answer：Okay, so I have this problem where I'm given two sets, A and B. Set A is {0, 1, 3}, and set B is {a+1, a² + 2}. The intersection of A and B is {1}, which means that the only common element between A and B is 1. My goal is to find the real number a that satisfies this condition.First, let me understand what the intersection means. If A ∩ B = {1}, that tells me that 1 is in both sets A and B, and nothing else is common between them. So, in set B, which has two elements: a+1 and a² + 2, one of these must be equal to 1, and the other should not be equal to any other elements in A, which are 0, 1, or 3.So, let me break it down. Since 1 is in both A and B, either a+1 = 1 or a² + 2 = 1. Let me consider both possibilities.First, let's suppose that a+1 = 1. If I solve for a, I subtract 1 from both sides:a + 1 = 1 a = 1 - 1 a = 0So, if a is 0, then set B becomes {0 + 1, 0² + 2} which is {1, 2}. Now, let's check the intersection of A and B. A is {0, 1, 3} and B is {1, 2}, so their intersection is {1}, which matches the given condition. That seems good.But wait, I should also check the other possibility where a² + 2 = 1. Let me solve that equation:a² + 2 = 1 a² = 1 - 2 a² = -1Hmm, a squared equals -1. But a is a real number, right? Because the problem specifies that a is a real number. So, a² can't be negative because any real number squared is non-negative. Therefore, this equation has no real solutions. So, a² + 2 can never be 1 for any real a. That means the only possibility is that a+1 = 1, which gives a = 0.But just to be thorough, let me make sure that when a = 0, the other element in set B, which is a² + 2, doesn't end up being another element in A. So, if a = 0, then a² + 2 is 0 + 2 = 2. 2 is not in set A, which is {0, 1, 3}, so that's fine. So, the intersection is only {1}, as required.Let me double-check if there's any other possible value of a that might satisfy the condition. Suppose a is some other number, say a = 1. Then, set B would be {1 + 1, 1² + 2} = {2, 3}. The intersection of A and B would be {3}, but the problem says the intersection is {1}, so a = 1 doesn't work.What if a = -1? Then, set B would be {-1 + 1, (-1)² + 2} = {0, 3}. The intersection of A and B would be {0, 3}, which is not just {1}, so that's not acceptable either.How about a = 2? Then, set B would be {2 + 1, 2² + 2} = {3, 6}. The intersection would be {3}, which again doesn't meet the condition.What if a is a fraction, like a = 0.5? Then, set B would be {0.5 + 1, (0.5)² + 2} = {1.5, 2.25}. The intersection with A is empty, which is not what we want.Wait, but the problem says that A ∩ B = {1}, so if a is such that neither a+1 nor a² + 2 is 1, then the intersection would be empty, which isn't the case here. So, we must have exactly one of them equal to 1, and the other not equal to any element in A.We already saw that a² + 2 can't be 1 because it would require a² = -1, which isn't possible for real a. So, the only way is a+1 = 1, leading to a = 0. Let me confirm once more:If a = 0, then B = {1, 2}. A = {0, 1, 3}. The intersection is {1}, which is exactly what we need. The other element in B is 2, which isn't in A, so that's perfect.I don't think there's any other value of a that would make the intersection exactly {1}. If a were something else, either the intersection would be empty or it would include more elements, which violates the given condition.So, after considering all possibilities and checking each case, I'm confident that the only real number a that satisfies the condition is 0.**Final Answer**The value of the real number ( a ) is boxed{0}.

answer：First, I need to determine how many slices of bread Tony used throughout the week. He made one sandwich each day from Monday to Friday, which totals 5 sandwiches. Since each sandwich requires 2 slices of bread, the total number of slices used during these 5 days is 5 sandwiches multiplied by 2 slices per sandwich, which equals 10 slices.On Saturday, Tony made two sandwiches because he was extra hungry. Each sandwich still requires 2 slices of bread, so the total number of slices used on Saturday is 2 sandwiches multiplied by 2 slices per sandwich, resulting in 4 slices.Adding the slices used from Monday to Friday and on Saturday gives a total of 10 slices plus 4 slices, which equals 14 slices used throughout the week.Finally, to find out how many slices of bread are left from the original 22-slice loaf, I subtract the total slices used from the initial number of slices. This calculation is 22 slices minus 14 slices used, which equals 8 slices remaining.

answer：First, identify the number of books Mr. Rodgers read, which is 10.Next, determine the number of chapters in each book, which is 24.To find the total number of chapters Mr. Rodgers read, multiply the number of books by the number of chapters per book.Calculating 10 multiplied by 24 gives 240.Therefore, Mr. Rodgers read a total of 240 chapters.

answer：Okay, so I have this problem where I need to find the equation of a tangent line ( l_2 ) to the curve ( y = x^2 + x - 2 ) such that ( l_2 ) is perpendicular to another tangent line ( l_1 ) at the point ( (1, 0) ). Then, I also need to find the area of the triangle formed by ( l_1 ), ( l_2 ), and the x-axis. Hmm, let me break this down step by step.First, I need to find the equation of ( l_1 ). Since ( l_1 ) is the tangent to the curve at ( (1, 0) ), I can find its slope by taking the derivative of the curve at that point. The derivative of ( y = x^2 + x - 2 ) is ( y' = 2x + 1 ). Plugging in ( x = 1 ), the slope ( m_1 ) is ( 2(1) + 1 = 3 ). So, the slope of ( l_1 ) is 3.Now, using the point-slope form of a line, which is ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is a point on the line. Here, the point is ( (1, 0) ), so plugging in, we get ( y - 0 = 3(x - 1) ), which simplifies to ( y = 3x - 3 ). So, that's the equation of ( l_1 ).Next, I need to find the equation of ( l_2 ), which is another tangent to the same curve and is perpendicular to ( l_1 ). Since ( l_1 ) has a slope of 3, the slope of ( l_2 ) must be the negative reciprocal, which is ( -1/3 ).To find ( l_2 ), I need another point on the curve where the tangent has a slope of ( -1/3 ). Let's denote this point as ( (b, b^2 + b - 2) ). The derivative at this point is ( y' = 2b + 1 ), which should equal ( -1/3 ). So, setting up the equation:( 2b + 1 = -1/3 )Solving for ( b ):( 2b = -1/3 - 1 )( 2b = -4/3 )( b = -2/3 )So, the x-coordinate of the point where ( l_2 ) is tangent to the curve is ( -2/3 ). Plugging this back into the curve equation to find the y-coordinate:( y = (-2/3)^2 + (-2/3) - 2 = 4/9 - 2/3 - 2 )Convert to ninths:( 4/9 - 6/9 - 18/9 = (4 - 6 - 18)/9 = (-20)/9 )So, the point is ( (-2/3, -20/9) ).Now, using the point-slope form again for ( l_2 ):( y - (-20/9) = (-1/3)(x - (-2/3)) )Simplify:( y + 20/9 = (-1/3)(x + 2/3) )Multiply out the right side:( y + 20/9 = (-1/3)x - 2/9 )Subtract 20/9 from both sides:( y = (-1/3)x - 2/9 - 20/9 )Combine like terms:( y = (-1/3)x - 22/9 )Wait, that doesn't seem right. Let me double-check my calculations. When I calculated the y-coordinate, I had ( (-2/3)^2 = 4/9 ), ( (-2/3) = -2/3 ), and then -2. So, 4/9 - 2/3 - 2. Converting to ninths:4/9 - 6/9 - 18/9 = (4 - 6 - 18)/9 = (-20)/9. That seems correct.Then, using point-slope:( y - (-20/9) = (-1/3)(x - (-2/3)) )Which is:( y + 20/9 = (-1/3)(x + 2/3) )Expanding the right side:( (-1/3)x - (2/3)(1/3) = (-1/3)x - 2/9 )So, subtracting 20/9:( y = (-1/3)x - 2/9 - 20/9 = (-1/3)x - 22/9 )Hmm, that seems correct. So, the equation of ( l_2 ) is ( y = (-1/3)x - 22/9 ).Wait, but when I think about the tangent line, it should pass through the point ( (-2/3, -20/9) ). Let me verify if plugging ( x = -2/3 ) into the equation gives ( y = -20/9 ):( y = (-1/3)(-2/3) - 22/9 = 2/9 - 22/9 = -20/9 ). Yes, that checks out.Okay, so part (1) is done. The equation of ( l_2 ) is ( y = (-1/3)x - 22/9 ).Now, moving on to part (2), finding the area of the triangle formed by ( l_1 ), ( l_2 ), and the x-axis.First, I need to find the points where ( l_1 ) and ( l_2 ) intersect the x-axis. For ( l_1 ), set ( y = 0 ):( 0 = 3x - 3 )( 3x = 3 )( x = 1 )So, the x-intercept is ( (1, 0) ), which is already given.For ( l_2 ), set ( y = 0 ):( 0 = (-1/3)x - 22/9 )Multiply both sides by 9 to eliminate denominators:( 0 = -3x - 22 )( 3x = -22 )( x = -22/3 )So, the x-intercept is ( (-22/3, 0) ).Next, I need the intersection point of ( l_1 ) and ( l_2 ). Let's set their equations equal:( 3x - 3 = (-1/3)x - 22/9 )Multiply both sides by 9 to eliminate denominators:( 27x - 27 = -3x - 22 )Bring all terms to one side:( 27x + 3x - 27 + 22 = 0 )( 30x - 5 = 0 )( 30x = 5 )( x = 5/30 = 1/6 )Now, plug ( x = 1/6 ) back into ( l_1 ) to find y:( y = 3(1/6) - 3 = 1/2 - 3 = -5/2 )So, the intersection point is ( (1/6, -5/2) ).Now, the triangle is formed by the points ( (1, 0) ), ( (-22/3, 0) ), and ( (1/6, -5/2) ). To find the area, I can use the formula for the area of a triangle given by three points:( text{Area} = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| )Plugging in the points ( (1, 0) ), ( (-22/3, 0) ), and ( (1/6, -5/2) ):( text{Area} = frac{1}{2} |1(0 - (-5/2)) + (-22/3)((-5/2) - 0) + (1/6)(0 - 0)| )Simplify each term:First term: ( 1*(5/2) = 5/2 )Second term: ( (-22/3)*(-5/2) = (110)/6 = 55/3 )Third term: ( (1/6)*(0) = 0 )So, adding them up:( 5/2 + 55/3 + 0 = (15/6 + 110/6) = 125/6 )Taking the absolute value (which is positive) and multiplying by 1/2:( text{Area} = frac{1}{2} * 125/6 = 125/12 )So, the area is ( 125/12 ) square units.Wait, let me double-check the area calculation. Alternatively, since two points are on the x-axis, the base of the triangle is the distance between ( (1, 0) ) and ( (-22/3, 0) ). The distance is ( |1 - (-22/3)| = |1 + 22/3| = |25/3| = 25/3 ). The height is the vertical distance from the intersection point ( (1/6, -5/2) ) to the x-axis, which is ( | -5/2 | = 5/2 ).So, area is ( frac{1}{2} * base * height = frac{1}{2} * (25/3) * (5/2) = (25/3)*(5/4) = 125/12 ). Yep, same result.So, the area is ( 125/12 ).**Final Answer**(1) The equation of line ( l_2 ) is boxed{y = -dfrac{1}{3}x - dfrac{22}{9}}.(2) The area of the triangle is boxed{dfrac{125}{12}}.

answer：Okay, so I have this problem where I'm given that the tangent of x is 3, and I need to find the value of two different expressions. Let me take this step by step because I want to make sure I understand each part correctly.Starting with the first expression:(1) (frac{2sin(pi - x) + 3cos(-x)}{sin(x + frac{pi}{2}) - sin(x + pi)})Hmm, I remember that there are some trigonometric identities that can simplify this expression. Let me recall them.First, (sin(pi - x)) is equal to (sin x). I think that's one of the co-function identities. Similarly, (cos(-x)) is equal to (cos x) because cosine is an even function. So, substituting these into the numerator:Numerator: (2sin x + 3cos x)Now, looking at the denominator: (sin(x + frac{pi}{2}) - sin(x + pi))I remember that (sin(x + frac{pi}{2})) is equal to (cos x), which is another co-function identity. And (sin(x + pi)) is equal to (-sin x) because adding pi to an angle reflects it across the x-axis, changing the sign. So substituting these into the denominator:Denominator: (cos x - (-sin x)) which simplifies to (cos x + sin x)So now, the entire expression becomes:(frac{2sin x + 3cos x}{cos x + sin x})Hmm, this looks like I can factor out a common term in the numerator and denominator. Let me see if I can express everything in terms of tangent x since I know that (tan x = 3). Remember that (tan x = frac{sin x}{cos x}), so maybe I can divide numerator and denominator by (cos x).Let's try that:Numerator: (2sin x + 3cos x = 2frac{sin x}{cos x} cdot cos x + 3cos x = 2tan x cdot cos x + 3cos x). Wait, maybe a better approach is to divide each term by (cos x):So, (frac{2sin x}{cos x} + frac{3cos x}{cos x} = 2tan x + 3)Similarly, the denominator: (cos x + sin x = frac{cos x}{cos x} + frac{sin x}{cos x} = 1 + tan x)So now, the expression becomes:(frac{2tan x + 3}{1 + tan x})Since (tan x = 3), substitute that in:(frac{2(3) + 3}{1 + 3} = frac{6 + 3}{4} = frac{9}{4})Alright, that seems straightforward. So the first part simplifies to (frac{9}{4}).Now, moving on to the second expression:(2) (2sin^2 x - sin 2x + cos^2 x)I need to simplify this expression as well. Let me recall some trigonometric identities. I know that (sin 2x = 2sin x cos x), so maybe I can substitute that in.First, let's rewrite the expression:(2sin^2 x - 2sin x cos x + cos^2 x)Hmm, I also remember that (sin^2 x + cos^2 x = 1). Maybe I can use that identity to simplify this expression.Looking at the expression, I have (2sin^2 x + cos^2 x - 2sin x cos x). Let me group the terms:( (2sin^2 x + cos^2 x) - 2sin x cos x )I can write (2sin^2 x) as (sin^2 x + sin^2 x), so:( (sin^2 x + sin^2 x + cos^2 x) - 2sin x cos x )Now, since (sin^2 x + cos^2 x = 1), this becomes:( (1 + sin^2 x) - 2sin x cos x )So, the expression simplifies to:(1 + sin^2 x - 2sin x cos x)Hmm, I wonder if I can express this in terms of (tan x) since I know that (tan x = 3). Let me try to express everything in terms of (sin x) and (cos x), then divide by (cos^2 x) to get (tan x).But before that, let me see if I can factor this expression differently. Alternatively, maybe I can express (sin^2 x) in terms of (tan x).Since (tan x = frac{sin x}{cos x}), let me denote (t = tan x = 3). Then, (sin x = t cos x). Also, since (sin^2 x + cos^2 x = 1), substituting (sin x) gives:(t^2 cos^2 x + cos^2 x = 1)(cos^2 x (t^2 + 1) = 1)So, (cos^2 x = frac{1}{t^2 + 1}), and (sin^2 x = t^2 cos^2 x = frac{t^2}{t^2 + 1})Alright, so let's substitute back into the expression:(1 + sin^2 x - 2sin x cos x = 1 + frac{t^2}{t^2 + 1} - 2(t cos x)(cos x))Wait, hold on, (sin x cos x = t cos^2 x), so:(1 + frac{t^2}{t^2 + 1} - 2t cos^2 x)But we already have (cos^2 x = frac{1}{t^2 + 1}), so substituting that in:(1 + frac{t^2}{t^2 + 1} - 2t cdot frac{1}{t^2 + 1})Combine the terms:(1 + frac{t^2 - 2t}{t^2 + 1})Now, let's write 1 as (frac{t^2 + 1}{t^2 + 1}):(frac{t^2 + 1}{t^2 + 1} + frac{t^2 - 2t}{t^2 + 1} = frac{t^2 + 1 + t^2 - 2t}{t^2 + 1})Combine like terms in the numerator:(2t^2 - 2t + 1) over (t^2 + 1)So, the expression becomes:(frac{2t^2 - 2t + 1}{t^2 + 1})Now, substitute (t = 3):(frac{2(9) - 2(3) + 1}{9 + 1} = frac{18 - 6 + 1}{10} = frac{13}{10})So, the second expression simplifies to (frac{13}{10}).Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the expression:(2sin^2 x - sin 2x + cos^2 x)Substituted (sin 2x = 2sin x cos x), so:(2sin^2 x - 2sin x cos x + cos^2 x)Then, grouped as:((2sin^2 x + cos^2 x) - 2sin x cos x)Expressed (2sin^2 x) as (sin^2 x + sin^2 x), so:((sin^2 x + sin^2 x + cos^2 x) - 2sin x cos x)Which simplifies to:((1 + sin^2 x) - 2sin x cos x)Then, expressed everything in terms of (t = tan x = 3):(1 + frac{t^2}{t^2 + 1} - 2t cdot frac{1}{t^2 + 1})Combined the terms:(1 + frac{t^2 - 2t}{t^2 + 1})Converted 1 to (frac{t^2 + 1}{t^2 + 1}):(frac{t^2 + 1 + t^2 - 2t}{t^2 + 1} = frac{2t^2 - 2t + 1}{t^2 + 1})Substituted (t = 3):(frac{18 - 6 + 1}{10} = frac{13}{10})Yes, that seems correct.So, summarizing:1. The first expression simplifies to (frac{9}{4}).2. The second expression simplifies to (frac{13}{10}).I think that's it. I don't see any mistakes in my reasoning, so I feel confident with these answers.

answer：First, I need to understand what ( g(z) ) represents. It is the product of all even integers from 2 up to ( z ). For example, ( g(14) = 2 times 4 times 6 times 8 times 10 times 12 times 14 ).The goal is to determine the exponent of 4 in ( g(z) ) when ( z = 24 ). Since 4 is ( 2^2 ), I need to count how many times 2 appears as a factor in the product of all even numbers up to 24.I'll list each even number up to 24 and determine the number of 2s in their prime factorizations:- 2: ( 2^1 ) (1 two)- 4: ( 2^2 ) (2 twos)- 6: ( 2^1 times 3 ) (1 two)- 8: ( 2^3 ) (3 twos)- 10: ( 2^1 times 5 ) (1 two)- 12: ( 2^2 times 3 ) (2 twos)- 14: ( 2^1 times 7 ) (1 two)- 16: ( 2^4 ) (4 twos)- 18: ( 2^1 times 9 ) (1 two)- 20: ( 2^2 times 5 ) (2 twos)- 22: ( 2^1 times 11 ) (1 two)- 24: ( 2^3 times 3 ) (3 twos)Next, I'll sum the number of 2s:[1 + 2 + 1 + 3 + 1 + 2 + 1 + 4 + 1 + 2 + 1 + 3 = 22]Since each 4 requires two 2s, I'll divide the total number of 2s by 2 to find the exponent of 4:[frac{22}{2} = 11]Therefore, the exponent of 4 in ( g(24) ) is 11.