💡First, I need to understand the costs and revenues involved in the company's operations. The company has a fixed daily maintenance fee of 600 and pays each worker 20 per hour. Each worker can produce 7 pens per hour, and these pens are sold at 2.80 each. The goal is to determine the minimum number of workers required for the company to make a profit over a 9-hour workday.To begin, I'll calculate the total daily cost, which includes both the fixed maintenance fee and the variable cost of paying the workers. The variable cost depends on the number of workers and the number of hours they work. Since each worker is paid 20 per hour and works for 9 hours, the daily wage per worker is 20 multiplied by 9.Next, I'll calculate the total revenue generated from selling the pens. Each worker produces 7 pens per hour, so over 9 hours, one worker produces 7 multiplied by 9 pens. With each pen sold at 2.80, the revenue per worker is the number of pens produced multiplied by the selling price.Once I have both the total cost and total revenue in terms of the number of workers, I can set up an inequality where revenue is greater than cost to find the minimum number of workers needed for the company to make a profit.Finally, I'll solve the inequality to determine the minimum number of workers required.

💡Alright, so I have this infinite series to solve: 1/5 + 2/5² + 3/5³ + ... where the nth term is n/(5^n). Hmm, okay, infinite series can be tricky, but I remember there are some formulas for specific types of series. Let me think about this step by step.First, I recognize that this series isn't a simple geometric series because the numerator increases by 1 each time, while the denominator is 5 raised to the power of n. So, it's more like a weighted geometric series where each term is multiplied by n. I think there's a formula for the sum of a series like this.I recall that for a geometric series, the sum S = a + ar + ar² + ar³ + ..., where |r| < 1, the sum is S = a / (1 - r). But in this case, each term is multiplied by n, so it's a bit different. Maybe there's a variation of the geometric series formula for this kind of series.Wait, yes! I think the formula for the sum of the series where each term is n times r^n is S = r / (1 - r)². Let me verify that. If I have S = r + 2r² + 3r³ + ..., then multiplying both sides by (1 - r) should give me something manageable.Let me try that. Let S = r + 2r² + 3r³ + ... Then, (1 - r)S = S - rS = (r + 2r² + 3r³ + ...) - (r² + 2r³ + 3r⁴ + ...) = r + (2r² - r²) + (3r³ - 2r³) + ... = r + r² + r³ + ... which is a geometric series with first term r and common ratio r. So, (1 - r)S = r / (1 - r), assuming |r| < 1.Therefore, S = [r / (1 - r)] / (1 - r) = r / (1 - r)². Okay, that checks out. So, the formula for the sum of n*r^n from n=1 to infinity is r / (1 - r)².In my problem, the nth term is n/(5^n), which can be written as n*(1/5)^n. So, here, r = 1/5. Plugging that into the formula, S = (1/5) / (1 - 1/5)².Let me compute that step by step. First, compute the denominator: 1 - 1/5 = 4/5. Then, square that: (4/5)² = 16/25. So, the denominator is 16/25. The numerator is 1/5. So, S = (1/5) / (16/25).Dividing by a fraction is the same as multiplying by its reciprocal, so S = (1/5) * (25/16) = (25/80) = 5/16.Let me double-check my calculations to make sure I didn't make a mistake. Starting with S = r / (1 - r)², r = 1/5. So, 1 - r = 4/5, squared is 16/25. Then, 1/5 divided by 16/25 is indeed (1/5)*(25/16) = 5/16. Yep, that seems correct.Just to be thorough, let me consider the first few terms and see if the sum is approaching 5/16. Let's compute the partial sums:First term: 1/5 = 0.2Second term: 1/5 + 2/25 = 0.2 + 0.08 = 0.28Third term: 0.28 + 3/125 ≈ 0.28 + 0.024 = 0.304Fourth term: 0.304 + 4/625 ≈ 0.304 + 0.0064 = 0.3104Fifth term: 0.3104 + 5/3125 ≈ 0.3104 + 0.0016 = 0.312Continuing this, the partial sums seem to be approaching around 0.3125, which is 5/16 because 5 divided by 16 is 0.3125. So, that matches up.Therefore, I'm confident that the sum of the series is 5/16.

💡To find the unknown number, let's denote it as ( x ).Starting with the equation:[ 45 - [ 28 - { 37 - ( x - 16 ) } ] = 55 ]First, simplify the innermost expression:[ 37 - ( x - 16 ) = 37 - x + 16 = 53 - x ]Substitute back into the equation:[ 45 - [ 28 - ( 53 - x ) ] = 55 ]Simplify the expression inside the square brackets:[ 28 - ( 53 - x ) = 28 - 53 + x = -25 + x ]Now, the equation becomes:[ 45 - ( -25 + x ) = 55 ]Simplify further:[ 45 + 25 - x = 55 ][ 70 - x = 55 ]Solve for ( x ):[ 70 - 55 = x ][ x = 15 ]Therefore, the unknown number is ( 15 ).

💡Alright, so I need to find the value of sin(-31π/6). Hmm, okay, let's break this down step by step. First, I remember that sine is a periodic function with a period of 2π, which means that sin(θ) = sin(θ + 2πk) for any integer k. Also, sine is an odd function, so sin(-θ) = -sin(θ). That might come in handy.So, starting with sin(-31π/6). Since sine is odd, I can rewrite this as -sin(31π/6). Now, 31π/6 is a bit more than 5π, which is more than 2π. That means I can subtract multiples of 2π to find an equivalent angle between 0 and 2π. Let's see how many times 2π fits into 31π/6.First, let's express 2π in terms of sixths to make it easier to subtract. 2π is equal to 12π/6. So, 31π/6 minus 12π/6 is 19π/6. That's still more than 2π, so I'll subtract another 12π/6. 19π/6 minus 12π/6 is 7π/6. Okay, 7π/6 is between π and 3π/2, so it's in the third quadrant where sine is negative.But wait, I already have a negative sign from the odd function property, so let me keep track of that. So, sin(-31π/6) = -sin(31π/6) = -sin(7π/6). Now, sin(7π/6) is equal to sin(π + π/6). I remember that sin(π + θ) = -sinθ, so sin(7π/6) = -sin(π/6). Sin(π/6) is 1/2, so sin(7π/6) is -1/2.Putting it all together: sin(-31π/6) = -sin(7π/6) = -(-1/2) = 1/2. Wait, that doesn't seem right because I thought sine in the third quadrant is negative, but I have a negative sign already from the odd function. Maybe I messed up somewhere.Let me double-check. Starting again: sin(-31π/6) = -sin(31π/6). 31π/6 divided by 2π is 31/12, which is approximately 2.583. So, subtracting 2π twice gives 31π/6 - 4π = 31π/6 - 24π/6 = 7π/6. So, sin(31π/6) = sin(7π/6) = -1/2. Therefore, sin(-31π/6) = -(-1/2) = 1/2.But wait, 7π/6 is in the third quadrant where sine is negative, so sin(7π/6) is -1/2. Then, sin(-31π/6) = -sin(31π/6) = -(-1/2) = 1/2. That seems correct. So, the answer should be 1/2, which is option A.Wait, but I initially thought it would be negative because of the negative angle, but the negative sign cancels out. Yeah, that makes sense. So, I think the correct answer is A: 1/2.

💡Okay, so I have this quadratic function, y equals negative x squared plus 2x plus 3. And I need to figure out the relationship between y1, y2, and y3 for the points A(-2, y1), B(2, y2), and C(-4, y3). The options are given as A, B, C, D, each with a different order of y1, y2, y3.First, I remember that quadratic functions graph as parabolas. Since the coefficient of x squared is negative (-1), the parabola opens downward. That means the vertex is the highest point on the graph.To understand where these points lie relative to the vertex, I think I need to find the vertex of this parabola. The general form of a quadratic function is y = ax² + bx + c, and the x-coordinate of the vertex is at -b/(2a). In this case, a is -1 and b is 2. So, plugging in, the x-coordinate of the vertex is -2/(2*(-1)) which is -2/-2, so that's 1. So the vertex is at x = 1.Now, since the parabola opens downward, the further a point is from the vertex along the x-axis, the lower its y-value will be. So, points closer to x = 1 will have higher y-values, and points further away will have lower y-values.Let me note the x-coordinates of the given points:- Point A is at x = -2- Point B is at x = 2- Point C is at x = -4I need to find how far each of these x-values is from the vertex at x = 1.For Point A: The distance from x = 1 to x = -2 is |1 - (-2)| = |1 + 2| = 3 units.For Point B: The distance from x = 1 to x = 2 is |1 - 2| = | -1 | = 1 unit.For Point C: The distance from x = 1 to x = -4 is |1 - (-4)| = |1 + 4| = 5 units.So, the distances are:- Point A: 3 units- Point B: 1 unit- Point C: 5 unitsSince the parabola opens downward, the point closest to the vertex (Point B at 1 unit) will have the highest y-value, followed by Point A at 3 units, and then Point C at 5 units, which is the farthest.Therefore, the order of y-values should be y2 > y1 > y3.Let me double-check by plugging in the x-values into the equation to find the actual y-values.For Point A (-2, y1):y1 = -(-2)² + 2*(-2) + 3 = -4 - 4 + 3 = -8 + 3 = -5For Point B (2, y2):y2 = -(2)² + 2*(2) + 3 = -4 + 4 + 3 = 0 + 3 = 3For Point C (-4, y3):y3 = -(-4)² + 2*(-4) + 3 = -16 - 8 + 3 = -24 + 3 = -21So, the actual y-values are:- y1 = -5- y2 = 3- y3 = -21Comparing these, y2 is the highest at 3, then y1 at -5, and y3 is the lowest at -21. So, the order is y2 > y1 > y3, which matches what I concluded earlier based on the distances from the vertex.Therefore, the correct answer should be option B: y2 > y1 > y3.**Final Answer**boxed{B}

💡Okay, so I have this problem about an acute triangle ABC. The sides a and b are given as 3 and 4 respectively, and the area of the triangle is 3 times the square root of 3. I need to find angle C. Hmm, let's see how to approach this.First, I remember that the area of a triangle can be calculated using the formula involving two sides and the sine of the included angle. The formula is: Area = (1/2) * a * b * sin(C)Where a and b are two sides, and C is the angle between them. So in this case, sides a and b are 3 and 4, and the area is 3√3. Let me plug these values into the formula.So, 3√3 = (1/2) * 3 * 4 * sin(C)Let me compute the right side step by step. First, multiply 1/2, 3, and 4. 1/2 * 3 = 1.5, and 1.5 * 4 = 6. So, the equation becomes:3√3 = 6 * sin(C)Now, I need to solve for sin(C). To do that, I'll divide both sides of the equation by 6.sin(C) = (3√3) / 6Simplify that fraction. 3 divided by 6 is 0.5, so:sin(C) = (√3)/2Okay, so sin(C) is √3 over 2. I remember that sin(60°) is √3/2, and sin(120°) is also √3/2. But wait, the triangle is acute, which means all angles are less than 90 degrees. So angle C must be 60 degrees, not 120 degrees.Let me just double-check my steps to make sure I didn't make a mistake. 1. I used the area formula correctly: (1/2)*a*b*sin(C).2. Plugged in a=3, b=4, and area=3√3.3. Calculated 1/2*3*4=6, so 3√3=6*sin(C).4. Divided both sides by 6 to get sin(C)=√3/2.5. Recognized that sin(60°)=√3/2 and since the triangle is acute, angle C is 60°, which is π/3 radians.Everything seems to check out. So, angle C is 60 degrees or π/3 radians. Just to be thorough, let me consider if there's any other possible angle. Since sine is positive in both the first and second quadrants, the angle could technically be 60° or 120°. But since the triangle is acute, all angles must be less than 90°, so 120° is out of the question. Therefore, angle C must indeed be 60°.I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**The measure of angle ( C ) is boxed{dfrac{pi}{3}}.