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Answer :
- Set up the equation $36.3 \tan^{-1}(0.8t) = 40$ to find the time $t$ when the 40th gallon is sold.
- Isolate the inverse tangent: $\tan^{-1}(0.8t) = \frac{40}{36.3}$.
- Take the tangent of both sides: $0.8t = \tan(\frac{40}{36.3})$.
- Solve for $t$: $t = \frac{\tan(\frac{40}{36.3})}{0.8} \approx 2.4677$.
- The 40th gallon of strawberries will be sold at $t \approx \boxed{2.4677}$ hours.
### Explanation
1. Problem Setup
We are given the function $b(t) = 36.3 \, \tan^{-1}(0.8t)$ that models the number of gallons of strawberries sold at a local farm, where $t$ is measured in hours and $0 \leq t \leq 8$. We want to find the time $t$ when the 40th gallon of strawberries is sold, which means we need to solve the equation $b(t) = 40$ for $t$.
2. Isolating the Inverse Tangent
We set $b(t) = 40$, so we have $36.3 \tan^{-1}(0.8t) = 40$. To isolate the inverse tangent function, we divide both sides by 36.3:$$\tan^{-1}(0.8t) = \frac{40}{36.3}$$
3. Taking the Tangent of Both Sides
Now, we take the tangent of both sides of the equation to get rid of the inverse tangent:$$0.8t = \tan\left(\frac{40}{36.3}\right)$$
4. Solving for t
To solve for $t$, we divide both sides by 0.8:$$t = \frac{\tan\left(\frac{40}{36.3}\right)}{0.8}$$
5. Calculating the Value of t
Now we calculate the value of $t$. The result of the calculation is approximately $t \approx 2.4677$. Since $0 \leq t \leq 8$, this value is within the given time interval.
6. Final Answer
Therefore, the 40th gallon of strawberries will be sold at approximately $t = 2.4677$ hours.
### Examples
Understanding inverse trigonometric functions and their applications, like in this problem, is crucial in many fields. For instance, in navigation, determining the angle of elevation to a landmark involves inverse trigonometric functions. Similarly, in physics, calculating the angle of refraction of light as it passes through different mediums uses these functions. This problem showcases how mathematical models can predict sales trends, aiding in inventory management and staffing decisions at local farms.
- Isolate the inverse tangent: $\tan^{-1}(0.8t) = \frac{40}{36.3}$.
- Take the tangent of both sides: $0.8t = \tan(\frac{40}{36.3})$.
- Solve for $t$: $t = \frac{\tan(\frac{40}{36.3})}{0.8} \approx 2.4677$.
- The 40th gallon of strawberries will be sold at $t \approx \boxed{2.4677}$ hours.
### Explanation
1. Problem Setup
We are given the function $b(t) = 36.3 \, \tan^{-1}(0.8t)$ that models the number of gallons of strawberries sold at a local farm, where $t$ is measured in hours and $0 \leq t \leq 8$. We want to find the time $t$ when the 40th gallon of strawberries is sold, which means we need to solve the equation $b(t) = 40$ for $t$.
2. Isolating the Inverse Tangent
We set $b(t) = 40$, so we have $36.3 \tan^{-1}(0.8t) = 40$. To isolate the inverse tangent function, we divide both sides by 36.3:$$\tan^{-1}(0.8t) = \frac{40}{36.3}$$
3. Taking the Tangent of Both Sides
Now, we take the tangent of both sides of the equation to get rid of the inverse tangent:$$0.8t = \tan\left(\frac{40}{36.3}\right)$$
4. Solving for t
To solve for $t$, we divide both sides by 0.8:$$t = \frac{\tan\left(\frac{40}{36.3}\right)}{0.8}$$
5. Calculating the Value of t
Now we calculate the value of $t$. The result of the calculation is approximately $t \approx 2.4677$. Since $0 \leq t \leq 8$, this value is within the given time interval.
6. Final Answer
Therefore, the 40th gallon of strawberries will be sold at approximately $t = 2.4677$ hours.
### Examples
Understanding inverse trigonometric functions and their applications, like in this problem, is crucial in many fields. For instance, in navigation, determining the angle of elevation to a landmark involves inverse trigonometric functions. Similarly, in physics, calculating the angle of refraction of light as it passes through different mediums uses these functions. This problem showcases how mathematical models can predict sales trends, aiding in inventory management and staffing decisions at local farms.
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