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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]$t \geq 0$[/tex].
The range of the function is [tex]$-5 \leq p(t) \leq 5$[/tex].



Answer :

GinnyAnswer
Sure, let's determine the correct values for the domain and range within the context of this situation.

### Domain:
The function [tex]\( p(t) \)[/tex] represents the change in pressure over time. Since time [tex]\( t \)[/tex] cannot be negative, the domain of the function is [tex]\( t \geq 0 \)[/tex].

### Range:
The function given is [tex]\( p(t) = 5 \sin(880 \pi t) \)[/tex].

The sine function, [tex]\( \sin(x) \)[/tex], oscillates between -1 and 1 for any real number [tex]\( x \)[/tex]. Therefore, when we multiply the sine function by 5, the range of [tex]\( 5 \sin(880 \pi t) \)[/tex] will be between [tex]\( -5 \)[/tex] and [tex]\( 5 \)[/tex].

So, the range of the function [tex]\( p(t) \)[/tex] is:
[tex]\[ -5 \leq p(t) \leq 5 \][/tex]

### Final Answer:
- The domain of the function is [tex]\( t \geq 0 \)[/tex].
- The range of the function is [tex]\( -5 \leq p(t) \leq 5 \)[/tex].

Therefore, the correct answers are:

The domain of the function is [tex]\( t \geq 0 \)[/tex].
The range of the function is [tex]\( -5 \leq p(t) \leq 5 \)[/tex].

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