answer：First, identify the cost of the pen, which is 4.Next, calculate the cost of the briefcase by multiplying the pen's price by five, resulting in 20.Finally, add the cost of the pen and the briefcase together to find the total amount Marcel paid, which is 24.

answer：Okay, so I have this arithmetic sequence problem. Let me try to figure it out step by step. First, part (I) asks for the general formula of the sequence {a_n}. They gave me that a₁ = 1 and a₃ = -3. Hmm, arithmetic sequences have a common difference, right? So, the nth term of an arithmetic sequence is usually given by a_n = a₁ + (n-1)d, where d is the common difference. Alright, I know a₁ is 1, so that's straightforward. I need to find d. Since a₃ is given as -3, I can use that to find d. Let me plug in n=3 into the formula: a₃ = a₁ + (3-1)d. So that's a₃ = 1 + 2d. But they told me a₃ is -3, so I can set up the equation: 1 + 2d = -3. Let me solve for d. Subtract 1 from both sides: 2d = -4. Then divide both sides by 2: d = -2. Okay, so the common difference is -2. That makes sense because starting from 1, subtracting 2 each time would get us to -3 on the third term. So now, plugging d back into the general formula: a_n = 1 + (n-1)(-2). Let me simplify that. Distribute the -2: a_n = 1 - 2(n - 1). Then, distribute the negative sign: a_n = 1 - 2n + 2. Combine like terms: 1 + 2 is 3, so a_n = 3 - 2n. Wait, let me check that. If n=1, a₁ should be 1. Plugging in n=1: 3 - 2(1) = 1. Good. For n=3: 3 - 2(3) = 3 - 6 = -3. Perfect, that matches a₃. So, I think that's correct. So the general formula is a_n = 3 - 2n.Moving on to part (II). It says that the sum of the first k terms, S_k, is -35. I need to find k. I remember that the sum of the first k terms of an arithmetic sequence is given by S_k = k/2 [2a₁ + (k - 1)d]. Alternatively, it can also be written as S_k = k(a₁ + a_k)/2. Maybe I can use either formula. Let me see which one is easier.Since I already have the general formula for a_n, maybe it's easier to use the second formula: S_k = k(a₁ + a_k)/2. I know a₁ is 1, and a_k is 3 - 2k, from part (I). So, plugging those in: S_k = k(1 + (3 - 2k))/2. Let me simplify inside the parentheses first: 1 + 3 - 2k = 4 - 2k. So, S_k = k(4 - 2k)/2. Simplify that: 4 - 2k divided by 2 is 2 - k. So, S_k = k(2 - k). Alternatively, that's 2k - k². But they told me that S_k = -35. So, set up the equation: 2k - k² = -35. Let me rearrange that: -k² + 2k + 35 = 0. Multiply both sides by -1 to make it a bit easier: k² - 2k - 35 = 0. Now, I have a quadratic equation: k² - 2k - 35 = 0. Let me try to factor this. Looking for two numbers that multiply to -35 and add to -2. Hmm, 5 and -7: 5 * (-7) = -35, and 5 + (-7) = -2. Perfect. So, the equation factors as (k - 7)(k + 5) = 0. Setting each factor equal to zero: k - 7 = 0 => k = 7, and k + 5 = 0 => k = -5. But since k represents the number of terms, it can't be negative. So, k = 7 is the solution. Let me verify that. If k=7, then S₇ should be -35. Using the sum formula: S₇ = 7/2 [2*1 + (7 - 1)(-2)]. Let's compute that. First, inside the brackets: 2*1 = 2, (7-1)=6, 6*(-2)=-12. So, 2 + (-12) = -10. Then, S₇ = 7/2 * (-10) = 7*(-5) = -35. Perfect, that checks out.Alternatively, using the other formula: S₇ = 7*(a₁ + a₇)/2. a₇ is 3 - 2*7 = 3 - 14 = -11. So, a₁ + a₇ = 1 + (-11) = -10. Then, S₇ = 7*(-10)/2 = 7*(-5) = -35. Same result. So, k=7 is correct.Just to make sure I didn't make any mistakes earlier, let me recap:For part (I), found the common difference d by using a₃ = a₁ + 2d, solved for d, got d=-2, then wrote the general term as a_n = 1 + (n-1)(-2) = 3 - 2n. Checked with n=1 and n=3, which worked.For part (II), used the sum formula, plugged in a₁ and a_k, simplified, set equal to -35, solved quadratic, got k=7, verified by plugging back in. All seems good.I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**(I) The general formula is boxed{a_n = 3 - 2n}.(II) The value of ( k ) is boxed{7}.

answer：To determine the number of valid schedules for the six lecturers with the constraint that Dr. Smith must speak after Dr. Jones, I start by considering the total number of possible orders without any restrictions. Since there are six lecturers, the total number of possible orders is 6 factorial, which is 720.Next, I recognize that without any constraints, Dr. Smith can either speak before or after Dr. Jones. These two scenarios are equally likely, meaning that half of the total permutations will have Dr. Smith speaking after Dr. Jones, and the other half will have him speaking before.Therefore, to find the number of valid schedules where Dr. Smith is scheduled after Dr. Jones, I divide the total number of permutations by 2. This calculation gives me 720 divided by 2, which equals 360.Thus, there are 360 possible valid schedules that satisfy the given condition.

answer：First, determine the total number of baby wipes Randy uses each day. Since he walks his dog twice a day and uses one wipe after each walk, he uses 2 wipes per day.Next, calculate the total number of wipes needed for 360 days by multiplying the daily usage by the number of days: 2 wipes/day * 360 days = 720 wipes.Each pack contains 120 wipes. To find out how many packs Randy needs, divide the total number of wipes by the number of wipes per pack: 720 wipes / 120 wipes/pack = 6 packs.Therefore, Randy will need 6 packs of baby wipes to have enough for 360 days.

answer：Okay, so I have this quadratic inequality: (m x^{2} - (m + 3)x - 1 < 0). The solution set is all real numbers, which means that this inequality holds true for every real number x. Hmm, I need to find the range of the real number m that satisfies this condition. First, I remember that for a quadratic inequality (ax^2 + bx + c < 0) to hold for all real x, the quadratic must open downwards and never touch the x-axis. That means two things: the leading coefficient a must be negative, and the discriminant must be negative. So, applying this to the given inequality.Let me write down the conditions:1. The coefficient of (x^2) must be negative. In this case, the coefficient is m. So, m < 0.2. The discriminant of the quadratic must be negative. The discriminant D is given by (D = b^2 - 4ac). Here, a is m, b is -(m + 3), and c is -1.Let me compute the discriminant step by step.First, b is -(m + 3), so (b^2 = [-(m + 3)]^2 = (m + 3)^2).Then, 4ac is 4 * m * (-1) = -4m.So, the discriminant D is:(D = (m + 3)^2 - 4 * m * (-1))Simplify that:(D = (m + 3)^2 + 4m)Let me expand ((m + 3)^2):(m^2 + 6m + 9 + 4m)Combine like terms:(m^2 + 10m + 9)So, the discriminant simplifies to (m^2 + 10m + 9). For the quadratic to never touch the x-axis, this discriminant must be less than zero:(m^2 + 10m + 9 < 0)Now, I need to solve this inequality. Let me factor the quadratic:Looking for two numbers that multiply to 9 and add to 10. Hmm, 1 and 9. Yes, 1 * 9 = 9 and 1 + 9 = 10.So, the quadratic factors as:((m + 1)(m + 9) < 0)Now, to find where this product is less than zero, I can analyze the intervals determined by the roots m = -1 and m = -9.The critical points are at m = -9 and m = -1. These divide the real number line into three intervals:1. (m < -9)2. (-9 < m < -1)3. (m > -1)I need to test each interval to see where the product ((m + 1)(m + 9)) is negative.Let me pick test points:1. For (m < -9), let's choose m = -10:((-10 + 1)(-10 + 9) = (-9)(-1) = 9), which is positive.2. For (-9 < m < -1), let's choose m = -5:((-5 + 1)(-5 + 9) = (-4)(4) = -16), which is negative.3. For (m > -1), let's choose m = 0:((0 + 1)(0 + 9) = (1)(9) = 9), which is positive.So, the inequality ((m + 1)(m + 9) < 0) holds true when (-9 < m < -1).But remember, we also have the condition that m < 0 from the leading coefficient. So, combining both conditions:From the discriminant, we have (-9 < m < -1), and from the leading coefficient, m < 0. Since (-9 < m < -1) already satisfies m < 0, the combined condition is just (-9 < m < -1).Wait, let me double-check. If m is between -9 and -1, it's definitely less than 0, so that's fine. So, the range of m is from -9 to -1.But hold on, let me make sure I didn't make a mistake in computing the discriminant.Original quadratic: (m x^2 - (m + 3)x - 1)So, a = m, b = -(m + 3), c = -1Discriminant D = b² - 4ac = [-(m + 3)]² - 4 * m * (-1) = (m + 3)² + 4mWhich expands to m² + 6m + 9 + 4m = m² + 10m + 9. That seems correct.Factored as (m + 1)(m + 9). Correct.So, the discriminant is negative when -9 < m < -1. So, that's correct.Therefore, combining with the leading coefficient condition, which is m < 0, but since -9 < m < -1 is already within m < 0, the range is -9 < m < -1.Wait, but let me think again. If m is between -9 and -1, then the quadratic opens downward because m is negative, and the discriminant is negative, so it doesn't cross the x-axis. Therefore, the quadratic is always negative for all x, which is exactly what we want.So, yes, the range of m is (-9, -1).But just to be thorough, let me test m = -5, which is in the interval.Plug into the quadratic: -5x² - (-5 + 3)x -1 = -5x² - (-2)x -1 = -5x² + 2x -1Compute discriminant: (2)^2 - 4*(-5)*(-1) = 4 - 20 = -16 < 0. So, it's always negative. Correct.What about m = -10, which is outside the interval.Quadratic becomes: -10x² - (-10 + 3)x -1 = -10x² - (-7)x -1 = -10x² +7x -1Discriminant: 7² -4*(-10)*(-1) = 49 - 40 = 9 > 0. So, it has real roots, meaning the quadratic is positive somewhere, which is not what we want. So, m = -10 is invalid.Similarly, m = 0 is invalid because the leading coefficient would be zero, making it a linear inequality, which isn't quadratic anymore. So, m must be strictly between -9 and -1.Therefore, the range of m is (-9, -1).**Final Answer**The range of the real number (m) is boxed{(-9, -1)}.

answer：Okay, so I have this problem here: There's a line given by the equation ( ax - by + 8 = 0 ), where ( a > 0 ) and ( b > 0 ). This line passes through the center of the circle defined by ( x^2 + y^2 + 4x - 4y = 0 ). I need to find the minimum value of ( frac{1}{a} + frac{1}{b} ). Hmm, okay, let me try to figure this out step by step.First, I remember that the general equation of a circle is ( x^2 + y^2 + Dx + Ey + F = 0 ), and its center is at ( (-D/2, -E/2) ). So, for the given circle ( x^2 + y^2 + 4x - 4y = 0 ), the coefficients D and E are 4 and -4, respectively. That means the center of the circle is at ( (-4/2, -(-4)/2) ), which simplifies to ( (-2, 2) ). Okay, so the center is at (-2, 2). Got that.Now, the line ( ax - by + 8 = 0 ) passes through this center point (-2, 2). So, if I plug x = -2 and y = 2 into the line equation, it should satisfy the equation. Let me do that:( a(-2) - b(2) + 8 = 0 )Simplifying that:( -2a - 2b + 8 = 0 )Hmm, let me rearrange this equation to make it a bit simpler. I can factor out a -2 from the first two terms:( -2(a + b) + 8 = 0 )Then, moving the -2(a + b) to the other side:( 8 = 2(a + b) )Divide both sides by 2:( 4 = a + b )So, ( a + b = 4 ). That's a key equation. So, the sum of a and b is 4. And since both a and b are positive, that gives me some constraints on their possible values.Now, the problem asks for the minimum value of ( frac{1}{a} + frac{1}{b} ). So, I need to minimize this expression given that ( a + b = 4 ) and ( a, b > 0 ).I think I can use some optimization techniques here. Maybe the method of Lagrange multipliers, but that might be overkill. Alternatively, since it's a problem involving two variables with a linear constraint, maybe I can express one variable in terms of the other and then find the minimum.Let me try that. Let's express b in terms of a. From ( a + b = 4 ), we have ( b = 4 - a ). Since both a and b are positive, a must be between 0 and 4.So, substituting ( b = 4 - a ) into ( frac{1}{a} + frac{1}{b} ), we get:( frac{1}{a} + frac{1}{4 - a} )Now, let me denote this function as ( f(a) = frac{1}{a} + frac{1}{4 - a} ). I need to find the minimum value of this function for ( 0 < a < 4 ).To find the minimum, I can take the derivative of f(a) with respect to a, set it equal to zero, and solve for a. That should give me the critical points, which could be minima or maxima.So, let's compute the derivative f'(a):First, ( f(a) = a^{-1} + (4 - a)^{-1} )Taking the derivative term by term:The derivative of ( a^{-1} ) is ( -a^{-2} ).The derivative of ( (4 - a)^{-1} ) is ( (-1)(4 - a)^{-2} times (-1) ), because of the chain rule. The derivative of (4 - a) is -1, so multiplying by that gives us ( (4 - a)^{-2} ).So, putting it all together:( f'(a) = -a^{-2} + (4 - a)^{-2} )Simplify that:( f'(a) = -frac{1}{a^2} + frac{1}{(4 - a)^2} )Now, set this equal to zero to find critical points:( -frac{1}{a^2} + frac{1}{(4 - a)^2} = 0 )Let me rearrange this equation:( frac{1}{(4 - a)^2} = frac{1}{a^2} )Taking reciprocals on both sides:( (4 - a)^2 = a^2 )Taking square roots on both sides, but since both sides are squared, we can consider both positive and negative roots. However, since a and (4 - a) are both positive (because a is between 0 and 4), we can just take the positive roots:( 4 - a = a )Solving for a:( 4 = 2a )( a = 2 )So, a is 2. Then, since ( a + b = 4 ), b is also 2.So, the critical point is at a = 2, b = 2.Now, I should check whether this critical point is indeed a minimum. Since the function ( f(a) ) is defined on the interval (0, 4), and as a approaches 0 or 4, ( f(a) ) tends to infinity, so the function must have a minimum somewhere in between. Since we found only one critical point, it must be the minimum.Therefore, the minimum occurs at a = 2, b = 2, and the minimum value is:( frac{1}{2} + frac{1}{2} = 1 )Wait, that seems straightforward, but let me double-check my calculations to make sure I didn't make a mistake.Starting from the derivative:( f'(a) = -frac{1}{a^2} + frac{1}{(4 - a)^2} )Setting it equal to zero:( -frac{1}{a^2} + frac{1}{(4 - a)^2} = 0 )Which simplifies to:( frac{1}{(4 - a)^2} = frac{1}{a^2} )Taking square roots:( frac{1}{4 - a} = frac{1}{a} ) or ( frac{1}{4 - a} = -frac{1}{a} )But since a and (4 - a) are positive, the second equation would lead to a negative denominator, which isn't possible because both a and (4 - a) are positive. So, we discard the negative solution.Thus, ( frac{1}{4 - a} = frac{1}{a} ), which implies ( 4 - a = a ), so ( 4 = 2a ), ( a = 2 ). So, that's correct.Therefore, the minimum value is indeed 1.Alternatively, I could have used the method of Lagrange multipliers or even the AM-HM inequality to solve this problem. Let me see if the AM-HM inequality applies here.The AM-HM inequality states that for positive real numbers, the arithmetic mean is always greater than or equal to the harmonic mean. Specifically, for two variables:( frac{a + b}{2} geq frac{2}{frac{1}{a} + frac{1}{b}} )Given that ( a + b = 4 ), the arithmetic mean is ( frac{4}{2} = 2 ). So,( 2 geq frac{2}{frac{1}{a} + frac{1}{b}} )Multiplying both sides by ( frac{1}{a} + frac{1}{b} ):( 2 left( frac{1}{a} + frac{1}{b} right) geq 2 )Dividing both sides by 2:( frac{1}{a} + frac{1}{b} geq 1 )So, the minimum value of ( frac{1}{a} + frac{1}{b} ) is 1, which occurs when ( a = b = 2 ). That's consistent with what I found earlier using calculus.Therefore, I can be confident that the minimum value is indeed 1.**Final Answer**The minimum value of ( frac{1}{a} + frac{1}{b} ) is boxed{1}.