The base of a triangle is decreasing at a rate of 13 millimeters per minute and the height of the triangle is increasing at a rate of 6 millimeters per minute. At a certain instant, the base is 5 millimeters and the height is 1 millimeter. What is the rate of change of the area of the triangle at that instant (in square millimeters per minute)? Choose 1 answer: (A) -8.5 (B) 8.5 (C) -21.5 (D) 21.5

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Area Formula: The area of a triangle is given by the formula A = 1/2 * base * height.Derivative Calculation: Let's denote the base by b and the height by h. The rate of change of the area with respect to time t can be found using the derivative dA/dt = 1/2 * (db/dt * h + b * dh/dt).Given Rates: Given db/dt = -13 mm/min (since the base is decreasing) and dh/dt = 6 mm/min (since the height is increasing).Instant Values: At the instant when the base b is 5 mm and the height h is 1 mm, we plug these values into the derivative formula to find dA/dt.Calculate dA/dt: So, dA/dt = 1/2 * (-13 * 1 + 5 * 6).Final Result: Calculating the above expression gives dA/dt = 1/2 * (-13 + 30).dA/dt = 1/2 * 17 = 8.5 square millimeters per minute.

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