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The average monthly temperature of the water in a mountain stream is modeled by the function , where T(x) is the temperature of the water in °F in month x (x = 1 → January).Solve algebraically to find which two months are most likely to give a temperature reading of 62°F.

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The average velocity of the stream is calculated as 5,000 m/s by dividing the given flow rate of 100,000 m³/s by the cross-sectional area of 20.0 m². However, this velocity is unreasonable for natural water systems, indicating a potential error in the data or premises. To find the two months with a temperature reading of 62°F, solve the equation T(x) = 62 for x.

To find which two months are most likely to give a temperature reading of 62°F, we need to solve the equation T(x) = 62, where T(x) is the temperature in °F in month x. In other words, we need to find the values of x for which T(x) = 62.

Given the function T(x), we can set it equal to 62 and solve for x algebraically.

Once we have the values of x, we can determine which months they correspond to. These two months are the most likely to have a temperature reading of 62°F.

To solve for the average velocity of the water in a mountain stream, we can use the formula of flow rate, which is the product of cross-sectional area (A) of the stream and the velocity (v) of the water: Q = A * v. Given a flow rate (Q) of 100,000 m³/s, a width of 10.0 m, and an average depth of 2.00 m, we can calculate the cross-sectional area as A = width * depth = 10.0 m * 2.00 m = 20.0 m².

Substituting the values into the formula, we get Q = 20.0 m² * v, which simplifies to v = Q / A. Plugging in the given flow rate, we have v = 100,000 m³/s / 20.0 m² = 5,000 m/s. However, this velocity is unreasonable because it is extremely high for any natural river, suggesting that there is either a mistake in the given data or the premise is not realistic. In reality, a velocity as high as 5,000 m/s would imply a catastrophic event or an incorrect flow rate measurement.

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