.74

Step-by-step explanation:

a) The following table shows the probability distribution of X:

X 0 1 2 3 4 or more

P(X) 0.58 0.23 0.11 0.05 0.03

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