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❓Let a and b be positive real numbers such that a^3 + b^3 = 3(a + b). Simplify the expression[frac{a}{b} + frac{b}{a} - frac{3}{ab}.]
💡Alright, so I have this problem here: Let ( a ) and ( b ) be positive real numbers such that ( a^3 + b^3 = 3(a + b) ). I need to simplify the expression ( frac{a}{b} + frac{b}{a} - frac{3}{ab} ). Hmm, okay, let's see how to approach this.First, I remember that ( a^3 + b^3 ) can be factored. The formula for factoring a sum of cubes is ( a^3 + b^3 = (a + b)(a^2 - ab + b^2) ). So, let me apply that here.Starting with the given equation:[ a^3 + b^3 = 3(a + b) ]Factor the left side:[ (a + b)(a^2 - ab + b^2) = 3(a + b) ]Now, since ( a ) and ( b ) are positive real numbers, ( a + b ) must be positive. That means I can safely divide both sides of the equation by ( a + b ) without worrying about dividing by zero. Let's do that:[ a^2 - ab + b^2 = 3 ]Okay, so now I have this equation:[ a^2 - ab + b^2 = 3 ]I need to simplify the expression ( frac{a}{b} + frac{b}{a} - frac{3}{ab} ). Let me write that expression down:[ frac{a}{b} + frac{b}{a} - frac{3}{ab} ]Hmm, I notice that all the terms have denominators involving ( ab ). Maybe I can combine them into a single fraction. Let me try that.First, combine ( frac{a}{b} ) and ( frac{b}{a} ):[ frac{a}{b} + frac{b}{a} = frac{a^2 + b^2}{ab} ]So now the expression becomes:[ frac{a^2 + b^2}{ab} - frac{3}{ab} ]Since both terms have the same denominator, I can combine them:[ frac{a^2 + b^2 - 3}{ab} ]Wait a second, from earlier, I have ( a^2 - ab + b^2 = 3 ). Maybe I can express ( a^2 + b^2 ) in terms of this equation. Let me see.Starting from ( a^2 - ab + b^2 = 3 ), if I add ( ab ) to both sides, I get:[ a^2 - ab + b^2 + ab = 3 + ab ]Simplifying the left side:[ a^2 + b^2 = 3 + ab ]Great! So ( a^2 + b^2 = 3 + ab ). Let me substitute this back into the expression I had earlier:[ frac{a^2 + b^2 - 3}{ab} = frac{(3 + ab) - 3}{ab} ]Simplify the numerator:[ frac{ab}{ab} ]Which simplifies to:[ 1 ]So, putting it all together, the original expression simplifies to 1. That seems straightforward, but let me double-check my steps to make sure I didn't make any mistakes.1. Factored ( a^3 + b^3 ) correctly into ( (a + b)(a^2 - ab + b^2) ).2. Divided both sides by ( a + b ) since ( a ) and ( b ) are positive, so ( a + b neq 0 ).3. Simplified the resulting equation to ( a^2 - ab + b^2 = 3 ).4. Expressed the given expression ( frac{a}{b} + frac{b}{a} - frac{3}{ab} ) as ( frac{a^2 + b^2 - 3}{ab} ).5. Substituted ( a^2 + b^2 = 3 + ab ) into the expression, leading to ( frac{ab}{ab} = 1 ).Everything seems to check out. I don't see any errors in my reasoning. So, I think the simplified expression is indeed 1.**Final Answer**boxed{1}
❓Mrs. Thompson's class uses a modified grading scale as shown below. The eighteen scores in her class were: [98, 73, 55, 100, 76, 93, 88, 72, 77, 65, 82, 79, 68, 85, 91, 56, 81, 89.] In Mrs. Thompson's class, what percent of the students received a grade outside the range of C?[boxed{begin{tabular}[t]{cc} A: & 94 - 100 B: & 86 - 93 C: & 76 - 85 D: & 65 - 75 F: & 0 - 64 end{tabular}}]A) 55.56%B) 61.11%C) 66.67%D) 72.22%
💡First, identify the scores that fall outside the C range (76 - 85) from the given list of scores. These scores are 98, 55, 100, 93, 88, 72, 65, 68, 91, 56, and 89.Next, count the number of students who received a grade outside of C. There are 11 such scores.Then, calculate the total number of students, which is 18.Finally, calculate the percentage of students who received a grade outside of C by dividing the number of students outside of C by the total number of students and multiplying by 100%. This gives:[text{Percentage} = left(frac{11}{18}right) times 100% approx 61.11%]Thus, the percentage of students who received a grade outside of C is approximately 61.11%.
❓after spending rs . 5000 on rent , rs . 1500 on milk , rs . 4500 on groceries , rs . 2500 on childrens education rs . 2000 on petrol and rs . 5200 on miscellaneous expenses , mr . kishore saved 10 % of his monthly salary . how much did he save in rs . ?
💡First, I need to calculate the total expenses Mr. Kishore incurred in a month. I'll add up all his expenditures:- Rent: ₹5,000- Milk: ₹1,500- Groceries: ₹4,500- Children's Education: ₹2,500- Petrol: ₹2,000- Miscellaneous Expenses: ₹5,200Adding these together: 5,000 + 1,500 + 4,500 + 2,500 + 2,000 + 5,200 = ₹18,700.Next, I know that Mr. Kishore saved 10% of his monthly salary. Let's denote his monthly salary as S. Therefore, his savings can be represented as 10% of S, which is 0.10 × S.The total amount Mr. Kishore has is the sum of his savings and his total expenses. So, I can set up the equation:Savings + Total Expenses = Monthly Salary0.10 × S + 18,700 = STo find the monthly salary (S), I'll rearrange the equation:0.10 × S + 18,700 = S18,700 = S - 0.10 × S18,700 = 0.90 × SNow, I'll solve for S by dividing both sides of the equation by 0.90:S = 18,700 / 0.90S = ₹20,777.78Finally, to find out how much Mr. Kishore saved, I'll calculate 10% of his monthly salary:Savings = 0.10 × 20,777.78Savings = ₹2,077.78Therefore, Mr. Kishore saved approximately ₹2,077.78.
❓Let A be one of the two points where the circles whose centers are the points M and N intersect. The tangents in A to such circles intersect them again in B and C , respectively. Let P be a point such that the quadrilateral AMPN is a parallelogram. Show that P is the circumcenter of triangle ABC .
💡Okay, so I have this geometry problem here, and I need to show that point P is the circumcenter of triangle ABC. Let me try to visualize and break down the problem step by step.First, there are two circles with centers at points M and N. These circles intersect at point A (and another point, but we're focusing on A). So, A is a common point on both circles. Now, the tangents at A to these circles intersect the circles again at points B and C, respectively. Hmm, tangents... so if I draw a tangent line at A for the circle centered at M, it will touch the circle only at A and then extend out, intersecting the circle again at B. Similarly, the tangent at A for the circle centered at N will intersect its circle again at C.Wait, hold on. If the tangent at A intersects the circle again, isn't that just the same point A? No, actually, the tangent line at A is just a line that touches the circle at A and doesn't cross it. So, maybe the problem means that the tangent line at A is extended until it intersects the other circle? Or perhaps it's a different circle? Hmm, the problem says "the tangents in A to such circles intersect them again in B and C, respectively." So, each tangent is to its respective circle, meaning the tangent to the circle centered at M at point A intersects the same circle again at B, and the tangent to the circle centered at N at point A intersects the same circle again at C.But wait, if it's a tangent at A, it should only touch the circle at A, right? So, how can it intersect the circle again? Maybe I'm misunderstanding. Perhaps it's not the tangent line at A, but another tangent from A to the other circle? That is, from point A, draw a tangent to the circle centered at M, which would touch the circle at some point B, and similarly, draw a tangent from A to the circle centered at N, which would touch at point C. That makes more sense because then B and C would be distinct points on each circle.Yes, that seems right. So, A is a common point, and from A, we draw tangents to each of the circles. The tangent to the circle centered at M (other than the tangent at A) touches the circle at B, and the tangent to the circle centered at N touches at C. So, AB is tangent to the circle at M, and AC is tangent to the circle at N.Now, quadrilateral AMPN is a parallelogram. So, points A, M, P, N form a parallelogram. In a parallelogram, opposite sides are equal and parallel. Therefore, AM is equal and parallel to PN, and AN is equal and parallel to PM.I need to show that P is the circumcenter of triangle ABC. The circumcenter is the intersection point of the perpendicular bisectors of the sides of the triangle. So, if I can show that P is equidistant from A, B, and C, or that it lies on the perpendicular bisectors of AB, BC, and AC, then I can conclude that P is the circumcenter.Let me think about the properties of tangents and circles. The tangent to a circle at a point is perpendicular to the radius at that point. So, AB is tangent to the circle centered at M, so AB is perpendicular to AM. Similarly, AC is tangent to the circle centered at N, so AC is perpendicular to AN.Therefore, AB ⊥ AM and AC ⊥ AN.Since AMPN is a parallelogram, as I noted earlier, AM is parallel to PN, and AN is parallel to PM. So, PN is parallel to AM, which is perpendicular to AB. Therefore, PN is also perpendicular to AB. Similarly, PM is parallel to AN, which is perpendicular to AC, so PM is also perpendicular to AC.So, PN is perpendicular to AB, and PM is perpendicular to AC. That suggests that PN and PM are the perpendicular bisectors of AB and AC, respectively. If I can show that P is on these perpendicular bisectors, then it would be the circumcenter.But wait, PN is perpendicular to AB, but is it the bisector? For PN to be the perpendicular bisector, it needs to pass through the midpoint of AB. Similarly, PM needs to pass through the midpoint of AC.Is that the case here? Let me think.Since AMPN is a parallelogram, vectorially, we can represent P as M + N - A, or something like that. Wait, maybe coordinate geometry would help here. Let me assign coordinates to the points to make this more concrete.Let me place point A at the origin (0,0) for simplicity. Let me denote the coordinates of M as (h, k) and N as (p, q). Since A is on both circles, the distance from A to M is the radius of the first circle, and the distance from A to N is the radius of the second circle.So, radius of circle centered at M is |AM| = sqrt(h² + k²), and radius of circle centered at N is |AN| = sqrt(p² + q²).Now, the tangent from A to circle M touches the circle at B. The tangent from A to circle N touches the circle at C.I know that the tangent from a point to a circle has a specific length, given by the formula sqrt(|AO|² - r²), where O is the center and r is the radius. But in this case, since A is on the circle, the tangent length would be zero, which doesn't make sense. Wait, no, if A is on the circle, then the tangent at A is just the line perpendicular to the radius at A. So, AB is the tangent at A to circle M, meaning AB is perpendicular to AM.Similarly, AC is the tangent at A to circle N, so AC is perpendicular to AN.So, AB is perpendicular to AM, and AC is perpendicular to AN.Since AMPN is a parallelogram, we have vectors:Vector AM = Vector PNVector AN = Vector PMSo, if I can express the coordinates of P in terms of M and N, maybe I can find the coordinates of P and then show that it's equidistant from A, B, and C.Alternatively, maybe using vectors would help. Let me denote vectors with their position vectors.Let me denote vector A as origin, so A = (0,0). Then, vector M is (h, k), vector N is (p, q). Since AMPN is a parallelogram, vector P = vector M + vector N.Wait, in a parallelogram, the position vector of P would be vector M + vector N, because starting from A, moving to M, then from M moving in the direction of N gives P. So, P = M + N.So, coordinates of P are (h + p, k + q).Now, I need to find the coordinates of B and C.Since AB is tangent to circle M at A, and AB is perpendicular to AM. So, the direction of AB is perpendicular to AM. Since AM is vector (h, k), the direction of AB is (-k, h) or (k, -h). Let's pick one direction, say (-k, h). So, the line AB has direction (-k, h).But AB is a tangent from A to circle M. Wait, but A is on the circle, so AB is just the tangent at A, which is a line perpendicular to AM. So, the line AB can be parametrized as t*(-k, h), where t is a scalar parameter.Similarly, AC is tangent at A to circle N, so AC is perpendicular to AN. Vector AN is (p, q), so the direction of AC is (-q, p) or (q, -p). Let's take (-q, p). So, line AC can be parametrized as s*(-q, p), where s is a scalar parameter.But points B and C are points where these tangents intersect the circles again. Wait, but since A is already on the circle, the tangent at A will only touch the circle at A. So, how can the tangent intersect the circle again? That seems contradictory.Wait, maybe I misinterpreted the problem. Perhaps the tangent is not at A, but from A to the other circle. So, from point A, draw a tangent to circle M (other than the tangent at A), which touches circle M at B, and similarly, draw a tangent from A to circle N (other than the tangent at A), which touches circle N at C.Yes, that makes more sense. So, AB is a tangent from A to circle M, touching at B, and AC is a tangent from A to circle N, touching at C.In that case, the lengths of AB and AC can be calculated using the tangent length formula.The length of tangent from A to circle M is sqrt(|AM|² - r_M²), but since A is on circle M, |AM| is equal to r_M, so the tangent length would be zero, which again doesn't make sense. Wait, no, if A is on circle M, then the tangent at A is just the line perpendicular to AM, as before. So, perhaps the problem is that the tangent at A is extended beyond A, but since it's already a tangent, it doesn't intersect the circle again.I'm getting confused here. Let me re-read the problem."Let A be one of the two points where the circles whose centers are the points M and N intersect. The tangents in A to such circles intersect them again in B and C, respectively."Hmm, so the tangents at A to the circles intersect the circles again at B and C. But if the tangent is at A, it should only touch the circle at A, right? So, how can it intersect again? Maybe the problem means that the tangent line at A intersects the other circle at B and C? That is, the tangent to circle M at A intersects circle N at B, and the tangent to circle N at A intersects circle M at C. That would make sense because then B and C are points on the other circle.Yes, that must be it. So, the tangent to circle M at A intersects circle N again at B, and the tangent to circle N at A intersects circle M again at C.Okay, that makes more sense. So, let's correct my earlier understanding. The tangent at A to circle M intersects circle N at B, and the tangent at A to circle N intersects circle M at C.So, AB is the tangent at A to circle M, extended to intersect circle N at B, and AC is the tangent at A to circle N, extended to intersect circle M at C.Got it. So, AB is tangent to circle M at A and intersects circle N at B, and AC is tangent to circle N at A and intersects circle M at C.Now, since AB is tangent to circle M at A, AB is perpendicular to AM. Similarly, AC is tangent to circle N at A, so AC is perpendicular to AN.So, AB ⊥ AM and AC ⊥ AN.Now, quadrilateral AMPN is a parallelogram. So, vectors AM and PN are equal and parallel, and vectors AN and PM are equal and parallel.So, if I can express the coordinates of P in terms of M and N, and then show that P is equidistant from A, B, and C, then P would be the circumcenter.Alternatively, since P is the result of the parallelogram, maybe I can find some relationships between the triangles or use perpendicular bisectors.Let me try to think about the perpendicular bisectors.Since AB is tangent to circle M at A, and AB is perpendicular to AM, then AM is the radius and AB is the tangent. Similarly, AN is the radius and AC is the tangent.Now, in the parallelogram AMPN, since AM is parallel to PN and AN is parallel to PM, we have that PN is parallel to AM and PM is parallel to AN.Since AM is perpendicular to AB, and PN is parallel to AM, then PN is also perpendicular to AB. Similarly, since AN is perpendicular to AC, and PM is parallel to AN, then PM is perpendicular to AC.So, PN is perpendicular to AB, and PM is perpendicular to AC.Now, in a triangle, the circumcenter lies at the intersection of the perpendicular bisectors. So, if I can show that P lies on the perpendicular bisectors of AB and AC, then it must be the circumcenter.But to be the perpendicular bisector, P must not only be perpendicular to AB and AC but also pass through their midpoints.Wait, so if PN is perpendicular to AB, does it pass through the midpoint of AB? Similarly, does PM pass through the midpoint of AC?Hmm, I'm not sure. Maybe I need to explore this further.Alternatively, since P is the result of the parallelogram, maybe I can express P in terms of M and N and then compute distances.Let me try coordinate geometry again, but this time with the corrected understanding.Let me place point A at (0,0). Let me denote the coordinates of M as (h, k) and N as (p, q). Since A is on both circles, the distance from A to M is the radius of circle M, which is sqrt(h² + k²), and the distance from A to N is sqrt(p² + q²), the radius of circle N.Now, the tangent at A to circle M is perpendicular to AM. So, the slope of AM is k/h, so the slope of AB (the tangent) is -h/k.Similarly, the slope of AN is q/p, so the slope of AC (the tangent) is -p/q.Now, the tangent line AB has slope -h/k and passes through A(0,0), so its equation is y = (-h/k)x.This line intersects circle N at point B. Circle N has center (p, q) and radius sqrt(p² + q²). So, the equation of circle N is (x - p)² + (y - q)² = p² + q².Substituting y = (-h/k)x into the equation of circle N:(x - p)² + ((-h/k)x - q)² = p² + q².Let me expand this:(x² - 2px + p²) + ((h²/k²)x² + (2hq/k)x + q²) = p² + q².Combine like terms:x² - 2px + p² + (h²/k²)x² + (2hq/k)x + q² = p² + q².Subtract p² + q² from both sides:x² - 2px + (h²/k²)x² + (2hq/k)x = 0.Factor x:x [x - 2p + (h²/k²)x + 2hq/k] = 0.So, x=0 is one solution (point A), and the other solution is:x - 2p + (h²/k²)x + 2hq/k = 0.Combine x terms:x(1 + h²/k²) + (-2p + 2hq/k) = 0.Solve for x:x = (2p - 2hq/k) / (1 + h²/k²).Multiply numerator and denominator by k² to eliminate denominators:x = (2p k² - 2hq k) / (k² + h²).Similarly, y = (-h/k)x = (-h/k) * (2p k² - 2hq k) / (k² + h²) = (-2p h k + 2h² q) / (k² + h²).So, coordinates of B are:B = ( (2p k² - 2h q k)/(k² + h²), (-2p h k + 2h² q)/(k² + h²) ).Similarly, let's find coordinates of C.The tangent at A to circle N has slope -p/q, so its equation is y = (-p/q)x.This line intersects circle M at point C. Circle M has center (h, k) and radius sqrt(h² + k²). Equation of circle M is (x - h)² + (y - k)² = h² + k².Substitute y = (-p/q)x into the equation:(x - h)² + ((-p/q)x - k)² = h² + k².Expand:(x² - 2hx + h²) + ((p²/q²)x² + (2pk/q)x + k²) = h² + k².Combine like terms:x² - 2hx + h² + (p²/q²)x² + (2pk/q)x + k² = h² + k².Subtract h² + k² from both sides:x² - 2hx + (p²/q²)x² + (2pk/q)x = 0.Factor x:x [x - 2h + (p²/q²)x + 2pk/q] = 0.So, x=0 is one solution (point A), and the other solution is:x - 2h + (p²/q²)x + 2pk/q = 0.Combine x terms:x(1 + p²/q²) + (-2h + 2pk/q) = 0.Solve for x:x = (2h - 2pk/q) / (1 + p²/q²).Multiply numerator and denominator by q²:x = (2h q² - 2pk q) / (q² + p²).Similarly, y = (-p/q)x = (-p/q)*(2h q² - 2pk q)/(q² + p²) = (-2h p q + 2p² k)/(q² + p²).So, coordinates of C are:C = ( (2h q² - 2p k q)/(q² + p²), (-2h p q + 2p² k)/(q² + p²) ).Now, we have coordinates for A, B, C, M, N, and P.Since AMPN is a parallelogram, P = M + N - A. Since A is at (0,0), P = M + N = (h + p, k + q).So, P = (h + p, k + q).Now, to show that P is the circumcenter of triangle ABC, we need to show that P is equidistant from A, B, and C.Let's compute the distances PA, PB, and PC.PA is the distance from P to A, which is sqrt((h + p - 0)^2 + (k + q - 0)^2) = sqrt((h + p)^2 + (k + q)^2).PB is the distance from P to B:B = ( (2p k² - 2h q k)/(k² + h²), (-2p h k + 2h² q)/(k² + h²) ).So, PB² = [ (h + p - (2p k² - 2h q k)/(k² + h²) )² + (k + q - (-2p h k + 2h² q)/(k² + h²) )² ].Similarly, PC² = [ (h + p - (2h q² - 2p k q)/(q² + p²) )² + (k + q - (-2h p q + 2p² k)/(q² + p²) )² ].This seems complicated, but maybe we can simplify.Alternatively, since we know that P lies on the perpendicular bisectors of AB and AC, and if we can show that P is equidistant from B and C as well, then it would be the circumcenter.But maybe there's a better approach.Let me consider the midpoint of AB and see if P lies on its perpendicular bisector.Midpoint of AB: Let's compute coordinates of midpoint.Coordinates of A: (0,0).Coordinates of B: ( (2p k² - 2h q k)/(k² + h²), (-2p h k + 2h² q)/(k² + h²) ).Midpoint M_AB = ( (2p k² - 2h q k)/(2(k² + h²)), (-2p h k + 2h² q)/(2(k² + h²)) ) = ( (p k² - h q k)/(k² + h²), (-p h k + h² q)/(k² + h²) ).Now, the slope of AB is (-h/k), as earlier.The perpendicular bisector of AB will have slope (k/h), since it's perpendicular to AB.So, the equation of the perpendicular bisector of AB is:y - [ (-p h k + h² q)/(k² + h²) ] = (k/h)(x - (p k² - h q k)/(k² + h²)).Now, does point P = (h + p, k + q) lie on this line?Let's plug in x = h + p, y = k + q into the equation.Left side: y - [ (-p h k + h² q)/(k² + h²) ] = (k + q) - [ (-p h k + h² q)/(k² + h²) ].Right side: (k/h)(x - (p k² - h q k)/(k² + h²)) = (k/h)(h + p - (p k² - h q k)/(k² + h²)).Let me compute both sides.Left side:(k + q) - [ (-p h k + h² q)/(k² + h²) ] = (k + q) + (p h k - h² q)/(k² + h²).Right side:(k/h)[ (h + p) - (p k² - h q k)/(k² + h²) ].Let me compute the expression inside the brackets:(h + p) - (p k² - h q k)/(k² + h²) = [ (h + p)(k² + h²) - (p k² - h q k) ] / (k² + h²).Expand numerator:(h + p)(k² + h²) = h k² + h³ + p k² + p h².Subtract (p k² - h q k):= h k² + h³ + p k² + p h² - p k² + h q k.Simplify:= h k² + h³ + p h² + h q k.Factor h:= h(k² + h² + p h + q k).So, the expression inside the brackets is [ h(k² + h² + p h + q k) ] / (k² + h²).Therefore, right side becomes:(k/h) * [ h(k² + h² + p h + q k) / (k² + h²) ] = k(k² + h² + p h + q k)/(k² + h²).Now, let's compute left side:(k + q) + (p h k - h² q)/(k² + h²).Let me write (k + q) as (k + q)(k² + h²)/(k² + h²) to have a common denominator.So, left side = [ (k + q)(k² + h²) + p h k - h² q ] / (k² + h²).Expand numerator:= k(k² + h²) + q(k² + h²) + p h k - h² q.= k³ + k h² + q k² + q h² + p h k - h² q.Simplify:= k³ + k h² + q k² + (q h² - h² q) + p h k.= k³ + k h² + q k² + p h k.Factor:= k(k² + h² + q k + p h).So, left side = k(k² + h² + q k + p h)/(k² + h²).Right side was k(k² + h² + p h + q k)/(k² + h²).So, both sides are equal. Therefore, point P lies on the perpendicular bisector of AB.Similarly, we can show that P lies on the perpendicular bisector of AC.Therefore, since P lies on both perpendicular bisectors of AB and AC, it must be the circumcenter of triangle ABC.Thus, P is the circumcenter of triangle ABC.
❓If proposition p: forall x in left(-frac{pi}{2}, frac{pi}{2}right), tan x > sin x, then the negation of proposition neg p is ( ).A: exists x_0 in left(-frac{pi}{2}, frac{pi}{2}right), tan x_0 geq sin x_0B: exists x_0 in left(-frac{pi}{2}, frac{pi}{2}right), tan x_0 > sin x_0C: exists x_0 in left(-frac{pi}{2}, frac{pi}{2}right), tan x_0 leq sin x_0D: exists x_0 notin left(-frac{pi}{2}, frac{pi}{2}right), tan x_0 leq sin x_0
💡Okay, so I have this proposition p: For all x in the interval (-π/2, π/2), tan x is greater than sin x. I need to find the negation of p, which is ¬p. Hmm, let me think about how to approach this.First, I remember that the negation of a universal statement ("for all") is an existential statement ("there exists"). So, if p is saying that something is true for every x in that interval, then ¬p should be saying that there exists at least one x in that interval where the statement is not true.So, p is: ∀x ∈ (-π/2, π/2), tan x > sin x.Therefore, ¬p should be: ∃x ∈ (-π/2, π/2) such that tan x ≤ sin x.Wait, let me make sure I'm doing this correctly. The original statement is tan x > sin x. The negation of "greater than" is "less than or equal to." So, yes, ¬p is that there exists some x in that interval where tan x is not greater than sin x, meaning tan x is less than or equal to sin x.Looking at the options:A: ∃x₀ ∈ (-π/2, π/2), tan x₀ ≥ sin x₀B: ∃x₀ ∈ (-π/2, π/2), tan x₀ > sin x₀C: ∃x₀ ∈ (-π/2, π/2), tan x₀ ≤ sin x₀D: ∃x₀ ∉ (-π/2, π/2), tan x₀ ≤ sin x₀So, based on my earlier reasoning, ¬p is option C because it states that there exists an x in the interval where tan x is less than or equal to sin x.Let me double-check to make sure I didn't make a mistake. The original statement is a universal one, so its negation must be an existential one. Also, the inequality flips from > to ≤ when negating. So, yes, option C seems correct.I think I got it right. The negation is about the existence of a counterexample within the same interval where the inequality doesn't hold. So, it's definitely option C.
❓36 men can complete a piece of work in 18 days. If a smaller group of men work on the same task, they can complete it in 72 days. How many men are in the smaller group?
💡First, I need to determine the total amount of work required to complete the task. This can be calculated by multiplying the number of men by the number of days they take to finish the work.Next, I'll set up an equation where the total work done by the larger group of men is equal to the total work done by the smaller group of men. This is because the amount of work remains the same regardless of the number of men working on it.By solving this equation, I can find the number of men in the smaller group.