💡First, I need to determine the initial number of songs Zoe has on her MP3 player, which is 15.Next, she deletes 8 old songs. To find out how many songs remain after the deletion, I subtract 8 from 15:15 - 8 = 7 songs.After deleting the old songs, Zoe adds 50 new songs to her MP3 player. To find the total number of songs now, I add 50 to the remaining 7 songs:7 + 50 = 57 songs.Therefore, Zoe has a total of 57 songs on her MP3 player.

💡Okay, so I have this problem about a parabola. Let me try to figure it out step by step. First, the problem says the parabola has its vertex at the origin, is symmetric about the x-axis, and passes through the point M(-2, 2√2). I need to find the equation of the parabola and the coordinates of its focus. Then, in part (2), a line passes through the focus with a slope of 45 degrees, and I have to find the length of the chord where this line intersects the parabola.Starting with part (1). Since the parabola is symmetric about the x-axis and has its vertex at the origin, I remember that the standard form of such a parabola is either y² = 4px or y² = -4px, depending on whether it opens to the right or left. Since the point M(-2, 2√2) is given, and the x-coordinate is negative, that suggests the parabola opens to the left, so the equation should be y² = -4px.Let me write that down: y² = -4px. Now, I need to find the value of p. To do this, I can plug in the coordinates of point M into the equation. So, substituting x = -2 and y = 2√2 into the equation:(2√2)² = -4p*(-2)Calculating the left side: (2√2)² = 4*2 = 8Right side: -4p*(-2) = 8pSo, 8 = 8p, which implies p = 1.Wait, hold on. If p = 1, then the equation is y² = -4x. Let me check that again. Plugging in M(-2, 2√2):(2√2)² = 8, and -4*(-2) = 8. So yes, 8 = 8. That works. So, the equation of the parabola is y² = -4x.Now, the focus of a parabola y² = -4px is at (-p, 0). Since p = 1, the focus F is at (-1, 0). That seems straightforward.Moving on to part (2). A line l passes through the focus F(-1, 0) and has a slope of 45 degrees. A slope of 45 degrees means the slope is 1 because tan(45°) = 1. So, the equation of line l is y = x + c. Since it passes through (-1, 0), I can plug that point into the equation to find c.0 = (-1) + c => c = 1. So, the equation of line l is y = x + 1.Now, I need to find the points where this line intersects the parabola y² = -4x. Let me substitute y from the line equation into the parabola equation.So, substituting y = x + 1 into y² = -4x:(x + 1)² = -4xExpanding the left side: x² + 2x + 1 = -4xBring all terms to one side: x² + 2x + 1 + 4x = 0 => x² + 6x + 1 = 0Hmm, that's a quadratic equation in x. Let me write it as x² + 6x + 1 = 0. To find the solutions, I can use the quadratic formula:x = [-b ± √(b² - 4ac)] / 2aHere, a = 1, b = 6, c = 1. Plugging these in:x = [-6 ± √(36 - 4)] / 2 = [-6 ± √32] / 2Simplify √32 as 4√2, so:x = [-6 ± 4√2] / 2 = -3 ± 2√2So, the x-coordinates of points A and B are x = -3 + 2√2 and x = -3 - 2√2.Now, let me find the corresponding y-coordinates using the line equation y = x + 1.For x = -3 + 2√2:y = (-3 + 2√2) + 1 = -2 + 2√2For x = -3 - 2√2:y = (-3 - 2√2) + 1 = -2 - 2√2So, the points A and B are:A: (-3 + 2√2, -2 + 2√2)B: (-3 - 2√2, -2 - 2√2)Now, I need to find the distance between points A and B, which is the length of chord AB.To find the distance between two points (x1, y1) and (x2, y2), the formula is:Distance = √[(x2 - x1)² + (y2 - y1)²]Let me compute the differences first.x2 - x1 = (-3 - 2√2) - (-3 + 2√2) = (-3 - 2√2 + 3 - 2√2) = (-4√2)Similarly, y2 - y1 = (-2 - 2√2) - (-2 + 2√2) = (-2 - 2√2 + 2 - 2√2) = (-4√2)So, both differences are -4√2. Now, squaring these:(-4√2)² = 16*2 = 32So, the distance is √(32 + 32) = √64 = 8Wait, that seems straightforward. So, the length of chord AB is 8 units.But let me double-check my calculations to make sure I didn't make any mistakes.First, the equation of the parabola: y² = -4x. Correct, since p = 1.Equation of the line: y = x + 1. Correct, passing through (-1, 0) with slope 1.Substituting into the parabola: (x + 1)² = -4x => x² + 2x + 1 = -4x => x² + 6x + 1 = 0. Correct.Solutions: x = [-6 ± √(36 - 4)] / 2 = [-6 ± √32]/2 = -3 ± 2√2. Correct.Points A and B: (-3 + 2√2, -2 + 2√2) and (-3 - 2√2, -2 - 2√2). Correct.Distance calculation: differences in x and y are both -4√2, squared gives 32 each, total 64, square root is 8. Correct.So, everything seems to check out. The length of chord AB is 8.**Final Answer**(1) The equation of the parabola is boxed{y^2 = -4x} and the focus is at boxed{(-1, 0)}.(2) The length of the chord (|AB|) is boxed{8}.

💡Alright, so I'm trying to understand this problem about the Golden Krosh Number System. It's based on the golden ratio, which is (varphi = frac{1 + sqrt{5}}{2}). I remember that the golden ratio has some interesting properties, especially related to its powers and how they can be expressed in terms of each other. The problem has four parts, and I need to tackle each one step by step. Let's start with part (a), which asks for at least five different decompositions of the number 1 in this system. First, I recall that in the golden ratio base system, numbers can be represented as sums of distinct powers of (varphi). For example, the number 1 can be represented in several ways because of the properties of (varphi). One key property is that (varphi^2 = varphi + 1). This means that higher powers of (varphi) can be broken down into sums of lower powers. So, starting with the simplest decomposition, 1 is just (varphi^0). That's straightforward. But how do I find other decompositions? Maybe by using the property (varphi^{-1} = varphi - 1). Let me see:1. (1 = varphi^0)2. Using (varphi^{-1} = varphi - 1), I can write (1 = varphi^{-1} + varphi^{-2}) because (varphi^{-2} = 1 - varphi). So, substituting, (1 = (varphi - 1) + (1 - varphi) = 0). Wait, that doesn't make sense. Maybe I made a mistake here.Let me double-check. If (varphi^{-1} = varphi - 1), then (varphi^{-2} = (varphi^{-1})^2 = (varphi - 1)^2 = varphi^2 - 2varphi + 1). But since (varphi^2 = varphi + 1), substituting that in, we get (varphi^{-2} = (varphi + 1) - 2varphi + 1 = -varphi + 2). Hmm, that's different from what I initially thought.Wait, maybe I should approach this differently. I know that in the golden ratio base, numbers can be represented without consecutive 1s in their expansions. This is similar to how binary numbers avoid consecutive 1s in certain representations. Maybe I can use this property to find different decompositions.Let me try to express 1 using negative powers of (varphi). Since (varphi^{-1} = varphi - 1), I can write 1 as (varphi^{-1} + varphi^{-2}) because:[varphi^{-1} + varphi^{-2} = (varphi - 1) + (varphi^{-2})]But I need to find what (varphi^{-2}) is. From (varphi^2 = varphi + 1), dividing both sides by (varphi^2), we get:[1 = varphi^{-1} + varphi^{-2}]So, indeed, (1 = varphi^{-1} + varphi^{-2}). That's the second decomposition.Now, for the third decomposition, maybe I can use higher negative powers. Let's see:[varphi^{-3} = varphi^{-2} cdot varphi^{-1} = (varphi^{-2}) cdot (varphi - 1)]But this might get complicated. Alternatively, I can use the fact that (varphi^{-n} = varphi^{-(n+1)} + varphi^{-(n+2)}). This recursive relation might help me break down higher negative powers into sums of lower ones.Using this, let's try to express 1 in another way:[1 = varphi^{-1} + varphi^{-3} + varphi^{-4}]Let me verify this:[varphi^{-1} + varphi^{-3} + varphi^{-4} = (varphi - 1) + varphi^{-3} + varphi^{-4}]But I need to express (varphi^{-3}) and (varphi^{-4}) in terms of (varphi). This might take some time, but I think it's doable.Continuing this process, I can find more decompositions by recursively applying the relation (varphi^{-n} = varphi^{-(n+1)} + varphi^{-(n+2)}). Each time I do this, I get a new decomposition with more terms.So, summarizing, I have:1. (1 = varphi^0)2. (1 = varphi^{-1} + varphi^{-2})3. (1 = varphi^{-1} + varphi^{-3} + varphi^{-4})4. (1 = varphi^{-1} + varphi^{-3} + varphi^{-5} + varphi^{-6})5. (1 = varphi^{-1} + varphi^{-3} + varphi^{-5} + varphi^{-7} + varphi^{-8})Each of these represents a different way to decompose 1 into sums of distinct powers of (varphi). I think this satisfies part (a).Moving on to part (b), which asks to prove that if a number can be decomposed into such a sum, it can be decomposed into a sum without adjacent powers of (varphi). I remember that in the golden ratio base system, representations avoid having two consecutive 1s, which corresponds to avoiding adjacent powers. So, if a number has a decomposition with adjacent powers, I can use the property (varphi^k + varphi^{k-1} = varphi^{k+1}) to eliminate the adjacent terms. For example, if I have (varphi^k + varphi^{k-1}), I can replace it with (varphi^{k+1}). This process can be repeated until there are no adjacent powers left. This ensures that any decomposition can be transformed into one without adjacent powers, which is what part (b) is asking for.For part (c), I need to prove that if (a) is decomposable, then (a + 1) is also decomposable. This seems like an inductive step. If I have a decomposition for (a), adding 1 should just involve adding (varphi^0) to the decomposition, provided that (varphi^0) is not already present. If it is present, I might need to carry over to higher powers, similar to how addition works in binary or decimal systems.But since the golden ratio base avoids consecutive 1s, adding 1 might not always be straightforward. However, the key idea is that adding 1 can be done by adjusting the decomposition appropriately, ensuring that no two adjacent powers are used. This should always be possible, maintaining the decomposability.Finally, part (d) asks to prove that numbers of the form (frac{m + n sqrt{5}}{2}), where (m) and (n) are integers of the same parity, are decomposable. I think this relates to the fact that such numbers can be expressed as linear combinations of powers of (varphi), which are themselves of the form (frac{m + n sqrt{5}}{2}). Since the golden ratio base system can represent any number that can be expressed as a sum of distinct powers of (varphi), and these numbers fit into that category, they should be decomposable. This likely involves showing that the given form can be broken down into sums of (varphi^k) terms, ensuring that each term is used at most once and no two are adjacent.Overall, each part builds on the properties of the golden ratio and its base system, ensuring that decompositions are possible and unique in a certain sense. I need to carefully apply these properties to construct the proofs for each part.

💡Okay, so I have this equation to solve: 3 - 7 cos^2 x sin x - 3 sin^3 x = 0Hmm, it looks a bit complicated with both sine and cosine terms. I remember that sometimes you can simplify equations by using trigonometric identities. Let me think... Oh, right! I can use the identity cos^2 x = 1 - sin^2 x to rewrite the equation in terms of sine only. That might make it easier to handle.So, substituting cos^2 x with 1 - sin^2 x, the equation becomes:3 - 7 (1 - sin^2 x) sin x - 3 sin^3 x = 0Let me expand that middle term:3 - 7 sin x + 7 sin^3 x - 3 sin^3 x = 0Wait, the 7 sin^3 x and -3 sin^3 x can be combined. Let me do that:3 - 7 sin x + (7 - 3) sin^3 x = 0Which simplifies to:3 - 7 sin x + 4 sin^3 x = 0Hmm, now it's a cubic equation in terms of sin x. Maybe I can let t = sin x to make it look simpler. Let's try that substitution:4 t^3 - 7 t + 3 = 0Alright, so now I have a cubic equation: 4t^3 - 7t + 3 = 0. I need to solve for t. I remember that for cubic equations, sometimes you can factor them or use the rational root theorem to find possible roots.Let me try the rational root theorem. The possible rational roots are factors of the constant term divided by factors of the leading coefficient. So, possible roots are pm1, pm3, pmfrac{1}{2}, pmfrac{3}{2}.Let me test t = 1:4(1)^3 - 7(1) + 3 = 4 - 7 + 3 = 0Oh, t = 1 is a root! That means (t - 1) is a factor of the cubic polynomial. Let me perform polynomial division or factor it out.Dividing 4t^3 - 7t + 3 by (t - 1):Using synthetic division:1 | 4 0 -7 3 | 4 4 -3 ---------------- 4 4 -3 0So, the cubic factors as:(t - 1)(4t^2 + 4t - 3) = 0Now, I can set each factor equal to zero:1. t - 1 = 0 gives t = 12. 4t^2 + 4t - 3 = 0Let me solve the quadratic equation 4t^2 + 4t - 3 = 0 using the quadratic formula:t = frac{-b pm sqrt{b^2 - 4ac}}{2a}Where a = 4, b = 4, and c = -3:t = frac{-4 pm sqrt{(4)^2 - 4(4)(-3)}}{2(4)} = frac{-4 pm sqrt{16 + 48}}{8} = frac{-4 pm sqrt{64}}{8} = frac{-4 pm 8}{8}So, two solutions:1. t = frac{-4 + 8}{8} = frac{4}{8} = frac{1}{2}2. t = frac{-4 - 8}{8} = frac{-12}{8} = -frac{3}{2}Wait, t = sin x, and the sine function only takes values between -1 and 1. So, t = -frac{3}{2} is not possible. So, we discard that.So, the valid solutions for t are t = 1 and t = frac{1}{2}.Now, let's translate these back to x.1. For t = 1:sin x = 1The solutions for this are:x = frac{pi}{2} + 2pi k quad text{where } k in mathbb{Z}2. For t = frac{1}{2}:sin x = frac{1}{2}The general solutions for this are:x = frac{pi}{6} + 2pi k quad text{and} quad x = frac{5pi}{6} + 2pi k quad text{where } k in mathbb{Z}Alternatively, this can be written as:x = (-1)^k frac{pi}{6} + pi k quad text{where } k in mathbb{Z}So, combining all solutions, the solutions to the original equation are:x = frac{pi}{2} + 2pi k quad text{and} quad x = (-1)^k frac{pi}{6} + pi k quad text{where } k in mathbb{Z}I think that's all the solutions. Let me just double-check if these satisfy the original equation.For x = frac{pi}{2}:sin x = 1, cos x = 0Plugging into the original equation:3 - 7(0)^2(1) - 3(1)^3 = 3 - 0 - 3 = 0Good.For x = frac{pi}{6}:sin x = frac{1}{2}, cos x = frac{sqrt{3}}{2}Plugging into the equation:3 - 7left(frac{sqrt{3}}{2}right)^2 left(frac{1}{2}right) - 3left(frac{1}{2}right)^3Calculating each term:7left(frac{3}{4}right)left(frac{1}{2}right) = 7 cdot frac{3}{8} = frac{21}{8}3left(frac{1}{8}right) = frac{3}{8}So, total:3 - frac{21}{8} - frac{3}{8} = 3 - frac{24}{8} = 3 - 3 = 0Perfect. Similarly, for x = frac{5pi}{6}:sin x = frac{1}{2}, cos x = -frac{sqrt{3}}{2}Plugging into the equation:3 - 7left(-frac{sqrt{3}}{2}right)^2 left(frac{1}{2}right) - 3left(frac{1}{2}right)^3Again, the square of cosine makes it positive:7left(frac{3}{4}right)left(frac{1}{2}right) = frac{21}{8}And the sine cubed term:3left(frac{1}{8}right) = frac{3}{8}So, total:3 - frac{21}{8} - frac{3}{8} = 0Which checks out.Therefore, I'm confident that the solutions are correct.

💡Okay, so I have this quadratic function: y = kx² - 4x - 2. The question is asking about the range of values for k such that the function has two distinct intersection points with the x-axis. Hmm, okay. Let me think about how to approach this.First, I remember that a quadratic function intersects the x-axis at points where y = 0. So, I need to set y to zero and solve for x. That gives me the equation: kx² - 4x - 2 = 0. Now, this is a quadratic equation in terms of x, and I want it to have two distinct real roots. I recall that for a quadratic equation ax² + bx + c = 0, the discriminant determines the nature of the roots. The discriminant is given by D = b² - 4ac. If D is greater than zero, there are two distinct real roots. If D is zero, there's exactly one real root (a repeated root), and if D is negative, the roots are complex and not real. So, in this case, I need the discriminant to be greater than zero.Let me identify the coefficients a, b, and c from my equation. Comparing kx² - 4x - 2 = 0 with ax² + bx + c = 0, I see that a = k, b = -4, and c = -2. Now, plugging these into the discriminant formula: D = (-4)² - 4 * k * (-2). Let me compute that step by step. (-4) squared is 16. Then, 4 times k times (-2) is -8k. But since it's subtracting that term, it becomes plus 8k. So, D = 16 + 8k.For two distinct real roots, I need D > 0. So, 16 + 8k > 0. Let me solve this inequality for k. Subtract 16 from both sides: 8k > -16. Then, divide both sides by 8: k > -2. Wait, so k has to be greater than -2. But hold on, I also remember that in a quadratic equation, the coefficient a (which is k in this case) cannot be zero because otherwise, it wouldn't be a quadratic equation anymore—it would become a linear equation. So, k cannot be zero. Therefore, combining these two conditions, k must be greater than -2 and not equal to zero. Let me double-check that. If k is greater than -2, the discriminant is positive, giving two distinct real roots, and k not equal to zero ensures that the equation remains quadratic. Looking back at the options provided:A: k > -2B: k > -2 and k ≠ 0C: k < 2D: k ≥ -2 and k ≠ 0So, option A is only considering k > -2, but it doesn't account for k ≠ 0. Option B includes both conditions, which seems correct. Option C is k < 2, which doesn't make sense because our discriminant condition gave us k > -2, not less than 2. Option D is k ≥ -2 and k ≠ 0, but since the discriminant requires k > -2, not just greater than or equal, D is not quite right either.Therefore, the correct answer should be option B: k > -2 and k ≠ 0.

💡First, I need to understand the problem. There is a square with vertices at (2, 0), (5, 0), (5, 3), and (2, 3). A line is drawn from the point (2, 3) to the point (5, 0), dividing the square into two parts. The goal is to determine what fraction of the square's area is above this line.To approach this, I'll start by finding the equation of the line that connects the points (2, 3) and (5, 0). Calculating the slope of the line will help in determining its equation. The slope (m) is calculated as the change in y divided by the change in x between the two points. Once I have the slope, I can use the point-slope form to write the equation of the line.Next, I'll identify where this line intersects the sides of the square. Since the line connects two vertices of the square, it already passes through (2, 3) and (5, 0). I need to check if it intersects any other sides of the square, but in this case, it seems it only connects these two vertices.After determining the equation of the line, I'll calculate the area of the square. The square has sides of length 3 units, so its area is 9 square units.To find the area above the line, I'll consider the shape formed above the line within the square. This shape is a triangle with vertices at (2, 3), (5, 3), and (5, 0). Calculating the area of this triangle will give me the area above the line.Finally, I'll determine the fraction of the square's area that is above the line by dividing the area of the triangle by the total area of the square.