answer：Okay, so I have this function f(n) which is the sum of reciprocals from (n+1) up to 2n. That is, f(n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n). I need to find the difference f(n+1) - f(n). Hmm, let's break this down step by step.First, let me write out what f(n+1) would look like. If I replace every n in f(n) with (n+1), then f(n+1) should be 1/((n+1)+1) + 1/((n+1)+2) + ... + 1/(2*(n+1)). Simplifying that, f(n+1) becomes 1/(n+2) + 1/(n+3) + ... + 1/(2n+2). Now, I need to subtract f(n) from f(n+1). So, f(n+1) - f(n) would be:[1/(n+2) + 1/(n+3) + ... + 1/(2n+2)] - [1/(n+1) + 1/(n+2) + ... + 1/(2n)].Looking at this, I notice that a lot of terms will cancel out. Specifically, the terms from 1/(n+2) up to 1/(2n) appear in both f(n+1) and f(n), so they'll subtract to zero. That leaves me with the terms that don't cancel. In f(n+1), the last two terms are 1/(2n+1) and 1/(2n+2), which aren't in f(n). In f(n), the first term is 1/(n+1), which isn't in f(n+1). So, after cancellation, the difference f(n+1) - f(n) should be:1/(2n+1) + 1/(2n+2) - 1/(n+1).Now, let me simplify this expression. I can combine the last two terms since they have a common denominator. 1/(2n+2) is the same as 1/(2(n+1)), which is (1/2)*(1/(n+1)). So, 1/(2n+2) - 1/(n+1) becomes (1/2)*(1/(n+1)) - 1/(n+1) = (-1/2)*(1/(n+1)).So, putting it all together, f(n+1) - f(n) = 1/(2n+1) - 1/(2(n+1)).Wait, let me check that again. I had 1/(2n+1) + 1/(2n+2) - 1/(n+1). Since 1/(2n+2) is 1/(2(n+1)), so 1/(2n+2) - 1/(n+1) is 1/(2(n+1)) - 2/(2(n+1)) = (-1)/(2(n+1)). So, yes, that part is correct.Therefore, the difference simplifies to 1/(2n+1) - 1/(2(n+1)). Alternatively, I can write this as 1/(2n+1) - 1/(2n+2) because 2(n+1) is 2n+2.Let me verify this with a small value of n to make sure. Let's take n=1.f(1) = 1/2.f(2) = 1/3 + 1/4 = 7/12.So, f(2) - f(1) = 7/12 - 1/2 = 7/12 - 6/12 = 1/12.Now, according to the formula I derived, 1/(2*1+1) - 1/(2*1+2) = 1/3 - 1/4 = 4/12 - 3/12 = 1/12. That matches, so it seems correct.Let me try n=2.f(2) = 1/3 + 1/4 = 7/12.f(3) = 1/4 + 1/5 + 1/6 = (15 + 12 + 10)/60 = 37/60.f(3) - f(2) = 37/60 - 7/12 = 37/60 - 35/60 = 2/60 = 1/30.Using the formula: 1/(2*2+1) - 1/(2*2+2) = 1/5 - 1/6 = 6/30 - 5/30 = 1/30. That also matches.Okay, so my formula seems to work for these small cases. Therefore, I think I've correctly found that f(n+1) - f(n) = 1/(2n+1) - 1/(2n+2).I can also factor this expression if needed. Let's see:1/(2n+1) - 1/(2n+2) = [ (2n+2) - (2n+1) ] / [ (2n+1)(2n+2) ] = (1) / [ (2n+1)(2n+2) ].So, another way to write it is 1 / [ (2n+1)(2n+2) ). But the original form is simpler, so I think that's the answer they're looking for.Just to recap, the process was:1. Write out f(n+1) by substituting n+1 into the original function.2. Subtract f(n) from f(n+1), which cancels out most terms except for the first term of f(n) and the last two terms of f(n+1).3. Simplify the remaining terms to get the final expression.This makes sense because when you increment n by 1, you're effectively adding two new terms to the sum (1/(2n+1) and 1/(2n+2)) and removing the first term (1/(n+1)). The difference between these additions and subtractions gives the change in the function value.I think I've got it. The difference f(n+1) - f(n) is simply 1/(2n+1) minus 1/(2n+2).

answer：Okay, so I have this piecewise function f(x) defined as follows: for x greater than 5, it's x squared minus 1, and for x less than or equal to 5, it's 3x plus some constant b. The question is asking me to find the value of b that makes this function continuous at x equals 5. Hmm, continuity at a point means that the left-hand limit and the right-hand limit at that point are equal, and they both equal the function's value at that point. So, I need to make sure that when x approaches 5 from the left and from the right, the function gives the same value, and that value should also be equal to f(5).Alright, let's break this down. First, I should find the limit as x approaches 5 from the right side. Since for x greater than 5, the function is x squared minus 1, I can plug in x equals 5 into that expression. So, 5 squared is 25, minus 1 is 24. So, the right-hand limit as x approaches 5 is 24.Next, I need to find the limit as x approaches 5 from the left side. For x less than or equal to 5, the function is 3x plus b. So, plugging in x equals 5 here, we get 3 times 5, which is 15, plus b. So, the left-hand limit is 15 plus b.Now, for the function to be continuous at x equals 5, these two limits must be equal. That means 24, which is the right-hand limit, must equal 15 plus b, the left-hand limit. So, setting them equal: 24 equals 15 plus b. To solve for b, I subtract 15 from both sides: 24 minus 15 is 9, so b equals 9.Wait, let me double-check that. If b is 9, then for x less than or equal to 5, the function is 3x plus 9. Plugging in x equals 5, that's 15 plus 9, which is 24. And for x greater than 5, plugging in x equals 5, it's 25 minus 1, which is also 24. So, both sides give 24 at x equals 5, which means the function is continuous there. Yeah, that makes sense.I think I got it right. So, b has to be 9 to make the function continuous at x equals 5.

answer：First, I need to determine the length of the blue line that Ezra drew. I know the length of the white line and how much longer it is compared to the blue line.The white line is 7.666666666666667 inches long, and it is 4.333333333333333 inches longer than the blue line. To find the length of the blue line, I can subtract the difference in length from the length of the white line.So, the calculation would be:7.666666666666667 inches (white line) minus 4.333333333333333 inches (difference) equals the length of the blue line.Performing the subtraction gives me the length of the blue line.

answer：Okay, so I've got these three math problems to solve, and I need to write the results in decimal form. Let me take them one at a time because I don't want to get confused. Starting with part (a): (7 times frac{2}{3} + 16 times frac{5}{12}). Hmm, fractions can be tricky, but I think I can handle this. First, I should probably calculate each multiplication separately and then add them together. Alright, let's do (7 times frac{2}{3}). Multiplying 7 by 2 gives me 14, and then I keep the denominator as 3, so that's (frac{14}{3}). I remember that fractions can be converted to decimals by dividing the numerator by the denominator, so 14 divided by 3. Let me do that: 3 goes into 14 four times with a remainder of 2, so it's 4.666... repeating. I'll note that down as approximately 4.6667.Next, (16 times frac{5}{12}). Multiplying 16 by 5 gives me 80, and the denominator remains 12, so that's (frac{80}{12}). I think I can simplify this fraction. Both 80 and 12 are divisible by 4. Dividing numerator and denominator by 4, I get (frac{20}{3}). Converting that to a decimal, 20 divided by 3 is the same as before, 6.666... repeating. So that's approximately 6.6667.Now, I need to add these two results together: 4.6667 + 6.6667. Let me add the whole numbers first: 4 + 6 is 10. Then the decimals: 0.6667 + 0.6667 is approximately 1.3334. Adding those together, 10 + 1.3334 gives me 11.3334. So, rounding that to four decimal places, it's 11.3333. I think that's the answer for part (a).Moving on to part (b): (5 - left(2 div frac{5}{3}right)). Okay, division of fractions. I remember that dividing by a fraction is the same as multiplying by its reciprocal. So, (2 div frac{5}{3}) is the same as (2 times frac{3}{5}). Let me calculate that: 2 times 3 is 6, and the denominator is 5, so that's (frac{6}{5}). Converting that to a decimal, 6 divided by 5 is 1.2.Now, subtracting that from 5: 5 - 1.2. That's straightforward. 5 minus 1 is 4, and 4 minus 0.2 is 3.8. So, the result for part (b) is 3.8.Finally, part (c): (1 + frac{2}{1 + frac{3}{1 + 4}}). This one looks a bit more complicated because of the nested fractions. I'll start from the innermost part to simplify it step by step.First, let's look at the innermost fraction: (1 + 4). That's easy, it's 5. So, replacing that, the expression becomes (1 + frac{3}{5}). Calculating that: 3 divided by 5 is 0.6, so 1 + 0.6 is 1.6.Now, substitute that back into the original expression: (1 + frac{2}{1.6}). Let me compute (frac{2}{1.6}). Dividing 2 by 1.6, I can think of it as 2 divided by 1.6. To make it easier, I'll multiply both numerator and denominator by 10 to eliminate the decimal: that gives me 20 divided by 16. Simplifying that, 20 divided by 16 is 1.25.So, now the expression is 1 + 1.25, which is 2.25. That seems straightforward. So, the result for part (c) is 2.25.Let me just double-check my calculations to make sure I didn't make any mistakes.For part (a), I had (7 times frac{2}{3}) which is (frac{14}{3}) or approximately 4.6667, and (16 times frac{5}{12}) which simplifies to (frac{20}{3}) or approximately 6.6667. Adding those together gives me 11.3334, which rounds to 11.3333. That seems correct.For part (b), dividing 2 by (frac{5}{3}) is the same as multiplying by (frac{3}{5}), giving me (frac{6}{5}) or 1.2. Subtracting that from 5 gives me 3.8. That checks out.For part (c), starting from the inside: (1 + 4 = 5), then (frac{3}{5} = 0.6), so (1 + 0.6 = 1.6). Then, (frac{2}{1.6} = 1.25), and finally, (1 + 1.25 = 2.25). That all makes sense.I think I've got all three parts correct. The decimals are straightforward once I break down the fractions step by step. It's important to handle each operation carefully, especially with fractions and nested expressions. Taking it one step at a time helps prevent mistakes.

answer：Alright, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, c respectively. I know that a is 1 and b is √3. Also, it's given that A + C = 2B. I need to find the value of c.First, I remember that in any triangle, the sum of the angles is π radians (or 180 degrees). So, A + B + C = π. But it's given that A + C = 2B. Hmm, maybe I can substitute that into the angle sum equation.Let me write that down:A + B + C = πBut A + C = 2B, so substituting that in:2B + B = πSo, 3B = πTherefore, B = π/3. That means angle B is 60 degrees. Okay, so angle B is 60 degrees. That might be useful later.Now, since I have sides a, b, c opposite angles A, B, C respectively, and I know two sides and one angle, maybe I can use the Law of Cosines or the Law of Sines to find the other sides or angles.Let me recall the Law of Cosines. It states that for any triangle,c² = a² + b² - 2ab cos(C)Similarly, for angle B, it would be:b² = a² + c² - 2ac cos(B)Wait, since I know angle B is 60 degrees, and I know sides a and b, maybe I can use the Law of Cosines to find side c.Let me write down what I know:a = 1b = √3B = 60 degrees, so cos(B) = cos(60°) = 0.5So, using the Law of Cosines for angle B:b² = a² + c² - 2ac cos(B)Plugging in the known values:(√3)² = 1² + c² - 2 * 1 * c * 0.5Simplify that:3 = 1 + c² - cSo, 3 = c² - c + 1Subtract 3 from both sides:0 = c² - c + 1 - 30 = c² - c - 2So, now I have a quadratic equation: c² - c - 2 = 0I can solve this quadratic equation for c. Let's use the quadratic formula:c = [1 ± √(1 + 8)] / 2Because the quadratic is c² - c - 2 = 0, so a=1, b=-1, c=-2.Wait, actually, in the quadratic formula, it's:c = [-b ± √(b² - 4ac)] / (2a)So, plugging in:c = [1 ± √(1 + 8)] / 2Because b² = (-1)² = 1, 4ac = 4*1*(-2) = -8, so discriminant is 1 - (-8) = 9.So, √9 = 3.Therefore, c = [1 ± 3] / 2So, two solutions:c = (1 + 3)/2 = 4/2 = 2c = (1 - 3)/2 = (-2)/2 = -1But since side lengths can't be negative, c = 2 is the only valid solution.Wait, but earlier I thought maybe using the Law of Sines. Let me check if that gives the same result.Law of Sines says that a/sin(A) = b/sin(B) = c/sin(C)I know a = 1, b = √3, and B = 60°, so sin(B) = sin(60°) = √3/2.So, 1/sin(A) = √3/(√3/2) = √3 / (√3/2) = 2So, 1/sin(A) = 2, which means sin(A) = 1/2Therefore, angle A is either 30° or 150°, since sine is positive in the first and second quadrants.But since A + C = 2B, and B is 60°, so A + C = 120°. If A were 150°, then C would have to be negative, which isn't possible. So, A must be 30°, and thus C = 120° - 30° = 90°.So, angle C is 90°, which is a right angle. Therefore, triangle ABC is a right-angled triangle at C.Wait, if angle C is 90°, then side c is the hypotenuse. But in a right-angled triangle, the hypotenuse is the longest side. Here, sides are a=1, b=√3, and c=?Wait, if c is the hypotenuse, then c should be longer than both a and b. But b is √3 ≈1.732, and c is 2, which is longer. So that makes sense.Alternatively, using Pythagoras theorem since it's a right-angled triangle:c² = a² + b²c² = 1 + 3 = 4c = 2So, that's consistent with what I found earlier.Therefore, c is 2.Wait, but in the initial problem, the user had a different approach where they got c=√3, but that was incorrect because they used angle C instead of angle B in the Law of Cosines. So, I think my approach is correct because I used angle B, which is known, and got c=2.Let me just recap:Given A + C = 2B, and A + B + C = π, so 3B = π, so B=π/3.Then, using Law of Cosines with angle B:b² = a² + c² - 2ac cos(B)Plugged in the values, solved for c, got c=2.Alternatively, using Law of Sines, found angle A=30°, angle C=90°, then used Pythagoras to get c=2.Both methods lead to c=2, so that must be the correct answer.I think the initial mistake in the problem's solution was using angle C instead of angle B in the Law of Cosines, which led to an incorrect value. So, it's important to make sure which angle corresponds to which side when applying the Law of Cosines.Another thing to note is that when solving for angles using the Law of Sines, we have to consider the possibility of the ambiguous case, but in this scenario, since A + C = 2B and B is 60°, angle A can't be 150°, so it must be 30°, making angle C 90°, which is a unique solution.So, all in all, c=2 is the correct answer.**Final Answer**The value of ( c ) is boxed{2}.

answer：Okay, so I have this problem about a tetrahedron SABC where the edges SA, SB, and SC are mutually perpendicular, and their lengths are 2, 3, and 6 respectively. I need to find the radius R of the circumscribed sphere around this tetrahedron. Hmm, okay, let me think about how to approach this.First, I remember that in a tetrahedron with three mutually perpendicular edges meeting at a vertex, we can place it in a coordinate system where that vertex is the origin, and the edges align with the coordinate axes. So, let me try to visualize this. Let's say point S is at (0, 0, 0). Then, point A would be at (2, 0, 0) because SA is 2 units long along the x-axis. Similarly, point B would be at (0, 3, 0) since SB is 3 units along the y-axis, and point C would be at (0, 0, 6) because SC is 6 units along the z-axis.Now, I need to find the radius of the circumscribed sphere around this tetrahedron. I recall that the circumscribed sphere (circum sphere) of a tetrahedron is the sphere that passes through all four vertices. The center of this sphere is equidistant from all four vertices. So, if I can find the coordinates of this center, I can then compute the distance from this center to any of the four points, which will be the radius R.Let me denote the center of the circum sphere as (x, y, z). Since this point is equidistant from S, A, B, and C, the distances from (x, y, z) to each of these points must be equal. So, I can set up equations based on the distance formula.First, the distance from (x, y, z) to S (0,0,0) is sqrt(x² + y² + z²). The distance to A (2,0,0) is sqrt((x-2)² + y² + z²). Similarly, the distance to B (0,3,0) is sqrt(x² + (y-3)² + z²), and the distance to C (0,0,6) is sqrt(x² + y² + (z-6)²).Since all these distances are equal, I can set up equations by equating them. Let me start by equating the distance to S and the distance to A:sqrt(x² + y² + z²) = sqrt((x-2)² + y² + z²)If I square both sides to eliminate the square roots:x² + y² + z² = (x - 2)² + y² + z²Simplifying the right side:(x - 2)² = x² - 4x + 4So, substituting back:x² + y² + z² = x² - 4x + 4 + y² + z²Subtracting x² + y² + z² from both sides:0 = -4x + 4So, 4x = 4, which gives x = 1.Okay, so the x-coordinate of the center is 1. That makes sense; it's halfway between 0 and 2, but since the edge SA is 2 units, maybe it's not exactly halfway, but in this case, it turned out to be 1.Now, let me equate the distance to S and the distance to B:sqrt(x² + y² + z²) = sqrt(x² + (y - 3)² + z²)Squaring both sides:x² + y² + z² = x² + (y - 3)² + z²Simplifying the right side:(y - 3)² = y² - 6y + 9So, substituting back:x² + y² + z² = x² + y² - 6y + 9 + z²Subtracting x² + y² + z² from both sides:0 = -6y + 9So, 6y = 9, which gives y = 9/6 = 3/2.Alright, so the y-coordinate is 3/2. That seems reasonable, as it's halfway between 0 and 3, but again, since the edge SB is 3 units, it's not exactly halfway, but it's 3/2.Now, let's equate the distance to S and the distance to C:sqrt(x² + y² + z²) = sqrt(x² + y² + (z - 6)²)Squaring both sides:x² + y² + z² = x² + y² + (z - 6)²Simplifying the right side:(z - 6)² = z² - 12z + 36So, substituting back:x² + y² + z² = x² + y² + z² - 12z + 36Subtracting x² + y² + z² from both sides:0 = -12z + 36So, 12z = 36, which gives z = 3.Okay, so the z-coordinate is 3. That's halfway between 0 and 6, which makes sense because SC is 6 units long.So, putting it all together, the center of the circumscribed sphere is at (1, 3/2, 3). Now, I need to find the radius R, which is the distance from this center to any of the four points. Let's choose point S (0,0,0) for simplicity.Using the distance formula:R = sqrt((1 - 0)² + (3/2 - 0)² + (3 - 0)²)= sqrt(1 + (9/4) + 9)= sqrt(1 + 2.25 + 9)= sqrt(12.25)= 3.5Wait, 12.25 is 49/4, so sqrt(49/4) is 7/2, which is 3.5. So, R is 7/2.Let me double-check this with another point, say point A (2,0,0):Distance from (1, 3/2, 3) to (2,0,0):sqrt((2 - 1)² + (0 - 3/2)² + (0 - 3)²)= sqrt(1 + (9/4) + 9)= sqrt(1 + 2.25 + 9)= sqrt(12.25)= 3.5Same result. Let me check point B (0,3,0):sqrt((0 - 1)² + (3 - 3/2)² + (0 - 3)²)= sqrt(1 + (3/2)² + 9)= sqrt(1 + 2.25 + 9)= sqrt(12.25)= 3.5And point C (0,0,6):sqrt((0 - 1)² + (0 - 3/2)² + (6 - 3)²)= sqrt(1 + (9/4) + 9)= sqrt(1 + 2.25 + 9)= sqrt(12.25)= 3.5Okay, so all distances are equal to 3.5, which is 7/2. So, R = 7/2.Now, for part b), the question is whether there exists a sphere with a radius smaller than R that contains the given tetrahedron. Hmm, so is there a sphere smaller than the circumscribed sphere that can still contain all four points of the tetrahedron?I think about the circumscribed sphere being the smallest sphere that passes through all four vertices. But maybe there's a sphere that doesn't pass through all four vertices but still contains them, possibly with a smaller radius. But wait, the circumscribed sphere is already the minimal one in terms of containing all the vertices because it's the smallest sphere that can pass through all four points. So, maybe the answer is no, but I'm not entirely sure.Wait, actually, in some cases, especially in non-regular tetrahedrons, there might be a sphere that contains all the points but doesn't necessarily pass through all four vertices, potentially with a smaller radius. But I'm not certain. Maybe I should think about the concept of the minimal enclosing sphere.I recall that the minimal enclosing sphere (also called the smallest enclosing sphere) of a set of points is the sphere with the smallest radius such that all points lie inside or on the sphere. For a tetrahedron, the minimal enclosing sphere is either the circumscribed sphere or a sphere whose center is somewhere else, possibly closer to the "center" of the tetrahedron.But in this case, since the tetrahedron is right-angled at S, maybe the minimal enclosing sphere is indeed the circumscribed sphere. Alternatively, perhaps the sphere with diameter as the longest edge or something like that.Wait, the longest edge in this tetrahedron is SC, which is 6 units. So, the sphere with diameter SC would have a radius of 3, which is less than 7/2 (which is 3.5). But does this sphere contain all the points?Let me check. If I take the sphere with diameter SC, its center would be at the midpoint of SC, which is at (0,0,3). The radius would be 3. Now, let's see if all four points lie inside or on this sphere.Point S (0,0,0): distance from (0,0,3) is sqrt(0 + 0 + 9) = 3, which is equal to the radius, so it's on the sphere.Point C (0,0,6): distance is also 3, so it's on the sphere.Point A (2,0,0): distance from (0,0,3) is sqrt(4 + 0 + 9) = sqrt(13) ≈ 3.605, which is greater than 3. So, point A is outside this sphere.Similarly, point B (0,3,0): distance is sqrt(0 + 9 + 9) = sqrt(18) ≈ 4.242, which is also greater than 3. So, points A and B are outside this sphere. Therefore, the sphere with diameter SC doesn't contain all the points.What about another sphere? Maybe centered somewhere else. Perhaps the centroid of the tetrahedron? The centroid is the average of the coordinates of the four points.Centroid (G) would be at ((0 + 2 + 0 + 0)/4, (0 + 0 + 3 + 0)/4, (0 + 0 + 0 + 6)/4) = (0.5, 0.75, 1.5). Let's compute the distance from G to each vertex.Distance to S (0,0,0):sqrt(0.5² + 0.75² + 1.5²) = sqrt(0.25 + 0.5625 + 2.25) = sqrt(3.0625) ≈ 1.75.Distance to A (2,0,0):sqrt((2 - 0.5)² + (0 - 0.75)² + (0 - 1.5)²) = sqrt(2.25 + 0.5625 + 2.25) = sqrt(5.0625) ≈ 2.25.Distance to B (0,3,0):sqrt((0 - 0.5)² + (3 - 0.75)² + (0 - 1.5)²) = sqrt(0.25 + 5.0625 + 2.25) = sqrt(7.5625) ≈ 2.75.Distance to C (0,0,6):sqrt((0 - 0.5)² + (0 - 0.75)² + (6 - 1.5)²) = sqrt(0.25 + 0.5625 + 20.25) = sqrt(21.0625) ≈ 4.59.So, the maximum distance from G to any vertex is approximately 4.59, which is greater than 3.5. Therefore, a sphere centered at G with radius 4.59 would contain all points, but that's larger than R = 3.5, so it's not helpful.Wait, but maybe there's another center where the maximum distance to the four points is less than 3.5. I'm not sure. Maybe I should think about the minimal enclosing sphere algorithm.I remember that for a set of points, the minimal enclosing sphere is either determined by two points (as diameter), three points on its surface, or four points on its surface. In our case, the circumscribed sphere is determined by four points, so it's already the minimal one. Therefore, I think the answer to part b) is no, there doesn't exist a sphere with a radius smaller than R that contains the given tetrahedron.Wait, but earlier I thought maybe the sphere with diameter SC could be smaller, but it didn't contain all points. So, perhaps the circumscribed sphere is indeed the minimal one. Therefore, the answer to part b) is no.But I'm a bit confused because sometimes minimal enclosing spheres can be different, but in this case, since the tetrahedron is right-angled, maybe the circumscribed sphere is indeed the minimal one.Wait, let me think again. The circumscribed sphere has radius 7/2 = 3.5. If I can find a sphere with a smaller radius that still contains all four points, then the answer is yes. Otherwise, no.I tried the sphere with diameter SC, which had radius 3, but it didn't contain points A and B. What about another sphere? Maybe centered somewhere else.Suppose I center the sphere at (1, 1.5, 3), which is the same as the circumcenter, but that's the same as the circumscribed sphere. So, that's radius 3.5.Alternatively, maybe center it closer to S. Let's say center it at (0.5, 0.75, 1.5), which is the centroid. But as I calculated earlier, the maximum distance from centroid is about 4.59, which is larger than 3.5, so that's worse.Alternatively, maybe center it somewhere else. Let me try to find a point (x, y, z) such that the maximum distance to S, A, B, C is less than 3.5.But this might be complicated. Alternatively, maybe the minimal enclosing sphere is indeed the circumscribed sphere because all four points are on its surface, so it's the minimal one.Wait, I think in general, for a tetrahedron, the circumscribed sphere is the minimal enclosing sphere if and only if the tetrahedron is acute. But I'm not sure if this tetrahedron is acute.Wait, in a right-angled tetrahedron, like this one, where three edges are mutually perpendicular, I think the circumscribed sphere is indeed the minimal enclosing sphere because all four vertices are on the sphere, and it's the smallest sphere that can contain them.Therefore, I think the answer to part b) is no, there does not exist a sphere with a radius smaller than R that contains the given tetrahedron.Wait, but I'm not entirely sure. Maybe I should look for another approach. Let me think about the coordinates again.The center of the circumscribed sphere is at (1, 1.5, 3), and the radius is 3.5. If I move the center slightly, can I get a smaller radius that still contains all points?Suppose I move the center towards S, say to (0.5, 0.75, 1.5). Then, the distance to S would be sqrt(0.5² + 0.75² + 1.5²) ≈ 1.75, which is less than 3.5. But the distance to C would be sqrt(0.5² + 0.75² + 4.5²) ≈ sqrt(0.25 + 0.5625 + 20.25) ≈ sqrt(21.0625) ≈ 4.59, which is larger than 3.5. So, that doesn't help.Alternatively, if I move the center towards C, say to (1, 1.5, 4.5). Then, the distance to C is sqrt(0 + 0 + 1.5²) = 1.5, which is less than 3.5. But the distance to S would be sqrt(1² + 1.5² + 4.5²) ≈ sqrt(1 + 2.25 + 20.25) ≈ sqrt(23.5) ≈ 4.85, which is larger than 3.5. So, again, not helpful.Alternatively, maybe move the center in some other direction. But it seems that moving the center in any direction either increases the distance to some points beyond 3.5 or doesn't reduce the maximum distance below 3.5.Therefore, I think the circumscribed sphere is indeed the minimal enclosing sphere, and there doesn't exist a smaller sphere that can contain the entire tetrahedron.So, to summarize:a) The radius R of the circumscribed sphere is 7/2.b) No, there does not exist a sphere with a radius smaller than R that contains the given tetrahedron.