💡Okay, so John bought a used car for 6500 to help with his pizza delivery job. He gets 12 for each pizza he delivers, but he also has to spend some money on gas and maintenance. Specifically, he spends 4 on gas and 1 on maintenance for each pizza. I need to figure out the minimum number of pizzas he needs to deliver to earn back the 6500 he spent on the car.First, let me break down his earnings and expenses. For each pizza, he earns 12. But then he has to subtract the costs. The gas is 4 per pizza, and maintenance is 1 per pizza. So, his net earnings per pizza would be 12 minus 4 minus 1. Let me write that down:Net earnings per pizza = 12 - 4 - 1 = 7.So, for each pizza he delivers, he makes a net profit of 7. Now, he needs to make back the 6500 he spent on the car. So, I need to find out how many pizzas he needs to deliver so that his total net earnings equal 6500.Let me denote the number of pizzas as 'p'. Then, his total net earnings would be 7 times p, which is 7p. We want this to be at least 6500, so:7p ≥ 6500.To find the minimum number of pizzas, I need to solve for p. So, I'll divide both sides of the inequality by 7:p ≥ 6500 / 7.Let me calculate that. 6500 divided by 7. Hmm, 7 times 900 is 6300, which is less than 6500. 7 times 928 is 6496, which is still less than 6500. 7 times 929 is 6503, which is just over 6500. So, p needs to be at least 929 pizzas.Wait, let me double-check that. 928 times 7 is 6496, which is 4 less than 6500. So, 928 pizzas would only give him 6496, which isn't enough. Therefore, he needs to deliver one more pizza, which is 929, to make 6503, which covers the 6500 cost.So, the minimum number of pizzas John must deliver is 929.

💡Okay, so I have this problem where I need to roll a rectangle with length 6 and width 4 into a cylinder and find the volume of that cylinder. Hmm, rolling a rectangle into a cylinder... I think that means I'm going to form the sides of the rectangle into the circular base of the cylinder. But wait, how exactly does that work?Let me visualize this. If I have a rectangle, and I roll it along one of its sides, the side that I roll along will become the circumference of the base of the cylinder, and the other side will become the height of the cylinder. So, there are two possibilities here: either I roll the rectangle along its length of 6, making the circumference 6 and the height 4, or I roll it along its width of 4, making the circumference 4 and the height 6.Alright, so I need to consider both cases and calculate the volume for each. The formula for the volume of a cylinder is V = πr²h, where r is the radius of the base and h is the height. But first, I need to find the radius from the circumference.The circumference of a circle is given by C = 2πr, so if I know the circumference, I can solve for the radius. Let's start with the first case where the circumference is 6.Case 1: Circumference C = 6So, C = 2πr = 6Solving for r: r = 6 / (2π) = 3 / πNow, the height h is 4. Plugging these into the volume formula:V = πr²h = π*(3/π)²*4First, square the radius: (3/π)² = 9 / π²Then multiply by π: π*(9 / π²) = 9 / πFinally, multiply by the height: (9 / π)*4 = 36 / πOkay, so the volume in this case is 36/π.Case 2: Circumference C = 4Again, using C = 2πr = 4Solving for r: r = 4 / (2π) = 2 / πThe height h is now 6. Plugging into the volume formula:V = πr²h = π*(2/π)²*6Square the radius: (2/π)² = 4 / π²Multiply by π: π*(4 / π²) = 4 / πMultiply by the height: (4 / π)*6 = 24 / πSo, the volume in this case is 24/π.Wait, so depending on which side I roll the rectangle along, I get two different volumes: 36/π and 24/π. The problem doesn't specify which side to roll, so I think both answers are possible. Therefore, the volume of the cylinder can be either 24/π or 36/π.Let me double-check my calculations to make sure I didn't make any mistakes.For Case 1:C = 6, so r = 3/πV = π*(3/π)²*4 = π*(9/π²)*4 = (9/π)*4 = 36/πThat seems correct.For Case 2:C = 4, so r = 2/πV = π*(2/π)²*6 = π*(4/π²)*6 = (4/π)*6 = 24/πThat also looks correct.I think I've covered both possibilities, so the volumes are indeed 24/π and 36/π.

💡Okay, so I have this problem about checking divisibility by 7. It's part a and part b. Let me start with part a.**Problem a: Check if 4,578 is divisible by 7 using the given method.**Alright, the method says: Take the last digit, double it, subtract it from the rest of the number, and see if the result is divisible by 7. If it is, then the original number is divisible by 7. If not, then it's not.So, let's apply this step by step.1. The number is 4,578. The last digit is 8.2. Double the last digit: 8 * 2 = 16.3. Subtract this from the remaining number: 457 - 16 = 441.Now, I need to check if 441 is divisible by 7. Hmm, 441 divided by 7... Let me do that division.441 ÷ 7 = 63. Oh, that's a whole number, so 441 is divisible by 7. Therefore, 4,578 should also be divisible by 7.Wait, just to make sure, maybe I should do the division to confirm.4,578 ÷ 7. Let's see:7 goes into 45 six times (6*7=42), remainder 3. Bring down the 7: 37.7 goes into 37 five times (5*7=35), remainder 2. Bring down the 8: 28.7 goes into 28 four times (4*7=28), no remainder.So, 4,578 ÷ 7 = 654, which is a whole number. Yep, so it is divisible by 7.**Problem b: How many three-digit numbers of the form AB5 are divisible by 7?**Okay, so AB5 means a three-digit number where A and B are digits, and the last digit is 5. So, numbers like 115, 125, 135, ..., up to 995.We need to find how many such numbers are divisible by 7. Using the same method as in part a.The method says: Take the last digit, double it, subtract from the rest, and check divisibility by 7.So, for a number AB5:1. Last digit is 5.2. Double it: 5*2 = 10.3. Subtract this from the remaining number AB: AB - 10.So, AB - 10 must be divisible by 7 for AB5 to be divisible by 7.Therefore, AB must be equal to 10 + 7k, where k is an integer, such that AB is a two-digit number (from 10 to 99).So, AB = 10 + 7k.Let me find all possible AB values.Start with k=0: AB=10. But AB is a two-digit number, so A can't be zero. Wait, AB=10 is allowed because A=1 and B=0. So, 10 is okay.k=1: AB=17k=2: AB=24k=3: AB=31k=4: AB=38k=5: AB=45k=6: AB=52k=7: AB=59k=8: AB=66k=9: AB=73k=10: AB=80k=11: AB=87k=12: AB=94k=13: AB=101. Wait, AB must be a two-digit number, so 101 is too big. So, k can go up to 12.Let me list all AB values:10, 17, 24, 31, 38, 45, 52, 59, 66, 73, 80, 87, 94.Let me count them: 10,17,24,31,38,45,52,59,66,73,80,87,94. That's 13 numbers.So, there are 13 such AB values, meaning 13 three-digit numbers of the form AB5 that are divisible by 7.Wait, let me double-check. For each AB, AB5 should be divisible by 7.Let me pick a few examples:- AB=10: 105 ÷7=15, which is whole.- AB=17: 175 ÷7=25, whole.- AB=24: 245 ÷7=35, whole.- AB=31: 315 ÷7=45, whole.- AB=38: 385 ÷7=55, whole.- AB=45: 455 ÷7=65, whole.- AB=52: 525 ÷7=75, whole.- AB=59: 595 ÷7=85, whole.- AB=66: 665 ÷7=95, whole.- AB=73: 735 ÷7=105, whole.- AB=80: 805 ÷7=115, whole.- AB=87: 875 ÷7=125, whole.- AB=94: 945 ÷7=135, whole.All of these divisions result in whole numbers, so it seems correct.Therefore, the answer is 13.**Final Answer**a) boxed{Yes}b) boxed{13}

💡Alright, so I have this problem here: Find cos(4 arccos(1/4)). Hmm, okay. Let me try to figure this out step by step. I remember that when dealing with expressions like cos(n arccos(x)), it often involves using multiple angle formulas or maybe some trigonometric identities. Let me recall what I know.First, let's break down the problem. I need to find the cosine of four times the arccosine of 1/4. So, if I let θ = arccos(1/4), then the problem becomes finding cos(4θ). That seems manageable because I can use multiple angle identities for cosine.So, θ = arccos(1/4) implies that cosθ = 1/4. That's straightforward. Now, I need to find cos(4θ). I remember that there are formulas for cos(2θ), cos(3θ), and so on. Since 4θ is a multiple of 2θ, maybe I can use the double angle formula twice. Let me try that.First, I'll find cos(2θ) using the double angle formula. The formula is cos(2θ) = 2cos²θ - 1. Plugging in cosθ = 1/4, I get:cos(2θ) = 2*(1/4)² - 1 = 2*(1/16) - 1 = 2/16 - 1 = 1/8 - 1 = -7/8.Okay, so cos(2θ) is -7/8. Now, I need to find cos(4θ), which is cos(2*(2θ)). I can apply the double angle formula again, this time using cos(2φ) where φ = 2θ. So, cos(4θ) = 2cos²(2θ) - 1.Plugging in cos(2θ) = -7/8, I get:cos(4θ) = 2*(-7/8)² - 1 = 2*(49/64) - 1 = 98/64 - 1.Hmm, 98/64 simplifies to 49/32. So, 49/32 - 1 is the same as 49/32 - 32/32 = 17/32.Wait, so cos(4θ) is 17/32? Let me double-check my steps to make sure I didn't make a mistake.Starting with θ = arccos(1/4), so cosθ = 1/4. Then, cos(2θ) = 2*(1/4)^2 - 1 = 2*(1/16) - 1 = 1/8 - 1 = -7/8. That seems right.Then, cos(4θ) = 2*(-7/8)^2 - 1 = 2*(49/64) - 1 = 98/64 - 64/64 = 34/64 = 17/32. Yeah, that checks out.Alternatively, I could use the multiple-angle formula for cos(4θ) directly. I think there's a formula that expresses cos(4θ) in terms of cosθ. Let me recall it.I believe it's cos(4θ) = 8cos⁴θ - 8cos²θ + 1. Let me verify that. Starting from cos(2θ) = 2cos²θ - 1, then cos(4θ) = 2cos²(2θ) - 1. Substituting cos(2θ) = 2cos²θ - 1, we get:cos(4θ) = 2*(2cos²θ - 1)^2 - 1.Expanding that, (2cos²θ - 1)^2 = 4cos⁴θ - 4cos²θ + 1. Then, multiplying by 2: 8cos⁴θ - 8cos²θ + 2. Subtracting 1 gives 8cos⁴θ - 8cos²θ + 1. Okay, so that's the correct formula.Using this formula with cosθ = 1/4, let's compute cos(4θ):cos(4θ) = 8*(1/4)^4 - 8*(1/4)^2 + 1.Calculating each term:(1/4)^4 = 1/256, so 8*(1/256) = 8/256 = 1/32.(1/4)^2 = 1/16, so 8*(1/16) = 8/16 = 1/2.Putting it all together:cos(4θ) = 1/32 - 1/2 + 1.Converting to a common denominator, which is 32:1/32 - 16/32 + 32/32 = (1 - 16 + 32)/32 = 17/32.Same result as before. So, that's reassuring. It seems like 17/32 is the correct answer.Just to make sure I haven't overlooked anything, let me think about another approach. Maybe using Euler's formula or something with complex numbers? Although that might be overcomplicating things.Euler's formula states that e^(iθ) = cosθ + i sinθ. So, if I can express cos(4θ) in terms of e^(iθ), maybe I can compute it. But that might not be necessary here since the double angle method worked fine.Alternatively, I could use the power-reduction formulas or other trigonometric identities, but I think the double angle approach is the most straightforward here.Wait, another thought: maybe using the cosine of multiple angles formula directly, like cos(4θ) = 4cosθ cos(2θ) - cos(2θ)? Hmm, no, that doesn't sound right. Let me recall the correct identity.Actually, the formula for cos(4θ) can be expressed as 2cos²(2θ) - 1, which is what I used earlier. So, I think I'm safe with that method.Just to recap:1. Let θ = arccos(1/4), so cosθ = 1/4.2. Compute cos(2θ) using the double angle formula: cos(2θ) = 2cos²θ - 1 = -7/8.3. Compute cos(4θ) using the double angle formula again: cos(4θ) = 2cos²(2θ) - 1 = 17/32.All steps seem logical and consistent. I don't see any calculation errors upon reviewing. Both methods—using the double angle formula twice and using the direct formula for cos(4θ)—give the same result, which adds confidence in the answer.Therefore, I conclude that cos(4 arccos(1/4)) is equal to 17/32.**Final Answer**boxed{dfrac{17}{32}}

💡Okay, so I have this integral to solve: the integral of (5x² - 2x³) with respect to x. Hmm, I remember from my calculus class that integrals can be split up, especially when there's addition or subtraction involved. So, maybe I can break this integral into two separate integrals and handle them individually. Let me think. The integral of a sum is the sum of the integrals, right? So, if I have ∫(a + b) dx, that's the same as ∫a dx + ∫b dx. In this case, my function inside the integral is 5x² minus 2x³. So, I can split this into two separate integrals: one for 5x² and another for -2x³. Wait, but there's a negative sign in front of the 2x³. Does that mean I can just factor that negative sign out of the integral? I think so. So, it would be ∫5x² dx minus ∫2x³ dx. That makes sense because the integral of a difference is the difference of the integrals. Now, I also remember that constants can be pulled out in front of the integral sign. So, for the first integral, I can take the 5 out, and for the second one, I can take the 2 out. That would give me 5 times the integral of x² dx minus 2 times the integral of x³ dx. Let me write that down to make sure I'm on the right track: 5∫x² dx - 2∫x³ dx. Yeah, that seems correct. I don't think I missed anything here. I wonder if I need to compute these integrals further or if just expressing them as the sum is enough. The problem says to express the integral as the algebraic sum of integrals, so I think I've done what's required. But just to double-check, let me recall the power rule for integration. The integral of xⁿ dx is (xⁿ⁺¹)/(n+1) plus the constant of integration, right? So, if I were to compute these integrals, I could apply that rule. For the first integral, ∫x² dx would be (x³)/3, and multiplying by 5 gives (5x³)/3. For the second integral, ∫x³ dx is (x⁴)/4, and multiplying by 2 gives (2x⁴)/4, which simplifies to (x⁴)/2. So, putting it all together, the integral of (5x² - 2x³) dx would be (5x³)/3 - (x⁴)/2 + C, where C is the constant of integration. But wait, the problem didn't ask me to compute the integral, just to express it as the sum of integrals. So, maybe I went a step further than necessary. However, going through the process of computing the integrals helped me confirm that splitting the original integral into two separate integrals was the correct approach. It also reinforced my understanding of how constants and coefficients work within integrals. I guess another way to think about it is that integration is a linear operator, which means it distributes over addition and subtraction, and constants can be factored out. So, applying these properties directly gives me the desired expression: 5∫x² dx - 2∫x³ dx. I don't see any mistakes in my reasoning, but let me think if there's another perspective. Maybe I can visualize the function 5x² - 2x³ and think about what its integral represents. The integral would give me the area under the curve of this function, and by splitting it into two parts, I'm essentially finding the areas under 5x² and -2x³ separately and then combining them. That makes sense because the integral of a function can be thought of as the accumulation of its values over an interval, and breaking it down into simpler parts makes the problem more manageable. So, to summarize, the integral ∫(5x² - 2x³) dx can be expressed as the algebraic sum of two integrals: 5 times the integral of x² dx minus 2 times the integral of x³ dx. This approach uses the linearity of integrals and the ability to factor out constants, which are fundamental properties in calculus. I think I've got it. Expressing the integral as the sum of simpler integrals not only simplifies the problem but also aligns with the principles I've learned in my calculus course. It's a good reminder of how breaking down complex expressions can make them easier to handle.

💡First, I'll define the variables for the rates of Machine A and Machine B. Let ( A ) represent the number of sprockets Machine A produces per hour, and ( B ) represent the number of sprockets Machine B produces per hour.According to the problem, Machine B produces 10% more sprockets per hour than Machine A. This means:[ B = 1.10A ]Next, I'll consider the time it takes for each machine to produce 220 sprockets. Let ( T_A ) be the time for Machine A and ( T_B ) be the time for Machine B. The relationship between their production times is:[ T_A = T_B + 10 ]Using the rate and time relationship for each machine:[ 220 = A cdot T_A ][ 220 = B cdot T_B ]Substituting ( B = 1.10A ) into the second equation:[ 220 = 1.10A cdot T_B ]Solving for ( T_B ):[ T_B = frac{220}{1.10A} = frac{200}{A} ]Now, substitute ( T_B ) back into the time difference equation:[ T_A = frac{200}{A} + 10 ]But from the first equation, ( T_A = frac{220}{A} ). Setting these equal:[ frac{220}{A} = frac{200}{A} + 10 ]Subtracting (frac{200}{A}) from both sides:[ frac{20}{A} = 10 ]Solving for ( A ):[ A = frac{20}{10} = 2 ]Therefore, Machine A produces 2 sprockets per hour.