Admission in india
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Determine whether each relation is a function.
1. {(2, 4), (1, 3), (–3, –1), (4, 6)} 2. {(2, 6), (–3, 1), (–2, 2)}
3. {(x, y)| x = 3} 4. {(x, y)| y = 8}
5. {(x, y)| x = y2} 6. {(x, y)| x2 + y2 = 36}
7. {(a, b)| a = b3} 8. {(w, z)| w = z – 36}
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1. {(2, 4), (1, 3), (–3, –1), (4, 6)}; yes, this is a function because each element of the domain is paired with exactly one element in the range.
2. {(2, 6), (–3, 1), (–2, 2)}; yes, this is a function because each element of the domain is paired with exactly one element in the range.
3. {(x, y)| x = 3}; no, this is not a function because it is a vertical line and fails the vertical line test.
4. {(x, y)| y = 8}; yes, this is a function because it is a horizontal line and passes the vertical line test.
Solutions
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5. {(x, y)| x = y2}; no, this is not a function because an element of the domain is paired with more than one element in the range.Example: 4 = 22 and 4 = (–2)2
6. {(x, y)| x2 + y2 = 36}; no, this is not a function because it is a circle and fails the vertical line test.
7. {(a, b)| a = b3}; yes, this is a function because each element of the domain is paired with exactly one element in the range.
8. {(w, z)| w = z – 36}; yes, this is a function because each element of the domain is paired with exactly one element in the range.
Solutions (continued)
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Periodic Function: a repeating pattern of y-values (outputs) at regular intervals. Cycle: One complete pattern of the function. A cycle can occur at any point on the graph of the functionPeriod: the horizontal length of one cycle of the function. Amplitude: half of the distance between the minimum and maximum values of the function.
Amplitude
Period Periodadmission.edhole.com
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Analyze this periodic function. Identify one cycle in two different ways.
Then determine the period of the function.
Each cycle is 7 units long. The period of the function is 7.
Begin at any point on the graph.Trace one complete pattern.
The beginning and ending x-values of each cycle determine the period of the function.
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Determine whether each function is or is not periodic. If it is, find the period.
The pattern of y-values in one section repeats exactly in other sections. The function is periodic.
a.
Find points at the beginning and end of one cycle.
Subtract the x-values of the points: 2 – 0 = 2.
The pattern of the graph repeats every 2 units, so the period is 2.
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(continued)
b.
The pattern of y-values in one section repeats exactly in other sections. The function is periodic.
Find points at the beginning and end of one cycle.
Subtract the x-values of the points: 3 – 0 = 3.
The pattern of the graph repeats every 3 units, so the period is 3.
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Find the amplitudes of the two functions in
Additional Example 2.
The amplitude of the function is 2.
a. amplitude = (maximum value – minimum value) Use definition of amplitude.
12
= [2 – (–2)] Substitute.12
= (4) = 2 Subtract within parentheses and simplify.12
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(continued)
The amplitude of the function is 3.
b. amplitude = (maximum value – minimum value) Use definition of amplitude.
12
= [6 – 0] Substitute.12
= (6) = 3 Subtract within parentheses and simplify.12
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The oscilloscope screen below shows the graph of the alternating
current electricity supplied to homes in the United States. Find the
period and amplitude.
1 unit on the t-axis = s1
360
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The maximum value of the function is 120, and the minimum is –120.
One cycle of the electric current occurs from 0 s to s.160
period = – 0 Use the definitions.
= Simplify.
160
160
amplitude = [120 – (–120)]
= (240) = 120
12
12
The amplitude is 120 volts.The period of the electric current is s.160
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