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❓A bowl contains 12 jellybeans (five red, three blue, and four white). If you randomly pick four jellybeans from the bowl without replacement, what is the probability that exactly three will be red? Express your answer as a common fraction.
💡To determine the probability of picking exactly three red jellybeans out of four selected from a bowl containing 12 jellybeans (five red, three blue, and four white), follow these steps:1. **Calculate the total number of ways to choose 4 jellybeans from 12:** [ {12 choose 4} = frac{12 times 11 times 10 times 9}{4 times 3 times 2 times 1} = 495 ]2. **Determine the number of ways to select exactly 3 red jellybeans from the 5 available:** [ {5 choose 3} = frac{5 times 4 times 3}{3 times 2 times 1} = 10 ]3. **Calculate the number of ways to choose 1 non-red jellybean from the remaining 7 (3 blue + 4 white):** [ {7 choose 1} = 7 ]4. **Find the total number of successful outcomes by multiplying the above two results:** [ 10 times 7 = 70 ]5. **Compute the probability by dividing the number of successful outcomes by the total number of possible outcomes:** [ frac{70}{495} ]6. **Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 5:** [ frac{70 div 5}{495 div 5} = frac{14}{99} ]
❓In 2000, there were 600,000 cases of a disease reported in a country. By 2020, this number had decreased to 2,000 cases. Based on a linear decrease, how many cases would have been reported in 2015?
💡First, I need to determine the time span between the years 2000 and 2020, which is 20 years.Next, I'll calculate the total decrease in disease cases during this period by subtracting the number of cases in 2020 from the number in 2000: 600,000 - 2,000 = 598,000 cases.To find the annual rate of decrease, I'll divide the total decrease by the number of years: 598,000 / 20 = 29,900 cases per year.Now, I'll determine how many years have passed between 2000 and 2015: 2015 - 2000 = 15 years.Using the annual rate of decrease, I'll calculate the total decrease in cases over these 15 years: 29,900 * 15 = 448,500 cases.Finally, I'll subtract this decrease from the initial number of cases in 2000 to find the number of cases in 2015: 600,000 - 448,500 = 151,500 cases.
❓Find the quartic polynomial p(x) such that p(1) = -3, p(2) = -5, p(3) = -11, p(4) = -27, and p(5) = -59.
💡Okay, so I need to find a quartic polynomial ( p(x) ) that satisfies the given conditions: ( p(1) = -3 ), ( p(2) = -5 ), ( p(3) = -11 ), ( p(4) = -27 ), and ( p(5) = -59 ). Hmm, quartic means it's a fourth-degree polynomial, so it should look something like ( p(x) = ax^4 + bx^3 + cx^2 + dx + e ). Alright, since I have five points, I can set up a system of equations by plugging each ( x ) value into the polynomial and setting it equal to the corresponding ( p(x) ) value. Let me write those out:1. For ( x = 1 ): ( a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e = -3 ) Simplifies to: ( a + b + c + d + e = -3 )2. For ( x = 2 ): ( a(2)^4 + b(2)^3 + c(2)^2 + d(2) + e = -5 ) Simplifies to: ( 16a + 8b + 4c + 2d + e = -5 )3. For ( x = 3 ): ( a(3)^4 + b(3)^3 + c(3)^2 + d(3) + e = -11 ) Simplifies to: ( 81a + 27b + 9c + 3d + e = -11 )4. For ( x = 4 ): ( a(4)^4 + b(4)^3 + c(4)^2 + d(4) + e = -27 ) Simplifies to: ( 256a + 64b + 16c + 4d + e = -27 )5. For ( x = 5 ): ( a(5)^4 + b(5)^3 + c(5)^2 + d(5) + e = -59 ) Simplifies to: ( 625a + 125b + 25c + 5d + e = -59 )So now I have a system of five equations:1. ( a + b + c + d + e = -3 ) 2. ( 16a + 8b + 4c + 2d + e = -5 ) 3. ( 81a + 27b + 9c + 3d + e = -11 ) 4. ( 256a + 64b + 16c + 4d + e = -27 ) 5. ( 625a + 125b + 25c + 5d + e = -59 )I need to solve this system to find the coefficients ( a, b, c, d, e ). Since it's a linear system, I can use elimination to solve for the variables step by step.Let me label the equations for clarity:Equation (1): ( a + b + c + d + e = -3 ) Equation (2): ( 16a + 8b + 4c + 2d + e = -5 ) Equation (3): ( 81a + 27b + 9c + 3d + e = -11 ) Equation (4): ( 256a + 64b + 16c + 4d + e = -27 ) Equation (5): ( 625a + 125b + 25c + 5d + e = -59 )First, I can subtract Equation (1) from Equation (2) to eliminate ( e ):Equation (2) - Equation (1): ( (16a - a) + (8b - b) + (4c - c) + (2d - d) + (e - e) = -5 - (-3) ) Simplifies to: ( 15a + 7b + 3c + d = -2 ) Let me call this Equation (6): ( 15a + 7b + 3c + d = -2 )Next, subtract Equation (2) from Equation (3):Equation (3) - Equation (2): ( (81a - 16a) + (27b - 8b) + (9c - 4c) + (3d - 2d) + (e - e) = -11 - (-5) ) Simplifies to: ( 65a + 19b + 5c + d = -6 ) Let me call this Equation (7): ( 65a + 19b + 5c + d = -6 )Similarly, subtract Equation (3) from Equation (4):Equation (4) - Equation (3): ( (256a - 81a) + (64b - 27b) + (16c - 9c) + (4d - 3d) + (e - e) = -27 - (-11) ) Simplifies to: ( 175a + 37b + 7c + d = -16 ) Equation (8): ( 175a + 37b + 7c + d = -16 )And subtract Equation (4) from Equation (5):Equation (5) - Equation (4): ( (625a - 256a) + (125b - 64b) + (25c - 16c) + (5d - 4d) + (e - e) = -59 - (-27) ) Simplifies to: ( 369a + 61b + 9c + d = -32 ) Equation (9): ( 369a + 61b + 9c + d = -32 )Now, I have four new equations (6, 7, 8, 9) with four variables: ( a, b, c, d ). Let me write them again:Equation (6): ( 15a + 7b + 3c + d = -2 ) Equation (7): ( 65a + 19b + 5c + d = -6 ) Equation (8): ( 175a + 37b + 7c + d = -16 ) Equation (9): ( 369a + 61b + 9c + d = -32 )I can now subtract Equation (6) from Equation (7) to eliminate ( d ):Equation (7) - Equation (6): ( (65a - 15a) + (19b - 7b) + (5c - 3c) + (d - d) = -6 - (-2) ) Simplifies to: ( 50a + 12b + 2c = -4 ) Equation (10): ( 50a + 12b + 2c = -4 )Next, subtract Equation (7) from Equation (8):Equation (8) - Equation (7): ( (175a - 65a) + (37b - 19b) + (7c - 5c) + (d - d) = -16 - (-6) ) Simplifies to: ( 110a + 18b + 2c = -10 ) Equation (11): ( 110a + 18b + 2c = -10 )Then, subtract Equation (8) from Equation (9):Equation (9) - Equation (8): ( (369a - 175a) + (61b - 37b) + (9c - 7c) + (d - d) = -32 - (-16) ) Simplifies to: ( 194a + 24b + 2c = -16 ) Equation (12): ( 194a + 24b + 2c = -16 )Now, I have three equations (10, 11, 12) with three variables: ( a, b, c ). Let me write them again:Equation (10): ( 50a + 12b + 2c = -4 ) Equation (11): ( 110a + 18b + 2c = -10 ) Equation (12): ( 194a + 24b + 2c = -16 )I can subtract Equation (10) from Equation (11) to eliminate ( c ):Equation (11) - Equation (10): ( (110a - 50a) + (18b - 12b) + (2c - 2c) = -10 - (-4) ) Simplifies to: ( 60a + 6b = -6 ) Equation (13): ( 60a + 6b = -6 )Similarly, subtract Equation (11) from Equation (12):Equation (12) - Equation (11): ( (194a - 110a) + (24b - 18b) + (2c - 2c) = -16 - (-10) ) Simplifies to: ( 84a + 6b = -6 ) Equation (14): ( 84a + 6b = -6 )Now, I have two equations (13 and 14) with two variables: ( a, b ). Let me write them again:Equation (13): ( 60a + 6b = -6 ) Equation (14): ( 84a + 6b = -6 )Subtract Equation (13) from Equation (14) to eliminate ( b ):Equation (14) - Equation (13): ( (84a - 60a) + (6b - 6b) = -6 - (-6) ) Simplifies to: ( 24a = 0 ) So, ( a = 0 ).Now that I have ( a = 0 ), I can plug this back into Equation (13) to find ( b ):Equation (13): ( 60(0) + 6b = -6 ) Simplifies to: ( 6b = -6 ) So, ( b = -1 ).Now, with ( a = 0 ) and ( b = -1 ), I can find ( c ) using Equation (10):Equation (10): ( 50(0) + 12(-1) + 2c = -4 ) Simplifies to: ( -12 + 2c = -4 ) Add 12 to both sides: ( 2c = 8 ) So, ( c = 4 ).Wait, hold on, that doesn't seem right because in the previous step, Equation (10) is ( 50a + 12b + 2c = -4 ). If ( a = 0 ) and ( b = -1 ), then:( 0 + 12(-1) + 2c = -4 ) Which is ( -12 + 2c = -4 ) Adding 12: ( 2c = 8 ) So, ( c = 4 ). Hmm, but in the initial solution, the coefficient for ( x^2 ) was 2. Maybe I made a mistake here.Wait, let me double-check. Maybe I substituted incorrectly.Wait, Equation (10): ( 50a + 12b + 2c = -4 ). If ( a = 0 ), ( b = -1 ), then:( 0 + 12(-1) + 2c = -4 ) Which is ( -12 + 2c = -4 ) So, ( 2c = 8 ) Thus, ( c = 4 ). Hmm, but in the initial solution, it was 2. Maybe I need to check my previous steps.Wait, let me go back. Maybe I made a mistake earlier.Wait, Equation (13) was ( 60a + 6b = -6 ), and Equation (14) was ( 84a + 6b = -6 ). Subtracting them gives ( 24a = 0 ), so ( a = 0 ). Then, plugging ( a = 0 ) into Equation (13): ( 6b = -6 ), so ( b = -1 ). Then, plugging ( a = 0 ) and ( b = -1 ) into Equation (10): ( 50(0) + 12(-1) + 2c = -4 ) Which is ( -12 + 2c = -4 ) So, ( 2c = 8 ) Thus, ( c = 4 ). Hmm, that's consistent. But in the initial solution, the coefficient ( c ) was 2. Maybe I made a mistake in the initial steps.Wait, perhaps I made a mistake in the earlier subtraction steps. Let me check the equations again.Wait, let's go back to Equation (10): ( 50a + 12b + 2c = -4 ). If ( a = 0 ) and ( b = -1 ), then:( 0 + 12(-1) + 2c = -4 ) Which is ( -12 + 2c = -4 ) So, ( 2c = 8 ) Thus, ( c = 4 ). Hmm, that seems correct.Wait, but in the initial solution, the polynomial was ( -x^4 + 2x^2 -5x +1 ). So, ( a = -1 ), but here I got ( a = 0 ). That's conflicting. Maybe I made a mistake in the earlier steps.Wait, let me check the initial equations. Maybe I made a mistake in setting up the equations.Wait, the initial equations were:1. ( a + b + c + d + e = -3 ) 2. ( 16a + 8b + 4c + 2d + e = -5 ) 3. ( 81a + 27b + 9c + 3d + e = -11 ) 4. ( 256a + 64b + 16c + 4d + e = -27 ) 5. ( 625a + 125b + 25c + 5d + e = -59 )Then, subtracting Equation (1) from Equation (2) gives Equation (6): ( 15a + 7b + 3c + d = -2 ). That seems correct.Subtracting Equation (2) from Equation (3) gives Equation (7): ( 65a + 19b + 5c + d = -6 ). Correct.Subtracting Equation (3) from Equation (4) gives Equation (8): ( 175a + 37b + 7c + d = -16 ). Correct.Subtracting Equation (4) from Equation (5) gives Equation (9): ( 369a + 61b + 9c + d = -32 ). Correct.Then, subtracting Equation (6) from Equation (7) gives Equation (10): ( 50a + 12b + 2c = -4 ). Correct.Subtracting Equation (7) from Equation (8) gives Equation (11): ( 110a + 18b + 2c = -10 ). Correct.Subtracting Equation (8) from Equation (9) gives Equation (12): ( 194a + 24b + 2c = -16 ). Correct.Then, subtracting Equation (10) from Equation (11) gives Equation (13): ( 60a + 6b = -6 ). Correct.Subtracting Equation (11) from Equation (12) gives Equation (14): ( 84a + 6b = -6 ). Correct.Subtracting Equation (13) from Equation (14) gives ( 24a = 0 ), so ( a = 0 ). Correct.Then, plugging ( a = 0 ) into Equation (13): ( 60(0) + 6b = -6 ) gives ( b = -1 ). Correct.Then, plugging ( a = 0 ) and ( b = -1 ) into Equation (10): ( 50(0) + 12(-1) + 2c = -4 ) gives ( -12 + 2c = -4 ), so ( 2c = 8 ), ( c = 4 ). Hmm, but in the initial solution, ( c = 2 ). So, where is the mistake?Wait, perhaps I made a mistake in the initial setup. Let me check the initial equations again.Wait, the initial equations were correct. So, maybe the initial solution was wrong. Let me proceed further.So, with ( a = 0 ), ( b = -1 ), ( c = 4 ), I can now find ( d ) using Equation (6): ( 15a + 7b + 3c + d = -2 ).Plugging in ( a = 0 ), ( b = -1 ), ( c = 4 ):( 15(0) + 7(-1) + 3(4) + d = -2 ) Simplifies to: ( 0 -7 + 12 + d = -2 ) Which is: ( 5 + d = -2 ) So, ( d = -7 ).Now, with ( a = 0 ), ( b = -1 ), ( c = 4 ), ( d = -7 ), I can find ( e ) using Equation (1): ( a + b + c + d + e = -3 ).Plugging in the known values:( 0 + (-1) + 4 + (-7) + e = -3 ) Simplifies to: ( -4 + e = -3 ) So, ( e = 1 ).Therefore, the polynomial is ( p(x) = 0x^4 -1x^3 +4x^2 -7x +1 ), which simplifies to ( p(x) = -x^3 +4x^2 -7x +1 ).Wait, but this is a cubic polynomial, not quartic. The initial problem asked for a quartic polynomial, which is degree 4. So, having ( a = 0 ) reduces it to a cubic. That's conflicting with the problem statement.Hmm, that suggests that maybe there was a mistake in the earlier steps. Let me double-check the calculations.Wait, when I subtracted Equation (13) from Equation (14), I got ( 24a = 0 ), so ( a = 0 ). But if ( a = 0 ), then the polynomial is indeed cubic, not quartic. But the problem states it's quartic, so perhaps I made a mistake in the calculations.Wait, let me check the subtraction steps again.Equation (13): ( 60a + 6b = -6 ) Equation (14): ( 84a + 6b = -6 ) Subtracting Equation (13) from Equation (14): ( (84a - 60a) + (6b - 6b) = (-6) - (-6) ) Which is ( 24a = 0 ), so ( a = 0 ). That seems correct.But if ( a = 0 ), then the polynomial is cubic, which contradicts the problem statement. So, perhaps there was a mistake in the earlier steps.Wait, let me check the initial equations again. Maybe I made a mistake in setting up the equations.Wait, the initial equations were:1. ( a + b + c + d + e = -3 ) 2. ( 16a + 8b + 4c + 2d + e = -5 ) 3. ( 81a + 27b + 9c + 3d + e = -11 ) 4. ( 256a + 64b + 16c + 4d + e = -27 ) 5. ( 625a + 125b + 25c + 5d + e = -59 )Let me verify these equations with the given points.For ( x = 1 ): ( a + b + c + d + e = -3 ). Correct.For ( x = 2 ): ( 16a + 8b + 4c + 2d + e = -5 ). Correct.For ( x = 3 ): ( 81a + 27b + 9c + 3d + e = -11 ). Correct.For ( x = 4 ): ( 256a + 64b + 16c + 4d + e = -27 ). Correct.For ( x = 5 ): ( 625a + 125b + 25c + 5d + e = -59 ). Correct.So, the equations are correct. Then, the subtraction steps leading to ( a = 0 ) must be correct, implying that the polynomial is cubic, not quartic. But the problem states it's quartic. That suggests that perhaps the initial assumption that the polynomial is quartic is incorrect, or there's a mistake in the problem statement.Alternatively, maybe I made a mistake in the calculations. Let me try a different approach.Another method to find the polynomial is to use finite differences. Let me try that.Given the points:( x ) | ( p(x) )---|---1 | -32 | -53 | -114 | -275 | -59Let me compute the finite differences.First, list the ( p(x) ) values: -3, -5, -11, -27, -59.First differences (Δp): -5 - (-3) = -2 -11 - (-5) = -6 -27 - (-11) = -16 -59 - (-27) = -32Second differences (Δ²p): -6 - (-2) = -4 -16 - (-6) = -10 -32 - (-16) = -16Third differences (Δ³p): -10 - (-4) = -6 -16 - (-10) = -6Fourth differences (Δ⁴p): -6 - (-6) = 0Wait, the fourth differences are zero, which suggests that the polynomial is of degree 3, not 4. Because for a polynomial of degree ( n ), the ( (n+1) )-th differences are zero. So, if the fourth differences are zero, it's a cubic polynomial. But the problem states it's quartic. That's conflicting.Wait, but in my earlier calculation, I found ( a = 0 ), which reduces it to a cubic. So, perhaps the problem is misstated, or perhaps I made a mistake in the finite difference calculation.Wait, let me recalculate the finite differences.Given ( p(1) = -3 ), ( p(2) = -5 ), ( p(3) = -11 ), ( p(4) = -27 ), ( p(5) = -59 ).First differences: -5 - (-3) = -2 -11 - (-5) = -6 -27 - (-11) = -16 -59 - (-27) = -32Second differences: -6 - (-2) = -4 -16 - (-6) = -10 -32 - (-16) = -16Third differences: -10 - (-4) = -6 -16 - (-10) = -6Fourth differences: -6 - (-6) = 0Yes, the fourth differences are zero, which suggests a cubic polynomial. So, perhaps the problem is misstated, or I made a mistake in the initial setup.Alternatively, maybe the polynomial is indeed quartic, but with ( a = 0 ), making it effectively cubic. So, perhaps the answer is a cubic polynomial, but the problem says quartic. That's confusing.Wait, let me check the initial solution provided. It was ( p(x) = -x^4 + 2x^2 -5x +1 ). Let me compute its values at x=1 to 5.Compute ( p(1) ): ( -1 + 2 -5 +1 = -3 ). Correct.( p(2) ): ( -16 + 8 -10 +1 = -17 ). Wait, but the given ( p(2) = -5 ). That's incorrect. So, the initial solution is wrong.Wait, that can't be. Let me compute ( p(2) ) correctly.( p(2) = - (2)^4 + 2*(2)^2 -5*(2) +1 = -16 + 8 -10 +1 = (-16 +8) + (-10 +1) = (-8) + (-9) = -17 ). But the given ( p(2) = -5 ). So, the initial solution is incorrect.Therefore, my earlier calculation leading to a cubic polynomial is correct, and the initial solution provided was wrong.So, the correct polynomial is cubic: ( p(x) = -x^3 +4x^2 -7x +1 ).Let me verify this polynomial at the given points.Compute ( p(1) ): ( -1 +4 -7 +1 = (-1 +4) + (-7 +1) = 3 -6 = -3 ). Correct.( p(2) ): ( -8 + 16 -14 +1 = (-8 +16) + (-14 +1) = 8 -13 = -5 ). Correct.( p(3) ): ( -27 + 36 -21 +1 = (-27 +36) + (-21 +1) = 9 -20 = -11 ). Correct.( p(4) ): ( -64 + 64 -28 +1 = (-64 +64) + (-28 +1) = 0 -27 = -27 ). Correct.( p(5) ): ( -125 + 100 -35 +1 = (-125 +100) + (-35 +1) = (-25) + (-34) = -59 ). Correct.So, the polynomial ( p(x) = -x^3 +4x^2 -7x +1 ) satisfies all the given conditions. Therefore, the initial solution provided was incorrect, and the correct polynomial is cubic, not quartic.But the problem statement says "quartic polynomial". So, perhaps there was a mistake in the problem statement, or perhaps I made a mistake in my calculations.Wait, but according to the finite differences, the fourth differences are zero, which suggests a cubic polynomial. So, the polynomial is indeed cubic, not quartic. Therefore, the problem statement might have an error.Alternatively, perhaps I made a mistake in the finite difference calculation. Let me check again.Given ( p(x) ) values: -3, -5, -11, -27, -59.First differences: -5 - (-3) = -2 -11 - (-5) = -6 -27 - (-11) = -16 -59 - (-27) = -32Second differences: -6 - (-2) = -4 -16 - (-6) = -10 -32 - (-16) = -16Third differences: -10 - (-4) = -6 -16 - (-10) = -6Fourth differences: -6 - (-6) = 0Yes, the fourth differences are zero, so it's a cubic polynomial. Therefore, the problem statement is incorrect in stating it's quartic. The correct polynomial is cubic.Therefore, the answer is ( p(x) = -x^3 +4x^2 -7x +1 ).But the initial solution provided was quartic, which was incorrect. So, I think the correct answer is the cubic polynomial I found.
❓In a company, there is a total of 450 employees with varying salaries and benefits. 150 employees earn a base salary of 2000 per month and receive a monthly bonus of 500, 200 employees earn a base salary of 2500 per month with a 400 monthly health benefit, and the remaining 100 employees earn a base salary of 3000 per month with an additional 600 monthly retirement benefit.Due to tough economic times, the company decided to lay off employees in two rounds and reduce some of the bonuses and benefits as well.In the first round, the company laid off 20% of the 2000 earners, 25% of the 2500 earners, and 15% of the 3000 earners. Additionally, they reduced the monthly bonus of the remaining 2000 earners to 400, and the monthly health benefits of the remaining 2500 earners to 300.After the first round, the company had to initiate a second round of layoffs where they let go of an additional 10% of the remaining 2000 earners, 15% of the remaining 2500 earners, and 5% of the remaining 3000 earners. They also decided to cut the monthly retirement benefit of the remaining 3000 earners by 20% to 480.Calculate the total amount of money the company paid to the remaining employees in terms of base salary, bonuses, and benefits after both rounds of layoffs.
💡First, I need to determine the number of employees remaining after each round of layoffs. In the first round, 20% of the 2000 earners, 25% of the 2500 earners, and 15% of the 3000 earners are laid off. This leaves 120 2000 earners, 150 2500 earners, and 85 3000 earners.Next, I'll adjust the bonuses and benefits after the first round. The 2000 earners' bonus is reduced to 400, and the 2500 earners' health benefit is reduced to 300. The 3000 earners' retirement benefit remains unchanged at 600.In the second round, an additional 10% of the remaining 2000 earners, 15% of the remaining 2500 earners, and 5% of the remaining 3000 earners are laid off. This results in 108 2000 earners, 127 2500 earners, and 80 3000 earners.Additionally, the 3000 earners' retirement benefit is reduced by 20%, bringing it down to 480.Finally, I'll calculate the total amount paid to the remaining employees by summing up the base salaries and the adjusted bonuses and benefits for each group.
❓The minimum value of the slope of the tangent line at the point (b, f(b)) on the graph of the function f(x) = ln x + x^2 - bx + a (b > 0, a in mathbb{R}) is ( )A: 2sqrt{2} B: sqrt{3} C: 1 D: 2
💡Okay, so I have this problem here where I need to find the minimum value of the slope of the tangent line at the point (b, f(b)) on the graph of the function f(x) = ln x + x² - b x + a. The options are given as A: 2√2, B: √3, C: 1, D: 2. Hmm, let me try to figure this out step by step.First, I remember that the slope of the tangent line at a particular point on a curve is given by the derivative of the function at that point. So, I need to find f'(x), the derivative of f(x), and then evaluate it at x = b.Let me write down the function again: f(x) = ln x + x² - b x + a. Okay, so to find the derivative, f'(x), I need to differentiate each term with respect to x.Starting with ln x, the derivative of ln x is 1/x. Then, the derivative of x² is 2x. The derivative of -b x is -b, since b is a constant. And finally, the derivative of a constant, a, is 0. So putting it all together, f'(x) = 1/x + 2x - b.Alright, so now I have f'(x) = 1/x + 2x - b. The next step is to evaluate this derivative at x = b, because we're interested in the slope at the point (b, f(b)). So, substituting x = b into f'(x), we get f'(b) = 1/b + 2b - b.Simplifying that, 2b - b is just b, so f'(b) = 1/b + b. So, the slope of the tangent line at x = b is 1/b + b. Now, the question is asking for the minimum value of this slope. So, I need to find the minimum value of the expression 1/b + b, where b > 0.Hmm, okay. So, I have to minimize the function g(b) = 1/b + b for b > 0. I think this is a classic optimization problem. Maybe I can use calculus to find the minimum, or perhaps there's an inequality that can help me here.Wait, I remember something called the AM-GM inequality, which states that for positive real numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe that applies here.Let me recall: For any positive real numbers a and b, (a + b)/2 ≥ √(ab), with equality when a = b. So, in this case, if I let a = 1/b and b = b, then the arithmetic mean would be (1/b + b)/2, and the geometric mean would be √(1/b * b) = √1 = 1.So, applying the AM-GM inequality: (1/b + b)/2 ≥ 1. Multiplying both sides by 2, we get 1/b + b ≥ 2. So, the minimum value of 1/b + b is 2, and this occurs when 1/b = b, which implies that b² = 1, so b = 1 (since b > 0).Therefore, the minimum value of the slope f'(b) is 2, which occurs when b = 1. So, looking back at the options, D is 2, which should be the correct answer.Wait, just to make sure I didn't make a mistake, let me verify this by using calculus. If I take g(b) = 1/b + b, then the derivative g'(b) would be -1/b² + 1. Setting this equal to zero to find critical points: -1/b² + 1 = 0, so 1 = 1/b², which implies b² = 1, so b = 1 (since b > 0). Then, plugging back into g(b), we get g(1) = 1/1 + 1 = 2. So, yes, that confirms it.Alternatively, I could have thought about the function g(b) = 1/b + b. As b approaches 0 from the right, 1/b tends to infinity, so g(b) tends to infinity. As b approaches infinity, b tends to infinity, so g(b) also tends to infinity. Therefore, the function must have a minimum somewhere in between. Since we found that at b = 1, g(b) = 2, and this is the only critical point, it must be the minimum.So, all methods point to the minimum slope being 2. Therefore, the correct answer is D: 2.**Final Answer**boxed{D}
❓Let U={triangle}, M={right triangle}, N={isosceles triangle}, then M∩N= .
💡Alright, so I have this problem here where I'm given a universal set U, which is the set of all triangles. Then there are two subsets, M and N. M is the set of right triangles, and N is the set of isosceles triangles. The question is asking for the intersection of M and N, which is M ∩ N. First, I need to understand what the intersection of two sets means. In set theory, the intersection of two sets is a set containing all elements that are common to both sets. So, in this case, M ∩ N would be all triangles that are both right triangles and isosceles triangles. Okay, so I need to think about what a right triangle is. A right triangle is a triangle that has one angle equal to 90 degrees. It can be either scalene or isosceles. A scalene triangle has all sides of different lengths, while an isosceles triangle has at least two sides that are equal in length. Now, an isosceles triangle has two equal sides and the angles opposite those sides are also equal. So, if I have a right triangle that is also isosceles, what does that look like? Well, in a right triangle, one angle is 90 degrees, and the other two angles must add up to 90 degrees because the total sum of angles in a triangle is 180 degrees. If it's also isosceles, then the two non-right angles must be equal. That means each of those angles is 45 degrees because 90 degrees divided by 2 is 45 degrees. So, this would be a 45-45-90 triangle. Wait, but is that the only possibility? Let me think. If a right triangle is isosceles, does it necessarily have two equal angles? Yes, because in an isosceles triangle, the base angles are equal. In a right triangle, the two non-right angles must be equal if it's isosceles, leading to each being 45 degrees. So, essentially, the intersection of M and N is the set of all triangles that are both right triangles and isosceles triangles, which are specifically the 45-45-90 triangles. Let me double-check to make sure I'm not missing anything. If a triangle is both right and isosceles, it must have one right angle and two equal angles, which are each 45 degrees. Therefore, it's a 45-45-90 triangle. Is there another type of triangle that could be both right and isosceles? I don't think so because if a right triangle has two equal sides, the angles opposite those sides must be equal, leading to the 45-45-90 configuration. So, I think I'm confident that M ∩ N is the set of isosceles right triangles, specifically the 45-45-90 triangles.