🧠:<think>Alright, so I've got this algebra problem here, and I need to figure it out step by step. Let me read it again carefully.Given:A = x² - 2xy + y²B = x² + 2xy + y²And the equation is 3A - 2B + C = 0.There are two parts to the problem:1. Find the expression for the algebraic term C.2. If x = 1/2 and y = -2, what is the value of C?Okay, starting with part (1). I need to find C in terms of x and y. The equation given is 3A - 2B + C = 0. So, I can rearrange this equation to solve for C.Let me write that down:3A - 2B + C = 0So, C = 2B - 3AAlright, now I need to substitute the expressions for A and B into this equation.Given:A = x² - 2xy + y²B = x² + 2xy + y²So, substituting these into C = 2B - 3A:C = 2*(x² + 2xy + y²) - 3*(x² - 2xy + y²)Now, I need to expand these terms. Let's do that step by step.First, expand 2*(x² + 2xy + y²):2*x² = 2x²2*2xy = 4xy2*y² = 2y²So, 2*(x² + 2xy + y²) = 2x² + 4xy + 2y²Next, expand -3*(x² - 2xy + y²):-3*x² = -3x²-3*(-2xy) = +6xy-3*y² = -3y²So, -3*(x² - 2xy + y²) = -3x² + 6xy - 3y²Now, combine these two results:C = (2x² + 4xy + 2y²) + (-3x² + 6xy - 3y²)Let me combine like terms.First, the x² terms:2x² - 3x² = -x²Next, the xy terms:4xy + 6xy = 10xyFinally, the y² terms:2y² - 3y² = -y²Putting it all together:C = -x² + 10xy - y²Okay, that seems to be the expression for C. Let me double-check my steps to make sure I didn't make a mistake.Starting from C = 2B - 3A, substituting A and B correctly, expanding each term, and then combining like terms. It looks correct. So, part (1) is done.Now, moving on to part (2). I need to find the value of C when x = 1/2 and y = -2.Given:x = 1/2y = -2And we have the expression for C:C = -x² + 10xy - y²Let me substitute x and y into this expression.First, calculate each term separately.1. -x²:x = 1/2, so x² = (1/2)² = 1/4So, -x² = -1/42. 10xy:x = 1/2, y = -2So, xy = (1/2)*(-2) = -1Then, 10xy = 10*(-1) = -103. -y²:y = -2, so y² = (-2)² = 4So, -y² = -4Now, add all these together:C = (-1/4) + (-10) + (-4)Let me compute this step by step.First, (-1/4) + (-10) = -10.25Then, -10.25 + (-4) = -14.25Wait, that's in decimal. But the question might prefer a fraction.Let me convert -14.25 to a fraction.0.25 is 1/4, so 14.25 is 14 and 1/4, which is 57/4.But since it's negative, it's -57/4.So, C = -57/4Let me verify my calculations again to make sure.Calculating each term:- x² = -(1/2)² = -1/410xy = 10*(1/2)*(-2) = 10*(-1) = -10-y² = -(-2)² = -4Adding them up: -1/4 -10 -4Convert all to fractions to add:-1/4 -10 is the same as -1/4 -40/4 = (-1 -40)/4 = -41/4Then, -41/4 -4 = -41/4 -16/4 = (-41 -16)/4 = -57/4Yes, that's correct.So, the value of C is -57/4 when x = 1/2 and y = -2.Just to make sure, let me plug x and y back into the original equation 3A - 2B + C = 0 and see if it holds.First, compute A and B with x = 1/2 and y = -2.A = x² - 2xy + y²= (1/2)² - 2*(1/2)*(-2) + (-2)²= 1/4 - 2*(-1) + 4= 1/4 + 2 + 4= 1/4 + 6= 6.25 or 25/4B = x² + 2xy + y²= (1/2)² + 2*(1/2)*(-2) + (-2)²= 1/4 + (-2) + 4= 1/4 + 2= 2.25 or 9/4Now, compute 3A - 2B + C.3A = 3*(25/4) = 75/42B = 2*(9/4) = 18/4 = 9/2C = -57/4So, 3A - 2B + C = 75/4 - 9/2 + (-57/4)Convert 9/2 to 18/4 to have a common denominator.So, 75/4 - 18/4 -57/4 = (75 - 18 -57)/4 = (75 -75)/4 = 0/4 = 0Yes, it satisfies the equation 3A - 2B + C = 0. So, my value for C is correct.Therefore, I'm confident that my answers are correct.</think>

🧠:<think>Alright, so I have this quadratic function ( f(x) = ax^2 + bx + 2 ) where ( a neq 0 ). The problem states that for any real number ( x ), the absolute value of ( f(x) ) is at least 2, meaning ( |f(x)| geq 2 ). I need to find the coordinates of the focus of this parabola.First, I remember that a quadratic function in standard form is ( f(x) = ax^2 + bx + c ), and its graph is a parabola. The focus of a parabola is a specific point that lies inside the parabola, and its coordinates can be found using the formula related to the vertex and the coefficient ( a ).Since the problem mentions that ( |f(x)| geq 2 ) for all real ( x ), this tells me something about the minimum or maximum value of the function. If the absolute value is always at least 2, then the function never goes below 2 or above -2. But since the constant term is 2, when ( x = 0 ), ( f(0) = 2 ). So, the point (0, 2) is on the graph.Now, if ( |f(x)| geq 2 ), that means the function doesn't dip below 2 or rise above -2. But since ( f(0) = 2 ), which is already at the lower bound, the function must be such that it doesn't go below 2. So, the minimum value of the function is 2. Therefore, the vertex of the parabola must be at (h, 2), where h is the x-coordinate of the vertex.Wait, but if the vertex is at (h, 2), and the parabola doesn't go below 2, that means the parabola opens upwards. So, ( a ) must be positive. If ( a ) were negative, the parabola would open downward, and the vertex would be a maximum, which would conflict with the condition ( |f(x)| geq 2 ) because the function would go below 2.So, ( a > 0 ). Now, to find the vertex, I know that the x-coordinate of the vertex is given by ( -frac{b}{2a} ). Let's denote the vertex as ( (h, k) ), so ( h = -frac{b}{2a} ) and ( k = f(h) ).Since the minimum value of the function is 2, ( k = 2 ). So, ( f(h) = 2 ). Let's compute ( f(h) ):( f(h) = a(h)^2 + b(h) + 2 )But ( h = -frac{b}{2a} ), so substituting:( f(h) = aleft(-frac{b}{2a}right)^2 + bleft(-frac{b}{2a}right) + 2 )Simplify this:First term: ( a times frac{b^2}{4a^2} = frac{b^2}{4a} )Second term: ( -frac{b^2}{2a} )Third term: 2So, putting it all together:( f(h) = frac{b^2}{4a} - frac{b^2}{2a} + 2 )Combine the first two terms:( frac{b^2}{4a} - frac{2b^2}{4a} = -frac{b^2}{4a} )So, ( f(h) = -frac{b^2}{4a} + 2 )But we know that ( f(h) = 2 ), so:( -frac{b^2}{4a} + 2 = 2 )Subtract 2 from both sides:( -frac{b^2}{4a} = 0 )Multiply both sides by ( -4a ):( b^2 = 0 )So, ( b = 0 )Ah, so the linear term ( bx ) is zero. Therefore, the quadratic function simplifies to ( f(x) = ax^2 + 2 ).Now, since ( b = 0 ), the vertex is at ( h = -frac{b}{2a} = 0 ). So, the vertex is at (0, 2).Now, to find the focus of the parabola. For a parabola in the form ( f(x) = ax^2 + c ), the focus is located at ( (h, k + frac{1}{4a}) ), where ( (h, k) ) is the vertex.In our case, the vertex is at (0, 2), so the focus is at ( (0, 2 + frac{1}{4a}) ).Wait, but I need to make sure about the formula for the focus. Let me recall: for a parabola ( y = ax^2 + bx + c ), the focus is at ( (h, k + frac{1}{4a}) ), where ( h = -frac{b}{2a} ) and ( k = c - frac{b^2}{4a} ).But in our case, ( b = 0 ), so ( h = 0 ) and ( k = 2 ). Therefore, the focus is at ( (0, 2 + frac{1}{4a}) ).But wait, I think I might have made a mistake here. Let me double-check the formula for the focus of a parabola.The standard form of a parabola that opens upward or downward is ( y = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. The focus is located at ( (h, k + frac{1}{4a}) ).Yes, that's correct. So, in our case, since the vertex is at (0, 2), the focus is at ( (0, 2 + frac{1}{4a}) ).But wait, the problem didn't specify whether the parabola opens upward or downward. However, since ( a > 0 ), it opens upward, so the focus is above the vertex.Therefore, the coordinates of the focus are ( (0, 2 + frac{1}{4a}) ).But let me think again. The problem states that ( |f(x)| geq 2 ) for all real ( x ). We concluded that ( b = 0 ) and the function is ( f(x) = ax^2 + 2 ). The vertex is at (0, 2), and since ( a > 0 ), the parabola opens upward, so the minimum value is 2, which satisfies ( |f(x)| geq 2 ).But wait, if the parabola opens upward, then ( f(x) ) can go to infinity as ( x ) increases or decreases, so ( |f(x)| ) will definitely be greater than or equal to 2. However, if the parabola opened downward, ( a < 0 ), then the function would have a maximum at the vertex, which is 2, and the function would go to negative infinity as ( x ) increases or decreases, which would violate the condition ( |f(x)| geq 2 ) because it would go below -2.Therefore, our conclusion that ( a > 0 ) and ( b = 0 ) seems correct.So, the focus is at ( (0, 2 + frac{1}{4a}) ).But let me check if there's another way to approach this problem, maybe by considering the discriminant or something else.Another approach: Since ( |f(x)| geq 2 ) for all ( x ), then either ( f(x) geq 2 ) for all ( x ) or ( f(x) leq -2 ) for all ( x ). But since ( f(0) = 2 ), it can't be that ( f(x) leq -2 ) for all ( x ), because at ( x = 0 ), it's exactly 2. Therefore, ( f(x) geq 2 ) for all ( x ).So, the quadratic function must be always above or equal to 2. That means the minimum value of the function is 2, which occurs at the vertex.Therefore, the vertex is at (0, 2), and the function is ( f(x) = ax^2 + 2 ).Now, to find the focus, we can write the function in vertex form: ( f(x) = a(x - 0)^2 + 2 ), so ( a ) is the same as before.The formula for the focus of a parabola ( y = a(x - h)^2 + k ) is ( (h, k + frac{1}{4a}) ).Therefore, the focus is at ( (0, 2 + frac{1}{4a}) ).So, I think that's the answer. But let me make sure I didn't miss anything.Wait, the problem didn't specify whether ( a ) is positive or negative, but from the condition ( |f(x)| geq 2 ) and ( f(0) = 2 ), we concluded that ( a ) must be positive. So, the focus is above the vertex.If ( a ) were negative, the parabola would open downward, and the vertex would be a maximum, which would mean the function can't satisfy ( |f(x)| geq 2 ) because it would go below 2 as ( x ) increases or decreases.Therefore, ( a > 0 ), and the focus is at ( (0, 2 + frac{1}{4a}) ).I think that's correct.</think>

🧠:<think>First, I need to determine how many times Jason fires his weapon in one minute. Since he fires every 15 seconds, I can calculate the number of fires per minute by dividing the total number of seconds in a minute (60 seconds) by the firing interval (15 seconds).Next, I know that Jason shoots flames for a total of 20 seconds per minute. To find out how long each individual flame lasts, I can divide the total flame duration (20 seconds) by the number of fires per minute (4 fires).By performing this calculation, I can determine the duration of each flame.</think>

🧠:<think>Alright, so I have this function f(x) = (1/2) sin 2x - sqrt(3) cos²x. I need to find its smallest positive period and its minimum value. Then, there's a second part where I have to stretch the x-coordinate of each point on the graph of f(x) to twice its original length to get g(x), and find the range of g(x) when x is between π/2 and π.Starting with part (1). I remember that for trigonometric functions, the period is the length after which the function repeats itself. The function f(x) has two terms: (1/2) sin 2x and -sqrt(3) cos²x. I need to find the periods of each term and then find the least common multiple (LCM) of those periods to get the smallest positive period of f(x).First, let's consider the term (1/2) sin 2x. The general form of a sine function is sin(Bx), and its period is 2π / |B|. Here, B is 2, so the period is 2π / 2 = π. So, the period for this term is π.Next, the term -sqrt(3) cos²x. The cosine squared function can be rewritten using a double-angle identity to make it easier to find the period. I recall that cos²x = (1 + cos 2x)/2. So, substituting that in, the term becomes -sqrt(3) * (1 + cos 2x)/2. Simplifying, that's -sqrt(3)/2 - (sqrt(3)/2) cos 2x.Now, the cos 2x term has a period of 2π / 2 = π, similar to the sine term. The constant term -sqrt(3)/2 doesn't affect the period. So, the period of the entire function f(x) is the LCM of π and π, which is just π. So, the smallest positive period of f(x) is π.Now, to find the minimum value of f(x). Let me rewrite f(x) using the identity I just applied. So, f(x) = (1/2) sin 2x - sqrt(3) cos²x becomes:f(x) = (1/2) sin 2x - sqrt(3) * (1 + cos 2x)/2Let me compute that step by step:First, expand the second term:= (1/2) sin 2x - (sqrt(3)/2)(1 + cos 2x)= (1/2) sin 2x - sqrt(3)/2 - (sqrt(3)/2) cos 2xSo, f(x) = (1/2) sin 2x - (sqrt(3)/2) cos 2x - sqrt(3)/2Now, I notice that the first two terms can be combined into a single sine function using the amplitude-phase form. The general form is A sin(2x + φ), where A is the amplitude and φ is the phase shift.Let me compute A:A = sqrt( (1/2)^2 + (sqrt(3)/2)^2 ) = sqrt(1/4 + 3/4) = sqrt(1) = 1So, the amplitude is 1. Now, let's find the phase shift φ.tan φ = (coefficient of cos term) / (coefficient of sin term) = (-sqrt(3)/2) / (1/2) = -sqrt(3)So, φ = arctan(-sqrt(3)). Since tan(π/3) = sqrt(3), and since the coefficient of cos is negative and sin is positive, the angle φ is in the fourth quadrant. So, φ = -π/3.Therefore, f(x) can be rewritten as:f(x) = sin(2x - π/3) - sqrt(3)/2So, f(x) = sin(2x - π/3) - sqrt(3)/2Now, the sine function has a range of [-1, 1]. Therefore, sin(2x - π/3) varies between -1 and 1. So, the minimum value of f(x) is when sin(2x - π/3) is -1, which gives:Minimum value = -1 - sqrt(3)/2To write that as a single fraction, it's (-2 - sqrt(3))/2.So, the minimum value is (-2 - sqrt(3))/2.Okay, so part (1) is done. The smallest positive period is π, and the minimum value is (-2 - sqrt(3))/2.Moving on to part (2). We need to stretch the x-coordinate of each point on the graph of f(x) to twice its original length while keeping the y-coordinate unchanged. This transformation affects the period of the function. Stretching the x-coordinate by a factor of 2 means that the period will also be stretched by 2. Since the original period of f(x) was π, the new period of g(x) will be 2π.But let me think about how the function changes. Stretching the x-coordinate by a factor of 2 is equivalent to replacing x with x/2 in the function. So, if f(x) = sin(2x - π/3) - sqrt(3)/2, then g(x) = sin(2*(x/2) - π/3) - sqrt(3)/2 = sin(x - π/3) - sqrt(3)/2.Wait, let me verify that. If you stretch the x-coordinate by a factor of 2, it's a horizontal stretch, which replaces x with x/k where k is the stretch factor. So, for a stretch factor of 2, it's x/2. So, yes, g(x) = f(x/2) = sin(2*(x/2) - π/3) - sqrt(3)/2 = sin(x - π/3) - sqrt(3)/2.So, g(x) = sin(x - π/3) - sqrt(3)/2.Now, we need to find the range of g(x) when x is in (π/2, π). So, let's analyze the function g(x) over this interval.First, let's find the range of the argument of the sine function, which is x - π/3. When x is in (π/2, π), then x - π/3 is in (π/2 - π/3, π - π/3) = (π/6, 2π/3).So, the argument of the sine function, x - π/3, varies from π/6 to 2π/3. Let's recall the behavior of the sine function in this interval.The sine function increases from π/6 to π/2, reaching its maximum at π/2, and then decreases from π/2 to 2π/3. So, the maximum value of sin(θ) in [π/6, 2π/3] is sin(π/2) = 1, and the minimum value is the smaller of sin(π/6) and sin(2π/3). Since sin(π/6) = 1/2 and sin(2π/3) = sqrt(3)/2 ≈ 0.866, so the minimum is 1/2.Therefore, sin(x - π/3) ranges from 1/2 to 1 as x varies from π/2 to π.Therefore, g(x) = sin(x - π/3) - sqrt(3)/2 ranges from (1/2 - sqrt(3)/2) to (1 - sqrt(3)/2).So, the range of g(x) is [ (1 - sqrt(3))/2 , (2 - sqrt(3))/2 ].Wait, let me compute that:Minimum value: 1/2 - sqrt(3)/2 = (1 - sqrt(3))/2Maximum value: 1 - sqrt(3)/2 = (2 - sqrt(3))/2Yes, that's correct.So, the range of g(x) when x is in (π/2, π) is [ (1 - sqrt(3))/2 , (2 - sqrt(3))/2 ].But let me double-check my steps to make sure I didn't make a mistake.1. I transformed f(x) into g(x) by replacing x with x/2, which is correct for a horizontal stretch by factor 2.2. Then, I found the argument of the sine function, x - π/3, and determined its range when x is in (π/2, π). That gives θ in (π/6, 2π/3), which is correct.3. Then, I considered the sine function over θ in (π/6, 2π/3). The sine function increases to π/2 and then decreases. So, the maximum is 1, and the minimum is 1/2. That seems correct.4. Therefore, subtracting sqrt(3)/2 from both ends gives the range of g(x). So, yes, the range is [ (1 - sqrt(3))/2 , (2 - sqrt(3))/2 ].I think that's correct. So, the final answer for part (2) is that the range of g(x) is [ (1 - sqrt(3))/2 , (2 - sqrt(3))/2 ].</think>