💡First, I need to identify all the numbers between 11 and 31 that are divisible by 5. The numbers in this range that meet the criteria are 15, 20, 25, and 30.Next, I will calculate the sum of these numbers: 15 + 20 + 25 + 30 equals 90.Then, I will determine the number of terms in this set, which is 4.Finally, to find the average, I will divide the sum by the number of terms: 90 divided by 4 equals 22.5.Therefore, the average of all the numbers between 11 and 31 that are divisible by 5 is 22.5.

💡Okay, so I have this problem here where f₀(x) is defined as sin(x). Then, each subsequent function f_{n+1}(x) is the derivative of the previous one, fₙ(x). I need to find what f₂₀₂₀(x) is. The options are sin(x), -sin(x), cos(x), and -cos(x). Hmm, let's see.First, I know that taking derivatives of sine and cosine functions cycles through each other. Let me write down the first few functions to see if there's a pattern.Starting with f₀(x) = sin(x). Then, f₁(x) would be the derivative of sin(x), which is cos(x). Okay, so f₁(x) = cos(x). Next, f₂(x) is the derivative of f₁(x), which is the derivative of cos(x). The derivative of cos(x) is -sin(x). So, f₂(x) = -sin(x).Continuing, f₃(x) is the derivative of f₂(x), which is the derivative of -sin(x). The derivative of -sin(x) is -cos(x). So, f₃(x) = -cos(x). Then, f₄(x) is the derivative of f₃(x), which is the derivative of -cos(x). The derivative of -cos(x) is sin(x). So, f₄(x) = sin(x).Wait a minute, that's the same as f₀(x). So, it seems like after every four derivatives, the functions cycle back to sin(x). So, the pattern repeats every four functions: sin(x), cos(x), -sin(x), -cos(x), and then back to sin(x), and so on.So, if the pattern repeats every four functions, then f₀, f₄, f₈, etc., are all sin(x). Similarly, f₁, f₅, f₉, etc., are cos(x), and so on.Now, I need to find f₂₀₂₀(x). Since the pattern repeats every four functions, I can figure out where 2020 falls in this cycle by dividing 2020 by 4 and looking at the remainder.Let me do that division. 2020 divided by 4. Well, 4 times 505 is 2020, so 2020 divided by 4 is exactly 505 with no remainder. That means 2020 is a multiple of 4, so it should correspond to f₀(x), which is sin(x).Wait, but let me double-check. If I consider that f₀ is 0, then f₁ is 1, f₂ is 2, up to f₂₀₂₀ being 2020. So, if I divide 2020 by 4, the remainder is 0, which would correspond to f₀(x). So, yes, f₂₀₂₀(x) should be sin(x).But just to make sure I didn't make a mistake, let me recount the cycle:f₀ = sin(x)f₁ = cos(x)f₂ = -sin(x)f₃ = -cos(x)f₄ = sin(x)Yes, it cycles every four functions. So, every time the index is a multiple of 4, it's sin(x). Since 2020 is 4 times 505, it's a multiple of 4, so f₂₀₂₀(x) is sin(x).I think that makes sense. So, the answer should be A: sin x.**Final Answer**boxed{A}

💡Okay, so I have this problem about complex numbers. It says that a complex number z is four-presentable if there's another complex number w with absolute value 4 such that z equals w minus 1 over w. Then, S is the set of all such z, and it forms a closed curve. I need to find the area inside this curve.Hmm, let me start by understanding what it means for z to be four-presentable. So, z = w - 1/w, and |w| = 4. That means w lies on a circle in the complex plane with radius 4 centered at the origin. I think I can represent w in polar form because it's on a circle. So, let me write w as 4e^{iθ}, where θ is the argument of w. Then, 1/w would be 1/(4e^{iθ}) which is (1/4)e^{-iθ}. So, substituting back into z, we have:z = 4e^{iθ} - (1/4)e^{-iθ}That's a parametric equation for z in terms of θ. So, if I can express this in terms of x and y coordinates, maybe I can figure out what shape this is.Let me write z as x + iy. So, expanding the expression:4e^{iθ} is 4cosθ + i4sinθSimilarly, (1/4)e^{-iθ} is (1/4)cosθ - i(1/4)sinθSo, subtracting these:z = [4cosθ - (1/4)cosθ] + i[4sinθ + (1/4)sinθ]Simplify the real and imaginary parts:Real part: (4 - 1/4)cosθ = (16/4 - 1/4)cosθ = (15/4)cosθImaginary part: (4 + 1/4)sinθ = (16/4 + 1/4)sinθ = (17/4)sinθSo, z can be written as:z = (15/4)cosθ + i(17/4)sinθHmm, this looks like the parametric equation of an ellipse. In general, an ellipse can be parametrized as x = a cosθ, y = b sinθ, where a and b are the semi-major and semi-minor axes.Comparing, here x = (15/4)cosθ and y = (17/4)sinθ. So, the semi-major axis is 17/4 and the semi-minor axis is 15/4. Wait, actually, which one is which? Since 17/4 is larger than 15/4, the major axis is along the y-axis with length 17/4, and the minor axis is along the x-axis with length 15/4.But wait, actually, in the parametrization, x is associated with cosθ and y with sinθ, so the major axis is along the y-axis if 17/4 > 15/4, which it is.So, the ellipse has semi-major axis a = 17/4 and semi-minor axis b = 15/4.The area of an ellipse is πab, so plugging in the values:Area = π * (17/4) * (15/4) = π * (255/16)Wait, but hold on. I think I made a mistake here. Because when I parametrized z, I considered w as 4e^{iθ}, but when I expressed z in terms of x and y, I directly got x = (15/4)cosθ and y = (17/4)sinθ. So, the ellipse has semi-axes 15/4 and 17/4.But let me think again. The original circle has radius 4, and we're transforming it via z = w - 1/w. So, is there another way to see this transformation?Alternatively, maybe I can think of this as a linear transformation applied to the circle. Let me write z = w - 1/w.But since |w| = 4, 1/w is the conjugate of w divided by |w|^2, which is 16. So, 1/w = (x - iy)/16 if w = x + iy.So, z = w - 1/w = (x + iy) - (x - iy)/16 = x - x/16 + iy + iy/16 = (15x/16) + i(17y/16)So, if w is on the circle of radius 4, then x^2 + y^2 = 16.So, substituting x = (16/15)Re(z) and y = (16/17)Im(z) into the circle equation:[(16/15)Re(z)]^2 + [(16/17)Im(z)]^2 = 16Which simplifies to:(256/225)(Re(z))^2 + (256/289)(Im(z))^2 = 16Dividing both sides by 16:(16/225)(Re(z))^2 + (16/289)(Im(z))^2 = 1Which is the equation of an ellipse with semi-major axis sqrt(225/16) = 15/4 and semi-minor axis sqrt(289/16) = 17/4. Wait, that seems contradictory to what I had earlier. Because in the parametrization, x was 15/4 cosθ and y was 17/4 sinθ, which would mean the semi-major axis is 17/4 along y, but in this equation, the denominators are 225 and 289, which are squares of 15 and 17, so the semi-axes are 15/4 and 17/4. So, actually, the semi-major axis is 17/4 along y, and semi-minor is 15/4 along x.Wait, but in the equation above, the coefficient for Re(z)^2 is (16/225) and for Im(z)^2 is (16/289). So, to write it in standard ellipse form:(Re(z))^2 / (225/16) + (Im(z))^2 / (289/16) = 1Which is:(Re(z))^2 / (15/4)^2 + (Im(z))^2 / (17/4)^2 = 1So, yes, the semi-major axis is 17/4 and semi-minor is 15/4.Therefore, the area is π * (15/4) * (17/4) = π * (255/16). So, 255/16 π.Wait, but earlier I thought it was 255π, but that was a miscalculation. Let me check.Wait, in the initial parametrization, I had z = (15/4)cosθ + i(17/4)sinθ, so the ellipse has semi-axes 15/4 and 17/4, so area is π*(15/4)*(17/4) = (255/16)π.But in the second approach, when I expressed x and y in terms of Re(z) and Im(z), I ended up with the same equation, leading to the same area.Wait, but in the initial thought process, I thought the area was 255π, but that was incorrect because I miscalculated the scaling factors.So, the correct area should be (255/16)π.But wait, let me double-check. When I wrote z = (15/4)cosθ + i(17/4)sinθ, that's the parametric equation of an ellipse with semi-axes 15/4 and 17/4. So, area is π*(15/4)*(17/4) = (255/16)π.Alternatively, when I did the substitution, I got the same result.So, maybe the initial thought process had a mistake in scaling, but the correct area is (255/16)π.Wait, but in the initial problem statement, the user wrote:"Let w be a complex number with absolute value 4. Then, z = w - frac{1}{w}. Let me write w = 4e^{itheta}. Then, 1/w = frac{1}{4}e^{-itheta}. So, z = 4e^{itheta} - frac{1}{4}e^{-itheta}. Expressing this in terms of sine and cosine, we get z = (4costheta - frac{1}{4}costheta) + i(4sintheta + frac{1}{4}sintheta) = frac{15}{4}costheta + ifrac{17}{4}sintheta."So, that's correct. Then, the parametric equation is x = (15/4)cosθ, y = (17/4)sinθ, which is an ellipse with semi-axes 15/4 and 17/4. So, area is π*(15/4)*(17/4) = 255/16 π.But in the initial thought process, the user wrote:"Let w = x + iy where x and y are real numbers. Then:z = x + iy - frac{x - iy}{16} = frac{15x}{16} + frac{17iy}{16}.This transformation corresponds to stretching the real part by frac{15}{16} and the imaginary part by frac{17}{16}. All complex numbers of absolute value 4 form a circle of radius 4. The resulting set S is an ellipse, formed by stretching this circle.The area inside S is given by: frac{15}{16} cdot frac{17}{16} cdot 4^2 pi = frac{15 cdot 17 cdot 16 pi}{16} = boxed{frac{4080}{16}pi} = boxed{255pi}.Wait, that's incorrect because the area scaling factor is (15/16)*(17/16), but the original area is π*(4)^2 = 16π. So, the transformed area should be 16π*(15/16)*(17/16) = (15*17/16)π = 255/16 π.But the user incorrectly multiplied 15/16 * 17/16 * 16π, which is (15*17)/(16*16) *16π = (15*17)/16 π = 255/16 π. Wait, actually, that's correct. Wait, let me compute:(15/16)*(17/16)*16π = (15*17)/(16*16)*16π = (15*17)/16 π = 255/16 π.So, actually, the user's calculation was correct, but the initial thought process in the problem statement was written as:"Let w = x + iy where x and y are real numbers. Then:z = x + iy - frac{x - iy}{16} = frac{15x}{16} + frac{17iy}{16}.This transformation corresponds to stretching the real part by frac{15}{16} and the imaginary part by frac{17}{16}. All complex numbers of absolute value 4 form a circle of radius 4. The resulting set S is an ellipse, formed by stretching this circle.The area inside S is given by: frac{15}{16} cdot frac{17}{16} cdot 4^2 pi = frac{15 cdot 17 cdot 16 pi}{16} = boxed{frac{4080}{16}pi} = boxed{255pi}.Wait, but 4^2 is 16, so 15/16 *17/16 *16π = (15*17)/(16*16)*16π = (15*17)/16 π = 255/16 π, not 255π.So, the user made a mistake in the calculation, writing 15*17*16π /16 = 255π, but actually, it's 15*17*16π / (16*16) = 255π/16.Wait, no. Wait, 15/16 *17/16 *16π = (15*17)/(16*16)*16π = (15*17)/16 π.So, the user incorrectly wrote 15*17*16π /16, which is 15*17*π, which is 255π, but that's wrong because the denominator is 16*16, not 16.So, the correct area is 255/16 π, not 255π.Therefore, the initial thought process had a miscalculation, leading to the wrong answer of 255π, but the correct area is 255/16 π.Wait, but in the parametrization approach, we saw that the semi-axes are 15/4 and 17/4, so area is π*(15/4)*(17/4) = 255/16 π.So, the correct answer should be 255/16 π.But in the initial problem statement, the user concluded 255π, which is incorrect.So, to summarize, the correct area is 255/16 π.But wait, let me think again. Maybe I'm missing something.When we have a linear transformation applied to a circle, the area scales by the determinant of the transformation matrix.In this case, the transformation is z = (15/16)x + i(17/16)y, where w = x + iy lies on the circle x^2 + y^2 = 16.So, the transformation matrix is diagonal with entries 15/16 and 17/16. The determinant is (15/16)*(17/16) = 255/256.The original area of the circle is π*(4)^2 = 16π.So, the area after transformation is 16π * (255/256) = (255/16)π.Yes, that's correct.Therefore, the area inside S is 255/16 π.So, the initial thought process had a mistake in the calculation, leading to 255π, but the correct answer is 255/16 π.Wait, but in the problem statement, the user wrote:"Let w = x + iy where x and y are real numbers. Then:z = x + iy - frac{x - iy}{16} = frac{15x}{16} + frac{17iy}{16}.This transformation corresponds to stretching the real part by frac{15}{16} and the imaginary part by frac{17}{16}. All complex numbers of absolute value 4 form a circle of radius 4. The resulting set S is an ellipse, formed by stretching this circle.The area inside S is given by: frac{15}{16} cdot frac{17}{16} cdot 4^2 pi = frac{15 cdot 17 cdot 16 pi}{16} = boxed{frac{4080}{16}pi} = boxed{255pi}.Wait, so the user wrote 15/16 *17/16 *16π, which is (15*17)/(16*16)*16π = (15*17)/16 π = 255/16 π, but they wrote it as 15*17*16π /16 = 255π. So, the user made a mistake in the calculation, thinking that 15/16 *17/16 *16π is 15*17*π, but actually, it's 15*17*π /16.Therefore, the correct area is 255/16 π.So, the initial thought process had a miscalculation, leading to the wrong answer of 255π, but the correct answer is 255/16 π.Therefore, the area inside S is 255/16 π.But wait, let me think again. Maybe I'm overcomplicating.Alternatively, using the parametrization, z = (15/4)cosθ + i(17/4)sinθ, which is an ellipse with semi-axes 15/4 and 17/4. The area is π*(15/4)*(17/4) = 255/16 π.Yes, that's correct.So, the correct answer is 255/16 π.But in the initial problem statement, the user concluded 255π, which is incorrect.Therefore, the correct area is 255/16 π.

💡Alright, so I've got this problem here with two lines, l₁ and l₂, and they intersect at a point P. I need to find two things: first, the equation of the line that goes through P and the origin, and second, the equation of another line l that passes through P and is perpendicular to a third line l₃. Hmm, okay, let me break this down step by step.First, I need to find the point P where l₁ and l₂ intersect. To do that, I should solve the system of equations given by l₁ and l₂. So, l₁ is 3x + 4y - 2 = 0, and l₂ is 2x + y + 2 = 0. I can use either substitution or elimination to solve for x and y. Let me try elimination because it might be straightforward here.Looking at l₂: 2x + y + 2 = 0. Maybe I can solve for y in terms of x to substitute into l₁. So, from l₂, y = -2x - 2. Now I can plug this expression for y into l₁.Substituting y into l₁: 3x + 4(-2x - 2) - 2 = 0. Let me compute that:3x + (-8x - 8) - 2 = 0 Combine like terms: 3x - 8x - 8 - 2 = 0 That simplifies to: -5x - 10 = 0 So, -5x = 10 Divide both sides by -5: x = -2Okay, so x is -2. Now, plug this back into the expression for y from l₂: y = -2(-2) - 2 = 4 - 2 = 2. So, y is 2. Therefore, point P is (-2, 2). Got that down.Now, moving on to part (1): finding the equation of the line passing through P and the origin. The origin is (0, 0), and P is (-2, 2). So, I need to find the equation of the line connecting (-2, 2) and (0, 0).To find the equation of a line, I can use the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. Since the line passes through the origin, when x = 0, y = 0, so b = 0. That simplifies things because the equation will just be y = mx.Now, let's find the slope m. The slope between two points (x₁, y₁) and (x₂, y₂) is given by (y₂ - y₁)/(x₂ - x₁). So, plugging in our points (-2, 2) and (0, 0):m = (0 - 2)/(0 - (-2)) = (-2)/(2) = -1So, the slope m is -1. Therefore, the equation of the line is y = -1x, which simplifies to y = -x. Alternatively, this can be written as x + y = 0. That seems straightforward.Let me just double-check. If I plug in the origin (0, 0) into x + y = 0, it works. And plugging in P (-2, 2): -2 + 2 = 0, which also works. Perfect, so part (1) is done.Now, part (2): finding the equation of the line l that passes through P and is perpendicular to line l₃: x - 2y - 1 = 0. Okay, so first, I need to find the slope of l₃ to determine the slope of the perpendicular line.Let's rewrite l₃ in slope-intercept form to find its slope. Starting with x - 2y - 1 = 0, I can rearrange it:x - 1 = 2y Divide both sides by 2: y = (1/2)x - 1/2So, the slope of l₃ is 1/2. Since line l is perpendicular to l₃, its slope will be the negative reciprocal of 1/2. The negative reciprocal of 1/2 is -2. So, the slope of line l is -2.Now, I have the slope of line l and a point it passes through, which is P (-2, 2). I can use the point-slope form of the equation of a line, which is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point.Plugging in the values: y - 2 = -2(x - (-2)) Simplify the equation: y - 2 = -2(x + 2) Distribute the -2: y - 2 = -2x - 4 Add 2 to both sides: y = -2x - 2Alternatively, I can write this in standard form. Starting from y = -2x - 2, I can add 2x to both sides: 2x + y = -2. So, the equation is 2x + y + 2 = 0.Let me verify this. The slope is -2, which is correct for being perpendicular to l₃. Plugging in point P (-2, 2): 2*(-2) + 2 + 2 = -4 + 2 + 2 = 0. Perfect, it satisfies the equation. So, that checks out.Just to recap, for part (1), I found the intersection point P by solving the system of equations, then calculated the slope between P and the origin to get the equation of the line. For part (2), I determined the slope of the given line l₃, found the negative reciprocal for the perpendicular slope, and used the point-slope form to write the equation of the desired line.I think I've covered all the steps and double-checked my work, so I feel confident with these answers.**Final Answer**(1) The equation of the line passing through point ( P ) and the origin is (boxed{x + y = 0}).(2) The equation of the line ( l ) that passes through point ( P ) and is perpendicular to line ( l_{3} ) is (boxed{2x + y + 2 = 0}).

💡Okay, so I have this problem about squares and their areas. Let me try to understand it step by step. First, it says that the diagonal of square I is 3 times (a + b), where a and b are the lengths of the sides of two smaller squares. Hmm, okay. So, square I has a diagonal of 3(a + b). I remember that in a square, the diagonal can be related to the side length using the Pythagorean theorem. For a square with side length s, the diagonal d is s√2. So, if the diagonal is 3(a + b), then the side length s of square I must be 3(a + b) divided by √2. Let me write that down:s = 3(a + b) / √2Alright, now that I have the side length of square I, I can find its area. The area of a square is side length squared, so:Area of square I = s² = [3(a + b) / √2]²Let me compute that. Squaring the numerator and the denominator separately:Numerator: [3(a + b)]² = 9(a + b)²Denominator: (√2)² = 2So, Area of square I = 9(a + b)² / 2Got that. Now, the problem says that square II is constructed such that its area is three times the area of square I. So, I need to find the area of square II, which is 3 times the area of square I.Area of square II = 3 * [Area of square I] = 3 * [9(a + b)² / 2]Let me compute that:3 * 9 = 27, so:Area of square II = 27(a + b)² / 2Hmm, looking at the answer choices, option B is 27(a + b)² / 2, which matches what I got. So, I think that's the answer.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Diagonal of square I is 3(a + b). Correct.2. Side length s = diagonal / √2 = 3(a + b) / √2. Correct.3. Area of square I = s² = [3(a + b) / √2]² = 9(a + b)² / 2. Correct.4. Area of square II = 3 * Area of square I = 3 * 9(a + b)² / 2 = 27(a + b)² / 2. Correct.Everything seems to add up. I don't see any errors in my calculations. So, yeah, the area of square II should be 27(a + b)² divided by 2, which is option B.

💡Okay, so I have two math problems to solve. The first one is an equation, and the second one is a system of inequalities. Let me tackle them one by one.Starting with the first problem: Solve the equation (frac{1}{x-3} = 3 + frac{x}{3 - x}). Hmm, this looks like a rational equation, meaning it has fractions with variables in the denominators. I remember that to solve such equations, I need to eliminate the denominators by finding a common denominator or multiplying both sides by the least common denominator (LCD). First, I should note the restrictions on the variable. The denominators (x - 3) and (3 - x) cannot be zero because division by zero is undefined. So, (x - 3 neq 0) implies (x neq 3), and (3 - x neq 0) also implies (x neq 3). So, (x) cannot be 3. I'll keep that in mind in case I get a solution that's 3, which would be extraneous.Looking at the equation again: (frac{1}{x - 3} = 3 + frac{x}{3 - x}). I notice that (3 - x) is the same as (-(x - 3)). So, (frac{x}{3 - x} = -frac{x}{x - 3}). That might be helpful to rewrite the equation so that both sides have the same denominator.Let me rewrite the equation:[frac{1}{x - 3} = 3 - frac{x}{x - 3}]Now, both terms on the right side have the same denominator, so I can combine them:[frac{1}{x - 3} = frac{3(x - 3) - x}{x - 3}]Wait, let me check that step. If I have 3 minus (frac{x}{x - 3}), I need to express 3 as a fraction with the same denominator. So, 3 is equal to (frac{3(x - 3)}{x - 3}). Therefore, the right side becomes:[frac{3(x - 3) - x}{x - 3}]Let me compute the numerator:[3(x - 3) - x = 3x - 9 - x = 2x - 9]So, the equation now is:[frac{1}{x - 3} = frac{2x - 9}{x - 3}]Since both sides have the same denominator and (x neq 3), I can set the numerators equal to each other:[1 = 2x - 9]Now, solving for (x):[1 + 9 = 2x 10 = 2x x = 5]Okay, so (x = 5). But I need to make sure this solution doesn't make any denominators zero in the original equation. Plugging (x = 5) into (x - 3) gives (5 - 3 = 2), which is not zero. So, (x = 5) is a valid solution.Wait, let me double-check by plugging (x = 5) back into the original equation:Left side: (frac{1}{5 - 3} = frac{1}{2}).Right side: (3 + frac{5}{3 - 5} = 3 + frac{5}{-2} = 3 - frac{5}{2} = frac{6}{2} - frac{5}{2} = frac{1}{2}).Yes, both sides equal (frac{1}{2}), so (x = 5) is indeed the correct solution.Alright, moving on to the second problem: Solve the system of inequalities:[begin{cases}frac{x - 1}{2} < frac{x + 1}{3} x - 3(x - 2) leq 4end{cases}]I need to solve each inequality separately and then find the intersection of the solutions.Starting with the first inequality: (frac{x - 1}{2} < frac{x + 1}{3}).To solve this, I can eliminate the denominators by finding a common multiple. The denominators are 2 and 3, so the least common multiple is 6. I'll multiply both sides of the inequality by 6 to eliminate the fractions:[6 cdot frac{x - 1}{2} < 6 cdot frac{x + 1}{3}]Simplifying both sides:[3(x - 1) < 2(x + 1)]Expanding both sides:[3x - 3 < 2x + 2]Now, subtract (2x) from both sides:[x - 3 < 2]Then, add 3 to both sides:[x < 5]So, the solution to the first inequality is (x < 5).Now, moving on to the second inequality: (x - 3(x - 2) leq 4).First, I'll simplify the left side by distributing the -3:[x - 3x + 6 leq 4]Combine like terms:[-2x + 6 leq 4]Subtract 6 from both sides:[-2x leq -2]Now, divide both sides by -2. Remember, when you divide or multiply both sides of an inequality by a negative number, the inequality sign flips.[x geq 1]So, the solution to the second inequality is (x geq 1).Now, combining both solutions:From the first inequality: (x < 5).From the second inequality: (x geq 1).So, the solution set is all real numbers (x) such that (1 leq x < 5).Let me write that in interval notation to visualize it better: ([1, 5)). This means (x) is greater than or equal to 1 and less than 5.Just to make sure I didn't make any mistakes, let me test a value in each interval defined by the critical points 1 and 5.For the first inequality, let's test (x = 0):(frac{0 - 1}{2} = -0.5) and (frac{0 + 1}{3} approx 0.333). So, (-0.5 < 0.333) is true.For the second inequality, testing (x = 0):(0 - 3(0 - 2) = 0 + 6 = 6), which is not less than or equal to 4. So, 0 is not in the solution set, which aligns with our solution (x geq 1).Testing (x = 2), which is within the solution interval:First inequality: (frac{2 - 1}{2} = 0.5) and (frac{2 + 1}{3} = 1). So, (0.5 < 1) is true.Second inequality: (2 - 3(2 - 2) = 2 - 0 = 2 leq 4), which is true.Testing (x = 5), which is the upper bound not included:First inequality: (frac{5 - 1}{2} = 2) and (frac{5 + 1}{3} = 2). So, (2 < 2) is false, which is why 5 is not included.Testing (x = 6), which is outside the solution set:First inequality: (frac{6 - 1}{2} = 2.5) and (frac{6 + 1}{3} approx 2.333). So, (2.5 < 2.333) is false.Second inequality: (6 - 3(6 - 2) = 6 - 12 = -6 leq 4), which is true, but since the first inequality is false, 6 is not in the solution set.Everything checks out. So, the solution to the system of inequalities is (1 leq x < 5).**Final Answer**(1) (boxed{5})(2) (boxed{[1, 5)})