🧠:<think>To determine the fewest number of times you must press the reciprocal key boxed{frac{1}{x}} to return the display to boxed{00050}, follow these steps:1. Understand the Reciprocal Function: The reciprocal of a number x is frac{1}{x}. Applying the reciprocal function twice returns the original number because frac{1}{frac{1}{x}} = x.2. Apply the Reciprocal Function: - Start with the number 50. - Press the reciprocal key once: frac{1}{50}. - Press the reciprocal key again: frac{1}{frac{1}{50}} = 50.3. Conclusion: After pressing the reciprocal key twice, the display returns to 50.Therefore, the fewest number of times you must press the reciprocal key is 2.</think>

🧠:<think>Okay, so I have this problem where I need to determine whether the condition m = -2 is sufficient, necessary, both, or neither for two lines to be perpendicular. The lines are given by the equations (m+2)x + my + 1 = 0 and (m-2)x + (m+2)y - 3 = 0. First, I remember that two lines are perpendicular if the product of their slopes is -1. So, I think I need to find the slopes of these two lines and then set up an equation where their product equals -1. Let me start by finding the slope of each line. For the first line, (m+2)x + my + 1 = 0, I can rewrite it in slope-intercept form (y = mx + b) to find the slope. So, rearranging the first equation:my = -(m+2)x - 1 Dividing both sides by m (assuming m ≠ 0):y = [-(m+2)/m]x - 1/m So, the slope (let's call it m1) is -(m+2)/m.Now, for the second line, (m-2)x + (m+2)y - 3 = 0. Similarly, I'll rearrange it into slope-intercept form.(m+2)y = -(m-2)x + 3 Dividing both sides by (m+2) (assuming m ≠ -2):y = [-(m-2)/(m+2)]x + 3/(m+2) So, the slope (let's call it m2) is -(m-2)/(m+2).Now, for the lines to be perpendicular, the product of their slopes should be -1. So, m1 * m2 = -1.Let's compute that:[-(m+2)/m] * [-(m-2)/(m+2)] = -1 Multiplying the numerators and denominators:[(m+2)(m-2)] / [m(m+2)] = -1 Simplify the numerator: (m+2)(m-2) is m² - 4 Denominator: m(m+2) So, (m² - 4)/(m(m+2)) = -1Now, let's solve this equation:(m² - 4)/(m(m+2)) = -1 Multiply both sides by m(m+2) to eliminate the denominator:m² - 4 = -m(m + 2) Expand the right side:m² - 4 = -m² - 2m Bring all terms to one side:m² - 4 + m² + 2m = 0 Combine like terms:2m² + 2m - 4 = 0 Divide the entire equation by 2 to simplify:m² + m - 2 = 0 Now, factor the quadratic:Looking for two numbers that multiply to -2 and add to 1. Those numbers are 2 and -1.So, (m + 2)(m - 1) = 0 Thus, m = -2 or m = 1.So, the solutions are m = -2 and m = 1. This means that for the lines to be perpendicular, m can be either -2 or 1. Now, the question is about the condition m = -2. Is this condition sufficient, necessary, both, or neither?Since m = -2 is one of the solutions that makes the lines perpendicular, it is a sufficient condition. That is, if m = -2, then the lines are definitely perpendicular. However, it's not the only solution because m = 1 also works. Therefore, m = -2 is not a necessary condition because there's another value of m that also satisfies the perpendicularity.So, m = -2 is sufficient but not necessary.Wait, let me double-check my calculations to make sure I didn't make any mistakes.Starting from the slopes:First line: y = [-(m+2)/m]x - 1/m, so slope m1 = -(m+2)/m Second line: y = [-(m-2)/(m+2)]x + 3/(m+2), so slope m2 = -(m-2)/(m+2)Product of slopes:[-(m+2)/m] * [-(m-2)/(m+2)] = [(m+2)(m-2)] / [m(m+2)] = (m² - 4)/(m² + 2m)Set equal to -1:(m² - 4)/(m² + 2m) = -1 Multiply both sides by denominator:m² - 4 = -m² - 2m Bring all terms to left:2m² + 2m - 4 = 0 Divide by 2:m² + m - 2 = 0 Factor:(m + 2)(m - 1) = 0 Solutions: m = -2, m = 1Yes, that seems correct. So, m = -2 is one solution, but there's another. Therefore, m = -2 is sufficient but not necessary.I think that's solid. So, the answer should be that m = -2 is a sufficient but not necessary condition.</think>

🧠:<think>First, I need to determine the total number of pears picked by Alyssa and Nancy.Alyssa picked 42 pears, and Nancy picked 17 pears.To find the total, I add the number of pears picked by each person: 42 + 17.Calculating the sum, 42 plus 17 equals 59.Therefore, the total number of pears picked is 59.</think>

🧠:<think>First, I need to understand the problem. Petya planned to run the entire distance at a constant speed ( V ). However, he decided to increase his speed by 25% for the first half of the distance and then decrease his speed by 20% for the second half. I need to determine if the actual time he took was more or less than the planned time.To start, I'll denote the total distance as ( a ). The planned time to complete the distance at speed ( V ) is straightforward:[T_{text{planned}} = frac{a}{V}]Next, for the first half of the distance, Petya increased his speed by 25%. This means his new speed is ( 1.25V ). The time taken to cover the first half of the distance at this increased speed is:[T_1 = frac{frac{a}{2}}{1.25V} = frac{a}{2 times 1.25V} = frac{a}{2.5V}]For the second half of the distance, Petya's speed decreased by 20% compared to his original speed ( V ). Therefore, his speed for the second half is ( 0.8V ). The time taken to cover the second half at this reduced speed is:[T_2 = frac{frac{a}{2}}{0.8V} = frac{a}{2 times 0.8V} = frac{a}{1.6V}]Now, I'll add the times for both halves to find the total actual time:[T_{text{real}} = T_1 + T_2 = frac{a}{2.5V} + frac{a}{1.6V}]To simplify this, I'll find a common denominator for the two fractions. The common denominator for 2.5 and 1.6 is 40. Converting both fractions:[frac{a}{2.5V} = frac{16a}{40V} quad text{and} quad frac{a}{1.6V} = frac{25a}{40V}]Adding these together:[T_{text{real}} = frac{16a}{40V} + frac{25a}{40V} = frac{41a}{40V}]Comparing this to the planned time:[T_{text{planned}} = frac{a}{V} = frac{40a}{40V}]Since ( frac{41a}{40V} > frac{40a}{40V} ), the actual time is greater than the planned time.Therefore, Petya's actual time was more than the planned time.</think>