💡First, I need to determine the total number of people who attended the baseball game. I'll denote this total as ( P ).From the problem, I know that 40% of the audience supports the first team and 34% supports the second team. Adding these percentages together gives me the total percentage of people who support either team: ( 40% + 34% = 74% ).This means that the remaining percentage of people who do not support either team is ( 100% - 74% = 26% ).The problem states that there are 3 people who did not support either team. This corresponds to the 26% of the total audience. I can set up the equation ( 0.26 times P = 3 ) to find the total number of attendees.Solving for ( P ), I divide both sides of the equation by 0.26:[P = frac{3}{0.26} = frac{300}{26} = frac{150}{13} approx 11.54]Since the number of people must be a whole number, I round up to the nearest whole number, which is 12.Therefore, the total number of people who attended the game is 12.

💡Okay, so I have this problem here: given that n is a positive integer and n is greater than 19, I need to express the product n(n-1)(n-2)...(n-19) using permutation notation. Hmm, permutations... I remember that permutations involve arranging objects where the order matters. Let me recall the formula for permutations. I think it's something like P(n, k) which represents the number of ways to arrange k objects out of n. The formula for that is n! divided by (n - k)!, right? So, P(n, k) = n! / (n - k)!.Now, looking at the expression n(n-1)(n-2)...(n-19), this is a product of 20 consecutive numbers starting from n and going down to n-19. So, how many terms are there in this product? Let me count: starting at n, then n-1, n-2, ..., down to n-19. That should be 20 terms because from 0 to 19 is 20 numbers.Wait, so if I have 20 terms, does that mean k is 20 in the permutation formula? Let me think. If I have P(n, k), that's n! / (n - k)!. So, if I set k = 20, then P(n, 20) would be n! / (n - 20)!.But in the given expression, it's n(n-1)...(n-19). Let me write that out as a product: n × (n-1) × (n-2) × ... × (n-19). How many terms is that? Let's see, starting at n and subtracting 0, 1, 2, ..., 19. So that's 20 terms in total.So, n(n-1)...(n-19) is equal to n! divided by (n - 20)! because it's the product of 20 consecutive numbers starting from n. Therefore, this should be equal to P(n, 20).Wait, let me verify that. If I take n! / (n - 20)!, that would indeed be the product of n × (n-1) × ... × (n - 19), which is exactly what we have here. So, yes, that makes sense.Is there another way to think about this? Maybe using factorial notation. The product n(n-1)...(n-19) can be written as n! / (n - 20)! because it's the same as n! divided by the factorial of (n - 20). And since permutation P(n, k) is defined as n! / (n - k)!, substituting k = 20 gives us P(n, 20).So, putting it all together, n(n-1)...(n-19) is equal to P(n, 20). Therefore, the expression in permutation notation is P(n, 20).I think that's it. It makes sense because permutations count the number of ways to arrange k items out of n, and the product n(n-1)...(n-19) is exactly the number of permutations of 20 items from n, which is P(n, 20).**Final Answer**The expression ( n(n-1)ldots(n-19) ) using permutation notation is boxed{P_n^{20}}.

💡Let the angle be ( x ) degrees.The complement of ( x ) is ( 90^circ - x ).The supplement of ( x ) is ( 180^circ - x ).According to the problem, the complement is 33.3% (or (frac{1}{3})) of the supplement:[90 - x = frac{1}{3}(180 - x)]Multiply both sides by 3 to eliminate the fraction:[3(90 - x) = 180 - x]Simplify:[270 - 3x = 180 - x]Bring like terms together:[270 - 180 = 3x - x][90 = 2x]Solve for ( x ):[x = frac{90}{2} = 45^circ]Thus, the degree measure of the angle is ( 45^circ ).

💡Alright, so I have this problem with two propositions, p and q, and I need to find the range of the real number a such that either p or q is true, but not both. Hmm, okay, let's break this down step by step.First, let's understand proposition p: ∃x ∈ [0, π/2], cos²x + 2cosx - a = 0. So, this means there exists some x in the interval [0, π/2] such that the equation cos²x + 2cosx - a = 0 holds. I need to find the values of a for which this is true.To make this easier, maybe I can substitute t = cosx. Since x is in [0, π/2], cosx will range from 0 to 1. So, t ∈ [0, 1]. Then, the equation becomes t² + 2t - a = 0. So, we can solve for a: a = t² + 2t. Now, since t is between 0 and 1, let's see what values a can take.Let me consider the function f(t) = t² + 2t. Since t is in [0,1], let's find the minimum and maximum of f(t) in this interval. The function f(t) is a quadratic that opens upwards because the coefficient of t² is positive. The vertex of this parabola is at t = -b/(2a) = -2/(2*1) = -1. But since our interval is [0,1], the minimum occurs at t=0 and the maximum at t=1.Calculating f(0) = 0² + 2*0 = 0, and f(1) = 1² + 2*1 = 1 + 2 = 3. So, as t increases from 0 to 1, a increases from 0 to 3. Therefore, the range of a for which p is true is [0, 3].Okay, so p is true when a is between 0 and 3, inclusive.Now, let's look at proposition q: ∀x ∈ ℝ, x² + 2ax - 8 + 6a ≥ 0. This means that for all real numbers x, the quadratic expression x² + 2ax - 8 + 6a is non-negative. For a quadratic to be non-negative for all x, its discriminant must be less than or equal to zero. Because if the discriminant is positive, the quadratic will have real roots and will dip below the x-axis between them, making it negative somewhere.The discriminant D of the quadratic equation x² + 2ax - 8 + 6a = 0 is D = (2a)² - 4*1*(-8 + 6a). Let's compute that:D = 4a² - 4*(-8 + 6a) = 4a² + 32 - 24a.For the quadratic to be non-negative for all x, we need D ≤ 0:4a² - 24a + 32 ≤ 0.Let me simplify this inequality. First, divide all terms by 4:a² - 6a + 8 ≤ 0.Now, factor the quadratic:a² - 6a + 8 = (a - 2)(a - 4).So, the inequality becomes (a - 2)(a - 4) ≤ 0.To solve this, let's find the critical points: a = 2 and a = 4. These divide the real line into three intervals:1. a < 2,2. 2 < a < 4,3. a > 4.We can test each interval:- For a < 2, say a = 1: (1 - 2)(1 - 4) = (-1)(-3) = 3 > 0.- For 2 < a < 4, say a = 3: (3 - 2)(3 - 4) = (1)(-1) = -1 < 0.- For a > 4, say a = 5: (5 - 2)(5 - 4) = (3)(1) = 3 > 0.Since the inequality is ≤ 0, the solution is the interval where the product is negative or zero, which is [2, 4].So, proposition q is true when a is between 2 and 4, inclusive.Now, the problem states that either p or q is true, but not both. That means we need the values of a where p is true and q is false, or p is false and q is true.Let's find these intervals.First, p is true when a ∈ [0, 3], and q is true when a ∈ [2, 4]. So, the overlap where both are true is [2, 3]. Since we need either p or q, but not both, we need to exclude this overlapping interval.So, let's consider two cases:1. p is true and q is false: This happens when a is in [0, 3] but not in [2, 4]. So, a ∈ [0, 2).2. p is false and q is true: This happens when a is not in [0, 3] but is in [2, 4]. So, a ∈ (3, 4].Therefore, combining these two intervals, the range of a is [0, 2) ∪ (3, 4].Wait, let me double-check. When p is true, a is in [0,3]. So, for p to be true and q to be false, a must be in [0,3] but not in [2,4]. Since [2,4] overlaps with [0,3] from [2,3], so excluding that, we get [0,2).Similarly, when q is true, a is in [2,4]. For q to be true and p to be false, a must be in [2,4] but not in [0,3]. The overlap is [2,3], so excluding that, we get (3,4].Therefore, the union of [0,2) and (3,4] is indeed [0,2) ∪ (3,4].Let me just visualize this on a number line to make sure.- From 0 to 2: p is true, q is false.- From 2 to 3: both p and q are true, which we don't want.- From 3 to 4: p is false, q is true.- Beyond 4: both p and q are false.So, the desired intervals are [0,2) and (3,4], which together make [0,2) ∪ (3,4].I think that makes sense. So, the range of a is [0,2) union (3,4].**Final Answer**The range of the real number ( a ) is boxed{[0, 2) cup (3, 4]}.

💡Okay, so I have this problem here with two lines, l₁ and l₂, and I need to find the value of m that makes them parallel. Let me see, the equations are:l₁: (m - 2)x - y + 5 = 0l₂: (m - 2)x + (3 - m)y + 2 = 0Hmm, I remember that for two lines to be parallel, their slopes must be equal. So, I need to find the slopes of these two lines and set them equal to each other.First, let me rewrite both equations in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. That should make it easier to compare the slopes.Starting with l₁:(m - 2)x - y + 5 = 0Let me solve for y:- y = -(m - 2)x - 5Multiply both sides by -1 to make it positive:y = (m - 2)x + 5So, the slope of l₁ is (m - 2).Now, moving on to l₂:(m - 2)x + (3 - m)y + 2 = 0Again, I'll solve for y:(3 - m)y = -(m - 2)x - 2Now, divide both sides by (3 - m):y = [-(m - 2)/(3 - m)]x - [2/(3 - m)]Simplify the slope term:-(m - 2)/(3 - m) can be rewritten as (2 - m)/(3 - m). Wait, is that correct? Let me check:-(m - 2) = -m + 2 = 2 - mSo, yes, the slope simplifies to (2 - m)/(3 - m).So, the slope of l₂ is (2 - m)/(3 - m).Now, since the lines are parallel, their slopes must be equal. Therefore, I can set the slope of l₁ equal to the slope of l₂:(m - 2) = (2 - m)/(3 - m)Hmm, okay, now I need to solve this equation for m.Let me write that out:(m - 2) = (2 - m)/(3 - m)To solve for m, I'll cross-multiply to eliminate the denominator:(m - 2)(3 - m) = (2 - m)Let me expand the left side:First, multiply m by (3 - m):m*3 - m*m = 3m - m²Then, multiply -2 by (3 - m):-2*3 + (-2)*(-m) = -6 + 2mSo, combining these, the left side becomes:3m - m² - 6 + 2mCombine like terms:3m + 2m = 5mSo, left side is:5m - m² - 6Now, the equation is:5m - m² - 6 = 2 - mLet me bring all terms to one side to set the equation to zero:5m - m² - 6 - 2 + m = 0Combine like terms:5m + m = 6m-6 - 2 = -8So, the equation becomes:- m² + 6m - 8 = 0I can rewrite this as:m² - 6m + 8 = 0Now, I have a quadratic equation: m² - 6m + 8 = 0Let me try to factor this.Looking for two numbers that multiply to 8 and add up to -6.Wait, 8 can be factored as 2 and 4, and 2 + 4 = 6. Since the middle term is -6m, both numbers should be negative.So, (m - 2)(m - 4) = 0Setting each factor equal to zero:m - 2 = 0 => m = 2m - 4 = 0 => m = 4So, the solutions are m = 2 and m = 4.But wait, I need to check if these values make the lines parallel or if they cause any issues.Let me check m = 2 first.If m = 2, then l₁ becomes:(2 - 2)x - y + 5 = 0 => 0x - y + 5 = 0 => y = 5And l₂ becomes:(2 - 2)x + (3 - 2)y + 2 = 0 => 0x + 1y + 2 = 0 => y = -2So, both lines are horizontal lines with slopes 0. Therefore, they are parallel.Now, let's check m = 4.If m = 4, then l₁ becomes:(4 - 2)x - y + 5 = 0 => 2x - y + 5 = 0 => y = 2x + 5And l₂ becomes:(4 - 2)x + (3 - 4)y + 2 = 0 => 2x - 1y + 2 = 0 => y = 2x + 2So, both lines have a slope of 2, which means they are parallel.Wait, but earlier when I set the slopes equal, I got m = 2 and m = 4, and both seem to work. So, both values are valid.But let me think again. When m = 2, both lines become horizontal lines, which are parallel. When m = 4, both lines have the same slope of 2, so they are also parallel.Therefore, both m = 2 and m = 4 satisfy the condition that the lines are parallel.Wait, but in the original problem, it says "find the value of the real number m". So, does that mean there are two possible values? Or is there something I missed?Let me double-check my calculations.Starting from the slopes:Slope of l₁: (m - 2)Slope of l₂: (2 - m)/(3 - m)Set them equal:(m - 2) = (2 - m)/(3 - m)Cross-multiplying:(m - 2)(3 - m) = 2 - mExpanding:3m - m² - 6 + 2m = 2 - mCombine like terms:5m - m² - 6 = 2 - mBring all terms to left:5m - m² - 6 - 2 + m = 06m - m² - 8 = 0Multiply by -1:m² - 6m + 8 = 0Factor:(m - 2)(m - 4) = 0Solutions: m = 2 and m = 4Yes, that seems correct.But wait, when m = 2, the lines are horizontal, which is fine. When m = 4, the lines have the same slope, which is also fine.So, both values are valid.Therefore, the real number m can be either 2 or 4.Wait, but in the initial problem, it says "find the value of the real number m". So, does that mean multiple values are possible? Or is there a restriction?Let me check the original equations again.l₁: (m - 2)x - y + 5 = 0l₂: (m - 2)x + (3 - m)y + 2 = 0If m = 2, then l₁ becomes -y + 5 = 0 => y = 5And l₂ becomes 0x + (3 - 2)y + 2 = 0 => y = -2So, both are horizontal lines, which are parallel.If m = 4, then l₁ becomes 2x - y + 5 = 0 => y = 2x + 5And l₂ becomes 2x - y + 2 = 0 => y = 2x + 2Both have slope 2, so they are parallel.Therefore, both m = 2 and m = 4 are valid solutions.So, the answer is m = 2 or m = 4.But in the initial problem, it says "find the value of the real number m". Maybe it's expecting both values.Alternatively, perhaps I made a mistake in the slope calculation.Wait, let me re-examine the slope of l₂.Original equation: (m - 2)x + (3 - m)y + 2 = 0Solving for y:(3 - m)y = -(m - 2)x - 2So, y = [-(m - 2)/(3 - m)]x - [2/(3 - m)]Simplify the slope:-(m - 2)/(3 - m) = (2 - m)/(3 - m)Wait, is that correct?Yes, because -(m - 2) = -m + 2 = 2 - m.So, slope is (2 - m)/(3 - m)Alternatively, I can write it as (m - 2)/(m - 3) by factoring out a negative from numerator and denominator.Wait, let's see:(2 - m)/(3 - m) = -(m - 2)/-(m - 3) = (m - 2)/(m - 3)Wait, no, that's not correct. Let me do it step by step.(2 - m)/(3 - m) = -(m - 2)/-(m - 3) = (m - 2)/(m - 3)Wait, actually, that's correct because both numerator and denominator have a negative sign, which cancels out.So, (2 - m)/(3 - m) = (m - 2)/(m - 3)Therefore, the slope of l₂ is (m - 2)/(m - 3)So, setting slope of l₁ equal to slope of l₂:(m - 2) = (m - 2)/(m - 3)Hmm, that's another way to write it.So, (m - 2) = (m - 2)/(m - 3)Now, if m - 2 ≠ 0, we can divide both sides by (m - 2):1 = 1/(m - 3)Which implies:m - 3 = 1So, m = 4Alternatively, if m - 2 = 0, then m = 2, which we already considered.So, this confirms that m = 2 and m = 4 are the solutions.Therefore, the real number m can be either 2 or 4.So, the final answer is m = 2 or m = 4.

💡Alright, so I've got this problem here with two propositions, p and q, and I need to figure out the range of the real number m based on the truth values of these propositions. Let me try to break it down step by step.First, let's understand what each proposition is saying.**Proposition p:** The equation (frac{x^2}{m+2} - frac{y^2}{m-3} = 1) represents a hyperbola.Okay, so I remember that for an equation to represent a hyperbola, the denominators of the squared terms need to be positive. That means both (m+2) and (m-3) should be positive. So, I can write that as:[(m + 2) > 0 quad text{and} quad (m - 3) > 0]Simplifying these inequalities:1. (m + 2 > 0 implies m > -2)2. (m - 3 > 0 implies m > 3)So, for both to be true, m has to be greater than 3. But wait, I think I might have made a mistake here. Let me think again. Actually, for a hyperbola, it's sufficient that the product of the denominators is positive. So, ((m + 2)(m - 3) > 0). That means either both are positive or both are negative.So, solving ((m + 2)(m - 3) > 0):- The critical points are m = -2 and m = 3.- Testing intervals: - For m < -2: Let's pick m = -3. Then, (-3 + 2)(-3 - 3) = (-1)(-6) = 6 > 0. - For -2 < m < 3: Let's pick m = 0. (0 + 2)(0 - 3) = 2*(-3) = -6 < 0. - For m > 3: Let's pick m = 4. (4 + 2)(4 - 3) = 6*1 = 6 > 0.So, the solution is m < -2 or m > 3. Got it. So, p is true when m < -2 or m > 3.**Proposition q:** The equation (mx^2 + (m + 3)x + 4 = 0) has no positive real roots.Alright, so I need to find when this quadratic equation has no positive real roots. Let's recall that for a quadratic equation (ax^2 + bx + c = 0), the roots can be found using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]For the equation to have no positive real roots, a few conditions must be satisfied. Let's consider different cases based on the value of m.**Case 1: m = 0**If m = 0, the equation becomes linear: (0x^2 + (0 + 3)x + 4 = 0 implies 3x + 4 = 0). Solving this gives x = -4/3, which is negative. So, there are no positive roots. Therefore, when m = 0, q is true.**Case 2: m ≠ 0**Here, it's a quadratic equation. For it to have no positive real roots, the following must be true:1. The discriminant must be negative (no real roots), or2. If there are real roots, both roots must be negative.Let's compute the discriminant:[Delta = (m + 3)^2 - 4 cdot m cdot 4 = m^2 + 6m + 9 - 16m = m^2 - 10m + 9]So, (Delta = m^2 - 10m + 9).**Subcase 2a: (Delta < 0)**We need to find m such that (m^2 - 10m + 9 < 0).Solving the inequality:First, find the roots of (m^2 - 10m + 9 = 0):[m = frac{10 pm sqrt{100 - 36}}{2} = frac{10 pm sqrt{64}}{2} = frac{10 pm 8}{2}]So, m = (10 + 8)/2 = 9 and m = (10 - 8)/2 = 1.Thus, the quadratic (m^2 - 10m + 9) is less than 0 between its roots:[1 < m < 9]So, for 1 < m < 9, the discriminant is negative, meaning no real roots, hence no positive roots. Therefore, q is true when 1 < m < 9.**Subcase 2b: (Delta geq 0)**When (Delta geq 0), the equation has real roots. To ensure no positive roots, both roots must be negative. For a quadratic equation (ax^2 + bx + c = 0), the conditions for both roots to be negative are:1. The sum of the roots, (-b/a), must be positive.2. The product of the roots, (c/a), must be positive.Let's compute these:1. Sum of roots: (- (m + 3)/m)2. Product of roots: (4/m)So, for both roots to be negative:1. (- (m + 3)/m > 0)2. (4/m > 0)Let's analyze these inequalities.**Condition 1: (- (m + 3)/m > 0)**This simplifies to:[frac{-(m + 3)}{m} > 0 implies frac{m + 3}{m} < 0]So, (frac{m + 3}{m} < 0). This occurs when numerator and denominator have opposite signs.- If m > 0, then m + 3 > 0, so (frac{m + 3}{m} > 0). Doesn't satisfy the inequality.- If m < 0, then m + 3 could be positive or negative depending on m.Let's find when m + 3 and m have opposite signs.- If m < -3, then m + 3 < 0 and m < 0, so both negative. Thus, (frac{m + 3}{m} > 0). Doesn't satisfy.- If -3 < m < 0, then m + 3 > 0 and m < 0, so (frac{m + 3}{m} < 0). This satisfies the inequality.So, Condition 1 is satisfied when -3 < m < 0.**Condition 2: (4/m > 0)**This is positive when m > 0.So, combining both conditions:- From Condition 1: -3 < m < 0- From Condition 2: m > 0There is no overlap between these two conditions. So, there is no m that satisfies both conditions simultaneously.Wait, that can't be right. If both roots need to be negative, but the conditions lead to no solution, does that mean that when (Delta geq 0), there are no m values where both roots are negative? Hmm, that seems odd.Let me double-check.Given the quadratic equation (mx^2 + (m + 3)x + 4 = 0), with m ≠ 0.For both roots to be negative:1. Sum of roots: (- (m + 3)/m > 0)2. Product of roots: (4/m > 0)From the product of roots, (4/m > 0) implies m > 0.From the sum of roots, (- (m + 3)/m > 0). Since m > 0, the denominator is positive, so the numerator must be negative:[-(m + 3) > 0 implies m + 3 < 0 implies m < -3]But we have m > 0 from the product condition. So, m must satisfy both m > 0 and m < -3, which is impossible. Therefore, there is no m > 0 for which both roots are negative.So, in the case where (Delta geq 0), there are no m values where the equation has no positive roots. Therefore, q is only true when (Delta < 0), which is 1 < m < 9, and when m = 0.Wait, but earlier, when m = 0, the equation is linear and has no positive roots. So, q is true for m = 0 as well.So, combining both cases:- When m = 0: q is true.- When 1 < m < 9: q is true.But what about when m ≤ 1 or m ≥ 9?For m ≤ 1 or m ≥ 9, (Delta geq 0), meaning real roots exist. But as we saw, for m > 0, the roots cannot both be negative because it's impossible. So, for m > 0, if there are real roots, at least one root is positive or both are positive.Wait, but for m > 0, if the product of roots is positive (since 4/m > 0), and the sum of roots is negative (since (- (m + 3)/m < 0) when m > 0). So, if the sum is negative and the product is positive, both roots are negative. Wait, that contradicts my earlier conclusion.Wait, let's re-examine.If m > 0:- Sum of roots: (- (m + 3)/m). Since m > 0, this is negative.- Product of roots: (4/m > 0).So, if the sum is negative and the product is positive, both roots are negative. Therefore, for m > 0, if (Delta geq 0), both roots are negative, meaning no positive roots.Wait, that's different from what I concluded earlier. So, perhaps I made a mistake earlier.Let me re-express the conditions:For a quadratic equation (ax^2 + bx + c = 0), with a ≠ 0.- If a > 0, and the sum of roots is negative, and the product is positive, then both roots are negative.- If a > 0, and the sum of roots is positive, and the product is positive, then both roots are positive.- If a > 0, and the product is negative, then one root is positive and one is negative.Similarly, if a < 0, the signs would flip.So, in our case, a = m.If m > 0:- Sum of roots: (- (m + 3)/m). Since m > 0, this is negative.- Product of roots: (4/m > 0).So, both roots are negative.Therefore, for m > 0, if (Delta geq 0), both roots are negative, so no positive roots.Similarly, if m < 0:- Sum of roots: (- (m + 3)/m). Let's see, m < 0, so denominator is negative.Sum of roots: (- (m + 3)/m = (-m - 3)/m = (- (m + 3))/m). Since m < 0, and m + 3 could be positive or negative.Product of roots: (4/m < 0), since m < 0.So, product is negative, meaning one root is positive and one is negative.Therefore, for m < 0, if (Delta geq 0), the equation has one positive and one negative root, which means there is a positive root. Therefore, q is false.So, summarizing:- When m = 0: q is true.- When m > 0: - If 1 < m < 9: (Delta < 0), so no real roots, hence no positive roots. q is true. - If m ≤ 1 or m ≥ 9: (Delta geq 0), but since m > 0, both roots are negative. So, no positive roots. q is true.- When m < 0: - If (Delta < 0): No real roots, so no positive roots. q is true. - If (Delta geq 0): One positive and one negative root. q is false.Wait, so earlier I thought that for m > 0, if (Delta geq 0), both roots are negative, so q is true. But earlier, I thought that for m > 0, when (Delta geq 0), it's impossible for both roots to be negative because of conflicting conditions. But now, upon re-examining, it seems that for m > 0, if (Delta geq 0), both roots are negative.But let's test with specific values.Let's pick m = 2, which is between 1 and 9.Equation: 2x² + (2 + 3)x + 4 = 0 → 2x² + 5x + 4 = 0Discriminant: 25 - 32 = -7 < 0. So, no real roots. q is true.Now, pick m = 10, which is greater than 9.Equation: 10x² + (10 + 3)x + 4 = 0 → 10x² + 13x + 4 = 0Discriminant: 169 - 160 = 9 > 0. So, real roots.Sum of roots: -13/10 < 0Product of roots: 4/10 > 0So, both roots are negative. Therefore, no positive roots. q is true.Similarly, pick m = 0.5, which is less than 1.Equation: 0.5x² + (0.5 + 3)x + 4 = 0 → 0.5x² + 3.5x + 4 = 0Discriminant: (3.5)^2 - 4*0.5*4 = 12.25 - 8 = 4.25 > 0Sum of roots: -3.5/0.5 = -7 < 0Product of roots: 4/0.5 = 8 > 0So, both roots are negative. q is true.Now, pick m = -1, which is less than 0.Equation: -1x² + (-1 + 3)x + 4 = 0 → -x² + 2x + 4 = 0Multiply both sides by -1: x² - 2x - 4 = 0Discriminant: 4 + 16 = 20 > 0Sum of roots: 2/1 = 2 > 0Product of roots: -4/1 = -4 < 0So, one positive and one negative root. Therefore, q is false.Similarly, pick m = -4.Equation: -4x² + (-4 + 3)x + 4 = 0 → -4x² - x + 4 = 0Multiply by -1: 4x² + x - 4 = 0Discriminant: 1 + 64 = 65 > 0Sum of roots: -1/4 < 0Product of roots: -4/4 = -1 < 0So, one positive and one negative root. q is false.Wait, but earlier, I thought that for m < 0, if (Delta geq 0), then one root is positive and one is negative, making q false. But for m = -4, the sum of roots is negative, and product is negative, so one positive and one negative. So, q is false.But what about m = -0.5?Equation: -0.5x² + (-0.5 + 3)x + 4 = 0 → -0.5x² + 2.5x + 4 = 0Multiply by -2: x² - 5x - 8 = 0Discriminant: 25 + 32 = 57 > 0Sum of roots: 5/1 = 5 > 0Product of roots: -8/1 = -8 < 0So, one positive and one negative root. q is false.Therefore, for m < 0, regardless of the discriminant, if (Delta geq 0), there is at least one positive root, making q false. If (Delta < 0), no real roots, so q is true.Wait, but for m < 0, when is (Delta < 0)?We have (Delta = m^2 - 10m + 9). Let's see when this is negative.We already found that (Delta < 0) when 1 < m < 9.But for m < 0, m is less than 1, so (Delta = m^2 - 10m + 9). Let's see:For m = -1, (Delta = 1 + 10 + 9 = 20 > 0)For m = -2, (Delta = 4 + 20 + 9 = 33 > 0)So, for m < 0, (Delta) is always positive because m^2 is positive, -10m is positive (since m is negative), and +9 is positive. So, (Delta) is always positive for m < 0.Therefore, for m < 0, (Delta geq 0), meaning real roots exist, and since m < 0, one root is positive and one is negative. Therefore, q is false for all m < 0.So, summarizing:- q is true when: - m = 0 - m > 0 (since for m > 0, regardless of (Delta), no positive roots)- q is false when m < 0Wait, but earlier, I thought that for m > 0, if (Delta geq 0), both roots are negative, so no positive roots. And if (Delta < 0), no real roots, so no positive roots. Therefore, for all m > 0, q is true.Similarly, for m = 0, q is true.For m < 0, q is false.So, q is true for m ≥ 0.Wait, that seems to be the case.Let me confirm with m = 10, which is greater than 9:Equation: 10x² + 13x + 4 = 0Discriminant: 169 - 160 = 9 > 0Sum of roots: -13/10 < 0Product of roots: 4/10 > 0So, both roots are negative. q is true.Similarly, m = 1:Equation: 1x² + 4x + 4 = 0Discriminant: 16 - 16 = 0So, repeated root at x = -2. Both roots are negative. q is true.m = 0.5:Equation: 0.5x² + 3.5x + 4 = 0Discriminant: 12.25 - 8 = 4.25 > 0Sum of roots: -3.5/0.5 = -7 < 0Product of roots: 4/0.5 = 8 > 0Both roots negative. q is true.m = 0:Equation: 3x + 4 = 0 → x = -4/3. No positive roots. q is true.m = -1:Equation: -x² + 2x + 4 = 0Discriminant: 4 + 16 = 20 > 0Sum of roots: 2/(-1) = -2 < 0Product of roots: 4/(-1) = -4 < 0So, one positive and one negative root. q is false.Therefore, q is true for m ≥ 0, and false for m < 0.So, to recap:- p is true when m < -2 or m > 3- q is true when m ≥ 0Now, the problem states that (p vee q) is true and (p wedge q) is false.Let's interpret this:- (p vee q) is true: At least one of p or q is true.- (p wedge q) is false: Both p and q cannot be true simultaneously.So, we need to find the values of m where either p is true or q is true, but not both.Let's analyze the truth values of p and q:1. When m < -2: - p is true - q is false (since m < 0) - So, p ∨ q is true (because p is true), and p ∧ q is false (because q is false). This satisfies both conditions.2. When -2 ≤ m < 0: - p is false (since m is not < -2 or > 3) - q is false (since m < 0) - So, p ∨ q is false, which does not satisfy the condition that p ∨ q is true. Therefore, this interval is excluded.3. When 0 ≤ m ≤ 3: - p is false (since m is not < -2 or > 3) - q is true (since m ≥ 0) - So, p ∨ q is true (because q is true), and p ∧ q is false (because p is false). This satisfies both conditions.4. When m > 3: - p is true - q is true (since m ≥ 0) - So, p ∨ q is true, but p ∧ q is also true, which violates the condition that p ∧ q is false. Therefore, this interval is excluded.Therefore, the valid intervals are:- m < -2- 0 ≤ m ≤ 3So, combining these, the range of m is:[m < -2 quad text{or} quad 0 leq m leq 3]I think that's the solution. Let me just double-check.For m < -2:- p is true, q is false. So, p ∨ q is true, p ∧ q is false. Good.For -2 ≤ m < 0:- p is false, q is false. So, p ∨ q is false. Doesn't satisfy. Correct.For 0 ≤ m ≤ 3:- p is false, q is true. So, p ∨ q is true, p ∧ q is false. Good.For m > 3:- p is true, q is true. So, p ∨ q is true, but p ∧ q is true. Doesn't satisfy. Correct.Yes, that seems consistent.