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To determine which equation models the distance [tex]\( d \)[/tex] of the weight from its equilibrium after [tex]\( t \)[/tex] seconds, we need to find an equation that satisfies the conditions provided: the weight is 9 inches below equilibrium at [tex]\( t = 0 \)[/tex] and returns to this position after [tex]\( t = 3 \)[/tex] seconds.
Given equations:
1. [tex]\( d = -9 \cos \left( \frac{\pi}{3} t \right) \)[/tex]
2. [tex]\( d = -9 \cos \left( \frac{2\pi}{3} t \right) \)[/tex]
3. [tex]\( d = -3 \cos \left( \frac{\pi}{9} t \right) \)[/tex]
4. [tex]\( d = -3 \cos \left( \frac{2\pi}{9} t \right) \)[/tex]
Let's test each equation separately:
### 1. [tex]\( d = -9 \cos \left( \frac{\pi}{3} t \right) \)[/tex]
At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{\pi}{3} \times 0 \right) = -9 \cos(0) = -9 \times 1 = -9 \][/tex]
This satisfies the condition at [tex]\( t = 0 \)[/tex].
At [tex]\( t = 3 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{\pi}{3} \times 3 \right) = -9 \cos(\pi) = -9 \times -1 = 9 \][/tex]
This does not match the condition at [tex]\( t = 3 \)[/tex] since we need [tex]\( d = -9 \)[/tex].
### 2. [tex]\( d = -9 \cos \left( \frac{2\pi}{3} t \right) \)[/tex]
At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{2\pi}{3} \times 0 \right) = -9 \cos(0) = -9 \times 1 = -9 \][/tex]
This satisfies the condition at [tex]\( t = 0 \)[/tex].
At [tex]\( t = 3 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{2\pi}{3} \times 3 \right) = -9 \cos(2\pi) = -9 \times 1 = -9 \][/tex]
This satisfies the condition at [tex]\( t = 3 \)[/tex].
### 3. [tex]\( d = -3 \cos \left( \frac{\pi}{9} t \right) \)[/tex]
At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -3 \cos \left( \frac{\pi}{9} \times 0 \right) = -3 \cos(0) = -3 \times 1 = -3 \][/tex]
This does not match the condition at [tex]\( t = 0 \)[/tex].
### 4. [tex]\( d = -3 \cos \left( \frac{2\pi}{9} t \right) \)[/tex]
At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -3 \cos \left( \frac{2\pi}{9} \times 0 \right) = -3 \cos(0) = -3 \times 1 = -3 \][/tex]
This does not match the condition at [tex]\( t = 0 \)[/tex].
Only the second equation [tex]\( d = -9 \cos \left( \frac{2\pi}{3} t \right) \)[/tex] satisfies the condition both at [tex]\( t = 0 \)[/tex] and [tex]\( t = 3 \)[/tex] seconds. Thus, the correct equation that models the distance [tex]\( d \)[/tex] of the weight from its equilibrium after [tex]\( t \)[/tex] seconds is:
[tex]\[ d = -9 \cos \left( \frac{2\pi}{3} t \right) \][/tex]
Given equations:
1. [tex]\( d = -9 \cos \left( \frac{\pi}{3} t \right) \)[/tex]
2. [tex]\( d = -9 \cos \left( \frac{2\pi}{3} t \right) \)[/tex]
3. [tex]\( d = -3 \cos \left( \frac{\pi}{9} t \right) \)[/tex]
4. [tex]\( d = -3 \cos \left( \frac{2\pi}{9} t \right) \)[/tex]
Let's test each equation separately:
### 1. [tex]\( d = -9 \cos \left( \frac{\pi}{3} t \right) \)[/tex]
At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{\pi}{3} \times 0 \right) = -9 \cos(0) = -9 \times 1 = -9 \][/tex]
This satisfies the condition at [tex]\( t = 0 \)[/tex].
At [tex]\( t = 3 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{\pi}{3} \times 3 \right) = -9 \cos(\pi) = -9 \times -1 = 9 \][/tex]
This does not match the condition at [tex]\( t = 3 \)[/tex] since we need [tex]\( d = -9 \)[/tex].
### 2. [tex]\( d = -9 \cos \left( \frac{2\pi}{3} t \right) \)[/tex]
At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{2\pi}{3} \times 0 \right) = -9 \cos(0) = -9 \times 1 = -9 \][/tex]
This satisfies the condition at [tex]\( t = 0 \)[/tex].
At [tex]\( t = 3 \)[/tex]:
[tex]\[ d = -9 \cos \left( \frac{2\pi}{3} \times 3 \right) = -9 \cos(2\pi) = -9 \times 1 = -9 \][/tex]
This satisfies the condition at [tex]\( t = 3 \)[/tex].
### 3. [tex]\( d = -3 \cos \left( \frac{\pi}{9} t \right) \)[/tex]
At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -3 \cos \left( \frac{\pi}{9} \times 0 \right) = -3 \cos(0) = -3 \times 1 = -3 \][/tex]
This does not match the condition at [tex]\( t = 0 \)[/tex].
### 4. [tex]\( d = -3 \cos \left( \frac{2\pi}{9} t \right) \)[/tex]
At [tex]\( t = 0 \)[/tex]:
[tex]\[ d = -3 \cos \left( \frac{2\pi}{9} \times 0 \right) = -3 \cos(0) = -3 \times 1 = -3 \][/tex]
This does not match the condition at [tex]\( t = 0 \)[/tex].
Only the second equation [tex]\( d = -9 \cos \left( \frac{2\pi}{3} t \right) \)[/tex] satisfies the condition both at [tex]\( t = 0 \)[/tex] and [tex]\( t = 3 \)[/tex] seconds. Thus, the correct equation that models the distance [tex]\( d \)[/tex] of the weight from its equilibrium after [tex]\( t \)[/tex] seconds is:
[tex]\[ d = -9 \cos \left( \frac{2\pi}{3} t \right) \][/tex]
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