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A tuning fork vibrates with a frequency of 440 hertz (cycles/second). When the tuning fork is struck, it produces a change in the normal air pressure in the room.

Function [tex] p [/tex] represents this situation, where [tex] p(t) [/tex] is the change in pressure, in pascals, relative to the normal air pressure in the room as a function of time, [tex] t [/tex], in seconds, after the tuning fork is struck.
[tex]
p(t) = 5 \sin (880 \pi t)
[/tex]

What are the domain and the range within the context of this situation?

The domain of the function is [tex] t \geq [/tex] [tex] \square [/tex]

The range of the function is [tex] \square \leq p(t) \leq [/tex] [tex] \square [/tex]

Sagot :

GinnyAnsweridn
To determine the domain and range of the function [tex]\( p(t) = 5 \sin (880 \pi t) \)[/tex]:

1. Domain:
- The domain of the function is related to the time [tex]\( t \)[/tex] after the tuning fork is struck. Since time cannot be negative, [tex]\( t \)[/tex] must be non-negative.
- Therefore, the domain of the function is [tex]\( t \geq 0 \)[/tex].

2. Range:
- The function [tex]\( \sin \)[/tex] oscillates between -1 and 1.
- When the amplitude of the sine wave is multiplied by 5, the pressure change [tex]\( p(t) \)[/tex] will oscillate between [tex]\( 5 \times (-1) \)[/tex] and [tex]\( 5 \times 1 \)[/tex], which are -5 and 5 Pascals, respectively.
- Therefore, the range of the function is [tex]\( -5 \leq p(t) \leq 5 \)[/tex].

So, the final answer is:

The domain of the function is [tex]\( t \geq \)[/tex] 0.

The range of the function is -5 [tex]\( \leq p(t) \leq \)[/tex] 5.
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