💡First, I recognize that the volume of a cylinder is calculated using the formula ( V = pi r^2 h ), where ( r ) is the radius and ( h ) is the height.When the radius is increased by 5 units, the new radius becomes ( r + 5 ). The new volume with the increased radius is ( V_{text{new radius}} = pi (r + 5)^2 cdot 3 ). The original volume is ( V_{text{original}} = pi r^2 cdot 3 ). The increase in volume due to the radius increase is:[V_{text{new radius}} - V_{text{original}} = pi (r + 5)^2 cdot 3 - pi r^2 cdot 3 = y]Simplifying this, we get:[3pi ((r + 5)^2 - r^2) = y][3pi (10r + 25) = y]Next, when the height is increased by 5 units, the new height becomes ( 3 + 5 = 8 ). The new volume with the increased height is ( V_{text{new height}} = pi r^2 cdot 8 ). The increase in volume due to the height increase is:[V_{text{new height}} - V_{text{original}} = pi r^2 cdot 8 - pi r^2 cdot 3 = y]Simplifying this, we get:[5pi r^2 = y]Since both expressions equal ( y ), we set them equal to each other:[3pi (10r + 25) = 5pi r^2]Dividing both sides by ( pi ) to simplify:[3(10r + 25) = 5r^2][30r + 75 = 5r^2]Rearranging the equation:[5r^2 - 30r - 75 = 0]Dividing the entire equation by 5:[r^2 - 6r - 15 = 0]To solve this quadratic equation, I can factor it:[(r - 7.5)(r + 2) = 0]This gives two potential solutions: ( r = 7.5 ) or ( r = -2 ). Since the radius cannot be negative, the original radius must be:[r = 7.5]

💡Okay, so I have this problem about an arithmetic sequence. The first term is 1, and the common difference is 2. I need to find the limit as n approaches infinity of S_n divided by a_n squared, where S_n is the sum of the first n terms. Hmm, let me break this down step by step.First, I remember that in an arithmetic sequence, each term is the previous term plus a common difference. So, the nth term, a_n, can be found using the formula a_n = a_1 + (n - 1)d. Here, a_1 is 1 and d is 2. Let me plug those values in.a_n = 1 + (n - 1)*2. Let me compute that: 1 + 2n - 2. Simplifying that, 1 - 2 is -1, so a_n = 2n - 1. Okay, that seems straightforward.Next, I need to find S_n, the sum of the first n terms. I recall that the formula for the sum of an arithmetic sequence is S_n = n/2 * (a_1 + a_n). Alternatively, it's also written as S_n = n/2 * [2a_1 + (n - 1)d]. Let me use the second formula because I already know a_1 and d.So, plugging in the values: S_n = n/2 * [2*1 + (n - 1)*2]. Calculating inside the brackets first: 2*1 is 2, and (n - 1)*2 is 2n - 2. So, adding those together: 2 + 2n - 2. The 2 and -2 cancel out, leaving just 2n. So, S_n = n/2 * 2n. The 2s cancel out, so S_n = n^2. That's nice and simple.Now, I need to find the limit as n approaches infinity of S_n divided by a_n squared. So, that's limit as n approaches infinity of (n^2) / (2n - 1)^2. Hmm, okay, let's write that out:lim(n→∞) [n^2 / (2n - 1)^2]I think I can simplify this by dividing numerator and denominator by n^2. Let me try that.Dividing numerator by n^2: n^2 / n^2 = 1.Dividing denominator by n^2: (2n - 1)^2 / n^2. Let me expand that denominator first. (2n - 1)^2 is (2n)^2 - 2*2n*1 + 1^2 = 4n^2 - 4n + 1.So, (4n^2 - 4n + 1)/n^2 = 4 - 4/n + 1/n^2.Putting it all together, the limit becomes:lim(n→∞) [1 / (4 - 4/n + 1/n^2)]Now, as n approaches infinity, the terms with 1/n and 1/n^2 will approach zero. So, the denominator simplifies to 4 - 0 + 0 = 4.Therefore, the limit is 1/4.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the beginning, a_n is 2n - 1, correct. S_n is n^2, that seems right because when you sum the arithmetic sequence, it's quadratic. Then, S_n / a_n squared is n^2 / (2n - 1)^2. Dividing numerator and denominator by n^2 gives 1 / (4 - 4/n + 1/n^2). As n approaches infinity, the fractions go to zero, so it's 1/4. Yeah, that seems consistent.I think I got it right. The limit is 1/4.**Final Answer**The value of the limit is boxed{dfrac{1}{4}}.

💡Sara used a total of (frac{1}{2}) of a roll of wrapping paper for five presents. To find out how much wrapping paper she used on each present, I need to divide the total amount of wrapping paper by the number of presents.First, I recognize that dividing by a number is the same as multiplying by its reciprocal. The reciprocal of 5 is (frac{1}{5}). So, I can rewrite the division as a multiplication:[frac{1}{2} div 5 = frac{1}{2} times frac{1}{5}]Next, I multiply the numerators together and the denominators together:[frac{1 times 1}{2 times 5} = frac{1}{10}]Therefore, Sara used (frac{1}{10}) of a roll of wrapping paper on each present.

💡First, I need to determine the additional distance the bus and truck need to cover to reach a total separation of 78 kilometers. Since they start 8 kilometers apart, the additional distance required is 78 minus 8, which equals 70 kilometers.Next, I'll calculate the combined speed of both vehicles. The truck travels at 60 kilometers per hour, and the bus travels at 40 kilometers per hour. Adding these together gives a combined speed of 100 kilometers per hour.Finally, to find out how long it takes for them to cover the additional 70 kilometers at a combined speed of 100 kilometers per hour, I'll divide the distance by the speed. This results in 70 divided by 100, which equals 0.7 hours.

💡Okay, so I have this problem where a line and a circle intersect at points A and B, forming a chord AB. The length of this chord is given as 2√3. I need to find the value of 'a' in the equation of the line, which is ax - y + 3 = 0. The circle's equation is (x - 1)² + (y - 2)² = 4. First, I should recall some basic concepts about circles and lines. The circle given has its center at (1, 2) and a radius of 2 because the standard form of a circle is (x - h)² + (y - k)² = r², so here h=1, k=2, and r²=4, so r=2.Now, the line intersects the circle at two points, A and B, forming a chord AB. The length of this chord is 2√3. I remember that the length of a chord in a circle can be related to the distance from the center of the circle to the chord. Specifically, there's a formula that connects the radius, the distance from the center to the chord, and half the length of the chord.The formula is: If a chord of length 2L is at a distance d from the center of a circle with radius r, then r² = d² + L². In this case, the length of the chord AB is 2√3, so half the length, L, is √3. The radius r is 2. So plugging into the formula: 2² = d² + (√3)². That simplifies to 4 = d² + 3. Solving for d, we get d² = 1, so d = 1. So the distance from the center of the circle (1, 2) to the line ax - y + 3 = 0 is 1. Now, I need to find the distance from the center (1, 2) to the line ax - y + 3 = 0. The formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / √(A² + B²). In this case, the line is ax - y + 3 = 0, so A = a, B = -1, and C = 3. The point is (1, 2). Plugging into the distance formula: |a*1 + (-1)*2 + 3| / √(a² + (-1)²) = |a - 2 + 3| / √(a² + 1) = |a + 1| / √(a² + 1). We already found that this distance must be 1, so we have |a + 1| / √(a² + 1) = 1. Now, solving for 'a'. Let's write the equation: |a + 1| = √(a² + 1). Since the square root is always non-negative, we can consider the absolute value as positive. So, we can drop the absolute value and write a + 1 = √(a² + 1). But wait, actually, |a + 1| equals √(a² + 1). So, squaring both sides to eliminate the square root: (a + 1)² = (√(a² + 1))², which simplifies to a² + 2a + 1 = a² + 1. Subtracting a² + 1 from both sides: 2a = 0, so a = 0. Hmm, that seems straightforward. But let me double-check. If a = 0, then the line equation becomes 0*x - y + 3 = 0, which simplifies to y = 3. Let me verify if the distance from (1, 2) to the line y = 3 is indeed 1. The distance from a point (x, y) to the line y = k is |y - k|. So, |2 - 3| = 1. Yes, that checks out. Also, let's confirm that the chord length is indeed 2√3. If the line is y = 3, then substituting into the circle equation: (x - 1)² + (3 - 2)² = 4. That simplifies to (x - 1)² + 1 = 4, so (x - 1)² = 3, which gives x = 1 ± √3. So the points of intersection are (1 + √3, 3) and (1 - √3, 3). The distance between these two points is the distance between x-coordinates since y is the same. So, (1 + √3) - (1 - √3) = 2√3. That's correct. Therefore, the value of 'a' is indeed 0.

💡Okay, so I have this problem with four parts about a function f(x) and another function g(x). Let me try to figure out each part step by step.Starting with part (a): I need to sketch the diagram of the function f(x) where f(x) = 4x(1 - |x|) for |x| ≤ 1. Hmm, okay. So, first, I know that |x| is the absolute value of x, which means it's V-shaped. So, 1 - |x| would be an upside-down V shifted up by 1. Then, multiplying by 4x... Hmm, that might make it a bit more complicated.Maybe I should break it down into two cases: when x is positive and when x is negative because of the absolute value. So, for x ≥ 0, |x| is just x, so f(x) becomes 4x(1 - x). Let me expand that: 4x - 4x². That's a quadratic function, and since the coefficient of x² is negative, it's a downward-opening parabola. Its vertex should be at x = -b/(2a) which is -4/(2*(-4)) = 0.5. So, the vertex is at x = 0.5, and plugging that back in, f(0.5) = 4*(0.5)*(1 - 0.5) = 4*(0.5)*(0.5) = 1. So, the vertex is at (0.5, 1). It crosses the x-axis at x = 0 and x = 1.For x < 0, |x| is -x, so f(x) becomes 4x(1 - (-x)) = 4x(1 + x). Expanding that: 4x + 4x². That's a quadratic function as well, but since the coefficient of x² is positive, it's an upward-opening parabola. The vertex is at x = -b/(2a) = -4/(2*4) = -0.5. Plugging that back in, f(-0.5) = 4*(-0.5)*(1 + (-0.5)) = 4*(-0.5)*(0.5) = -1. So, the vertex is at (-0.5, -1). It crosses the x-axis at x = 0 and x = -1.So, putting it all together, the graph of f(x) is a piecewise function. From x = -1 to x = 0, it's an upward-opening parabola with vertex at (-0.5, -1), and from x = 0 to x = 1, it's a downward-opening parabola with vertex at (0.5, 1). It's symmetric about the origin because if I plug in -x, I get f(-x) = -f(x), so it's an odd function.Alright, that should be the sketch for part (a).Moving on to part (b): Does there exist a derivative of f at x = 0?Hmm, derivatives at a point require the function to be smooth there, meaning the left-hand derivative and the right-hand derivative should be equal. Since f(x) is defined piecewise, I should check both sides.First, let's find the derivative from the right (x > 0). For x > 0, f(x) = 4x - 4x², so f'(x) = 4 - 8x. At x = 0, the right-hand derivative is 4 - 8*0 = 4.Now, the derivative from the left (x < 0). For x < 0, f(x) = 4x + 4x², so f'(x) = 4 + 8x. At x = 0, the left-hand derivative is 4 + 8*0 = 4.Since both the left-hand and right-hand derivatives are equal to 4, the derivative exists at x = 0 and is equal to 4.Okay, so part (b) is yes, the derivative exists at x = 0 and is 4.Part (c): Let g be a function defined as f(x)/x when x ≠ 0 and 4 when x = 0. Is g continuous at x = 0?Continuity at a point requires that the limit as x approaches 0 of g(x) equals g(0). So, let's compute the limit of g(x) as x approaches 0.For x ≠ 0, g(x) = f(x)/x. So, let's compute the limit as x approaches 0 of f(x)/x.From part (a), we know f(x) is 4x(1 - |x|). So, f(x)/x = 4(1 - |x|) when x ≠ 0.Therefore, as x approaches 0, |x| approaches 0, so 4(1 - |x|) approaches 4*(1 - 0) = 4.Since g(0) is defined as 4, the limit as x approaches 0 of g(x) is 4, which equals g(0). Therefore, g is continuous at x = 0.Alright, so part (c) is yes, g is continuous at x = 0.Finally, part (d): Sketch the diagram of g.From part (c), we know that for x ≠ 0, g(x) = f(x)/x = 4(1 - |x|). So, g(x) is 4(1 - |x|) for all x ≠ 0, and g(0) = 4.But wait, 4(1 - |x|) is a V-shaped graph centered at x = 0, peaking at 4 when x = 0, and decreasing linearly to 0 as x approaches 1 or -1.But hold on, when x ≠ 0, g(x) = 4(1 - |x|). So, it's a V-shaped graph with the peak at (0,4), going down to (1,0) and (-1,0). But at x = 0, it's also defined as 4, so it's continuous there.So, the graph of g(x) is a V-shape with vertices at (1,0) and (-1,0), peaking at (0,4). It's a linear function on both sides of x = 0.Wait, but let me double-check. For x > 0, g(x) = 4(1 - x), which is a line with slope -4 starting at (0,4) and going down to (1,0). For x < 0, g(x) = 4(1 + x), which is a line with slope 4 starting at (0,4) and going down to (-1,0). So, yes, it's a V-shape.Therefore, the sketch of g is a V-shaped graph with the vertex at (0,4), crossing the x-axis at x = 1 and x = -1.Wait, but hold on a second. When x approaches 0 from the right, g(x) approaches 4, and from the left, it also approaches 4. So, it's continuous at x = 0, as we found in part (c). So, the graph is just a V-shape without any breaks.Yes, that makes sense. So, g(x) is a continuous V-shaped function peaking at (0,4) and crossing the x-axis at x = 1 and x = -1.I think that's all. Let me just recap:(a) f(x) is a piecewise function with two parabolas, one opening downward from x=0 to x=1, and one opening upward from x=-1 to x=0.(b) The derivative at x=0 exists and is 4.(c) g(x) is continuous at x=0.(d) g(x) is a V-shaped graph peaking at (0,4) and crossing the x-axis at x=1 and x=-1.Yeah, that seems right. I don't think I made any mistakes here, but let me just quickly verify part (c) again because sometimes limits can be tricky.We have g(x) = f(x)/x for x ≠ 0, which simplifies to 4(1 - |x|). So, as x approaches 0, |x| approaches 0, so 4(1 - |x|) approaches 4. And g(0) is defined as 4, so yes, the limit equals the function value, hence continuous.Alright, I think I'm confident with these answers.**Final Answer**(a) The graph of ( f(x) ) consists of two parabolas meeting at the origin. For ( 0 leq x leq 1 ), it is a downward-opening parabola with vertex at ( (0.5, 1) ). For ( -1 leq x < 0 ), it is an upward-opening parabola with vertex at ( (-0.5, -1) ).(b) The derivative of ( f ) at ( x = 0 ) exists and is equal to ( boxed{4} ).(c) The function ( g ) is continuous at ( x = 0 ).(d) The graph of ( g(x) ) is a V-shaped graph peaking at ( (0, 4) ) and crossing the x-axis at ( x = 1 ) and ( x = -1 ).boxed{4}