💡To determine the height of the scale model, we start by noting the actual height of the Empire State Building, which is 1454 feet, and the scale ratio of 1:50. This means that every 1 unit of measurement on the model corresponds to 50 units on the actual building.First, we calculate the height of the model by dividing the actual height by the scale ratio:[text{Height of model} = frac{1454}{50}]Performing the division:[frac{1454}{50} = 29.08]Next, we round 29.08 to the nearest whole number, which is 29.Therefore, the height of the model is (boxed{textbf{(C) }29}) feet.

💡Okay, so I have this math problem here with four options, A through D, and I need to figure out which one is correct. Let me take it step by step.Starting with Option A: 3a + 2a = 5a. Hmm, that seems straightforward. If I have 3 of something and add 2 more of the same thing, I should have 5 of that thing. So, 3a plus 2a should indeed be 5a. That makes sense. So, A looks correct.Moving on to Option B: 3a - 2a = 1. Wait a minute, if I have 3a and I subtract 2a, I should get a, right? Because 3 minus 2 is 1, and the a stays the same. So, 3a minus 2a should be a, not just 1. So, B doesn't seem right because it's missing the a.Next is Option C: 3a + 2a = 6a. Okay, so adding 3a and 2a again. I thought that was 5a, not 6a. Maybe I'm missing something here. Let me double-check. 3 plus 2 is 5, so 3a plus 2a should be 5a. So, C is definitely wrong because it says 6a.Finally, Option D: (3a) divided by (2a) equals (3/2)a. Hmm, dividing 3a by 2a. Well, when you divide like terms, the a's should cancel out, right? So, 3 divided by 2 is 3/2, and the a's cancel, so it should just be 3/2, not 3/2 times a. So, D is also incorrect because it still has an a in the answer.So, after going through each option, A is the only one that seems correct. B, C, and D all have mistakes in their calculations.

💡First, I need to determine the cost per bagel when purchasing individually and when buying a dozen.The cost per bagel when bought individually is €3.50.For the dozen purchase, the total cost is €38 for 12 bagels. To find the cost per bagel in a dozen, I divide the total cost by the number of bagels:Cost per bagel in a dozen = €38 / 12 ≈ €3.1667Next, I calculate the savings per bagel by subtracting the cost per bagel in a dozen from the individual cost:Savings per bagel = €3.50 - €3.1667 ≈ €0.3333Finally, to express the savings in Euro cents, I multiply the savings per bagel by 100:Savings per bagel in cents = €0.3333 * 100 ≈ 33.33 centsTherefore, buying a dozen bagels at a time saves approximately 33.33 Euro cents per bagel.

💡First, I recognize that the arithmetic mean of four numbers being 20 means the total sum of these numbers is 80.Given two of the numbers are 12 and 22, I subtract these from the total sum to find the combined value of the remaining two equal numbers, which is 46.Since the two numbers are equal, each number must be 23.Finally, I calculate the product of these two equal numbers by multiplying 23 by 23, resulting in 529.

💡Okay, so I have this problem about an ellipse and a parabola. It's divided into two parts. Let me try to understand each part step by step.Starting with part (1): I need to find the equation of the ellipse C. The given ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) where (a > b > 0). The left and right foci are (F_1) and (F_2), respectively. There's a point (P(1, frac{3}{2})) on the ellipse, and the line connecting P and the right focus (F_2) is perpendicular to the x-axis. Hmm, if the line connecting P and (F_2) is perpendicular to the x-axis, that means it's a vertical line. So, the x-coordinate of (F_2) must be the same as that of P. Since (F_2) is the right focus, its coordinates are ((c, 0)) where (c) is the distance from the center to the focus. Therefore, (c = 1) because the line is vertical and passes through (x = 1). So, (c = 1). For an ellipse, (c^2 = a^2 - b^2). So, (1 = a^2 - b^2). That's one equation.Also, since point P lies on the ellipse, substituting (x = 1) and (y = frac{3}{2}) into the ellipse equation gives:[frac{1^2}{a^2} + frac{left(frac{3}{2}right)^2}{b^2} = 1]Simplifying that:[frac{1}{a^2} + frac{9}{4b^2} = 1]So now, I have two equations:1. (a^2 - b^2 = 1)2. (frac{1}{a^2} + frac{9}{4b^2} = 1)I need to solve these simultaneously to find (a^2) and (b^2).Let me denote (a^2 = A) and (b^2 = B) for simplicity. Then the equations become:1. (A - B = 1)2. (frac{1}{A} + frac{9}{4B} = 1)From the first equation, (A = B + 1). Substitute this into the second equation:[frac{1}{B + 1} + frac{9}{4B} = 1]Let me find a common denominator for the left side. The denominators are (B + 1) and (4B), so the common denominator is (4B(B + 1)). Multiplying each term:[frac{4B}{4B(B + 1)} + frac{9(B + 1)}{4B(B + 1)} = 1]Combine the numerators:[frac{4B + 9B + 9}{4B(B + 1)} = 1]Simplify the numerator:[frac{13B + 9}{4B(B + 1)} = 1]Multiply both sides by (4B(B + 1)):[13B + 9 = 4B(B + 1)]Expand the right side:[13B + 9 = 4B^2 + 4B]Bring all terms to one side:[4B^2 + 4B - 13B - 9 = 0]Simplify:[4B^2 - 9B - 9 = 0]Now, solve this quadratic equation for B. Using the quadratic formula:[B = frac{9 pm sqrt{81 + 144}}{8} = frac{9 pm sqrt{225}}{8} = frac{9 pm 15}{8}]So, two solutions:1. (B = frac{9 + 15}{8} = frac{24}{8} = 3)2. (B = frac{9 - 15}{8} = frac{-6}{8} = -frac{3}{4})Since (B = b^2) must be positive, we discard the negative solution. So, (B = 3). Then, from (A = B + 1), (A = 4).Therefore, (a^2 = 4) and (b^2 = 3). So, the equation of the ellipse is:[frac{x^2}{4} + frac{y^2}{3} = 1]Alright, that seems solid. Let me just double-check by plugging point P into this equation:[frac{1^2}{4} + frac{left(frac{3}{2}right)^2}{3} = frac{1}{4} + frac{9/4}{3} = frac{1}{4} + frac{3}{4} = 1]Yep, that works. And since (c = 1), (a^2 - b^2 = 4 - 3 = 1), which also checks out. So part (1) is done.Moving on to part (2): I need to find the minimum value of the slope of line MN. The setup is as follows:- Line l is tangent to the parabola (y^2 = 4x) in the first quadrant.- This line l intersects the ellipse C at points A and B.- It intersects the x-axis at point M.- The perpendicular bisector of segment AB intersects the y-axis at point N.- We need to find the minimum value of the slope of line MN.Okay, that's a bit involved. Let me break it down step by step.First, let's recall that the parabola (y^2 = 4x) has its vertex at the origin and opens to the right. The tangent lines to this parabola in the first quadrant will have positive slopes.The general equation of a tangent to the parabola (y^2 = 4x) can be written as (y = mx + frac{1}{m}), where m is the slope. This is because for the parabola (y^2 = 4ax), the tangent is (y = mx + frac{a}{m}), and here (a = 1). So, the tangent is (y = mx + frac{1}{m}).But wait, in the problem, it's mentioned that the tangent is in the first quadrant. So, both the x and y intercepts should be positive. Since the tangent is (y = mx + frac{1}{m}), the y-intercept is (frac{1}{m}), which is positive if m is positive. The x-intercept can be found by setting y = 0:[0 = mx + frac{1}{m} implies x = -frac{1}{m^2}]Wait, that's negative. But the tangent intersects the x-axis at point M, which is supposed to be in the first quadrant? Wait, no. The tangent line is in the first quadrant, but the x-intercept is negative. Hmm, that seems contradictory. Maybe I made a mistake.Wait, no, actually, the tangent line touches the parabola in the first quadrant, but it can still intersect the x-axis at a negative x-coordinate. So, point M is on the x-axis, but it's not necessarily in the first quadrant. So, that's okay.But let me think again. The tangent line touches the parabola at a point in the first quadrant, so the tangent line must have a positive slope because it's going from the origin (ish) upwards. So, m is positive.So, the tangent line is (y = mx + frac{1}{m}), with m > 0.Alternatively, another way to represent the tangent line is by using a parameter. For the parabola (y^2 = 4x), the parametric equations are (x = at^2), (y = 2at), where a = 1, so (x = t^2), (y = 2t). The equation of the tangent at parameter t is (ty = x + t^2). So, rearranged, it's (x = ty - t^2). But maybe using the slope form is easier here.So, let's stick with (y = mx + frac{1}{m}). So, m is the slope. Let me denote this as line l.Now, this line l intersects the ellipse ( frac{x^2}{4} + frac{y^2}{3} = 1 ) at points A and B. So, to find points A and B, I need to solve the system of equations:1. (y = mx + frac{1}{m})2. ( frac{x^2}{4} + frac{y^2}{3} = 1 )Substituting equation 1 into equation 2:[frac{x^2}{4} + frac{(mx + frac{1}{m})^2}{3} = 1]Let me expand this:First, square the y-term:[(mx + frac{1}{m})^2 = m^2x^2 + 2 cdot mx cdot frac{1}{m} + frac{1}{m^2} = m^2x^2 + 2x + frac{1}{m^2}]So, plug that into the ellipse equation:[frac{x^2}{4} + frac{m^2x^2 + 2x + frac{1}{m^2}}{3} = 1]Multiply through by 12 to eliminate denominators:[3x^2 + 4(m^2x^2 + 2x + frac{1}{m^2}) = 12]Expand:[3x^2 + 4m^2x^2 + 8x + frac{4}{m^2} = 12]Combine like terms:[(3 + 4m^2)x^2 + 8x + left( frac{4}{m^2} - 12 right) = 0]So, that's a quadratic in x:[(4m^2 + 3)x^2 + 8x + left( frac{4}{m^2} - 12 right) = 0]Let me denote this as:[A x^2 + B x + C = 0]Where:- (A = 4m^2 + 3)- (B = 8)- (C = frac{4}{m^2} - 12)Now, for this quadratic to have real solutions (since the line intersects the ellipse), the discriminant must be non-negative.Discriminant (D = B^2 - 4AC):[D = 64 - 4(4m^2 + 3)left( frac{4}{m^2} - 12 right)]Let me compute this:First, compute (4AC):[4(4m^2 + 3)left( frac{4}{m^2} - 12 right)]Let me compute ((4m^2 + 3)(frac{4}{m^2} - 12)):Multiply term by term:- (4m^2 cdot frac{4}{m^2} = 16)- (4m^2 cdot (-12) = -48m^2)- (3 cdot frac{4}{m^2} = frac{12}{m^2})- (3 cdot (-12) = -36)So, adding these together:[16 - 48m^2 + frac{12}{m^2} - 36 = (-48m^2) + frac{12}{m^2} - 20]Therefore, (4AC = 4(-48m^2 + frac{12}{m^2} - 20) = -192m^2 + frac{48}{m^2} - 80)So, discriminant (D = 64 - (-192m^2 + frac{48}{m^2} - 80))Simplify:[D = 64 + 192m^2 - frac{48}{m^2} + 80 = 192m^2 - frac{48}{m^2} + 144]Factor out 48:[D = 48(4m^2 - frac{1}{m^2} + 3)]Hmm, not sure if that helps. Maybe I can write it as:[D = 192m^2 - frac{48}{m^2} + 144]But regardless, since the line is tangent to the parabola, it should intersect the ellipse at two points, so D must be positive. So, I can note that (D > 0), but maybe I don't need to delve deeper into this right now.Next, I need to find the coordinates of points A and B. Let me denote them as (A(x_1, y_1)) and (B(x_2, y_2)). From the quadratic equation above, I can find expressions for (x_1 + x_2) and (x_1 x_2).From quadratic equation:- Sum of roots: (x_1 + x_2 = -frac{B}{A} = -frac{8}{4m^2 + 3})- Product of roots: (x_1 x_2 = frac{C}{A} = frac{frac{4}{m^2} - 12}{4m^2 + 3})So, the midpoint of AB would be (left( frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2} right)). Let me compute that.First, the x-coordinate of the midpoint:[frac{x_1 + x_2}{2} = -frac{4}{4m^2 + 3}]Now, the y-coordinate. Since (y = mx + frac{1}{m}), so:[y_1 = m x_1 + frac{1}{m}, quad y_2 = m x_2 + frac{1}{m}]So,[frac{y_1 + y_2}{2} = frac{m(x_1 + x_2) + frac{2}{m}}{2} = frac{m(-frac{8}{4m^2 + 3}) + frac{2}{m}}{2}]Simplify:[= frac{ -frac{8m}{4m^2 + 3} + frac{2}{m} }{2 } = frac{ -frac{8m}{4m^2 + 3} + frac{2}{m} }{2 }]To combine the terms, find a common denominator, which is (m(4m^2 + 3)):[= frac{ -8m^2 + 2(4m^2 + 3) }{2m(4m^2 + 3)} = frac{ -8m^2 + 8m^2 + 6 }{2m(4m^2 + 3)} = frac{6}{2m(4m^2 + 3)} = frac{3}{m(4m^2 + 3)}]So, the midpoint of AB is:[left( -frac{4}{4m^2 + 3}, frac{3}{m(4m^2 + 3)} right)]Now, I need the perpendicular bisector of AB. Since AB has a slope of m, the perpendicular bisector will have a slope of (-frac{1}{m}).So, the equation of the perpendicular bisector is:[y - frac{3}{m(4m^2 + 3)} = -frac{1}{m} left( x + frac{4}{4m^2 + 3} right )]Simplify this equation:Multiply both sides by m to eliminate the denominator:[m left( y - frac{3}{m(4m^2 + 3)} right ) = - left( x + frac{4}{4m^2 + 3} right )]Simplify left side:[my - frac{3}{4m^2 + 3} = -x - frac{4}{4m^2 + 3}]Bring all terms to one side:[my + x - frac{3}{4m^2 + 3} + frac{4}{4m^2 + 3} = 0]Simplify constants:[my + x + frac{1}{4m^2 + 3} = 0]Alternatively, solving for y:[y = -frac{1}{m}x - frac{1}{m(4m^2 + 3)}]Wait, let me double-check that. Starting from:[my - frac{3}{4m^2 + 3} = -x - frac{4}{4m^2 + 3}]Bring the x term to the left:[my + x = frac{3}{4m^2 + 3} - frac{4}{4m^2 + 3}]Simplify the right side:[my + x = frac{-1}{4m^2 + 3}]Then, solving for y:[y = -frac{1}{m}x - frac{1}{m(4m^2 + 3)}]Yes, that's correct.Now, this perpendicular bisector intersects the y-axis at point N. The y-axis is where x = 0, so plug x = 0 into the equation:[y = -frac{1}{m}(0) - frac{1}{m(4m^2 + 3)} = -frac{1}{m(4m^2 + 3)}]So, point N is (left( 0, -frac{1}{m(4m^2 + 3)} right )).Earlier, we found point M, which is where line l intersects the x-axis. Line l is (y = mx + frac{1}{m}). Setting y = 0:[0 = mx + frac{1}{m} implies x = -frac{1}{m^2}]So, point M is (left( -frac{1}{m^2}, 0 right )).Now, we need to find the slope of line MN. Points M and N are:- M: (left( -frac{1}{m^2}, 0 right ))- N: (left( 0, -frac{1}{m(4m^2 + 3)} right ))The slope (k_{MN}) is:[k_{MN} = frac{y_N - y_M}{x_N - x_M} = frac{ -frac{1}{m(4m^2 + 3)} - 0 }{0 - left( -frac{1}{m^2} right )} = frac{ -frac{1}{m(4m^2 + 3)} }{ frac{1}{m^2} } = -frac{1}{m(4m^2 + 3)} cdot m^2 = -frac{m}{4m^2 + 3}]So, (k_{MN} = -frac{m}{4m^2 + 3}).Now, we need to find the minimum value of this slope. Since m > 0 (as the tangent is in the first quadrant), the slope is negative. So, we need to find the minimum (most negative) value of (k_{MN}).Alternatively, since we can think of it as finding the maximum of (frac{m}{4m^2 + 3}), because the negative of that will give the slope.Let me denote (f(m) = frac{m}{4m^2 + 3}). We need to find the maximum value of (f(m)) for m > 0.To find the maximum, take the derivative of (f(m)) with respect to m and set it equal to zero.Compute (f'(m)):[f'(m) = frac{(1)(4m^2 + 3) - m(8m)}{(4m^2 + 3)^2} = frac{4m^2 + 3 - 8m^2}{(4m^2 + 3)^2} = frac{-4m^2 + 3}{(4m^2 + 3)^2}]Set (f'(m) = 0):[-4m^2 + 3 = 0 implies 4m^2 = 3 implies m^2 = frac{3}{4} implies m = frac{sqrt{3}}{2}]Since m > 0, we take the positive root.Now, check if this is a maximum. The second derivative test or sign changes can be used, but intuitively, since the function (f(m)) tends to 0 as m approaches 0 and as m approaches infinity, the critical point at (m = frac{sqrt{3}}{2}) must be a maximum.Compute (fleft( frac{sqrt{3}}{2} right )):[fleft( frac{sqrt{3}}{2} right ) = frac{ frac{sqrt{3}}{2} }{ 4 left( frac{3}{4} right ) + 3 } = frac{ frac{sqrt{3}}{2} }{ 3 + 3 } = frac{ frac{sqrt{3}}{2} }{6} = frac{sqrt{3}}{12}]Therefore, the maximum value of (f(m)) is (frac{sqrt{3}}{12}), so the minimum value of (k_{MN}) is (-frac{sqrt{3}}{12}).Let me just verify this result.We found that the slope (k_{MN} = -frac{m}{4m^2 + 3}), and by maximizing (frac{m}{4m^2 + 3}), we found the minimum slope. So, yes, that makes sense.Alternatively, another way to approach this is by using substitution. Let me set (t = m^2), so (t > 0). Then, (f(m) = frac{sqrt{t}}{4t + 3}). To maximize this, square it to make it easier:[f(m)^2 = frac{t}{(4t + 3)^2}]Let (g(t) = frac{t}{(4t + 3)^2}). Take derivative with respect to t:[g'(t) = frac{(1)(4t + 3)^2 - t(2)(4t + 3)(4)}{(4t + 3)^4} = frac{(4t + 3) - 8t}{(4t + 3)^3} = frac{-4t + 3}{(4t + 3)^3}]Set (g'(t) = 0):[-4t + 3 = 0 implies t = frac{3}{4}]So, (t = frac{3}{4}), which means (m = sqrt{frac{3}{4}} = frac{sqrt{3}}{2}), same as before.Thus, the maximum of (f(m)) is indeed (frac{sqrt{3}}{12}), so the minimum slope is (-frac{sqrt{3}}{12}).Therefore, the minimum value of the slope of line MN is (-frac{sqrt{3}}{12}).**Final Answer**(1) The equation of ellipse C is (boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}).(2) The minimum value of the slope of line MN is (boxed{-dfrac{sqrt{3}}{12}}).

💡Alright, so I've got this problem about geometric sequences, and I need to figure out which of the given conclusions is correct. Let me take it step by step.First, let's recall what a geometric sequence is. A geometric sequence is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as ( q ). So, if the first term is ( a_1 ), then the second term ( a_2 = a_1 times q ), the third term ( a_3 = a_2 times q = a_1 times q^2 ), and so on.Now, let's look at each option one by one.**Option A: ( a_1 + a_3 geqslant 2a_2 )**Hmm, let's express ( a_3 ) and ( a_2 ) in terms of ( a_1 ) and ( q ). So, ( a_2 = a_1 q ) and ( a_3 = a_1 q^2 ). Substituting these into the inequality:( a_1 + a_1 q^2 geqslant 2 a_1 q )Let's factor out ( a_1 ):( a_1 (1 + q^2) geqslant 2 a_1 q )Now, assuming ( a_1 ) is positive, we can divide both sides by ( a_1 ):( 1 + q^2 geqslant 2 q )This simplifies to:( q^2 - 2 q + 1 geqslant 0 )Which is:( (q - 1)^2 geqslant 0 )Since a square is always non-negative, this inequality holds true for all real ( q ). But wait, what if ( a_1 ) is negative? If ( a_1 ) is negative, then dividing both sides by ( a_1 ) would reverse the inequality:( 1 + q^2 leqslant 2 q )Which would mean:( (q - 1)^2 leqslant 0 )But a square is always non-negative, so the only solution is ( q = 1 ). So, if ( a_1 ) is negative, the inequality ( a_1 + a_3 geqslant 2a_2 ) only holds if ( q = 1 ). Otherwise, it might not hold. Therefore, Option A isn't always true.**Option B: ( {a}_{1}^{2} + {a}_{3}^{2} geqslant 2{a}_{2}^{2} )**Again, let's express everything in terms of ( a_1 ) and ( q ):( {a}_{1}^{2} + {a}_{1}^{2} q^4 geqslant 2 {a}_{1}^{2} q^2 )Factor out ( {a}_{1}^{2} ):( {a}_{1}^{2} (1 + q^4) geqslant 2 {a}_{1}^{2} q^2 )Assuming ( a_1 neq 0 ), we can divide both sides by ( {a}_{1}^{2} ):( 1 + q^4 geqslant 2 q^2 )This simplifies to:( q^4 - 2 q^2 + 1 geqslant 0 )Which factors as:( (q^2 - 1)^2 geqslant 0 )Again, a square is always non-negative, so this inequality holds true for all real ( q ). Therefore, Option B is always correct.**Option C: If ( a_{1} = a_{3} ), then ( a_{1} = a_{2} )**Let's see. If ( a_1 = a_3 ), then:( a_1 = a_1 q^2 )Assuming ( a_1 neq 0 ), we can divide both sides by ( a_1 ):( 1 = q^2 )So, ( q = 1 ) or ( q = -1 ).If ( q = 1 ), then ( a_2 = a_1 times 1 = a_1 ), so ( a_1 = a_2 ).If ( q = -1 ), then ( a_2 = a_1 times (-1) = -a_1 ), so ( a_1 neq a_2 ) unless ( a_1 = 0 ).But the problem doesn't specify that ( a_1 ) is non-zero. If ( a_1 = 0 ), then all terms are zero, so ( a_1 = a_2 ) trivially. However, if ( a_1 neq 0 ), then ( q = -1 ) would mean ( a_1 neq a_2 ). Therefore, the conclusion ( a_1 = a_2 ) isn't necessarily true. So, Option C is incorrect.**Option D: If ( a_{3} > a_{1} ), then ( a_{4} > a_{2} )**Let's express ( a_3 ) and ( a_4 ) in terms of ( a_1 ) and ( q ):Given ( a_3 = a_1 q^2 > a_1 ), so:( a_1 q^2 > a_1 )Assuming ( a_1 > 0 ), we can divide both sides by ( a_1 ):( q^2 > 1 )Which implies ( q > 1 ) or ( q < -1 ).Now, ( a_4 = a_1 q^3 ) and ( a_2 = a_1 q ).We need to check if ( a_4 > a_2 ):( a_1 q^3 > a_1 q )Again, assuming ( a_1 > 0 ), divide both sides by ( a_1 ):( q^3 > q )Which simplifies to:( q^3 - q > 0 )Factor out ( q ):( q (q^2 - 1) > 0 )We already know that ( q^2 > 1 ), so ( q^2 - 1 > 0 ).Thus, the inequality becomes:( q (positive) > 0 )So, the sign of the expression depends on ( q ).If ( q > 1 ), then ( q ) is positive, so ( q (positive) > 0 ) is true.If ( q < -1 ), then ( q ) is negative, so ( q (positive) < 0 ), which is false.Therefore, if ( q < -1 ), ( a_4 < a_2 ), which contradicts the conclusion. So, Option D isn't always true.**Final Thoughts**After analyzing all options:- **Option A**: Not always true.- **Option B**: Always true.- **Option C**: Not necessarily true.- **Option D**: Not always true.So, the correct conclusion is Option B.