answer：First, I need to solve the inequality (-x^{2} - x + 2 geqslant 0). To make it easier, I'll start by multiplying both sides of the inequality by (-1). This will reverse the inequality sign, so the inequality becomes (x^{2} + x - 2 leqslant 0).Next, I'll factor the quadratic expression (x^{2} + x - 2). I'm looking for two numbers that multiply to (-2) and add up to (1). The numbers (2) and (-1) satisfy these conditions, so the factored form of the quadratic is ((x - 1)(x + 2)).Now, I have the inequality ((x - 1)(x + 2) leqslant 0). To find the solution set, I'll determine the intervals where the product of ((x - 1)) and ((x + 2)) is less than or equal to zero. The critical points are (x = 1) and (x = -2), which divide the number line into three intervals: (x < -2), (-2 leqslant x leqslant 1), and (x > 1).I'll test each interval to see where the product is negative or zero:1. For (x < -2), both ((x - 1)) and ((x + 2)) are negative, so their product is positive.2. For (-2 leqslant x leqslant 1), ((x - 1)) is negative and ((x + 2)) is positive, so their product is negative.3. For (x > 1), both ((x - 1)) and ((x + 2)) are positive, so their product is positive.Since we're looking for where the product is less than or equal to zero, the solution is the interval where the product is negative or zero, which is (-2 leqslant x leqslant 1).Therefore, the solution set of the inequality is ({x | -2 leqslant x leqslant 1}).

answer：Alright, let's tackle this problem step by step. I have an odd function ( f(x) ) defined for all real numbers. I know that ( f(x+2) ) is an even function, and also that ( f(-1) = -1 ). I need to find ( f(2017) + f(2016) ). Hmm, okay, let's break this down.First, let's recall what it means for a function to be odd or even. An odd function satisfies ( f(-x) = -f(x) ) for all ( x ), and an even function satisfies ( f(-x) = f(x) ) for all ( x ). So, ( f(x) ) is odd, which means ( f(-x) = -f(x) ). Also, ( f(x+2) ) is even, so ( f(-x+2) = f(x+2) ).Let me write down these properties:1. ( f(-x) = -f(x) ) (since ( f ) is odd)2. ( f(-x + 2) = f(x + 2) ) (since ( f(x+2) ) is even)I need to find ( f(2017) + f(2016) ). These are pretty large numbers, so maybe there's a periodicity or some pattern I can exploit.Let me see if I can find a relationship or a recurrence for ( f(x) ). Starting from the second property:( f(-x + 2) = f(x + 2) )Let me substitute ( x ) with ( x + 2 ) to see if that helps:( f(-(x + 2) + 2) = f((x + 2) + 2) )Simplify:( f(-x - 2 + 2) = f(x + 4) )Which simplifies to:( f(-x) = f(x + 4) )But from the first property, ( f(-x) = -f(x) ). So, substituting that in:( -f(x) = f(x + 4) )So, ( f(x + 4) = -f(x) ). That's interesting. Let's see what happens if we shift ( x ) by 4 again:( f(x + 8) = -f(x + 4) )But from the previous equation, ( f(x + 4) = -f(x) ), so:( f(x + 8) = -(-f(x)) = f(x) )Ah! So, ( f(x + 8) = f(x) ). This means the function ( f ) is periodic with period 8. That's a crucial piece of information.So, the function repeats every 8 units. Therefore, to find ( f(2017) ) and ( f(2016) ), I can reduce these numbers modulo 8.Let's compute ( 2017 mod 8 ) and ( 2016 mod 8 ).First, ( 2016 div 8 = 252 ) exactly, since ( 252 times 8 = 2016 ). So, ( 2016 mod 8 = 0 ).Next, ( 2017 div 8 ) is 252 with a remainder of 1, because ( 252 times 8 = 2016 ), so ( 2017 = 2016 + 1 ). Therefore, ( 2017 mod 8 = 1 ).So, ( f(2016) = f(0) ) and ( f(2017) = f(1) ).Now, I need to find ( f(0) ) and ( f(1) ).Since ( f ) is an odd function, ( f(-0) = -f(0) ), which simplifies to ( f(0) = -f(0) ). The only number that is equal to its own negative is zero. Therefore, ( f(0) = 0 ).So, ( f(2016) = 0 ).Next, let's find ( f(1) ). I know that ( f(-1) = -1 ) from the problem statement. Since ( f ) is odd, ( f(-1) = -f(1) ). Therefore:( f(-1) = -f(1) )( -1 = -f(1) )Multiply both sides by -1:( 1 = f(1) )So, ( f(1) = 1 ).Therefore, ( f(2017) = f(1) = 1 ).Putting it all together:( f(2017) + f(2016) = 1 + 0 = 1 )Wait, but looking back at the options, D is 1. So, that should be the answer. But just to make sure I didn't make any mistakes, let me verify.We established that ( f ) is periodic with period 8, so ( f(2016) = f(0) = 0 ) and ( f(2017) = f(1) = 1 ). Adding them gives 1. That seems correct.But just to double-check, let's see if the function's properties hold.We know ( f(x + 4) = -f(x) ). So, ( f(1 + 4) = f(5) = -f(1) = -1 ). Then, ( f(5 + 4) = f(9) = -f(5) = -(-1) = 1 ). And since the period is 8, ( f(9) = f(1) = 1 ), which is consistent.Similarly, ( f(2016) = f(0) = 0 ), which is consistent with the function being odd.So, yeah, I think the answer is D: 1.**Final Answer**boxed{D}

answer：First, I need to determine how many hours the first mechanic worked. Since the total combined hours are 20 and the second mechanic worked for 5 hours, the first mechanic must have worked for 15 hours.Next, I'll calculate the total charge by the first mechanic by multiplying their hourly rate by the number of hours they worked: 45 dollars/hour multiplied by 15 hours equals 675 dollars.To find out how much the second mechanic charged in total, I'll subtract the first mechanic's total charge from the overall total charge: 1100 dollars minus 675 dollars equals 425 dollars.Finally, to determine the second mechanic's hourly rate, I'll divide their total charge by the number of hours they worked: 425 dollars divided by 5 hours equals 85 dollars per hour.

answer：Alright, so I've got this problem here with five pieces, and I need to figure out which of the five given figures can't be formed using these pieces. The pieces are all made up of squares, and from the Asymptote code, it looks like they're arranged in specific configurations. Let me try to visualize them based on the description.First, the pieces are shown in an Asymptote figure, which I can't see directly, but from the code, it seems like there are five distinct pieces, each made up of multiple squares. The first piece is just a single square. The second piece is two squares stacked vertically. The third piece is three squares in a sort of L-shape. The fourth piece is four squares arranged in a 2x2 square. The fifth piece is five squares in a straight line. So, in total, we have 1 + 2 + 3 + 4 + 5 = 15 squares.Now, the figures we need to check are also made up of these squares. Each figure is a different arrangement, and four of them can be formed using the five pieces, but one cannot. The options are labeled (A) through (E), and I need to figure out which one is impossible.Let me think about how to approach this. Since all the pieces are made up of squares, and the total number of squares is 15, each figure must also consist of 15 squares. So, first, I should check if all the given figures have 15 squares. If any of them don't, that would be the answer. But I think they all have 15 squares because the problem states that four can be formed, implying they all have the same number of squares.Next, I need to consider the shapes of the pieces. The pieces have different configurations, so the key is whether the arrangement of squares in each figure can be covered exactly by the given pieces without overlapping or leaving gaps.Let me try to analyze each figure one by one.Starting with figure (A). From the Asymptote code, it seems like figure (A) is a 5x3 rectangle. Wait, no, actually, looking at the code, it's a bit more complex. It has multiple rows with varying numbers of squares. Let me parse the code:For figure (A), the coordinates are:{(0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (2,0), (2,1), (3,0), (3,1), (3,2), (3,3), (3,4)}So, plotting these points, it seems like figure (A) has a sort of irregular shape. The first column has 5 squares, the second column has 3 squares, the third column has 2 squares, the fourth column has 4 squares, and the fifth column has 1 square. Hmm, that seems a bit complicated, but maybe it can be formed by the given pieces.Moving on to figure (B). Its coordinates are:{(5,3), (5,4), (6,2), (6,3), (6,4), (7,1), (7,2), (7,3), (7,4), (8,0), (8,1), (8,2), (9,0), (9,1), (9,2)}Plotting these, figure (B) seems to have a kind of stepped shape. The first column has 2 squares, the second column has 3 squares, the third column has 4 squares, the fourth column has 3 squares, and the fifth column has 2 squares. This also seems a bit tricky, but perhaps manageable.Figure (C) has coordinates:{(11,0), (11,1), (11,2), (11,3), (11,4), (12,1), (12,2), (12,3), (12,4), (13,2), (13,3), (13,4), (14,3), (14,4), (15,4)}This looks like a sort of pyramid or stepped shape as well. The first column has 5 squares, the second column has 4 squares, the third column has 3 squares, the fourth column has 2 squares, and the fifth column has 1 square. This might be a bit challenging, but maybe possible.Figure (D) has coordinates:{(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)}This seems like a 3x5 rectangle. So, three rows and five columns, each with 5 squares. That should be straightforward to form with the given pieces, especially since we have a 2x2 square and a straight line of five squares.Figure (E) has coordinates:{(21,4), (22,1), (22,2), (22,3), (22,4), (23,0), (23,1), (23,2), (23,3), (23,4), (24,1), (24,2), (24,3), (24,4), (25,4)}This figure seems to have a kind of hook shape. The first column has 1 square, the second column has 4 squares, the third column has 5 squares, the fourth column has 4 squares, and the fifth column has 1 square. This might be a bit complex, but perhaps doable.Now, considering the pieces we have: a single square, two squares vertical, three squares L-shape, four squares 2x2, and five squares straight. Let's see how these can fit into each figure.Starting with figure (D), which is a 3x5 rectangle. This should be straightforward because we have a straight line of five squares, which can form the base. Then, the 2x2 square can fit into the middle, and the L-shape and vertical domino can fill in the remaining spaces. So, figure (D) is likely possible.Figure (A) is a bit irregular, but let's see. It has a column of five, which can be formed by the straight line of five squares. Then, the adjacent columns have fewer squares, which can be filled with the other pieces. The L-shape and the vertical domino can fit into the gaps. So, figure (A) might be possible.Figure (C) is a stepped pyramid. The first column has five squares, which can be the straight line. Then, the next columns have decreasing numbers, which can be filled with the L-shape, the vertical domino, and the 2x2 square. It might require some careful placement, but it seems feasible.Figure (E) has a hook shape. The long straight line can form part of the hook, and the 2x2 square can fill in the corner. The L-shape and vertical domino can fill the remaining parts. It might be a bit tricky, but possible.Now, figure (B) is the one that stands out. It has a stepped shape with columns of 2, 3, 4, 3, 2 squares. The issue here might be the central column with four squares. We have a straight line of five squares, which is longer than needed, and the 2x2 square, which is 2x2. The L-shape is three squares, and the vertical domino is two squares. Wait, the central column has four squares. If we try to place the straight line of five squares vertically, it would extend beyond the figure. If we place it horizontally, it might not fit because the figure is only five columns wide. The 2x2 square can cover four squares, but it's a square, so it might not fit into the linear column of four squares.Alternatively, maybe we can combine pieces. The straight line of five can be broken down, but no, we have to use the pieces as they are. So, we can't break the straight line into smaller pieces; we have to use it as a single unit.Similarly, the 2x2 square is a block, so it can't be split. The L-shape and vertical domino are also fixed. So, trying to fit these into figure (B)'s central column of four squares might not work because none of the pieces can cover four squares in a straight line except the straight line of five, which is too long, or the 2x2 square, which is a block and doesn't fit into a straight line.Therefore, figure (B) might not be possible to form with the given pieces because we can't cover the central column of four squares without either overlapping or leaving gaps.Wait, let me double-check. Maybe there's a way to arrange the pieces differently. If I place the 2x2 square in the center, it would cover four squares, but it's a square, so it would cover a 2x2 area, which might not align with the linear column. The straight line of five is too long, so it can't be used vertically. Maybe horizontally, but the figure is only five columns wide, and the straight line would fit, but then the other pieces might not fit around it.Alternatively, maybe the L-shape and vertical domino can be arranged to cover the four squares in the center. Let's see. The L-shape covers three squares, and the vertical domino covers two. If we place the L-shape in the center, it would leave one square uncovered, which the vertical domino can't cover because it's two squares. Similarly, placing the vertical domino first would leave two squares, which the L-shape can't cover because it needs three.So, it seems like the central column of four squares can't be covered by any combination of the given pieces without overlapping or leaving gaps. Therefore, figure (B) is the one that cannot be formed.I think I've considered all possibilities, and figure (B) is the only one that can't be formed with the given pieces.

answer：Okay, so I have this problem where ( a ) varies inversely with ( b^2 ). Hmm, I remember that inverse variation means that as one quantity increases, the other decreases proportionally. But let me think more carefully about what that exactly means in terms of an equation.I think when two quantities vary inversely, their product is a constant. So, if ( a ) varies inversely with ( b^2 ), then ( a times b^2 = k ) for some constant ( k ). Yeah, that sounds right. So, the equation would be ( a = frac{k}{b^2} ).Alright, the problem gives me that ( a = 4 ) when ( b = 2 ). I can use these values to find the constant ( k ). Let me plug those numbers into the equation.So, substituting ( a = 4 ) and ( b = 2 ) into ( a = frac{k}{b^2} ), I get:[4 = frac{k}{2^2}]Simplifying ( 2^2 ) gives me 4, so:[4 = frac{k}{4}]To solve for ( k ), I can multiply both sides by 4:[4 times 4 = k]Which means:[k = 16]Okay, so now I know that the constant ( k ) is 16. That means the relationship between ( a ) and ( b ) is:[a = frac{16}{b^2}]Now, the question asks me to find the value of ( a ) when ( b = 8 ). Let me substitute ( b = 8 ) into the equation.So, plugging in ( b = 8 ):[a = frac{16}{8^2}]Calculating ( 8^2 ) gives me 64, so:[a = frac{16}{64}]Simplifying that fraction, I divide both numerator and denominator by 16:[a = frac{1}{4}]Wait, let me double-check that. If I have 16 divided by 64, that's the same as 1 divided by 4, which is 0.25. Yeah, that seems right.So, when ( b ) increases from 2 to 8, ( a ) decreases from 4 to ( frac{1}{4} ). That makes sense because ( a ) varies inversely with ( b^2 ), so as ( b ) gets larger, ( a ) gets smaller, and since ( b ) is squared, the effect is more pronounced.Just to make sure I didn't make any mistakes, let me go through the steps again.1. Start with the inverse variation formula: ( a = frac{k}{b^2} ).2. Plug in ( a = 4 ) and ( b = 2 ) to find ( k ): [ 4 = frac{k}{4} implies k = 16 ]3. Now, use ( k = 16 ) and ( b = 8 ) to find ( a ): [ a = frac{16}{64} = frac{1}{4} ]Everything checks out. I think I've got it right.Another way to think about it is by using the ratio of the two situations. Since ( a ) varies inversely with ( b^2 ), the ratio of the new ( a ) to the original ( a ) should be equal to the inverse ratio of the squares of the new ( b ) to the original ( b ).So, mathematically, that would be:[frac{a_{text{new}}}{a_{text{original}}} = frac{b_{text{original}}^2}{b_{text{new}}^2}]Plugging in the values:[frac{a_{text{new}}}{4} = frac{2^2}{8^2} = frac{4}{64} = frac{1}{16}]Then, solving for ( a_{text{new}} ):[a_{text{new}} = 4 times frac{1}{16} = frac{4}{16} = frac{1}{4}]Yep, that's the same answer as before. This method also confirms that ( a ) becomes ( frac{1}{4} ) when ( b ) is 8.I guess another thing to consider is the units or context, but since no units are given, I don't think that affects the calculation here. It's purely a mathematical relationship.Just to visualize it, if I plot ( a ) against ( b^2 ), it should be a straight line passing through the origin because it's an inverse variation. The constant ( k ) would be the product ( a times b^2 ), which we found to be 16. So, any point on that line would satisfy ( a times b^2 = 16 ).When ( b = 2 ), ( b^2 = 4 ), and ( a = 4 ), so ( 4 times 4 = 16 ). When ( b = 8 ), ( b^2 = 64 ), and ( a = frac{1}{4} ), so ( frac{1}{4} times 64 = 16 ). Both points lie on the same line, confirming the consistency of the solution.I think I've covered all the bases here. I've used two different methods to arrive at the same answer, checked my work, and even thought about the graphical representation to ensure everything makes sense. I feel confident that ( a = frac{1}{4} ) when ( b = 8 ).**Final Answer**The value of ( a ) when ( b = 8 ) is boxed{dfrac{1}{4}}.

answer：Okay, so I need to figure out the probability of getting all heads when flipping a fair coin three times. Hmm, let's see. I remember that when dealing with probabilities, especially with multiple events, I need to consider whether the events are independent or not. In this case, each coin flip is independent because the outcome of one flip doesn't affect the others.First, I should recall that a fair coin has two possible outcomes: heads or tails, each with an equal probability of 1/2. So, for each flip, the chance of getting heads is 1/2. Now, since the flips are independent, the probability of getting heads on all three flips should be the product of the probabilities of getting heads on each individual flip.Let me write that down to make it clearer. The probability of getting heads on the first flip is 1/2. The probability of getting heads on the second flip is also 1/2. Similarly, the probability of getting heads on the third flip is 1/2. To find the probability of all three events happening together, I need to multiply these probabilities.So, that would be 1/2 multiplied by 1/2 multiplied by 1/2. Let me calculate that: 1/2 times 1/2 is 1/4, and then 1/4 times 1/2 is 1/8. Therefore, the probability of getting all heads in three flips is 1/8.Wait, let me double-check that. Maybe I can list out all possible outcomes to make sure. When flipping a coin three times, the possible outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. That's eight possible outcomes in total. Out of these, only one outcome is HHH, which is all heads. So, the probability is indeed 1 out of 8, which is 1/8.I think that makes sense. Each flip is independent, and the probability of each specific outcome is the product of the probabilities of each individual event. So, multiplying 1/2 three times gives me 1/8. Yeah, that seems right.Another way to think about it is using exponents. Since each flip has 2 possible outcomes, the total number of possible outcomes for three flips is 2^3, which is 8. And since only one of those outcomes is all heads, the probability is 1/8.Okay, I'm pretty confident now that the probability is 1/8.