💡Okay, so I have this problem about a department store selling shirts. They currently sell an average of 20 shirts a day with a profit of 40 per shirt. They want to increase sales and profits by reducing the price. For every 1 reduction in price, they can sell 2 more shirts per day. There are two parts to the problem: first, finding out how much they should reduce the price to make an average daily profit of 1200, and second, determining the price reduction that would maximize their average daily profit.Alright, let's start with part (1). I need to figure out how much to reduce the price so that the daily profit is 1200. Let me denote the price reduction per shirt as ( x ) dollars. So, if they reduce the price by ( x ) dollars, the new profit per shirt would be ( 40 - x ) dollars. Now, for each 1 reduction, they sell 2 more shirts. So, if they reduce the price by ( x ) dollars, they would sell ( 20 + 2x ) shirts per day. That makes sense because for every dollar they reduce, they get two more sales. So, the total daily profit ( P ) would be the new profit per shirt multiplied by the new number of shirts sold. That gives me the equation:[P = (40 - x)(20 + 2x)]They want this profit to be 1200, so I set up the equation:[(40 - x)(20 + 2x) = 1200]Let me expand this equation to solve for ( x ). First, I'll multiply out the terms:[40 times 20 + 40 times 2x - x times 20 - x times 2x = 1200]Calculating each term:[800 + 80x - 20x - 2x^2 = 1200]Combine like terms:[800 + 60x - 2x^2 = 1200]Now, let's move all terms to one side to form a quadratic equation:[-2x^2 + 60x + 800 - 1200 = 0]Simplify:[-2x^2 + 60x - 400 = 0]I can divide the entire equation by -2 to make it simpler:[x^2 - 30x + 200 = 0]Now, I need to solve this quadratic equation. Let's try factoring. I'm looking for two numbers that multiply to 200 and add up to -30. Hmm, 10 and 20 multiply to 200, and 10 + 20 = 30. Since the middle term is -30x, both numbers should be negative:[(x - 10)(x - 20) = 0]So, the solutions are ( x = 10 ) and ( x = 20 ). Wait, so does that mean they can reduce the price by either 10 or 20 to reach a profit of 1200? Let me check both solutions to make sure.First, if ( x = 10 ):- New profit per shirt: ( 40 - 10 = 30 ) dollars- New number of shirts sold: ( 20 + 2 times 10 = 40 ) shirts- Total profit: ( 30 times 40 = 1200 ) dollarsThat works.Now, if ( x = 20 ):- New profit per shirt: ( 40 - 20 = 20 ) dollars- New number of shirts sold: ( 20 + 2 times 20 = 60 ) shirts- Total profit: ( 20 times 60 = 1200 ) dollarsThat also works.So, both price reductions of 10 and 20 result in a daily profit of 1200. But the problem mentions wanting to "reduce inventory quickly." Selling more shirts would help reduce inventory faster. So, reducing the price by 20 would lead to selling 60 shirts a day, which is more than selling 40 shirts a day with a 10 reduction. Therefore, to reduce inventory quickly, they should reduce the price by 20.Alright, that seems to make sense. So, for part (1), the price should be reduced by 20.Now, moving on to part (2): finding the price reduction that would maximize the average daily profit. From part (1), we have the profit equation:[P = (40 - x)(20 + 2x)]Which simplifies to:[P = -2x^2 + 60x + 800]This is a quadratic equation in the form of ( P = ax^2 + bx + c ), where ( a = -2 ), ( b = 60 ), and ( c = 800 ). Since the coefficient of ( x^2 ) is negative, the parabola opens downward, meaning the vertex is the maximum point.The vertex of a parabola given by ( ax^2 + bx + c ) is at ( x = -frac{b}{2a} ). Plugging in the values:[x = -frac{60}{2 times -2} = -frac{60}{-4} = 15]So, the price reduction that would maximize the profit is 15.Let me verify this by plugging ( x = 15 ) back into the profit equation:[P = -2(15)^2 + 60(15) + 800 = -2(225) + 900 + 800 = -450 + 900 + 800 = 1250]So, the maximum profit is 1250 when the price is reduced by 15.Just to be thorough, let me check the profits around 15 to ensure it's indeed the maximum.If ( x = 14 ):[P = -2(14)^2 + 60(14) + 800 = -2(196) + 840 + 800 = -392 + 840 + 800 = 1248]If ( x = 16 ):[P = -2(16)^2 + 60(16) + 800 = -2(256) + 960 + 800 = -512 + 960 + 800 = 1248]So, at both 14 and 16 reductions, the profit is 1248, which is less than 1250. This confirms that 15 is indeed the price reduction that maximizes the profit.Therefore, for part (2), the price should be reduced by 15 to achieve the maximum average daily profit of 1250.**Final Answer**(1) The price should be reduced by boxed{20} dollars.(2) The price should be reduced by boxed{15} dollars to achieve the maximum average daily profit.

💡First, I need to determine the actual dimensions of Mr. Green's garden in feet. Since his stride length is 30 inches, which is equivalent to 2.5 feet, I can calculate the garden's length and width in feet by multiplying the number of steps by the stride length.Next, I'll calculate the area of the garden by multiplying its length and width in feet. This will give me the total area in square feet.After finding the area, I'll compare it to the threshold of 1200 square feet to determine the appropriate yield rate. If the area is more than 1200 square feet, the yield rate is 0.6 pounds per square foot; otherwise, it's 0.4 pounds per square foot.Finally, I'll multiply the area by the applicable yield rate to find the total expected potato yield from the garden.

💡First, I need to divide 54 into two parts, which I'll call ( x ) and ( y ).From the problem, I know that:[ x + y = 54 ]I also know that:[ 10x + 22y = 780 ]To solve this system of equations, I'll express ( x ) in terms of ( y ) from the first equation:[ x = 54 - y ]Next, I'll substitute this expression for ( x ) into the second equation:[ 10(54 - y) + 22y = 780 ]Expanding and simplifying:[ 540 - 10y + 22y = 780 ][ 540 + 12y = 780 ][ 12y = 240 ][ y = 20 ]Now, I'll find ( x ) using the value of ( y ):[ x = 54 - 20 = 34 ]The two parts are 34 and 20, with 34 being the bigger part.

💡Okay, so I have this problem where there's an arithmetic sequence {a_n} and a geometric sequence {b_n}. The arithmetic sequence starts with 1 and has a common difference of 2. The geometric sequence also starts with 1 but has a common ratio of 2. I need to find the sum of the products of their corresponding terms from the first term up to the tenth term. That is, compute a₁b₁ + a₂b₂ + … + a₁₀b₁₀.First, let me recall what arithmetic and geometric sequences are. An arithmetic sequence is one where each term increases by a constant difference. In this case, the first term is 1, and each subsequent term is 2 more than the previous one. So, the nth term of an arithmetic sequence can be written as a_n = a₁ + (n - 1)d, where d is the common difference. Plugging in the values, a_n = 1 + (n - 1)*2. Simplifying that, a_n = 1 + 2n - 2, which is 2n - 1. So, a_n = 2n - 1.On the other hand, a geometric sequence is one where each term is multiplied by a constant ratio. Here, the first term is 1, and each subsequent term is twice the previous one. The nth term of a geometric sequence is given by b_n = b₁ * r^(n - 1), where r is the common ratio. So, b_n = 1 * 2^(n - 1), which simplifies to b_n = 2^(n - 1).Now, I need to compute the sum S = a₁b₁ + a₂b₂ + … + a₁₀b₁₀. Let's write out the terms to see if I can spot a pattern or find a way to compute this sum efficiently.First, let's compute a few terms:- a₁ = 2*1 - 1 = 1, b₁ = 2^(1 - 1) = 1, so a₁b₁ = 1*1 = 1- a₂ = 2*2 - 1 = 3, b₂ = 2^(2 - 1) = 2, so a₂b₂ = 3*2 = 6- a₃ = 2*3 - 1 = 5, b₃ = 2^(3 - 1) = 4, so a₃b₃ = 5*4 = 20- a₄ = 2*4 - 1 = 7, b₄ = 2^(4 - 1) = 8, so a₄b₄ = 7*8 = 56- a₅ = 2*5 - 1 = 9, b₅ = 2^(5 - 1) = 16, so a₅b₅ = 9*16 = 144- a₆ = 2*6 - 1 = 11, b₆ = 2^(6 - 1) = 32, so a₆b₆ = 11*32 = 352- a₇ = 2*7 - 1 = 13, b₇ = 2^(7 - 1) = 64, so a₇b₇ = 13*64 = 832- a₈ = 2*8 - 1 = 15, b₈ = 2^(8 - 1) = 128, so a₈b₈ = 15*128 = 1920- a₉ = 2*9 - 1 = 17, b₉ = 2^(9 - 1) = 256, so a₉b₉ = 17*256 = 4352- a₁₀ = 2*10 - 1 = 19, b₁₀ = 2^(10 - 1) = 512, so a₁₀b₁₀ = 19*512 = 9728Now, if I add all these up:1 + 6 = 77 + 20 = 2727 + 56 = 8383 + 144 = 227227 + 352 = 579579 + 832 = 14111411 + 1920 = 33313331 + 4352 = 76837683 + 9728 = 17411So, the sum is 17,411. But computing each term individually and adding them up seems tedious, especially for larger n. There must be a smarter way to compute this sum without having to calculate each term.Let me think about the general term. The nth term of the sum is a_n * b_n = (2n - 1) * 2^(n - 1). So, the sum S = Σ (from n=1 to 10) (2n - 1) * 2^(n - 1).This looks like a sum that can be tackled using the method of generating functions or perhaps by recognizing it as a telescoping series. Alternatively, I might recall that there's a formula for sums of the form Σ n * r^n, which is similar to what we have here.Let me try to manipulate the sum S. Let's write S as:S = Σ (2n - 1) * 2^(n - 1) from n=1 to 10.I can split this into two separate sums:S = 2 Σ n * 2^(n - 1) - Σ 2^(n - 1) from n=1 to 10.So, S = 2S₁ - S₂, where S₁ = Σ n * 2^(n - 1) and S₂ = Σ 2^(n - 1).Let me compute S₂ first because it's a geometric series. The sum S₂ is Σ 2^(n - 1) from n=1 to 10.The sum of a geometric series Σ r^(k) from k=0 to m is (r^(m + 1) - 1)/(r - 1). In this case, S₂ is Σ 2^(n - 1) from n=1 to 10, which is the same as Σ 2^k from k=0 to 9. So, S₂ = (2^10 - 1)/(2 - 1) = (1024 - 1)/1 = 1023.Now, S₂ = 1023.Next, let's compute S₁ = Σ n * 2^(n - 1) from n=1 to 10.This is similar to the sum Σ n * r^(n), which has a known formula. Let me recall that formula.The sum Σ n * r^(n) from n=1 to m is r*(1 - (m + 1)*r^m + m*r^(m + 1))/(1 - r)^2.But in our case, S₁ is Σ n * 2^(n - 1). Let me adjust the formula accordingly.Let me make a substitution: let k = n - 1. Then, when n=1, k=0, and when n=10, k=9. So, S₁ = Σ (k + 1) * 2^k from k=0 to 9.So, S₁ = Σ (k + 1) * 2^k from k=0 to 9.This can be split into two sums: Σ k * 2^k + Σ 2^k from k=0 to 9.We already know Σ 2^k from k=0 to 9 is 2^10 - 1 = 1023.Now, let's compute Σ k * 2^k from k=0 to 9.I remember that Σ k * r^k from k=0 to m is r*(1 - (m + 1)*r^m + m*r^(m + 1))/(1 - r)^2.Plugging in r = 2 and m = 9:Σ k * 2^k from k=0 to 9 = 2*(1 - 10*2^9 + 9*2^10)/(1 - 2)^2.First, compute the denominator: (1 - 2)^2 = (-1)^2 = 1.So, it's 2*(1 - 10*512 + 9*1024).Compute 10*512 = 5120.Compute 9*1024 = 9216.So, inside the brackets: 1 - 5120 + 9216 = 1 + (9216 - 5120) = 1 + 4096 = 4097.Multiply by 2: 2*4097 = 8194.Therefore, Σ k * 2^k from k=0 to 9 = 8194.So, going back to S₁:S₁ = Σ (k + 1)*2^k from k=0 to 9 = Σ k*2^k + Σ 2^k = 8194 + 1023 = 9217.Therefore, S₁ = 9217.Now, going back to S:S = 2*S₁ - S₂ = 2*9217 - 1023.Compute 2*9217: 9217*2 = 18434.Subtract 1023: 18434 - 1023 = 17411.So, the sum S is 17,411.Wait, that's the same result I got earlier by computing each term individually. So, that's reassuring. But I wanted to find a smarter way without computing each term, and this method worked.Alternatively, I remember that sometimes for such sums, you can use the technique of subtracting the sum multiplied by the common ratio to telescope the series. Let me try that approach as well to verify.Let me denote S = Σ (2n - 1)*2^(n - 1) from n=1 to 10.So, S = 1*1 + 3*2 + 5*4 + 7*8 + ... + 19*512.Let me write this out:S = 1 + 6 + 20 + 56 + 144 + 352 + 832 + 1920 + 4352 + 9728.Now, if I multiply both sides by 2:2S = 2 + 12 + 40 + 112 + 288 + 704 + 1664 + 3840 + 8704 + 19456.Now, subtract the original S from this 2S:2S - S = (2 + 12 + 40 + 112 + 288 + 704 + 1664 + 3840 + 8704 + 19456) - (1 + 6 + 20 + 56 + 144 + 352 + 832 + 1920 + 4352 + 9728).Let's compute this term by term:2 - 1 = 112 - 6 = 640 - 20 = 20112 - 56 = 56288 - 144 = 144704 - 352 = 3521664 - 832 = 8323840 - 1920 = 19208704 - 4352 = 435219456 - 9728 = 9728So, the result is:1 + 6 + 20 + 56 + 144 + 352 + 832 + 1920 + 4352 + 9728.Wait, that's exactly the same as the original S. So, 2S - S = S = 1 + 6 + 20 + 56 + 144 + 352 + 832 + 1920 + 4352 + 9728.But this doesn't help me because it just gives me S = S. That approach didn't telescope anything. Maybe I need to adjust the method.Perhaps I should consider the general term and find a way to express it as a difference of two terms so that when I sum them up, most terms cancel out. Let me think about that.Given that a_n = 2n - 1 and b_n = 2^(n - 1), the term a_n*b_n = (2n - 1)*2^(n - 1).Let me try to express this as a difference. Suppose I can write (2n - 1)*2^(n - 1) as something like c_n - c_{n+1}, where c_n is some expression involving n and 2^n.Let me assume that c_n = (An + B) * 2^n. Then, c_n - c_{n+1} = (An + B)2^n - (A(n + 1) + B)2^{n + 1}.Simplify this:= (An + B)2^n - (An + A + B)2^{n + 1}= (An + B)2^n - (An + A + B)2*2^n= (An + B)2^n - 2(An + A + B)2^n= [An + B - 2An - 2A - 2B]2^n= [-An - A - B]2^nWe want this to equal (2n - 1)2^{n - 1}.But let's note that (2n - 1)2^{n - 1} = (2n - 1)/2 * 2^n = (n - 0.5)2^n.So, we have:- An - A - B = n - 0.5.Therefore, equating coefficients:- A = 1 (coefficient of n)- A - B = -0.5 (constant term)From the first equation, A = -1.Plugging into the second equation:-(-1) - B = -0.5 => 1 - B = -0.5 => -B = -1.5 => B = 1.5.So, c_n = (An + B)2^n = (-n + 1.5)2^n.Therefore, (2n - 1)2^{n - 1} = c_n - c_{n+1} = (-n + 1.5)2^n - (-(n + 1) + 1.5)2^{n + 1}.Simplify c_{n+1}:= (-n -1 + 1.5)2^{n + 1} = (-n + 0.5)2^{n + 1}.So, c_n - c_{n+1} = (-n + 1.5)2^n - (-n + 0.5)2^{n + 1}.Factor out 2^n:= 2^n [ (-n + 1.5) - 2*(-n + 0.5) ]= 2^n [ (-n + 1.5) + 2n - 1 ]= 2^n [ n + 0.5 ]= (n + 0.5)2^n.But wait, earlier we had (2n - 1)2^{n - 1} = (n - 0.5)2^n.But according to this, c_n - c_{n+1} = (n + 0.5)2^n, which is not equal to (n - 0.5)2^n. Hmm, seems like a discrepancy.Wait, maybe I made a mistake in the calculation. Let me check again.We have:c_n = (-n + 1.5)2^nc_{n+1} = (-(n + 1) + 1.5)2^{n + 1} = (-n -1 + 1.5)2^{n + 1} = (-n + 0.5)2^{n + 1}.So, c_n - c_{n+1} = (-n + 1.5)2^n - (-n + 0.5)2^{n + 1}.Factor out 2^n:= 2^n [ (-n + 1.5) - 2*(-n + 0.5) ]= 2^n [ (-n + 1.5) + 2n - 1 ]= 2^n [ n + 0.5 ]So, c_n - c_{n+1} = (n + 0.5)2^n.But we wanted this to be equal to (2n - 1)2^{n - 1} = (n - 0.5)2^n.So, (n + 0.5)2^n vs (n - 0.5)2^n. They are not equal. So, my initial assumption might be incorrect, or perhaps I need to adjust the coefficients.Alternatively, maybe I need to consider a different form for c_n. Perhaps c_n = (An + B)2^{n + 1} instead of 2^n.Let me try that. Let c_n = (An + B)2^{n + 1}.Then, c_n - c_{n+1} = (An + B)2^{n + 1} - (A(n + 1) + B)2^{n + 2}.Factor out 2^{n + 1}:= 2^{n + 1} [ (An + B) - 2(A(n + 1) + B) ]= 2^{n + 1} [ An + B - 2An - 2A - 2B ]= 2^{n + 1} [ -An - 2A - B ]We want this to equal (2n - 1)2^{n - 1}.Note that (2n - 1)2^{n - 1} = (2n - 1)/2 * 2^n = (n - 0.5)2^n.But c_n - c_{n+1} = 2^{n + 1}(-An - 2A - B) = 2*(-An - 2A - B)2^n.So, 2*(-An - 2A - B)2^n = (n - 0.5)2^n.Therefore, equate the coefficients:-2A = 1 (coefficient of n)-4A - 2B = -0.5 (constant term)From the first equation: -2A = 1 => A = -0.5.From the second equation: -4*(-0.5) - 2B = -0.5 => 2 - 2B = -0.5 => -2B = -2.5 => B = 1.25.So, c_n = (An + B)2^{n + 1} = (-0.5n + 1.25)2^{n + 1}.Therefore, (2n - 1)2^{n - 1} = c_n - c_{n+1}.Let me verify this:c_n - c_{n+1} = (-0.5n + 1.25)2^{n + 1} - (-0.5(n + 1) + 1.25)2^{n + 2}.Simplify c_{n+1}:= (-0.5n - 0.5 + 1.25)2^{n + 2} = (-0.5n + 0.75)2^{n + 2}.So, c_n - c_{n+1} = (-0.5n + 1.25)2^{n + 1} - (-0.5n + 0.75)2^{n + 2}.Factor out 2^{n + 1}:= 2^{n + 1} [ (-0.5n + 1.25) - 2*(-0.5n + 0.75) ]= 2^{n + 1} [ (-0.5n + 1.25) + n - 1.5 ]= 2^{n + 1} [ 0.5n - 0.25 ]= 2^{n + 1}*(0.5n - 0.25)= 2^n*(n - 0.5).Which is equal to (2n - 1)2^{n - 1} because:(2n - 1)2^{n - 1} = (2n - 1)/2 * 2^n = (n - 0.5)2^n.Yes, that works. So, we have:(2n - 1)2^{n - 1} = c_n - c_{n+1}, where c_n = (-0.5n + 1.25)2^{n + 1}.Therefore, the sum S = Σ (from n=1 to 10) (2n - 1)2^{n - 1} = Σ (c_n - c_{n+1}) from n=1 to 10.This telescopes to c₁ - c_{11}.Compute c₁:c₁ = (-0.5*1 + 1.25)2^{1 + 1} = (-0.5 + 1.25)2^2 = (0.75)*4 = 3.Compute c_{11}:c_{11} = (-0.5*11 + 1.25)2^{11 + 1} = (-5.5 + 1.25)2^12 = (-4.25)*4096.Compute -4.25*4096:First, 4*4096 = 16384.0.25*4096 = 1024.So, 4.25*4096 = 16384 + 1024 = 17408.Therefore, c_{11} = -17408.Thus, S = c₁ - c_{11} = 3 - (-17408) = 3 + 17408 = 17411.So, again, we get S = 17,411. This confirms the result using the telescoping series method.Alternatively, another approach is to recognize that the sum S can be expressed in terms of known sums. Let me try that as well.We have S = Σ (2n - 1)2^{n - 1} from n=1 to 10.Let me rewrite this as:S = Σ (2n*2^{n - 1} - 2^{n - 1}) from n=1 to 10.= 2 Σ n*2^{n - 1} - Σ 2^{n - 1}.Let me denote S₁ = Σ n*2^{n - 1} from n=1 to 10, and S₂ = Σ 2^{n - 1} from n=1 to 10.We already computed S₂ earlier as 1023.Now, let's compute S₁.Note that S₁ = Σ n*2^{n - 1} from n=1 to 10.Let me consider the sum Σ n*r^{n} from n=1 to m, which is known to be r*(1 - (m + 1)*r^m + m*r^{m + 1}) / (1 - r)^2.In our case, r = 2, and the sum is Σ n*2^{n - 1} from n=1 to 10.Let me make a substitution: let k = n - 1. Then, when n=1, k=0; when n=10, k=9.So, S₁ = Σ (k + 1)*2^k from k=0 to 9.Which is Σ k*2^k + Σ 2^k from k=0 to 9.We already know Σ 2^k from k=0 to 9 is 2^10 - 1 = 1023.Now, compute Σ k*2^k from k=0 to 9.Using the formula for Σ k*r^k from k=0 to m:Σ k*r^k = r*(1 - (m + 1)*r^m + m*r^{m + 1}) / (1 - r)^2.Plugging in r = 2, m = 9:Σ k*2^k = 2*(1 - 10*2^9 + 9*2^10) / (1 - 2)^2.Compute denominator: (1 - 2)^2 = 1.So, numerator: 2*(1 - 10*512 + 9*1024).Compute 10*512 = 5120.Compute 9*1024 = 9216.So, inside the brackets: 1 - 5120 + 9216 = 1 + (9216 - 5120) = 1 + 4096 = 4097.Multiply by 2: 2*4097 = 8194.Therefore, Σ k*2^k from k=0 to 9 = 8194.Thus, S₁ = 8194 + 1023 = 9217.Therefore, S = 2*S₁ - S₂ = 2*9217 - 1023 = 18434 - 1023 = 17411.Again, the same result. So, no matter which method I use—direct computation, telescoping series, or known summation formulas—I consistently get S = 17,411.I think I've verified this result thoroughly using multiple methods, so I'm confident that the answer is correct.

💡Okay, so I have this problem where Caleb visited a rectangular park and counted a certain type of angles. Then he went to a square field and counted the same type of angles. The sum of these two numbers is 8. I need to figure out what type of angles Caleb counted.First, I should recall the properties of rectangles and squares in terms of their angles. A rectangle has four right angles, each measuring 90 degrees. Similarly, a square, which is a special type of rectangle with all sides equal, also has four right angles. So, both shapes have four right angles each.If Caleb counted the number of right angles in both the rectangle and the square, he would have counted 4 in the rectangle and 4 in the square. Adding these together, 4 + 4 equals 8, which matches the sum given in the problem.But wait, could there be another type of angle that also sums up to 8? Let me think about other types of angles. There are acute angles (less than 90 degrees), obtuse angles (more than 90 degrees but less than 180 degrees), and reflex angles (more than 180 degrees but less than 360 degrees). In a rectangle, all angles are right angles, so there are no acute or obtuse angles. In a square, the same applies—all angles are right angles. Therefore, if Caleb counted any other type of angle besides right angles, the sum wouldn't make sense because both shapes don't have those types of angles.Another thought: maybe Caleb counted the number of angles in total, not just the right angles. But a rectangle and a square each have four angles, so counting all angles would also give 4 + 4 = 8. But the problem specifies that he counted a certain type of angles, not all angles. So, it's more likely that he counted the number of right angles.To double-check, let's consider if there's any other angle type that could add up to 8. If he counted acute angles, but since both shapes have right angles, there are no acute angles to count. Similarly, there are no obtuse or reflex angles in these shapes. Therefore, the only plausible type of angle Caleb counted is the right angle.So, putting it all together, Caleb must have counted the number of right angles in both the rectangular park and the square field. Each location has four right angles, and together they sum up to eight.

💡First, I need to identify the pattern of the sequence. The sequence starts at -48 and each subsequent term increases by 5. This means the common difference ( d ) is 5.Next, I'll use the formula for the ( n )-th term of an arithmetic sequence:[a_n = a_1 + (n-1)d]where ( a_1 = -48 ), ( d = 5 ), and ( a_n = 72 ).Plugging in the known values:[72 = -48 + (n-1) times 5]Solving for ( n ):[72 + 48 = (n-1) times 5][120 = (n-1) times 5][frac{120}{5} = n - 1][24 = n - 1][n = 25]Therefore, there are 25 numbers in the sequence.