💡Okay, so I have this problem here: "The value range of ( x ) for the set ( {x^2 + x, 2x} ) is ( )". The options are A, B, C, and D, with different intervals. Hmm, let me try to figure this out step by step.First, I need to understand what the question is asking. It's about the range of values that ( x ) can take such that the set ( {x^2 + x, 2x} ) is valid. I think that means the set should have two distinct elements, right? Because if both elements were the same, it wouldn't be a set with two elements anymore—it would just have one element. So, I guess the key here is to make sure that ( x^2 + x ) is not equal to ( 2x ).Alright, so let me write that down as an equation:( x^2 + x neq 2x )Now, I need to solve this inequality to find the values of ( x ) that satisfy it. Let me subtract ( 2x ) from both sides to get all terms on one side:( x^2 + x - 2x neq 0 )Simplifying that, I get:( x^2 - x neq 0 )Hmm, okay, so ( x^2 - x ) is not equal to zero. I can factor this expression to make it easier to solve. Let's factor out an ( x ):( x(x - 1) neq 0 )So, this product is not equal to zero. For a product of two terms to be zero, at least one of the terms has to be zero. Therefore, for the product to not be zero, neither of the terms can be zero. That means:( x neq 0 ) and ( x - 1 neq 0 )Which simplifies to:( x neq 0 ) and ( x neq 1 )So, ( x ) cannot be 0 or 1. Therefore, the values of ( x ) must be all real numbers except 0 and 1. Now, I need to express this in interval notation.Let me recall how interval notation works. All real numbers except 0 and 1 would be split into three intervals:1. From negative infinity up to 0, not including 0.2. From 0 up to 1, not including 0 and 1.3. From 1 to positive infinity, not including 1.So, in interval notation, that would be:( (-infty, 0) cup (0, 1) cup (1, +infty) )Looking back at the options given:A: ( (-infty, +infty) )B: ( (-infty, 0) cup (0, +infty) )C: ( (-infty, 1) cup (1, +infty) )D: ( (-infty, 0) cup (0, 1) cup (1, +infty) )So, option D matches the interval notation I came up with. Therefore, the correct answer should be D.Wait, let me double-check to make sure I didn't make any mistakes. The original condition was that ( x^2 + x neq 2x ). Solving that, I subtracted ( 2x ) and factored to find ( x neq 0 ) and ( x neq 1 ). So, yes, ( x ) can be any real number except 0 and 1, which is exactly what option D says. I think that's correct.Just to be thorough, let me plug in some numbers to test. If ( x = 0 ), then the set becomes ( {0, 0} ), which is just ( {0} ), so it's not a set with two elements. Similarly, if ( x = 1 ), the set becomes ( {1 + 1, 2} = {2, 2} = {2} ), again only one element. So, 0 and 1 are excluded, which confirms my earlier conclusion.What about other numbers? Let's say ( x = 2 ). Then the set is ( {4 + 2, 4} = {6, 4} ), which has two distinct elements. Similarly, ( x = -1 ) gives ( {1 - 1, -2} = {0, -2} ), which is also two distinct elements. So, it seems like for any ( x ) not equal to 0 or 1, the set has two distinct elements, which is what we want.Therefore, I'm confident that the correct answer is D.

💡First, I need to determine the total number of oranges in the crates. There are 12 crates, and each crate contains 150 oranges. By multiplying the number of crates by the number of oranges per crate, I can find the total number of oranges.Next, I will calculate the total number of nectarines in the boxes. There are 16 boxes, and each box holds 30 nectarines. Multiplying the number of boxes by the number of nectarines per box will give me the total number of nectarines.Finally, I will add the total number of oranges and nectarines together to find the overall total number of pieces of fruit.

💡Alright, so I have this sequence problem here. The sequence starts with 2000, then y, and each subsequent term is obtained by subtracting the preceding term from the one before that. The last term is the first negative term encountered. I need to find the positive integer y that produces a sequence of maximum length. Hmm, okay, let me try to break this down.First, let me write out the sequence as per the given rule. The first term is 2000, the second term is y. Then each term after that is the previous term subtracted from the one before that. So, term 3 would be 2000 - y, term 4 would be y - (2000 - y) which simplifies to 2y - 2000, term 5 would be (2000 - y) - (2y - 2000) which is 4000 - 3y, and so on.Let me write these out step by step:- b₁ = 2000- b₂ = y- b₃ = b₁ - b₂ = 2000 - y- b₄ = b₂ - b₃ = y - (2000 - y) = 2y - 2000- b₅ = b₃ - b₄ = (2000 - y) - (2y - 2000) = 2000 - y - 2y + 2000 = 4000 - 3y- b₆ = b₄ - b₅ = (2y - 2000) - (4000 - 3y) = 2y - 2000 - 4000 + 3y = 5y - 6000- b₇ = b₅ - b₆ = (4000 - 3y) - (5y - 6000) = 4000 - 3y - 5y + 6000 = 10000 - 8y- b₈ = b₆ - b₇ = (5y - 6000) - (10000 - 8y) = 5y - 6000 - 10000 + 8y = 13y - 16000- b₉ = b₇ - b₈ = (10000 - 8y) - (13y - 16000) = 10000 - 8y - 13y + 16000 = 26000 - 21y- b₁₀ = b₈ - b₉ = (13y - 16000) - (26000 - 21y) = 13y - 16000 - 26000 + 21y = 34y - 42000- b₁₁ = b₉ - b₁₀ = (26000 - 21y) - (34y - 42000) = 26000 - 21y - 34y + 42000 = 68000 - 55yOkay, so up to term 11, I have these expressions. Now, the sequence stops when it encounters the first negative term. So, I need to find the y such that the sequence continues as long as possible without hitting a negative term, and then the next term is negative.Looking at the terms, each term alternates between expressions with positive coefficients for y and negative coefficients. So, for the terms to be positive, certain inequalities must hold.For example, term 3: 2000 - y > 0 ⇒ y < 2000Term 4: 2y - 2000 > 0 ⇒ y > 1000Term 5: 4000 - 3y > 0 ⇒ y < 4000/3 ≈ 1333.33Term 6: 5y - 6000 > 0 ⇒ y > 1200Term 7: 10000 - 8y > 0 ⇒ y < 1250Term 8: 13y - 16000 > 0 ⇒ y > 16000/13 ≈ 1230.77Term 9: 26000 - 21y > 0 ⇒ y < 26000/21 ≈ 1238.095Term 10: 34y - 42000 > 0 ⇒ y > 42000/34 ≈ 1235.29Term 11: 68000 - 55y > 0 ⇒ y < 68000/55 ≈ 1236.36So, each term imposes a condition on y. To have the sequence as long as possible, all these conditions must be satisfied. So, y must be greater than the maximum of the lower bounds and less than the minimum of the upper bounds.Looking at the lower bounds:From term 4: y > 1000From term 6: y > 1200From term 8: y > ~1230.77From term 10: y > ~1235.29So, the most restrictive lower bound is y > ~1235.29Upper bounds:From term 3: y < 2000From term 5: y < ~1333.33From term 7: y < 1250From term 9: y < ~1238.095From term 11: y < ~1236.36So, the most restrictive upper bound is y < ~1236.36Therefore, y must satisfy:~1235.29 < y < ~1236.36Since y is a positive integer, the only integer in this interval is 1236.So, y = 1236.Let me verify this.If y = 1236, let's compute each term:b₁ = 2000b₂ = 1236b₃ = 2000 - 1236 = 764b₄ = 1236 - 764 = 472b₅ = 764 - 472 = 292b₆ = 472 - 292 = 180b₇ = 292 - 180 = 112b₈ = 180 - 112 = 68b₉ = 112 - 68 = 44b₁₀ = 68 - 44 = 24b₁₁ = 44 - 24 = 20b₁₂ = 24 - 20 = 4b₁₃ = 20 - 4 = 16b₁₄ = 4 - 16 = -12So, the 14th term is negative, which is the first negative term. So, the sequence has 14 terms before hitting a negative number.Wait, but in the initial calculation, I only went up to term 11, which was 68000 - 55y. So, if y=1236, term 11 is 68000 - 55*1236.Let me compute that:55*1236 = 55*(1200 + 36) = 55*1200 + 55*36 = 66000 + 1980 = 67980So, term 11: 68000 - 67980 = 20Term 12: term 10 - term 11 = (34y - 42000) - (68000 - 55y) = (34*1236 - 42000) - (68000 - 55*1236)Wait, but maybe it's easier to compute step by step.Wait, term 12 is term 10 - term 11, which is 24 - 20 = 4Term 13 is term 11 - term 12 = 20 - 4 = 16Term 14 is term 12 - term 13 = 4 - 16 = -12So, yes, term 14 is the first negative term. So, the sequence length is 14 terms.Is this the maximum possible? Let's check if y=1235 would give a longer sequence.If y=1235, let's compute:b₁ = 2000b₂ = 1235b₃ = 2000 - 1235 = 765b₄ = 1235 - 765 = 470b₅ = 765 - 470 = 295b₆ = 470 - 295 = 175b₇ = 295 - 175 = 120b₈ = 175 - 120 = 55b₉ = 120 - 55 = 65b₁₀ = 55 - 65 = -10So, term 10 is negative. So, the sequence length is 10 terms, which is shorter than 14.Similarly, if y=1237, let's compute:b₁ = 2000b₂ = 1237b₃ = 2000 - 1237 = 763b₄ = 1237 - 763 = 474b₅ = 763 - 474 = 289b₆ = 474 - 289 = 185b₇ = 289 - 185 = 104b₈ = 185 - 104 = 81b₉ = 104 - 81 = 23b₁₀ = 81 - 23 = 58b₁₁ = 23 - 58 = -35So, term 11 is negative. So, the sequence length is 11 terms, which is still shorter than 14.Therefore, y=1236 gives the maximum sequence length of 14 terms before hitting a negative number.Wait, but in the initial problem, the assistant's solution only went up to term 11 and concluded that y must be between ~1235.29 and ~1236.36, so y=1236. But when I computed with y=1236, I got 14 terms, which is longer than the assistant's initial calculation.Wait, maybe the assistant only went up to term 11, but in reality, the sequence can continue beyond that. So, perhaps the assistant's solution was incomplete.But regardless, through my step-by-step computation, I found that y=1236 gives a sequence length of 14 terms, which is longer than y=1235 or y=1237.Therefore, the answer is y=1236.**Final Answer**The positive integer ( y ) that produces a sequence of maximum length is (boxed{1236}).

💡Okay, so I need to find the slope angle of the line AB given points A(0,1) and B(1,0). Hmm, slope angle... I think that's the angle that the line makes with the positive direction of the x-axis. Right? So, to find that, I probably need to find the slope of the line first.Alright, the formula for slope is (y2 - y1)/(x2 - x1). Let me assign A as (x1, y1) which is (0,1) and B as (x2, y2) which is (1,0). So plugging into the formula, it would be (0 - 1)/(1 - 0). That simplifies to (-1)/1, which is -1. So the slope is -1.Now, the slope is equal to the tangent of the angle that the line makes with the x-axis. So, tan(theta) = -1. I need to find theta such that tan(theta) = -1. I remember that tan(theta) is negative in the second and fourth quadrants. But since we're talking about the angle of the line with the x-axis, it should be between 0 and pi, right? So, we're looking for an angle in the second quadrant where tan(theta) is -1.I also recall that tan(pi/4) = 1, so in the second quadrant, the angle would be pi - pi/4, which is 3pi/4. So, theta is 3pi/4. Let me check that: tan(3pi/4) is indeed -1 because in the second quadrant, tangent is negative, and 3pi/4 is 135 degrees, which has a reference angle of 45 degrees where tan is 1, so tan(135 degrees) is -1. That makes sense.So, the slope angle of line AB is 3pi/4. Looking back at the options, A is 3pi/4, so that should be the correct answer.Wait, just to make sure I didn't make any mistakes. Let me visualize the points. Point A is at (0,1), which is on the y-axis one unit up, and point B is at (1,0), which is on the x-axis one unit to the right. So, the line connecting these two points goes from the upper left to the lower right, which means it's a negative slope, which aligns with the slope being -1. The angle it makes with the positive x-axis should be obtuse, more than 90 degrees but less than 180 degrees, which 3pi/4 is, since pi is 180 degrees, so 3pi/4 is 135 degrees. Yep, that seems right.I think I'm confident now that the answer is A: 3pi/4.

💡First, I need to determine the price of each brownie that Jack sold. I'll start by identifying all the given information.Jack has sold:- 4 brownies at an unknown price each.- 5 lemon squares at 2 each.- 7 cookies at 4 each.His total goal is to make 50.Next, I'll calculate the total amount earned from the lemon squares and cookies:- Lemon squares: 5 × 2 = 10- Cookies: 7 × 4 = 28Adding these together gives 10 + 28 = 38.Now, I'll set up an equation to represent the total amount earned:4 × (price of one brownie) + 38 = 50Let ( B ) represent the price of one brownie:4B + 38 = 50To solve for ( B ), I'll subtract 38 from both sides of the equation:4B = 12Finally, I'll divide both sides by 4 to find the price of one brownie:B = 3Therefore, the price of each brownie is 3.

💡First, I need to determine the total number of medals displayed, which is 67.Next, I'll identify the number of gold and silver medals. There are 19 gold medals and 32 silver medals.To find out how many bronze medals there are, I'll subtract the number of gold and silver medals from the total number of medals.Adding the gold and silver medals together: 19 + 32 = 51.Subtracting this sum from the total number of medals: 67 - 51 = 16.Therefore, there are 16 bronze medals displayed.