💡Okay, so I have this problem with two parts. Let me try to figure out each step carefully.First, the function given is ( f(x) = cos^2x - frac{sqrt{3}}{2}sin2x - frac{1}{2} ). I need to find the symmetry center and the interval where it's monotonically increasing.Hmm, I remember that symmetry center for a function usually refers to a point around which the function is symmetric. For trigonometric functions, especially cosine and sine, they have certain symmetries. Maybe I can rewrite this function in a simpler form.Looking at ( f(x) ), I see ( cos^2x ) and ( sin2x ). I recall that ( cos^2x ) can be written using a double-angle identity: ( cos^2x = frac{1 + cos2x}{2} ). Let me substitute that in.So, substituting, we get:[f(x) = frac{1 + cos2x}{2} - frac{sqrt{3}}{2}sin2x - frac{1}{2}]Simplify the constants:[f(x) = frac{1}{2} + frac{cos2x}{2} - frac{sqrt{3}}{2}sin2x - frac{1}{2}]The ( frac{1}{2} ) and ( -frac{1}{2} ) cancel out, so:[f(x) = frac{cos2x}{2} - frac{sqrt{3}}{2}sin2x]Hmm, this looks like a combination of cosine and sine terms with the same argument ( 2x ). Maybe I can write this as a single cosine function using the amplitude-phase form.I remember that ( Acostheta + Bsintheta = Ccos(theta - phi) ), where ( C = sqrt{A^2 + B^2} ) and ( tanphi = frac{B}{A} ).In this case, ( A = frac{1}{2} ) and ( B = -frac{sqrt{3}}{2} ). Let me compute ( C ):[C = sqrt{left(frac{1}{2}right)^2 + left(-frac{sqrt{3}}{2}right)^2} = sqrt{frac{1}{4} + frac{3}{4}} = sqrt{1} = 1]Okay, so the amplitude is 1. Now, the phase shift ( phi ):[tanphi = frac{B}{A} = frac{-sqrt{3}/2}{1/2} = -sqrt{3}]So, ( phi ) is an angle whose tangent is ( -sqrt{3} ). The principal value for ( tan^{-1}(-sqrt{3}) ) is ( -frac{pi}{3} ), but since tangent is periodic with period ( pi ), we can also express this as ( frac{2pi}{3} ).Wait, but let me think. Since ( A ) is positive and ( B ) is negative, the angle ( phi ) is in the fourth quadrant. So, ( phi = -frac{pi}{3} ).Therefore, the function can be rewritten as:[f(x) = cosleft(2x - (-frac{pi}{3})right) = cosleft(2x + frac{pi}{3}right)]Wait, hold on. Let me double-check. The formula is ( Acostheta + Bsintheta = Ccos(theta - phi) ). So, if ( tanphi = B/A ), then ( phi = tan^{-1}(B/A) ). But since ( B ) is negative, ( phi ) is negative.But in our case, ( A = frac{1}{2} ) and ( B = -frac{sqrt{3}}{2} ), so ( tanphi = -sqrt{3} ), which gives ( phi = -frac{pi}{3} ). So, substituting back:[f(x) = cosleft(2x - (-frac{pi}{3})right) = cosleft(2x + frac{pi}{3}right)]Yes, that seems correct.So, ( f(x) = cosleft(2x + frac{pi}{3}right) ).Now, for part (1), we need the coordinates of the symmetry center of the graph of ( f(x) ). I think for a cosine function, which is symmetric about its maxima and minima, the symmetry center would be the points where it crosses the x-axis, i.e., the points of inflection.Wait, but actually, for a standard cosine function ( cos(kx + phi) ), it's symmetric about its maxima and minima. However, the symmetry center, if it's a point, would be the center of symmetry for the graph. For a cosine function, it's symmetric about its peaks and troughs, but those are points of maxima and minima, not centers.Wait, maybe I'm confusing with sine functions. Sine functions are symmetric about the origin, but cosine functions are symmetric about the y-axis. Hmm, perhaps for a transformed cosine function, the symmetry center would be the point around which the function is symmetric.Alternatively, maybe the function has a point of symmetry, like a center of symmetry, which is a point such that the function is symmetric with respect to that point. For example, if a function is odd around a point ( (h, k) ), then ( f(h + x) + f(h - x) = 2k ).In this case, since ( f(x) = cos(2x + frac{pi}{3}) ), let's see if it's symmetric about a certain point.Let me set ( f(h + x) + f(h - x) = 2k ).Compute ( f(h + x) = cos(2(h + x) + frac{pi}{3}) = cos(2h + 2x + frac{pi}{3}) )Compute ( f(h - x) = cos(2(h - x) + frac{pi}{3}) = cos(2h - 2x + frac{pi}{3}) )So, ( f(h + x) + f(h - x) = cos(2h + 2x + frac{pi}{3}) + cos(2h - 2x + frac{pi}{3}) )Using the identity ( cos(A + B) + cos(A - B) = 2cos A cos B ), so:( 2cos(2h + frac{pi}{3})cos(2x) )We want this to equal ( 2k ), regardless of ( x ). So, ( 2cos(2h + frac{pi}{3})cos(2x) = 2k ).This must hold for all ( x ), which implies that ( cos(2x) ) must be constant, which is only possible if ( cos(2h + frac{pi}{3}) = 0 ) and ( k = 0 ).So, ( cos(2h + frac{pi}{3}) = 0 ), which implies ( 2h + frac{pi}{3} = frac{pi}{2} + npi ), where ( n ) is an integer.Solving for ( h ):[2h = frac{pi}{2} - frac{pi}{3} + npi = frac{3pi}{6} - frac{2pi}{6} + npi = frac{pi}{6} + npi]Thus,[h = frac{pi}{12} + frac{npi}{2}]Therefore, the function ( f(x) ) is symmetric about the points ( left( frac{pi}{12} + frac{npi}{2}, 0 right) ) for any integer ( n ).So, the symmetry centers are at ( left( frac{pi}{12} + frac{npi}{2}, 0 right) ).Now, moving on to the interval where ( f(x) ) is monotonically increasing.Since ( f(x) = cos(2x + frac{pi}{3}) ), its derivative is:[f'(x) = -2sin(2x + frac{pi}{3})]We need to find where ( f'(x) > 0 ), which means ( -2sin(2x + frac{pi}{3}) > 0 ), so ( sin(2x + frac{pi}{3}) < 0 ).The sine function is negative in the intervals ( pi < 2x + frac{pi}{3} < 2pi ), modulo ( 2pi ).So, solving ( pi < 2x + frac{pi}{3} < 2pi ):Subtract ( frac{pi}{3} ):[pi - frac{pi}{3} < 2x < 2pi - frac{pi}{3}]Simplify:[frac{2pi}{3} < 2x < frac{5pi}{3}]Divide by 2:[frac{pi}{3} < x < frac{5pi}{6}]But since sine is periodic with period ( 2pi ), this interval repeats every ( pi ) for ( x ). So, the general solution is:[frac{pi}{3} + kpi < x < frac{5pi}{6} + kpi quad text{for integer } k]Wait, but in the original problem, the solution given was ( left[-frac{2pi}{3} + kpi, -frac{pi}{6} + kpiright] ). Hmm, that seems different from what I got.Let me check my steps again.We have ( f'(x) = -2sin(2x + frac{pi}{3}) ). So, ( f'(x) > 0 ) when ( sin(2x + frac{pi}{3}) < 0 ).The sine function is negative in intervals ( pi < theta < 2pi ), so:[pi < 2x + frac{pi}{3} < 2pi]Subtract ( frac{pi}{3} ):[frac{2pi}{3} < 2x < frac{5pi}{3}]Divide by 2:[frac{pi}{3} < x < frac{5pi}{6}]But this is just one interval. Since sine has a period of ( 2pi ), the inequality ( sin(2x + frac{pi}{3}) < 0 ) will hold in intervals of length ( pi ) for ( x ).Wait, actually, since the argument is ( 2x + frac{pi}{3} ), the period for ( x ) is ( pi ). So, the solution should be:[frac{pi}{3} + kpi < x < frac{5pi}{6} + kpi quad text{for integer } k]But the solution given was ( left[-frac{2pi}{3} + kpi, -frac{pi}{6} + kpiright] ). Hmm, perhaps I made a miscalculation.Wait, let's consider another approach. Let me solve ( sin(2x + frac{pi}{3}) < 0 ).The sine function is negative in the third and fourth quadrants, so:[pi < 2x + frac{pi}{3} < 2pi]But also, considering the periodicity, this can be written as:[pi + 2kpi < 2x + frac{pi}{3} < 2pi + 2kpi]Subtract ( frac{pi}{3} ):[frac{2pi}{3} + 2kpi < 2x < frac{5pi}{3} + 2kpi]Divide by 2:[frac{pi}{3} + kpi < x < frac{5pi}{6} + kpi]So, that's consistent with what I had before. So, the intervals where ( f(x) ) is increasing are ( left( frac{pi}{3} + kpi, frac{5pi}{6} + kpi right) ).But the solution given was ( left[-frac{2pi}{3} + kpi, -frac{pi}{6} + kpiright] ). Wait, maybe they shifted the interval by ( pi ). Let me check.If I take ( k = -1 ), then:[frac{pi}{3} - pi = -frac{2pi}{3}, quad frac{5pi}{6} - pi = -frac{pi}{6}]Ah, so the interval ( left[-frac{2pi}{3} + kpi, -frac{pi}{6} + kpiright] ) is just shifting the interval I found by ( kpi ). So, both are correct, just expressed differently.Therefore, the intervals where ( f(x) ) is monotonically increasing are ( left[-frac{2pi}{3} + kpi, -frac{pi}{6} + kpiright] ) for integer ( k ).Okay, so part (1) is done. The symmetry centers are ( left( frac{pi}{12} + frac{kpi}{2}, 0 right) ) and the increasing intervals are ( left[-frac{2pi}{3} + kpi, -frac{pi}{6} + kpiright] ).Now, moving on to part (2). In triangle ( triangle ABC ), sides opposite angles ( A, B, C ) are ( a, b, c ) respectively. Given ( f(A) + 1 = 0 ) and ( b + c = 2 ). Need to find the minimum value of ( a ).First, let's find ( A ). Since ( f(A) + 1 = 0 ), so ( f(A) = -1 ).From part (1), we have ( f(x) = cos(2x + frac{pi}{3}) ). So,[cos(2A + frac{pi}{3}) = -1]The cosine function equals -1 at ( pi + 2kpi ). So,[2A + frac{pi}{3} = pi + 2kpi]Solving for ( A ):[2A = pi - frac{pi}{3} + 2kpi = frac{2pi}{3} + 2kpi]Thus,[A = frac{pi}{3} + kpi]But in a triangle, angles are between 0 and ( pi ). So, ( A = frac{pi}{3} ) when ( k = 0 ), and ( A = frac{pi}{3} + pi = frac{4pi}{3} ), which is more than ( pi ), so invalid. Therefore, ( A = frac{pi}{3} ).So, angle ( A = 60^circ ).Now, in triangle ( ABC ), with ( A = 60^circ ), sides ( a, b, c ) opposite angles ( A, B, C ) respectively, and ( b + c = 2 ). Need to find the minimum value of ( a ).I remember that in a triangle, the sides are related by the Law of Cosines:[a^2 = b^2 + c^2 - 2bccos A]Since ( A = 60^circ ), ( cos A = frac{1}{2} ), so:[a^2 = b^2 + c^2 - 2bc cdot frac{1}{2} = b^2 + c^2 - bc]We also know that ( b + c = 2 ). Let me express ( b^2 + c^2 ) in terms of ( (b + c)^2 ):[b^2 + c^2 = (b + c)^2 - 2bc = 4 - 2bc]Therefore, substituting back into ( a^2 ):[a^2 = (4 - 2bc) - bc = 4 - 3bc]So, ( a^2 = 4 - 3bc ). To minimize ( a ), we need to maximize ( bc ).Given that ( b + c = 2 ), the product ( bc ) is maximized when ( b = c ). Because for a fixed sum, the product is maximum when the numbers are equal.So, when ( b = c = 1 ), ( bc = 1 times 1 = 1 ).Therefore, the maximum value of ( bc ) is 1, so the minimum value of ( a^2 ) is ( 4 - 3(1) = 1 ), hence ( a = 1 ).Therefore, the minimum value of ( a ) is 1.Wait, let me verify this. If ( b = c = 1 ), then triangle sides are ( a, 1, 1 ), with angle ( A = 60^circ ). Using the Law of Cosines:[a^2 = 1^2 + 1^2 - 2 times 1 times 1 times cos 60^circ = 1 + 1 - 2 times frac{1}{2} = 2 - 1 = 1]So, ( a = 1 ). That seems correct.Alternatively, if ( b ) and ( c ) are not equal, say ( b = 2 - c ), then ( bc = c(2 - c) = 2c - c^2 ). To find the maximum of this quadratic, take derivative or note that it's a downward parabola with vertex at ( c = 1 ), so maximum at ( c = 1 ), giving ( bc = 1 ).Therefore, the minimum ( a ) is indeed 1.**Final Answer**(1) The symmetry center is (left(frac{pi}{12} + frac{kpi}{2}, 0right)) and the function is increasing on (left[-frac{2pi}{3} + kpi, -frac{pi}{6} + kpiright]). (2) The minimum value of (a) is (boxed{1}).

💡Okay, so I need to figure out the molecular weight of Al(OH)3 under specific conditions: temperature at 35°C, pressure at 1.55 atm, and humidity of 70%. Hmm, molecular weight... I remember that molecular weight is the sum of the atomic weights of all the atoms in a molecule. But wait, does the molecular weight change with temperature, pressure, or humidity? I'm not sure. Maybe it does, maybe it doesn't. I think molecular weight is a constant value because it's based on the atomic weights of the elements, which don't change with environmental conditions. But I'm not entirely certain. Maybe I should double-check that.Alright, let's break it down. Al(OH)3 is aluminum hydroxide. So, it has one aluminum atom, three oxygen atoms, and three hydrogen atoms. I need to find the atomic weights of aluminum, oxygen, and hydrogen. I recall that aluminum has an atomic weight of about 27 g/mol, oxygen is around 16 g/mol, and hydrogen is approximately 1 g/mol. So, if I add those up: 27 (for Al) + 3*16 (for O) + 3*1 (for H). That would be 27 + 48 + 3, which totals 78 g/mol. So, the molecular weight of Al(OH)3 is 78 g/mol.But wait, the question mentions specific conditions: 35°C, 1.55 atm, and 70% humidity. Does that affect the molecular weight? I think molecular weight is an intrinsic property of the compound and doesn't change with external conditions like temperature or pressure. It's more about the composition of the molecule itself. So, even if the temperature or pressure changes, the molecular weight should remain the same because it's based on the number and type of atoms in the molecule.However, I'm a bit confused because sometimes people talk about molar mass depending on conditions, but I think that's more about the mass of a mole of the substance under certain conditions, not the molecular weight itself. Molecular weight is a constant value for a given compound.To be thorough, I'll look up the exact atomic weights to make sure I'm accurate. Aluminum's atomic weight is approximately 26.98 g/mol, oxygen is about 16.00 g/mol, and hydrogen is roughly 1.008 g/mol. So, recalculating with these precise values: 26.98 + 3*(16.00) + 3*(1.008) = 26.98 + 48 + 3.024 = 78.004 g/mol. That's very close to my initial estimate of 78 g/mol.So, even with the exact atomic weights, the molecular weight of Al(OH)3 is approximately 78 g/mol. And since molecular weight doesn't change with temperature, pressure, or humidity, the specific conditions provided in the question don't affect the calculation.I think I've got it. The molecular weight of Al(OH)3 is 78 g/mol, and it remains the same regardless of the environmental conditions mentioned.

💡Alright, so I have this problem about a circle and some lines. Let me try to figure it out step by step. The circle is given by the equation (x^2 + y^2 = 4), which I know is a circle centered at the origin (0,0) with a radius of 2. The first part asks for the equation of the tangent line to the circle that passes through the point P(1,2). Hmm, okay. I remember that a tangent line to a circle is perpendicular to the radius at the point of contact. So, if I can find the slope of the radius from the center to the point of tangency, the tangent line will have a slope that's the negative reciprocal of that.But wait, I don't know the point of tangency yet. Maybe there's another way. I think I can use the formula for the distance from a point to a line. Since the tangent line touches the circle at exactly one point, the distance from the center of the circle to the tangent line must be equal to the radius.Let me denote the equation of the tangent line as (y = kx + b). Since it passes through P(1,2), substituting that in gives (2 = k(1) + b), so (b = 2 - k). Therefore, the equation of the line is (y = kx + (2 - k)).Now, the distance from the center (0,0) to this line should be equal to the radius, which is 2. The formula for the distance from a point ((x_0, y_0)) to the line (Ax + By + C = 0) is (frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}). Let me rewrite the equation of the line in standard form: (kx - y + (2 - k) = 0). So, A = k, B = -1, and C = 2 - k. Plugging into the distance formula:[frac{|k(0) + (-1)(0) + (2 - k)|}{sqrt{k^2 + (-1)^2}} = 2]Simplifying the numerator:[frac{|2 - k|}{sqrt{k^2 + 1}} = 2]Now, I can square both sides to eliminate the square root:[(2 - k)^2 = 4(k^2 + 1)]Expanding the left side:[4 - 4k + k^2 = 4k^2 + 4]Subtracting (4k^2 + 4) from both sides:[4 - 4k + k^2 - 4k^2 - 4 = 0]Simplifying:[-3k^2 - 4k = 0]Factor out a -k:[-k(3k + 4) = 0]So, either (k = 0) or (3k + 4 = 0), which gives (k = -frac{4}{3}).Therefore, there are two possible tangent lines. 1. When (k = 0), the equation is (y = 0x + (2 - 0)), which simplifies to (y = 2).2. When (k = -frac{4}{3}), the equation is (y = -frac{4}{3}x + (2 + frac{4}{3}) = -frac{4}{3}x + frac{10}{3}). Multiplying through by 3 to eliminate fractions: (3y = -4x + 10), so (4x + 3y - 10 = 0).Okay, that seems to make sense. I can double-check by plugging in the point P(1,2) into both equations to ensure they pass through it.For (y = 2): Plugging in x=1, y=2, which works.For (4x + 3y - 10 = 0): Plugging in x=1 and y=2: (4(1) + 3(2) - 10 = 4 + 6 - 10 = 0), which also works.Great, so part (1) is done. The tangent lines are (y = 2) and (4x + 3y - 10 = 0).Now, moving on to part (2). It says that line l passes through P(1,2) and intersects the circle at points A and B, with the distance between A and B being (2sqrt{3}). I need to find the equation of line l.Hmm, okay. So, unlike part (1), this line isn't necessarily tangent; it's a secant that intersects the circle at two points A and B, and the distance between them is given.I think I can use the formula that relates the length of the chord to the distance from the center of the circle to the line. The formula is:[|AB| = 2sqrt{r^2 - d^2}]Where (r) is the radius of the circle, and (d) is the distance from the center to the line. Here, (|AB| = 2sqrt{3}) and (r = 2), so plugging in:[2sqrt{3} = 2sqrt{2^2 - d^2}]Divide both sides by 2:[sqrt{3} = sqrt{4 - d^2}]Square both sides:[3 = 4 - d^2]So,[d^2 = 1 implies d = 1]Therefore, the distance from the center (0,0) to the line l must be 1.Now, let me denote the equation of line l. Since it passes through P(1,2), I can write it in the form (y - 2 = k(x - 1)), where k is the slope. So, the equation is (y = kx + (2 - k)). In standard form, that's (kx - y + (2 - k) = 0).The distance from (0,0) to this line is:[frac{|0 - 0 + (2 - k)|}{sqrt{k^2 + 1}}} = 1]So,[frac{|2 - k|}{sqrt{k^2 + 1}} = 1]Square both sides:[(2 - k)^2 = k^2 + 1]Expanding the left side:[4 - 4k + k^2 = k^2 + 1]Subtract (k^2 + 1) from both sides:[4 - 4k + k^2 - k^2 - 1 = 0]Simplify:[3 - 4k = 0 implies 4k = 3 implies k = frac{3}{4}]So, the slope is (frac{3}{4}), and the equation of the line is:[y - 2 = frac{3}{4}(x - 1)]Multiplying through by 4 to eliminate fractions:[4(y - 2) = 3(x - 1) implies 4y - 8 = 3x - 3 implies 3x - 4y + 5 = 0]Wait, but I should also consider if the line is vertical. Because sometimes, when dealing with slopes, we might miss vertical lines since their slope is undefined.A vertical line passing through P(1,2) would have the equation (x = 1). Let's check if this line intersects the circle and what the chord length would be.Substituting (x = 1) into the circle equation:[1^2 + y^2 = 4 implies y^2 = 3 implies y = sqrt{3} text{ or } y = -sqrt{3}]So, the points of intersection are (1, (sqrt{3})) and (1, (-sqrt{3})). The distance between these two points is:[sqrt{(1 - 1)^2 + (sqrt{3} - (-sqrt{3}))^2} = sqrt{0 + (2sqrt{3})^2} = sqrt{12} = 2sqrt{3}]Which matches the given chord length. So, (x = 1) is also a valid solution.Therefore, there are two possible lines: one with slope (frac{3}{4}) and equation (3x - 4y + 5 = 0), and the vertical line (x = 1).Let me just recap to make sure I didn't miss anything. For part (2), I found the distance from the center to the line must be 1, leading to the slope (k = frac{3}{4}). Then, I also considered the vertical line case, which also satisfies the chord length condition. So, both lines are valid solutions.I think that's it. I don't see any mistakes in my reasoning, but let me just verify the chord length for the non-vertical line.For the line (3x - 4y + 5 = 0), let's find the points of intersection with the circle (x^2 + y^2 = 4).First, solve for y:[3x + 5 = 4y implies y = frac{3x + 5}{4}]Substitute into the circle equation:[x^2 + left(frac{3x + 5}{4}right)^2 = 4]Multiply through by 16 to eliminate denominators:[16x^2 + (3x + 5)^2 = 64]Expand ((3x + 5)^2):[9x^2 + 30x + 25]So,[16x^2 + 9x^2 + 30x + 25 = 64]Combine like terms:[25x^2 + 30x + 25 - 64 = 0 implies 25x^2 + 30x - 39 = 0]Use quadratic formula:[x = frac{-30 pm sqrt{30^2 - 4 cdot 25 cdot (-39)}}{2 cdot 25} = frac{-30 pm sqrt{900 + 3900}}{50} = frac{-30 pm sqrt{4800}}{50}]Simplify (sqrt{4800}):[sqrt{4800} = sqrt{16 cdot 300} = 4sqrt{300} = 4 cdot 10 sqrt{3} = 40sqrt{3}]So,[x = frac{-30 pm 40sqrt{3}}{50} = frac{-3 pm 4sqrt{3}}{5}]Thus, the x-coordinates are (frac{-3 + 4sqrt{3}}{5}) and (frac{-3 - 4sqrt{3}}{5}). Plugging back into (y = frac{3x + 5}{4}):For (x = frac{-3 + 4sqrt{3}}{5}):[y = frac{3left(frac{-3 + 4sqrt{3}}{5}right) + 5}{4} = frac{frac{-9 + 12sqrt{3}}{5} + frac{25}{5}}{4} = frac{frac{16 + 12sqrt{3}}{5}}{4} = frac{16 + 12sqrt{3}}{20} = frac{4 + 3sqrt{3}}{5}]Similarly, for (x = frac{-3 - 4sqrt{3}}{5}):[y = frac{3left(frac{-3 - 4sqrt{3}}{5}right) + 5}{4} = frac{frac{-9 - 12sqrt{3}}{5} + frac{25}{5}}{4} = frac{frac{16 - 12sqrt{3}}{5}}{4} = frac{16 - 12sqrt{3}}{20} = frac{4 - 3sqrt{3}}{5}]So, the points A and B are (left(frac{-3 + 4sqrt{3}}{5}, frac{4 + 3sqrt{3}}{5}right)) and (left(frac{-3 - 4sqrt{3}}{5}, frac{4 - 3sqrt{3}}{5}right)).Now, let's compute the distance between these two points.The distance formula is:[sqrt{left(frac{-3 + 4sqrt{3}}{5} - frac{-3 - 4sqrt{3}}{5}right)^2 + left(frac{4 + 3sqrt{3}}{5} - frac{4 - 3sqrt{3}}{5}right)^2}]Simplify the differences:For x-coordinates:[frac{-3 + 4sqrt{3} + 3 + 4sqrt{3}}{5} = frac{8sqrt{3}}{5}]For y-coordinates:[frac{4 + 3sqrt{3} - 4 + 3sqrt{3}}{5} = frac{6sqrt{3}}{5}]So, the distance is:[sqrt{left(frac{8sqrt{3}}{5}right)^2 + left(frac{6sqrt{3}}{5}right)^2} = sqrt{frac{64 cdot 3}{25} + frac{36 cdot 3}{25}} = sqrt{frac{192 + 108}{25}} = sqrt{frac{300}{25}} = sqrt{12} = 2sqrt{3}]Which matches the given condition. So, the line (3x - 4y + 5 = 0) is indeed a valid solution.Therefore, both lines (x = 1) and (3x - 4y + 5 = 0) satisfy the conditions of part (2). I think I've covered all possibilities and double-checked my work, so I'm confident in these answers.**Final Answer**(1) The equations of the tangent lines are (boxed{y = 2}) and (boxed{4x + 3y - 10 = 0}).(2) The equations of the line are (boxed{3x - 4y + 5 = 0}) and (boxed{x = 1}).

💡Okay, so I need to figure out how many hours David will spend on his data analytics course. Let's see, the course lasts for 24 weeks. Each week, there are 2 three-hour classes and 1 four-hour class. Plus, he has to spend 4 hours each week on small group homework assignments. Hmm, that sounds like a lot, but let's break it down step by step.First, let's think about the classes. There are two types of classes each week: the three-hour classes and the four-hour class. So, for the three-hour classes, there are 2 of them each week. That means 2 times 3 hours, which is 6 hours. Then, there's one four-hour class each week, so that's 4 hours. So, in total, for classes each week, it's 6 hours plus 4 hours, which equals 10 hours.Now, besides the classes, David also has to work on small group homework assignments. He spends 4 hours each week on that. So, adding that to the class time, it's 10 hours for classes plus 4 hours for homework, which gives us 14 hours per week.Okay, so each week, David spends 14 hours on this course. But the course lasts for 24 weeks. So, to find out the total number of hours he'll spend, I need to multiply the weekly hours by the number of weeks. That would be 14 hours per week times 24 weeks.Let me do that multiplication. 14 times 24. Hmm, 14 times 20 is 280, and 14 times 4 is 56. So, adding those together, 280 plus 56 equals 336. So, David will spend a total of 336 hours on this course.Wait, let me double-check to make sure I didn't make any mistakes. So, 2 three-hour classes are 6 hours, plus 1 four-hour class is 4 hours, totaling 10 hours for classes. Then, 4 hours for homework makes it 14 hours per week. Multiplying 14 by 24 gives 336 hours. Yeah, that seems right. I think I got it.

💡Okay, so I have this problem where a receipt shows that 72 identical trophies were purchased for 99.9, but the first and last digits are illegible. I need to figure out how much each trophy cost. Hmm, let's see. First, I need to understand what the total cost could be. The receipt shows 99.9, but the first and last digits are missing. So, it's something like a99.9b, where 'a' is the first missing digit and 'b' is the last missing digit. That means the total cost is a five-digit number where the first digit is 'a', followed by two 9s, a decimal point, another 9, and then the last digit 'b'. So, it's like a99.9b.Now, since the total cost is a99.9b, and we need to find 'a' and 'b', I think we can use some divisibility rules to figure this out. Because the total cost must be divisible by 72, right? Since 72 trophies were bought, and the cost per trophy would be the total cost divided by 72. So, the total cost must be a multiple of 72.I remember that 72 is 8 times 9, so if a number is divisible by both 8 and 9, it's divisible by 72. That gives me two rules to apply here: the divisibility rule for 8 and the divisibility rule for 9.Starting with the divisibility rule for 8: A number is divisible by 8 if the last three digits form a number that's divisible by 8. In this case, the last three digits of the total cost are '99b'. So, I need to find a digit 'b' such that 99b is divisible by 8.Let me list the possible values for 'b' from 0 to 9 and check which one makes 99b divisible by 8:- 990 ÷ 8 = 123.75 → Not divisible- 991 ÷ 8 = 123.875 → Not divisible- 992 ÷ 8 = 124 → Divisible- 993 ÷ 8 = 124.125 → Not divisible- 994 ÷ 8 = 124.25 → Not divisible- 995 ÷ 8 = 124.375 → Not divisible- 996 ÷ 8 = 124.5 → Not divisible- 997 ÷ 8 = 124.625 → Not divisible- 998 ÷ 8 = 124.75 → Not divisible- 999 ÷ 8 = 124.875 → Not divisibleSo, the only value that works is when b = 2, because 992 is divisible by 8. Okay, so now I know that the last digit 'b' is 2. So, the total cost is now a99.92.Next, I need to apply the divisibility rule for 9. A number is divisible by 9 if the sum of its digits is divisible by 9. So, let's add up the digits of the total cost:The total cost is a99.92, which is a five-digit number: a, 9, 9, 9, 2.So, the sum of the digits is a + 9 + 9 + 9 + 2 = a + 29.This sum must be divisible by 9. So, a + 29 ≡ 0 mod 9.Let's find the value of 'a' that satisfies this condition. First, let's find what 29 is modulo 9:29 ÷ 9 = 3 with a remainder of 2. So, 29 ≡ 2 mod 9.Therefore, a + 2 ≡ 0 mod 9.This means that a ≡ -2 mod 9, which is the same as a ≡ 7 mod 9 (since -2 + 9 = 7).Since 'a' is a single digit (0-9), the only possible value is a = 7.So, the total cost is 799.92.Now, to find the cost per trophy, I need to divide the total cost by the number of trophies, which is 72.So, 799.92 ÷ 72.Let me do this division step by step:First, I can write this as 799.92 ÷ 72.Let's see how many times 72 goes into 799.92.Alternatively, I can think of it as 79992 ÷ 7200 to eliminate the decimal.But maybe it's easier to do it directly:72 goes into 799.92 how many times?Let's see:72 × 11 = 79272 × 11.1 = 799.272 × 11.11 = 799.92So, 72 × 11.11 = 799.92Therefore, each trophy costs 11.11.So, putting it all together, the total cost is 799.92, and dividing that by 72 gives us 11.11 per trophy.I think that makes sense. I used the divisibility rules for 8 and 9 to find the missing digits, then performed the division to find the cost per trophy. It all checks out.

💡Alright, so I have this problem here: I need to determine the value for which the function (x⁴ + 1)/(x² + 1) is continuous at x = 1. The options are A) 0, B) 1, C) 2, and D) 3. Hmm, okay. Let me think about this step by step.First, I remember that a function is continuous at a point if the limit as x approaches that point equals the function's value at that point. So, for the function to be continuous at x = 1, the limit as x approaches 1 should be equal to f(1). Let me write that down:limₓ→1 (x⁴ + 1)/(x² + 1) = f(1)So, I need to find the limit as x approaches 1 of (x⁴ + 1)/(x² + 1) and then check if it equals f(1). If it does, then the function is continuous at x = 1.But wait, the question is asking for the value for which the function is continuous at x = 1. That makes me think that maybe the function isn't already continuous there, and I need to adjust something to make it continuous. Maybe there's a hole or a removable discontinuity?Let me check if the function is defined at x = 1. Plugging in x = 1:f(1) = (1⁴ + 1)/(1² + 1) = (1 + 1)/(1 + 1) = 2/2 = 1So, f(1) is 1. Now, let's find the limit as x approaches 1:limₓ→1 (x⁴ + 1)/(x² + 1)Since both the numerator and the denominator are polynomials, and the denominator at x = 1 is 2, which is not zero, the function is defined and continuous at x = 1. So, the limit should just be f(1), which is 1.Wait, but the question is asking for the value for which the function is continuous at x = 1. If it's already continuous, then maybe the value is just 1? But let me double-check.Alternatively, maybe the function isn't continuous because of some other reason, like a jump or an asymptote. But looking at the function, both numerator and denominator are polynomials, and the denominator doesn't become zero at x = 1, so there shouldn't be any discontinuity there.Hmm, maybe I'm overcomplicating it. Since f(1) is 1 and the limit as x approaches 1 is also 1, the function is already continuous at x = 1 without any adjustments. Therefore, the value for which it's continuous is just 1.But let me think again. The options are A) 0, B) 1, C) 2, D) 3. So, 1 is an option, which is B. So, I think the answer is B) 1.Wait, but just to be thorough, let me consider if there's any way the function could be made continuous by redefining it at x = 1. But since it's already continuous there, redefining it wouldn't change anything. So, the value is already 1.Yeah, I think I'm confident now. The function is continuous at x = 1, and the value is 1.