💡First, I recognize that the average of 7 numbers is 20. To find the total sum of these numbers, I multiply the average by the number of numbers: 20 multiplied by 7 equals 140.Next, each number in the set is multiplied by 5. This means the total sum of the new set of numbers will also be multiplied by 5. Therefore, I multiply the original sum of 140 by 5 to get 700.Finally, to find the new average, I divide the new total sum of 700 by the number of numbers, which is still 7. Dividing 700 by 7 gives me the new average of 100.

💡First, I need to determine the total number of ways five people can line up without any restrictions. This is simply the number of permutations of five distinct individuals, which is 5! = 120.Next, I need to calculate the number of arrangements where at least one of the two youngest individuals is in the first position. To do this, I'll use the principle of inclusion-exclusion.1. **Calculate the number of arrangements where the first youngest person is in the first position:** - If the first position is fixed with the youngest person, there are 4 remaining positions to arrange the other four people. - The number of such arrangements is 4! = 24.2. **Calculate the number of arrangements where the second youngest person is in the first position:** - Similarly, if the second youngest person is in the first position, there are 4 remaining positions to arrange the other four people. - The number of such arrangements is also 4! = 24.3. **Adjust for double-counting:** - If both the youngest and the second youngest are considered in the first position, there's an overlap where both scenarios are counted twice. - However, since only one person can occupy the first position at a time, there is no actual overlap, so no adjustment is needed.Finally, subtract the number of restricted arrangements from the total number of arrangements to find the number of valid arrangements where neither of the two youngest individuals is in the first position:120 (total) - 24 (youngest first) - 24 (second youngest first) = 72 valid arrangements.

💡Okay, so I have this problem here about a function f(x) = ax³ + x², where a is a real number. It says that this function has an extremum at x = -4/3. I need to find the value of a, and then for the second part, I have to discuss the monotonicity of another function g(x) which is f(x) multiplied by e^x.Alright, let's start with the first part. I remember that an extremum of a function occurs where the first derivative is zero. So, I need to find the derivative of f(x) and set it equal to zero at x = -4/3. That should give me an equation to solve for a.So, f(x) = ax³ + x². The derivative f'(x) would be 3ax² + 2x, right? Yeah, that makes sense because the derivative of ax³ is 3ax² and the derivative of x² is 2x.Now, since there's an extremum at x = -4/3, plugging that into f'(x) should give zero. Let me write that out:f'(-4/3) = 3a*(-4/3)² + 2*(-4/3) = 0.Let me compute each part step by step. First, (-4/3) squared is (16/9). So, 3a*(16/9) is (48a/9) which simplifies to (16a/3). Then, 2*(-4/3) is (-8/3). So, putting it all together:16a/3 - 8/3 = 0.Hmm, so combining these terms, I get (16a - 8)/3 = 0. Since the denominator is 3, which isn't zero, the numerator must be zero. So, 16a - 8 = 0. Solving for a, I add 8 to both sides:16a = 8.Then, divide both sides by 16:a = 8/16 = 1/2.Okay, so a is 1/2. That seems straightforward. Let me just double-check my calculations to make sure I didn't make a mistake.Starting again, f'(x) = 3ax² + 2x. Plugging in x = -4/3:3a*(16/9) + 2*(-4/3) = 0.Calculates to (48a/9) - (8/3) = 0. Simplify 48a/9 to 16a/3, so 16a/3 - 8/3 = 0. Combine terms: (16a - 8)/3 = 0. Multiply both sides by 3: 16a - 8 = 0. So, 16a = 8, a = 1/2. Yep, that checks out.Alright, so part 1 is done. Now, moving on to part 2. We have g(x) = f(x)e^x, and we need to discuss its monotonicity. Monotonicity refers to where the function is increasing or decreasing, which is determined by the sign of its derivative.So, first, let's write out g(x):g(x) = f(x)e^x = ( (1/2)x³ + x² )e^x.To find where g(x) is increasing or decreasing, I need to compute its derivative g'(x). I remember that the derivative of a product of two functions is the derivative of the first times the second plus the first times the derivative of the second. So, using the product rule:g'(x) = f'(x)e^x + f(x)e^x.We already know f'(x) from earlier, which is 3ax² + 2x. Since we found a = 1/2, let's plug that in:f'(x) = 3*(1/2)x² + 2x = (3/2)x² + 2x.So, f'(x) = (3/2)x² + 2x.Therefore, g'(x) = [ (3/2)x² + 2x ]e^x + [ (1/2)x³ + x² ]e^x.Hmm, let's factor out e^x since it's common to both terms:g'(x) = e^x [ (3/2)x² + 2x + (1/2)x³ + x² ].Now, let's combine like terms inside the brackets:First, let's write all the terms:(3/2)x² + 2x + (1/2)x³ + x².Combine the x³ term: (1/2)x³.Combine the x² terms: (3/2)x² + x² = (3/2 + 2/2)x² = (5/2)x².And the x term: 2x.So, putting it all together:g'(x) = e^x [ (1/2)x³ + (5/2)x² + 2x ].Hmm, this is a cubic polynomial multiplied by e^x. Since e^x is always positive, the sign of g'(x) depends entirely on the cubic polynomial:(1/2)x³ + (5/2)x² + 2x.Let me factor this polynomial to make it easier to analyze.First, factor out an x:x [ (1/2)x² + (5/2)x + 2 ].So, g'(x) = e^x * x [ (1/2)x² + (5/2)x + 2 ].Now, let's factor the quadratic inside the brackets:(1/2)x² + (5/2)x + 2.To factor this, I can multiply all terms by 2 to eliminate the fractions:x² + 5x + 4.Now, factor this quadratic:Looking for two numbers that multiply to 4 and add to 5. That would be 1 and 4.So, x² + 5x + 4 = (x + 1)(x + 4).Therefore, going back to the original expression:(1/2)x² + (5/2)x + 2 = (1/2)(x² + 5x + 4) = (1/2)(x + 1)(x + 4).So, putting it all together, the derivative g'(x) is:g'(x) = e^x * x * (1/2)(x + 1)(x + 4).Simplify the constants:(1/2) is a constant factor, so we can write:g'(x) = (1/2)e^x * x(x + 1)(x + 4).Since e^x is always positive, and 1/2 is positive, the sign of g'(x) depends on the product x(x + 1)(x + 4).So, to find where g'(x) is positive or negative, we need to analyze the sign of x(x + 1)(x + 4).Let's find the critical points where the expression equals zero:x = 0, x = -1, x = -4.These points divide the real number line into intervals. We'll test each interval to determine the sign of the product.The intervals are:1. (-∞, -4)2. (-4, -1)3. (-1, 0)4. (0, ∞)Let's pick a test point in each interval and plug it into x(x + 1)(x + 4):1. For (-∞, -4), let's choose x = -5:(-5)(-5 + 1)(-5 + 4) = (-5)(-4)(-1) = (-5)(4) = -20. Negative.2. For (-4, -1), let's choose x = -2:(-2)(-2 + 1)(-2 + 4) = (-2)(-1)(2) = (-2)(-2) = 4. Positive.3. For (-1, 0), let's choose x = -0.5:(-0.5)(-0.5 + 1)(-0.5 + 4) = (-0.5)(0.5)(3.5) = (-0.5)(1.75) = -0.875. Negative.4. For (0, ∞), let's choose x = 1:(1)(1 + 1)(1 + 4) = (1)(2)(5) = 10. Positive.So, summarizing:- On (-∞, -4): g'(x) < 0, so g(x) is decreasing.- On (-4, -1): g'(x) > 0, so g(x) is increasing.- On (-1, 0): g'(x) < 0, so g(x) is decreasing.- On (0, ∞): g'(x) > 0, so g(x) is increasing.Therefore, the function g(x) decreases on (-∞, -4) and (-1, 0), and increases on (-4, -1) and (0, ∞).Wait, let me double-check the test points to make sure I didn't make a mistake.For x = -5: (-5)(-4)(-1) = (-5)(4) = -20. Correct, negative.For x = -2: (-2)(-1)(2) = 4. Correct, positive.For x = -0.5: (-0.5)(0.5)(3.5) = (-0.5)(1.75) = -0.875. Correct, negative.For x = 1: (1)(2)(5) = 10. Correct, positive.So, the sign analysis seems correct.Therefore, the monotonicity of g(x) is as follows:- Decreasing on (-∞, -4)- Increasing on (-4, -1)- Decreasing on (-1, 0)- Increasing on (0, ∞)I think that's all for part 2. Let me just recap:1. Found a by setting the derivative of f(x) to zero at x = -4/3, which gave a = 1/2.2. For g(x) = f(x)e^x, computed its derivative, factored it, found critical points, and analyzed the sign of the derivative in each interval to determine where g(x) is increasing or decreasing.Everything seems to check out. I don't see any mistakes in my calculations or reasoning.**Final Answer**1. The value of ( a ) is (boxed{dfrac{1}{2}}).2. The function ( g(x) ) is decreasing on the intervals ( (-infty, -4) ) and ( (-1, 0) ), and increasing on the intervals ( (-4, -1) ) and ( (0, +infty) ).

💡Alright, so I've got this problem about Alyna, Bohdan, Vika, and Grysha coming to a physical education lesson wearing shorts and t-shirts, and each piece of clothing is either blue or red. I need to figure out the color of each child's t-shirt and shorts based on the given information.First, let's list out all the information provided:1. Alyna and Bohdan have red t-shirts, but their shorts are of different colors.2. Vika and Grysha have t-shirts of different colors, but both have blue shorts.3. The girls (Alyna and Vika) have t-shirts of different colors, and their shorts are also different in color.Okay, let's break this down step by step.Starting with point 1: Alyna and Bohdan have red t-shirts, but their shorts are different. So, Alyna has a red t-shirt, and Bohdan also has a red t-shirt. Their shorts, however, are different colors. Since the only colors available are blue and red, one of them must have blue shorts, and the other must have red shorts. But I don't know which one yet.Moving on to point 2: Vika and Grysha have t-shirts of different colors, but both have blue shorts. So, Vika and Grysha both have blue shorts. Their t-shirts are different, meaning one has a red t-shirt and the other has a blue t-shirt. Again, I don't know which is which yet.Point 3: The girls (Alyna and Vika) have t-shirts of different colors, and their shorts are also different in color. Since Alyna has a red t-shirt, Vika must have a blue t-shirt to make their t-shirts different. Also, their shorts are different, so if Alyna has red shorts, Vika must have blue shorts, and vice versa.Wait a minute, but from point 2, we already know that both Vika and Grysha have blue shorts. So, if Vika has blue shorts, then according to point 3, Alyna must have red shorts because their shorts are different.So, let's summarize what we have so far:- Alyna: Red t-shirt, Red shorts- Bohdan: Red t-shirt, Blue shorts (since Alyna has red shorts, Bohdan must have blue shorts to make their shorts different)- Vika: Blue t-shirt, Blue shorts- Grysha: Red t-shirt, Blue shorts (since Vika has a blue t-shirt, Grysha must have a red t-shirt)Wait, but hold on. If Grysha has a red t-shirt and blue shorts, and Bohdan also has a red t-shirt and blue shorts, that means both Grysha and Bohdan have the same combination of red t-shirt and blue shorts. Is that allowed? The problem doesn't specify that each child must have a unique combination, just that the girls have different t-shirts and shorts.But let's double-check. The girls (Alyna and Vika) have different t-shirts and different shorts, which is satisfied because Alyna has red t-shirt and red shorts, while Vika has blue t-shirt and blue shorts. So that's good.Bohdan and Grysha are both boys, and they both have red t-shirts, but their shorts are different? Wait, no, from point 1, Alyna and Bohdan have different shorts, but there's no information about Grysha's shorts relative to Bohdan's. However, from point 2, both Vika and Grysha have blue shorts, so Grysha has blue shorts, and Bohdan has blue shorts as well. So both Grysha and Bohdan have blue shorts, which is okay because they're both boys, and the problem doesn't specify that boys need to have different shorts.But wait, the problem says that Alyna and Bohdan have different shorts. So if Alyna has red shorts, Bohdan has blue shorts. And since Vika and Grysha both have blue shorts, that's consistent.So, putting it all together:- Alyna: Red t-shirt, Red shorts- Bohdan: Red t-shirt, Blue shorts- Vika: Blue t-shirt, Blue shorts- Grysha: Red t-shirt, Blue shortsIs there any conflict here? Let's see.- Alyna and Bohdan: Red t-shirts, different shorts (Alyna red, Bohdan blue) – checks out.- Vika and Grysha: Different t-shirts (Vika blue, Grysha red), both blue shorts – checks out.- Girls (Alyna and Vika): Different t-shirts (red and blue), different shorts (red and blue) – checks out.I think that works.

💡Alright, so I have this equation to consider:[(1 + a)x^4 + x^3 - (3a + 2)x^2 - 4a = 0]And I need to answer two questions about it. First, does this equation always have a real solution for any real value of ( a )? Second, is there a real number ( x_0 ) that is never a solution, no matter what ( a ) is?Okay, let's start with the first question. I need to check if there's always at least one real solution for ( x ) regardless of the value of ( a ). Hmm, since this is a quartic equation (degree 4), it can have up to four real roots, but it's not guaranteed to have any. However, I remember that for polynomials, the number of real roots depends on the behavior of the function as ( x ) approaches positive and negative infinity and whether it crosses the x-axis.Let me think about the leading term. The leading term is ( (1 + a)x^4 ). If ( 1 + a ) is positive, then as ( x ) approaches both positive and negative infinity, the function will go to positive infinity. If ( 1 + a ) is negative, it will go to negative infinity on both ends. If ( 1 + a = 0 ), then the leading term disappears, and the equation becomes a cubic.Wait, if ( 1 + a = 0 ), that means ( a = -1 ). Let me check what the equation becomes in that case:[(1 - 1)x^4 + x^3 - (3(-1) + 2)x^2 - 4(-1) = 0]Simplify:[0x^4 + x^3 - (-3 + 2)x^2 + 4 = 0]Which is:[x^3 + x^2 + 4 = 0]So, when ( a = -1 ), the equation reduces to a cubic. Cubic equations always have at least one real root, so in this case, there is at least one real solution. That's good to know.But what about when ( a neq -1 )? Then it's a quartic equation. As I thought earlier, the leading term is ( (1 + a)x^4 ). So, if ( 1 + a > 0 ), the function tends to positive infinity as ( x ) approaches both infinities. If ( 1 + a < 0 ), it tends to negative infinity on both ends.Now, for the function to have real roots, it needs to cross the x-axis. If the function tends to positive infinity on both ends, it must dip below the x-axis somewhere in between to have real roots. Similarly, if it tends to negative infinity on both ends, it must rise above the x-axis somewhere in between.But is that always the case? I mean, could the function just stay entirely above or below the x-axis without crossing it? For quartic functions, it's possible, but I need to check if that's the case here.Alternatively, maybe I can factor the equation or find a specific root that works for any ( a ). Let me try plugging in some simple values for ( x ) and see if they satisfy the equation regardless of ( a ).Let's try ( x = 1 ):[(1 + a)(1)^4 + (1)^3 - (3a + 2)(1)^2 - 4a = (1 + a) + 1 - (3a + 2) - 4a]Simplify:[1 + a + 1 - 3a - 2 - 4a = (1 + 1 - 2) + (a - 3a - 4a) = 0 - 6a]So, ( 0 - 6a = 0 ) implies ( a = 0 ). So, ( x = 1 ) is a solution only when ( a = 0 ). Not helpful for all ( a ).How about ( x = -1 ):[(1 + a)(-1)^4 + (-1)^3 - (3a + 2)(-1)^2 - 4a = (1 + a) - 1 - (3a + 2) - 4a]Simplify:[1 + a - 1 - 3a - 2 - 4a = (1 - 1 - 2) + (a - 3a - 4a) = (-2) + (-6a)]So, ( -2 - 6a = 0 ) implies ( a = -frac{1}{3} ). Again, only a specific ( a ).Hmm, maybe ( x = 2 ):[(1 + a)(16) + 8 - (3a + 2)(4) - 4a = 16(1 + a) + 8 - 4(3a + 2) - 4a]Simplify:[16 + 16a + 8 - 12a - 8 - 4a = (16 + 8 - 8) + (16a - 12a - 4a) = 16 + 0 = 16]So, ( 16 = 0 ) is never true. So, ( x = 2 ) is never a solution, regardless of ( a ). Interesting, that might be the answer to part 2.But let's get back to part 1. Maybe ( x = -2 ):[(1 + a)(16) + (-8) - (3a + 2)(4) - 4a = 16(1 + a) - 8 - 4(3a + 2) - 4a]Simplify:[16 + 16a - 8 - 12a - 8 - 4a = (16 - 8 - 8) + (16a - 12a - 4a) = 0 + 0 = 0]Wow, so ( x = -2 ) is always a solution, regardless of ( a ). That's great! So, for any real ( a ), ( x = -2 ) is a real solution. Therefore, the answer to part 1 is yes.Now, for part 2, as I saw earlier, ( x = 2 ) is never a solution because plugging it in gives 16 = 0, which is impossible. So, ( x_0 = 2 ) is a real number that is never a solution, no matter what ( a ) is.Wait, let me double-check that. If I plug ( x = 2 ) into the original equation:[(1 + a)(16) + 8 - (3a + 2)(4) - 4a = 16(1 + a) + 8 - 4(3a + 2) - 4a]Expanding:[16 + 16a + 8 - 12a - 8 - 4a = 16 + (16a - 12a - 4a) + (8 - 8) = 16 + 0 + 0 = 16]Yes, it's always 16, which is never zero. So, ( x = 2 ) is indeed never a solution.I think that's solid. So, to recap:1. The equation always has a real solution, specifically ( x = -2 ), for any real ( a ).2. There exists a real number ( x_0 = 2 ) that is never a solution, regardless of ( a ).I don't see any mistakes in my reasoning. I tested specific values, factored where possible, and confirmed the results. It seems consistent.

💡Okay, so I have this quadratic equation: 3x² + bx + 6x + 4 = 0. The problem is asking for the sum of all values of b that make this equation have exactly one solution. Hmm, I remember that a quadratic equation has exactly one solution when its discriminant is zero. Let me try to recall how that works.First, I think I should simplify the equation by combining like terms. The equation has two terms with x: bx and 6x. So, I can combine those to get (b + 6)x. That means the equation becomes 3x² + (b + 6)x + 4 = 0. Okay, that looks better.Now, for a quadratic equation in the form ax² + cx + d = 0, the discriminant is given by c² - 4ad. If the discriminant is zero, the equation has exactly one real solution. So, in this case, a is 3, c is (b + 6), and d is 4. Let me write that down:Discriminant = (b + 6)² - 4 * 3 * 4.Simplifying that, it becomes (b + 6)² - 48. And since we want exactly one solution, we set this equal to zero:(b + 6)² - 48 = 0.Alright, now I need to solve for b. Let me expand (b + 6)² first. That would be b² + 12b + 36. So, substituting back into the equation:b² + 12b + 36 - 48 = 0.Simplifying that, 36 - 48 is -12, so the equation becomes:b² + 12b - 12 = 0.Now, I have a quadratic equation in terms of b: b² + 12b - 12 = 0. I need to find the values of b that satisfy this equation. I can use the quadratic formula here, which is (-c ± √(c² - 4ad)) / (2a). In this case, a is 1, c is 12, and d is -12.Plugging those values into the quadratic formula:b = [-12 ± √(12² - 4 * 1 * (-12))] / (2 * 1).Calculating inside the square root:12² is 144, and 4 * 1 * (-12) is -48. So, it becomes:√(144 - (-48)) = √(144 + 48) = √192.Hmm, √192 can be simplified. Let me see, 192 divided by 64 is 3, so √192 is √(64 * 3) which is 8√3.So, substituting back into the formula:b = [-12 ± 8√3] / 2.I can simplify this by dividing both terms in the numerator by 2:b = (-12/2) ± (8√3)/2 = -6 ± 4√3.So, the two values of b are -6 + 4√3 and -6 - 4√3. The problem asks for the sum of all values of b. So, I need to add these two solutions together.Let me write that out:(-6 + 4√3) + (-6 - 4√3).Combining like terms, the -6 and -6 add up to -12, and the 4√3 and -4√3 cancel each other out, leaving 0. So, the sum is -12.Wait, that's interesting. I remember something about Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots. For a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a. In this case, our quadratic in b is b² + 12b - 12 = 0. So, the sum of the roots should be -12/1, which is -12. That matches what I got by adding the two solutions. So, that's a good check.I think I'm confident now that the sum of all values of b is -12. Let me just recap the steps to make sure I didn't skip anything:1. Combine like terms in the original equation to get 3x² + (b + 6)x + 4 = 0.2. Set the discriminant equal to zero for exactly one solution: (b + 6)² - 48 = 0.3. Solve the resulting quadratic equation for b, which gives two solutions: -6 + 4√3 and -6 - 4√3.4. Add these two solutions together to get the sum, which is -12.5. Verified using Vieta's formula, which also gives -12.I don't see any mistakes in my reasoning, so I think the answer is correct.**Final Answer**The sum of all values of ( b ) is boxed{-12}.