💡Alright, so I have this triangle problem here, and I need to solve two parts. Let me try to figure it out step by step.First, part (1) asks me to find the measure of angle A. The given equation is 2 cos 2A + 4 cos(B + C) + 3 = 0. Hmm, okay. I remember that in a triangle, the sum of angles is π radians, so B + C = π - A. That might be useful.Let me write that down: B + C = π - A. So, cos(B + C) is cos(π - A). I also recall that cos(π - A) is equal to -cos A. So, cos(B + C) = -cos A. That simplifies the equation a bit.Now, substituting that into the equation: 2 cos 2A + 4*(-cos A) + 3 = 0. So, that's 2 cos 2A - 4 cos A + 3 = 0.I also remember the double-angle formula for cosine: cos 2A = 2 cos² A - 1. Let me substitute that in. So, 2*(2 cos² A - 1) - 4 cos A + 3 = 0.Expanding that: 4 cos² A - 2 - 4 cos A + 3 = 0. Combining like terms: 4 cos² A - 4 cos A + 1 = 0.Hmm, this looks like a quadratic equation in terms of cos A. Let me set x = cos A for simplicity. Then the equation becomes 4x² - 4x + 1 = 0.Let me solve this quadratic equation. The quadratic formula is x = [4 ± sqrt(16 - 16)] / 8, since the discriminant is b² - 4ac = 16 - 16 = 0. So, x = [4 ± 0]/8 = 4/8 = 0.5.So, cos A = 0.5. Now, what angle A has cosine 0.5? I remember that cos(π/3) = 0.5, so A = π/3 radians, which is 60 degrees.Okay, that seems straightforward. So, part (1) is solved: angle A is π/3.Now, moving on to part (2). We're given that a = sqrt(3) and b + c = 3. We need to find the values of b and c.First, let's recall the Law of Cosines. It states that a² = b² + c² - 2bc cos A. We know a, which is sqrt(3), so a² = 3. We also know angle A is π/3, so cos A = 0.5.Substituting these into the Law of Cosines: 3 = b² + c² - 2bc*(0.5). Simplifying that: 3 = b² + c² - bc.We also know that b + c = 3. Maybe I can express one variable in terms of the other. Let's say c = 3 - b. Then, substitute c into the equation.So, 3 = b² + (3 - b)² - b*(3 - b). Let's expand that.First, (3 - b)² is 9 - 6b + b². So, substituting back: 3 = b² + 9 - 6b + b² - [3b - b²].Wait, let me make sure I expand that correctly. The term -b*(3 - b) is -3b + b². So, putting it all together:3 = b² + 9 - 6b + b² - 3b + b².Wait, that seems a bit off. Let me double-check.Original equation after substitution: 3 = b² + (3 - b)² - b*(3 - b).Expanding (3 - b)²: 9 - 6b + b².So, 3 = b² + 9 - 6b + b² - (3b - b²).Wait, no, the last term is -b*(3 - b) = -3b + b². So, putting it all together:3 = b² + 9 - 6b + b² - 3b + b².Combine like terms:b² + b² + b² = 3b².-6b - 3b = -9b.And the constant term is +9.So, the equation becomes: 3 = 3b² - 9b + 9.Let me subtract 3 from both sides to set it to zero: 0 = 3b² - 9b + 6.I can factor out a 3: 0 = 3(b² - 3b + 2).Divide both sides by 3: 0 = b² - 3b + 2.Now, factor the quadratic: b² - 3b + 2 = (b - 1)(b - 2) = 0.So, the solutions are b = 1 and b = 2.Since b + c = 3, if b = 1, then c = 2. If b = 2, then c = 1.Therefore, the possible pairs for (b, c) are (1, 2) and (2, 1).Let me just verify that these satisfy the original equation.First, take b = 1 and c = 2.Using the Law of Cosines: a² = 1² + 2² - 2*1*2*cos(π/3) = 1 + 4 - 4*(0.5) = 5 - 2 = 3. Which matches a² = 3.Similarly, if b = 2 and c = 1, the calculation is the same: 4 + 1 - 4*(0.5) = 5 - 2 = 3. So, that also works.Therefore, both solutions are valid.So, summarizing:1. Angle A is π/3.2. The sides b and c are 1 and 2, in some order.I think that's it. I didn't encounter any contradictions, so I believe my solutions are correct.

💡Okay, so I have this problem here: "The opposite of -2020 is" with options A: 2020, B: -1/2020, C: 1/2020, and D: -2020. Hmm, I need to figure out what the opposite of -2020 is. First, I remember that the opposite of a number is also called its additive inverse. That means if I add the number and its opposite together, the result should be zero. So, if I take -2020 and add its opposite, I should get zero. Let me write that down: -2020 + opposite = 0. To find the opposite, I can rearrange this equation. If I add 2020 to both sides, I get opposite = 2020. So, the opposite of -2020 is 2020. Looking back at the options, A is 2020, which matches what I just found. So, A should be the correct answer. Wait, but let me double-check to make sure I didn't make a mistake. Sometimes, people confuse opposites with reciprocals. The reciprocal of a number is 1 divided by that number. So, the reciprocal of -2020 would be -1/2020, which is option B. But that's not the opposite; that's the reciprocal. So, the opposite is definitely 2020, not the reciprocal. Therefore, I'm confident that A is the right choice. Just to be thorough, let's consider what the other options mean. Option C is 1/2020, which is the reciprocal of 2020, not related to the opposite of -2020. Option D is -2020 itself, which is the original number, not its opposite. So, yeah, A is definitely the correct answer. It all lines up with what I know about additive inverses and opposites. I don't think I made any mistakes in my reasoning. I guess another way to think about it is on the number line. If -2020 is 2020 units to the left of zero, then its opposite should be 2020 units to the right of zero, which is 2020. That makes sense. Alright, I think I've got it figured out. The opposite of -2020 is 2020, which is option A.

💡To find the value of ( star ) (which is ( y )) in the system of equations:[begin{cases}2x + y = bullet 2x - y = 12end{cases}]given that ( x = 5 ), follow these steps:1. **Substitute ( x = 5 ) into the second equation:** [ 2(5) - y = 12 ]2. **Simplify the equation:** [ 10 - y = 12 ]3. **Solve for ( y ):** [ -y = 12 - 10 ] [ -y = 2 ] [ y = -2 ]Therefore, the value of ( star ) is ( -2 ).

💡First, I need to simplify the expression (120r - 68r + 15r).All the terms in the expression have the same variable (r), so they are like terms and can be combined.I'll start by subtracting (68r) from (120r):[120r - 68r = 52r]Next, I'll add (15r) to the result:[52r + 15r = 67r]Therefore, the simplified expression is (67r).

💡Okay, so I need to find the equation of the tangent line with the smallest slope for the curve given by y = x³ + 3x² + 6x - 1. Hmm, let's see. I remember that the slope of the tangent line to a curve at a particular point is given by the derivative of the function at that point. So, first, I should find the derivative of y with respect to x.Alright, let me compute that. The derivative of x³ is 3x², the derivative of 3x² is 6x, the derivative of 6x is 6, and the derivative of -1 is 0. So putting it all together, the derivative y' should be 3x² + 6x + 6. Got that down.Now, I need to find the tangent line with the smallest slope. That means I need to find the minimum value of y'. Since y' is a quadratic function in terms of x, I can think of it as a parabola. Quadratic functions have either a minimum or maximum value depending on the coefficient of x². In this case, the coefficient is 3, which is positive, so the parabola opens upwards, meaning it has a minimum point.To find the minimum value of y', I can complete the square or use the vertex formula. I think completing the square might be straightforward here. Let me try that. So, starting with y' = 3x² + 6x + 6. I can factor out the 3 from the first two terms: y' = 3(x² + 2x) + 6.Now, to complete the square inside the parentheses, I take the coefficient of x, which is 2, divide it by 2 to get 1, and then square it to get 1. So, I add and subtract 1 inside the parentheses: y' = 3[(x² + 2x + 1 - 1)] + 6. This simplifies to y' = 3[(x + 1)² - 1] + 6.Distributing the 3, I get y' = 3(x + 1)² - 3 + 6, which simplifies further to y' = 3(x + 1)² + 3. Okay, so now it's clear that the minimum value of y' occurs when (x + 1)² is zero because squares are always non-negative. So, when x + 1 = 0, which is x = -1, the minimum slope occurs.Plugging x = -1 into y', we get y' = 3(0) + 3 = 3. So, the smallest slope of the tangent line is 3. Now, I need to find the equation of the tangent line at this point.To write the equation of the tangent line, I need a point on the curve and the slope. I already have the slope, which is 3. Now, I need the corresponding y-value when x = -1. Let's plug x = -1 into the original function y = x³ + 3x² + 6x - 1.Calculating that: y = (-1)³ + 3(-1)² + 6(-1) - 1. Let's compute each term step by step. (-1)³ is -1, 3(-1)² is 3(1) which is 3, 6(-1) is -6, and then -1. So adding them up: -1 + 3 is 2, 2 - 6 is -4, and -4 -1 is -5. So, the point is (-1, -5).Now, using the point-slope form of a line, which is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point. Plugging in the values, we get y - (-5) = 3(x - (-1)), which simplifies to y + 5 = 3(x + 1).Let me simplify this further. Distribute the 3 on the right side: y + 5 = 3x + 3. Then, subtract 5 from both sides: y = 3x + 3 - 5, which simplifies to y = 3x - 2.Wait, but the question asks for the equation of the tangent line in a standard form, I think. So, maybe I should rearrange it to the form Ax + By + C = 0. Starting from y = 3x - 2, subtract 3x and add 2 to both sides: -3x + y + 2 = 0. Multiplying both sides by -1 to make the coefficient of x positive, we get 3x - y - 2 = 0.So, the equation of the tangent line with the smallest slope is 3x - y - 2 = 0. Let me just double-check my steps to make sure I didn't make any mistakes. Found the derivative correctly, completed the square correctly, found the minimum slope at x = -1, computed the y-value correctly, used point-slope form, and rearranged properly. Seems solid to me.**Final Answer**The equation of the tangent line with the smallest slope is boxed{3x - y - 2 = 0}.

💡Okay, so I have this equation of a circle: 2x² = -2y² + 16x - 8y + 40. Hmm, I need to find the area of a square that's inscribed around this circle, with one pair of sides parallel to the x-axis. Alright, let me break this down step by step.First, I remember that the general equation of a circle is usually written in the form (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. But the given equation doesn't look like that. It has coefficients in front of the x² and y² terms, which isn't standard. So, I think my first step should be to rearrange and simplify this equation to get it into the standard circle form.Let me write down the given equation again:2x² = -2y² + 16x - 8y + 40.Hmm, I notice that both x² and y² have coefficients, which makes it a bit tricky. Maybe I can move all terms to one side to see if I can factor out the coefficients. Let me subtract 2x² from both sides to get:0 = -2y² + 16x - 8y + 40 - 2x².Wait, that's a bit messy. Maybe I should bring all terms to the left side:2x² + 2y² - 16x + 8y - 40 = 0.Ah, that looks better. Now, I can factor out the 2 from all the terms:2(x² + y² - 8x + 4y - 20) = 0.Dividing both sides by 2 gives:x² + y² - 8x + 4y - 20 = 0.Okay, now this looks more like the standard circle equation, except it's not completed the square yet. I need to complete the square for both x and y terms.Let's group the x terms and y terms:(x² - 8x) + (y² + 4y) = 20.Now, completing the square for x:Take the coefficient of x, which is -8, divide by 2 to get -4, then square it to get 16. So, add and subtract 16 inside the equation.Similarly, for y:Coefficient of y is 4, divide by 2 to get 2, square it to get 4. So, add and subtract 4.Putting it all together:(x² - 8x + 16 - 16) + (y² + 4y + 4 - 4) = 20.Simplify:(x - 4)² - 16 + (y + 2)² - 4 = 20.Combine constants:(x - 4)² + (y + 2)² - 20 = 20.Add 20 to both sides:(x - 4)² + (y + 2)² = 40.Alright, so now the equation is in standard form. The center of the circle is at (4, -2), and the radius is √40. Simplifying √40, that's 2√10.Now, the problem asks for the area of a square inscribed around the circle, with one pair of sides parallel to the x-axis. Hmm, inscribed around the circle... So, the circle is inscribed within the square, meaning the square touches the circle at exactly four points, one on each side.In that case, the diameter of the circle should be equal to the side length of the square. Because the circle touches the square exactly at the midpoints of the square's sides. So, the diameter of the circle is equal to the side length of the square.The radius of the circle is 2√10, so the diameter is twice that, which is 4√10.Therefore, the side length of the square is 4√10. To find the area of the square, I square the side length:Area = (4√10)² = 16 * 10 = 160.So, the area of the square is 160 square units.Wait, let me double-check to make sure I didn't make any mistakes. Starting from the equation:2x² = -2y² + 16x - 8y + 40.I moved all terms to the left:2x² + 2y² - 16x + 8y - 40 = 0.Factored out a 2:2(x² + y² - 8x + 4y - 20) = 0.Divided by 2:x² + y² - 8x + 4y - 20 = 0.Completed the square:(x - 4)² + (y + 2)² = 40.Radius is √40, diameter is 2√40 = 4√10.Side of square is 4√10, area is (4√10)² = 160.Yes, that seems correct. I think I got it right.