answer：Okay, so Sarah needs to fence a rectangular area, and she wants to use the least amount of material possible. That means she wants to minimize the perimeter of the rectangle. But there are some conditions: the area has to be at least 150 square feet, and the length has to be 20 feet longer than the width. Hmm, okay, let's break this down.First, let's define the variables. Let's say the width of the rectangle is ( w ) feet. Then, since the length is 20 feet longer than the width, the length would be ( w + 20 ) feet.Now, the area of a rectangle is given by the formula:[text{Area} = text{length} times text{width}]So, plugging in our variables, the area ( A ) would be:[A = w times (w + 20)]And we know that this area has to be at least 150 square feet:[w(w + 20) geq 150]Let's expand this equation:[w^2 + 20w geq 150]To make it easier to solve, let's subtract 150 from both sides:[w^2 + 20w - 150 geq 0]Now, we have a quadratic inequality. To solve this, we can first find the roots of the quadratic equation ( w^2 + 20w - 150 = 0 ). We can use the quadratic formula:[w = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Here, ( a = 1 ), ( b = 20 ), and ( c = -150 ). Plugging these values in:[w = frac{-20 pm sqrt{(20)^2 - 4(1)(-150)}}{2(1)}][w = frac{-20 pm sqrt{400 + 600}}{2}][w = frac{-20 pm sqrt{1000}}{2}]Simplifying the square root:[sqrt{1000} = sqrt{100 times 10} = 10sqrt{10} approx 31.62]So, the roots are:[w = frac{-20 + 31.62}{2} approx frac{11.62}{2} approx 5.81]and[w = frac{-20 - 31.62}{2} approx frac{-51.62}{2} approx -25.81]Since width cannot be negative, we discard the negative root. So, the critical point is approximately ( w = 5.81 ) feet.Now, we need to determine where the quadratic expression ( w^2 + 20w - 150 ) is greater than or equal to zero. Since the coefficient of ( w^2 ) is positive, the parabola opens upwards. This means the expression is greater than or equal to zero when ( w leq -25.81 ) or ( w geq 5.81 ). Again, since width cannot be negative, we only consider ( w geq 5.81 ).However, Sarah wants the area to be at least 150 square feet, so the smallest width that satisfies this condition is approximately 5.81 feet. But let's check if this makes sense.If ( w = 5.81 ) feet, then the length ( l = w + 20 = 5.81 + 20 = 25.81 ) feet. The area would be:[A = 5.81 times 25.81 approx 150 text{ square feet}]That's exactly the minimum area required. But Sarah wants to use the least material possible, which means minimizing the perimeter.The perimeter ( P ) of a rectangle is given by:[P = 2(text{length} + text{width}) = 2(l + w)]Substituting ( l = w + 20 ):[P = 2(w + 20 + w) = 2(2w + 20) = 4w + 40]To minimize the perimeter, we need to minimize ( w ). But we already found that the smallest ( w ) that satisfies the area requirement is approximately 5.81 feet. So, using this width would give the minimum perimeter.But let's verify if this is indeed the case. If we choose a width slightly larger than 5.81 feet, say 6 feet, then the length would be 26 feet, and the area would be:[A = 6 times 26 = 156 text{ square feet}]Which is more than 150 square feet. The perimeter would be:[P = 2(6 + 26) = 2(32) = 64 text{ feet}]If we choose the width exactly at 5.81 feet, the perimeter would be:[P = 4(5.81) + 40 = 23.24 + 40 = 63.24 text{ feet}]Which is slightly less than 64 feet. So, indeed, the perimeter is minimized when the width is approximately 5.81 feet.However, in practical terms, it's unlikely that Sarah can have a width of exactly 5.81 feet. She might need to use a whole number or a more standard measurement. If we round up to the nearest whole number, the width would be 6 feet, and the length would be 26 feet, giving an area of 156 square feet and a perimeter of 64 feet.But since the problem asks for the width that minimizes the material used, and we've determined that the exact width is approximately 5.81 feet, we can express this as an exact value rather than an approximation.Recall that:[w = frac{-20 + sqrt{1000}}{2} = frac{-20 + 10sqrt{10}}{2} = -10 + 5sqrt{10}]So, the exact width is ( -10 + 5sqrt{10} ) feet. Since width cannot be negative, and ( 5sqrt{10} ) is approximately 15.81, subtracting 10 gives us approximately 5.81 feet, which matches our earlier approximation.Therefore, the width that minimizes the material used while satisfying the area requirement is ( -10 + 5sqrt{10} ) feet. However, since negative width doesn't make sense, we take the positive solution.But let's double-check our work to ensure there are no mistakes. We set up the area correctly, solved the quadratic equation accurately, and interpreted the roots properly. We also considered the practical aspect of using whole numbers but concluded that the exact value is more precise.In conclusion, the width should be ( -10 + 5sqrt{10} ) feet to minimize the material used while ensuring the area is at least 150 square feet.

answer：Alright, so I have this problem about complex numbers, and I need to figure out which statement is false. Let me take it step by step because I'm still getting the hang of complex numbers and their properties.First, let's restate the problem to make sure I understand it correctly. We have two complex numbers, z₁ and z₂, and there are four statements labeled A to D. I need to determine which one is false. Let me list out the statements again:A: If |z₁| = |z₂|, then z₁² = z₂² B: If z₁ = conjugate(z₂), then conjugate(z₁) = z₂ C: If |z₁| = |z₂|, then z₁ · conjugate(z₁) = z₂ · conjugate(z₂) D: If |z₁ - z₂| = 0, then conjugate(z₁) = conjugate(z₂)Okay, so I need to analyze each statement one by one.Starting with statement A: If the magnitudes of z₁ and z₂ are equal, does that mean their squares are equal? Hmm, I'm not sure. Let me think about what squaring a complex number does. If z = a + bi, then z² = (a² - b²) + 2abi. So, squaring a complex number not only affects the magnitude but also the angle or argument of the complex number. Wait, the magnitude of z² is |z|², which is the square of the magnitude of z. So, if |z₁| = |z₂|, then |z₁|² = |z₂|², which would mean |z₁²| = |z₂²|. But does that mean z₁² = z₂²? Not necessarily, because two complex numbers can have the same magnitude but different arguments, so their squares would not be equal unless their arguments are also related in a specific way.Let me test this with an example. Suppose z₁ = 1 + i and z₂ = 1 - i. Both have the same magnitude: |z₁| = sqrt(1² + 1²) = sqrt(2), and |z₂| is also sqrt(2). Now, let's compute z₁² and z₂².z₁² = (1 + i)² = 1² + 2*1*i + i² = 1 + 2i - 1 = 2i z₂² = (1 - i)² = 1² - 2*1*i + i² = 1 - 2i - 1 = -2iSo, z₁² = 2i and z₂² = -2i. These are not equal. Therefore, even though |z₁| = |z₂|, their squares are not equal. So statement A is false. But wait, before I jump to conclusions, let me check the other statements too, just to be thorough.Moving on to statement B: If z₁ is the conjugate of z₂, then the conjugate of z₁ is z₂. Let me recall that the conjugate of a complex number z = a + bi is conjugate(z) = a - bi. So, if z₁ = conjugate(z₂), then z₁ = a - bi, and z₂ = a + bi. Taking the conjugate of z₁ would give us a + bi, which is exactly z₂. So yes, statement B is true.Statement C: If |z₁| = |z₂|, then z₁ · conjugate(z₁) = z₂ · conjugate(z₂). Let me remember that for any complex number z, z · conjugate(z) = |z|². So, if |z₁| = |z₂|, then |z₁|² = |z₂|², which means z₁ · conjugate(z₁) = z₂ · conjugate(z₂). So, statement C is definitely true.Statement D: If |z₁ - z₂| = 0, then conjugate(z₁) = conjugate(z₂). The modulus |z₁ - z₂| being zero implies that z₁ - z₂ = 0, so z₁ = z₂. If z₁ equals z₂, then their conjugates must also be equal. Therefore, statement D is true.So, after checking all the statements, only statement A is false. My initial thought was correct. But just to make sure I didn't make a mistake with statement A, let me think of another example. Suppose z₁ = 2 and z₂ = 2i. Both have magnitude 2. Now, z₁² = 4 and z₂² = (2i)² = -4. Again, z₁² ≠ z₂². So, this further confirms that statement A is false.Another way to think about it is that squaring a complex number is not just about scaling; it also involves rotation in the complex plane. So, even if two complex numbers have the same magnitude, their angles could be different, and squaring them would rotate them by twice their original angles, potentially making them different.Therefore, I'm confident that statement A is the false one.**Final Answer**The false statement is boxed{A}.

answer：Okay, so I have this problem about folding a square piece of paper. Let me try to visualize it first. The square is ABCD with each side being 8 cm. So, if I imagine the square, let's say A is at the bottom-left corner, B at the bottom-right, C at the top-right, and D at the top-left. So, it's like a standard square on a coordinate plane.Now, the problem says that corner C is folded over so that it coincides with point E, which is one-third of the way along AD from D. Hmm, okay. So, AD is the left side of the square, right? From D to A. So, E is closer to D, one-third of the way. Since AD is 8 cm, one-third would be 8/3 cm. So, E is 8/3 cm away from D towards A.So, when we fold corner C to E, the crease formed is GF, where F is on CD. I need to find the length of FD. So, FD is the segment from F to D on side CD.Alright, let me try to think about what happens when we fold C to E. In folding, the crease GF will be the perpendicular bisector of the segment connecting C and E. So, maybe I can use some coordinate geometry here.Let me assign coordinates to the square. Let's place point A at (0,0), so B would be at (8,0), C at (8,8), and D at (0,8). Then, point E is one-third along AD from D, so since AD goes from (0,8) to (0,0), one-third from D would be at (0, 8 - 8/3) = (0, 16/3). So, E is at (0, 16/3).Now, when we fold C(8,8) over to E(0,16/3), the crease GF will be the perpendicular bisector of the segment CE. So, first, let's find the midpoint of CE. The midpoint M would have coordinates ((8+0)/2, (8 + 16/3)/2) = (4, (24/3 + 16/3)/2) = (4, (40/3)/2) = (4, 20/3).So, the midpoint M is at (4, 20/3). Now, the crease GF is the perpendicular bisector of CE. So, I need the slope of CE to find the slope of GF.The slope of CE is (16/3 - 8)/(0 - 8) = (16/3 - 24/3)/(-8) = (-8/3)/(-8) = (1/3). So, the slope of CE is 1/3. Therefore, the slope of the perpendicular bisector GF is the negative reciprocal, which is -3.So, the crease GF has a slope of -3 and passes through the midpoint M(4, 20/3). So, the equation of GF is y - 20/3 = -3(x - 4). Let me write that out:y - 20/3 = -3x + 12So, y = -3x + 12 + 20/3Convert 12 to thirds: 12 = 36/3, so y = -3x + 36/3 + 20/3 = -3x + 56/3.So, the equation of GF is y = -3x + 56/3.Now, point F is on CD. CD is the top side of the square from C(8,8) to D(0,8). Wait, no, CD is actually from C(8,8) to D(0,8)? Wait, no, in the square ABCD, the sides are AB, BC, CD, DA. So, CD is from C(8,8) to D(0,8). So, CD is the top side, horizontal line at y=8.Wait, but in the problem statement, it says F is on CD. So, CD is from C(8,8) to D(0,8). So, F is somewhere along y=8.But the crease GF intersects CD at F. So, to find F, we can find the intersection of GF with CD.Since CD is y=8, let's set y=8 in the equation of GF:8 = -3x + 56/3Let me solve for x:-3x = 8 - 56/3Convert 8 to thirds: 8 = 24/3So, -3x = 24/3 - 56/3 = (-32)/3So, x = (-32/3)/(-3) = (32/3)/3 = 32/9.So, x = 32/9. Therefore, point F is at (32/9, 8).Now, FD is the distance from F(32/9, 8) to D(0,8). Since both are on y=8, it's just the horizontal distance.So, FD = |32/9 - 0| = 32/9 cm.Wait, that seems straightforward. Let me double-check my steps.1. Assigned coordinates correctly: A(0,0), B(8,0), C(8,8), D(0,8). Correct.2. Point E is one-third along AD from D. AD is from (0,8) to (0,0). One-third from D is 8/3 cm down, so E is at (0, 8 - 8/3) = (0, 16/3). Correct.3. Fold C(8,8) to E(0,16/3). Midpoint M is ((8+0)/2, (8 + 16/3)/2) = (4, 20/3). Correct.4. Slope of CE: (16/3 - 8)/(0 - 8) = (-8/3)/(-8) = 1/3. So, slope of perpendicular bisector is -3. Correct.5. Equation of GF: y - 20/3 = -3(x - 4). Simplify to y = -3x + 56/3. Correct.6. Intersection with CD (y=8): 8 = -3x + 56/3. Solve for x: x=32/9. Correct.7. FD is the distance from F(32/9,8) to D(0,8): 32/9 cm. Correct.So, I think that's solid. The length of FD is 32/9 cm.

answer：First, I recognize that the question is asking for the thickness of 100 volumes of a primary school mathematics textbook. The options provided are in different units: millimeters, centimeters, meters, and decimeters.I start by estimating the thickness of a single textbook. Based on common knowledge, a typical textbook might be around 7 millimeters thick. This seems reasonable for a primary school mathematics book.Next, I need to calculate the total thickness for 100 such books. Multiplying the thickness of one book by 100 gives me:7 millimeters/book × 100 books = 700 millimetersNow, I convert millimeters to a more appropriate unit for the total thickness. Since 1 decimeter equals 100 millimeters, I can convert 700 millimeters to decimeters:700 millimeters ÷ 100 = 7 decimetersFinally, I compare this result with the provided options. The closest match is 7 decimeters, which corresponds to option D.

answer：Okay, so I have this problem here with two propositions, P and Q, and I need to figure out which one is true or false. Let me start by understanding each proposition separately.**Proposition P:**The solution set for the inequality (lg(x(1 - x) + 1) > 0) is ({x | 0 < x < 1}).Hmm, okay. So, this is a logarithmic inequality. I remember that (lg) is the logarithm base 10. The inequality is saying that the logarithm of (x(1 - x) + 1) is greater than 0. First, let me recall that (lg(a) > 0) implies that (a > 10^0), which is (a > 1). So, I can rewrite the inequality as:[ x(1 - x) + 1 > 1 ]Simplify that:[ x(1 - x) + 1 > 1 ]Subtract 1 from both sides:[ x(1 - x) > 0 ]Now, let's solve (x(1 - x) > 0). This is a quadratic inequality. The expression (x(1 - x)) can be rewritten as (-x^2 + x). To find where this is positive, I can find the roots and test intervals.The roots are at (x = 0) and (x = 1). So, the critical points are 0 and 1. The quadratic opens downward because the coefficient of (x^2) is negative. Therefore, the expression is positive between the roots.So, the solution is (0 < x < 1). That matches the given solution set for proposition P. So, P seems to be true.Wait, but let me double-check. The original inequality was (lg(x(1 - x) + 1) > 0). So, I need to make sure that the argument of the logarithm is positive as well, because logarithm is only defined for positive numbers.So, (x(1 - x) + 1 > 0). Let's see:[ x(1 - x) + 1 = -x^2 + x + 1 ]This is a quadratic equation. Let me find its discriminant to see if it ever becomes zero or negative.Discriminant (D = b^2 - 4ac = 1^2 - 4*(-1)*1 = 1 + 4 = 5). Since the discriminant is positive, the quadratic has two real roots. The quadratic opens downward, so it will be positive between its roots.But wait, the quadratic ( -x^2 + x + 1 ) is positive for all x except between its roots. Wait, no, since it opens downward, it's positive between its two roots. Let me find the roots:[ -x^2 + x + 1 = 0 ]Multiply both sides by -1:[ x^2 - x - 1 = 0 ]Using quadratic formula:[ x = frac{1 pm sqrt{1 + 4}}{2} = frac{1 pm sqrt{5}}{2} ]So, the roots are (frac{1 + sqrt{5}}{2}) and (frac{1 - sqrt{5}}{2}). Approximately, (sqrt{5}) is about 2.236, so the roots are roughly (1.618) and (-0.618).Therefore, the quadratic ( -x^2 + x + 1 ) is positive for ( -0.618 < x < 1.618 ). But in the original inequality, we have (x(1 - x) + 1 > 1), which simplifies to (x(1 - x) > 0), giving (0 < x < 1). So, the domain where the logarithm is defined is ( -0.618 < x < 1.618 ), but the inequality ( lg(x(1 - x) + 1) > 0 ) only holds when (x(1 - x) + 1 > 1), which is (0 < x < 1). So, the solution set is indeed (0 < x < 1). Therefore, proposition P is true.**Proposition Q:**In triangle ABC, (angle A > angle B) is a necessary but not sufficient condition for (cos^2left(frac{A}{2}right) < cos^2left(frac{B}{2}right)) to hold.Okay, so this is about triangle angles and cosine squared of half-angles. Let me recall some trigonometric identities and properties of triangles.First, in a triangle, the larger angle is opposite the longer side. So, if (angle A > angle B), then side a (opposite angle A) is longer than side b (opposite angle B).Now, we need to relate this to (cos^2left(frac{A}{2}right)) and (cos^2left(frac{B}{2}right)). Let me recall that (cos^2left(frac{theta}{2}right)) can be expressed using the double-angle identity:[ cos^2left(frac{theta}{2}right) = frac{1 + costheta}{2} ]So, let's rewrite the inequality:[ cos^2left(frac{A}{2}right) < cos^2left(frac{B}{2}right) ][ frac{1 + cos A}{2} < frac{1 + cos B}{2} ]Multiply both sides by 2:[ 1 + cos A < 1 + cos B ]Subtract 1:[ cos A < cos B ]So, the inequality (cos^2left(frac{A}{2}right) < cos^2left(frac{B}{2}right)) simplifies to (cos A < cos B).Now, in the range of angles in a triangle, each angle is between 0 and 180 degrees (0 and π radians). The cosine function is decreasing in the interval [0, π]. That means, if angle A is larger than angle B, then (cos A) is less than (cos B). So, if (angle A > angle B), then (cos A < cos B), which implies (cos^2left(frac{A}{2}right) < cos^2left(frac{B}{2}right)).Wait, so if (angle A > angle B), then (cos A < cos B), which leads to (cos^2left(frac{A}{2}right) < cos^2left(frac{B}{2}right)). So, does that mean that (angle A > angle B) is both necessary and sufficient?But the proposition says it's necessary but not sufficient. Hmm, that seems contradictory to what I just thought.Wait, maybe I made a mistake. Let me think again.We have (cos A < cos B) if and only if (A > B) because cosine is decreasing in [0, π]. So, (angle A > angle B) is equivalent to (cos A < cos B), which is equivalent to (cos^2left(frac{A}{2}right) < cos^2left(frac{B}{2}right)).Therefore, (angle A > angle B) is both necessary and sufficient for (cos^2left(frac{A}{2}right) < cos^2left(frac{B}{2}right)). So, the proposition Q says it's necessary but not sufficient, which would be false.But wait, maybe I'm missing something. Let me consider if there are cases where (angle A > angle B) doesn't necessarily lead to (cos^2left(frac{A}{2}right) < cos^2left(frac{B}{2}right)).Wait, but as I derived, it's an equivalence. So, if (angle A > angle B), then (cos A < cos B), which implies (cos^2left(frac{A}{2}right) < cos^2left(frac{B}{2}right)), and vice versa.Therefore, (angle A > angle B) is both necessary and sufficient. So, proposition Q is incorrect in saying it's only necessary but not sufficient.Wait, but maybe in some cases, even if (angle A > angle B), (cos^2left(frac{A}{2}right)) might not be less than (cos^2left(frac{B}{2}right)). Let me test with specific angles.Let me take a triangle where angle A is 120 degrees and angle B is 60 degrees. So, A > B.Compute (cos^2left(frac{A}{2}right)):[ cos^2left(frac{120}{2}right) = cos^2(60^circ) = (0.5)^2 = 0.25 ]Compute (cos^2left(frac{B}{2}right)):[ cos^2left(frac{60}{2}right) = cos^2(30^circ) = left(frac{sqrt{3}}{2}right)^2 = 0.75 ]So, 0.25 < 0.75, which holds.Another example: angle A = 90 degrees, angle B = 45 degrees.(cos^2(45^circ/2) = cos^2(22.5^circ) ≈ (0.924)^2 ≈ 0.853)(cos^2(45^circ/2) = same as above ≈ 0.853)Wait, no, angle A is 90, so (cos^2(45^circ)) is 0.5, and angle B is 45, so (cos^2(22.5^circ) ≈ 0.853). So, 0.5 < 0.853, which holds.Wait, but in this case, A > B, and (cos^2(A/2) < cos^2(B/2)).Wait, another example: angle A = 100 degrees, angle B = 80 degrees.(cos^2(50^circ) ≈ (0.6428)^2 ≈ 0.4132)(cos^2(40^circ) ≈ (0.7660)^2 ≈ 0.5868)So, 0.4132 < 0.5868, which holds.Wait, so in all these cases, when A > B, (cos^2(A/2) < cos^2(B/2)). So, it seems that it's always true.But wait, what if angle A is greater than 90 degrees and angle B is less than 90 degrees?Wait, in the first example, A was 120, B was 60, and it held.Wait, what if angle A is 150 degrees, angle B is 30 degrees.(cos^2(75^circ) ≈ (0.2588)^2 ≈ 0.06699)(cos^2(15^circ) ≈ (0.9659)^2 ≈ 0.9330)So, 0.06699 < 0.9330, which holds.Wait, so in all these cases, when A > B, (cos^2(A/2) < cos^2(B/2)). So, it seems that (angle A > angle B) is both necessary and sufficient.Therefore, proposition Q is incorrect in stating that it's only necessary but not sufficient. It is actually both necessary and sufficient.Wait, but maybe I'm missing something. Let me think about the definitions.A necessary condition is one that must be true for the statement to hold, but it alone doesn't guarantee the statement. A sufficient condition is one that, if met, guarantees the statement, but there might be other ways for the statement to hold.In this case, since (angle A > angle B) is equivalent to (cos^2(A/2) < cos^2(B/2)), they are equivalent statements. So, (angle A > angle B) is both necessary and sufficient.Therefore, proposition Q is false because it claims that (angle A > angle B) is necessary but not sufficient, which is not the case.Wait, but the original proposition Q says: "In triangle ABC, (angle A > angle B) is a necessary but not sufficient condition for (cos^2(frac{A}{2}) < cos^2(frac{B}{2})) to hold."But from my analysis, (angle A > angle B) is both necessary and sufficient. Therefore, Q is false.Wait, but maybe I'm misunderstanding the proposition. Let me read it again."In triangle ABC, (angle A > angle B) is a necessary but not sufficient condition for (cos^2(frac{A}{2}) < cos^2(frac{B}{2})) to hold."So, it's saying that for (cos^2(A/2) < cos^2(B/2)) to hold, it's necessary that (angle A > angle B), but (angle A > angle B) alone is not sufficient to guarantee (cos^2(A/2) < cos^2(B/2)).But from my analysis, (angle A > angle B) is both necessary and sufficient. So, the proposition is incorrect in saying it's only necessary but not sufficient.Therefore, proposition Q is false.Wait, but maybe I'm missing something. Let me think about the cosine half-angle formula again.(cos^2(theta/2) = frac{1 + costheta}{2}). So, if (angle A > angle B), then (cos A < cos B), which implies (frac{1 + cos A}{2} < frac{1 + cos B}{2}), which implies (cos^2(A/2) < cos^2(B/2)).Conversely, if (cos^2(A/2) < cos^2(B/2)), then (frac{1 + cos A}{2} < frac{1 + cos B}{2}), so (cos A < cos B), which implies (angle A > angle B).Therefore, they are equivalent. So, (angle A > angle B) is both necessary and sufficient for (cos^2(A/2) < cos^2(B/2)).Therefore, proposition Q is false because it claims it's only necessary but not sufficient, which is incorrect.So, summarizing:- Proposition P is true.- Proposition Q is false.Therefore, the correct answer should be A: P is true, Q is false.Wait, but let me check if there's any case where (angle A > angle B) doesn't imply (cos^2(A/2) < cos^2(B/2)). Suppose angle A is 100 degrees, angle B is 80 degrees.(cos^2(50) ≈ 0.4132), (cos^2(40) ≈ 0.5868). So, 0.4132 < 0.5868, which holds.Another case: angle A = 170 degrees, angle B = 10 degrees.(cos^2(85) ≈ (0.08716)^2 ≈ 0.0076), (cos^2(5) ≈ (0.9962)^2 ≈ 0.9924). So, 0.0076 < 0.9924, which holds.Wait, what if angle A is 90 degrees, angle B is 80 degrees.(cos^2(45) = 0.5), (cos^2(40) ≈ 0.5868). So, 0.5 < 0.5868, which holds.Wait, another case: angle A = 80 degrees, angle B = 100 degrees.Then, (cos^2(40) ≈ 0.5868), (cos^2(50) ≈ 0.4132). So, 0.5868 > 0.4132, which is consistent with angle A < angle B.So, in all these cases, the relationship holds. Therefore, (angle A > angle B) is equivalent to (cos^2(A/2) < cos^2(B/2)). Therefore, Q is false.So, the answer is A: P is true, Q is false.

answer：First, I need to evaluate each expression separately to understand their values.Starting with (-left(-frac{1}{9}right)), the double negative cancels out, so this simplifies to (frac{1}{9}).Next, for (-left|-frac{1}{9}right|), the absolute value of (-frac{1}{9}) is (frac{1}{9}), and applying the negative sign in front gives (-frac{1}{9}).Now, I compare the two results: (frac{1}{9}) and (-frac{1}{9}). Since any positive number is greater than its negative counterpart, (frac{1}{9}) is greater than (-frac{1}{9}).Therefore, the correct symbol to fill in the blank is (>).