answer：Okay, so I have this problem here where y varies inversely as the square root of x. Hmm, I remember that inverse variation means that as one thing increases, the other decreases proportionally. But let me think about how exactly that works.Alright, if y varies inversely as the square root of x, that should mean that y is equal to some constant divided by the square root of x. So, I can write that as y = k / sqrt(x), where k is the constant of proportionality. Yeah, that sounds right.The problem gives me a specific case where when x is 4, y is 3. I can use this information to find the value of k. Let me plug those numbers into the equation. So, 3 = k / sqrt(4). Since sqrt(4) is 2, that simplifies to 3 = k / 2. To solve for k, I can multiply both sides by 2, which gives me k = 6. Okay, so now I know that the constant k is 6.Now, the question asks me to find the value of y when x is 18. Using the same formula, y = k / sqrt(x), and substituting k with 6 and x with 18, I get y = 6 / sqrt(18). Hmm, sqrt(18) is a bit tricky. Let me see, 18 can be broken down into 9 times 2, so sqrt(18) is sqrt(9*2) which is sqrt(9) times sqrt(2), which is 3*sqrt(2). So, sqrt(18) is 3*sqrt(2).Substituting that back into the equation, y = 6 / (3*sqrt(2)). I can simplify this by dividing 6 by 3, which is 2. So now, y = 2 / sqrt(2). Wait, that can be simplified further. If I rationalize the denominator, I can multiply numerator and denominator by sqrt(2), which gives me (2*sqrt(2)) / (sqrt(2)*sqrt(2)). The denominator simplifies to 2 because sqrt(2)*sqrt(2) is 2. So, now I have (2*sqrt(2)) / 2, which simplifies to sqrt(2).So, putting it all together, when x is 18, y is sqrt(2). Let me just double-check my steps to make sure I didn't make any mistakes. I started with the inverse variation formula, found the constant using the given values, substituted the new x value, simplified the square root, and then rationalized the denominator. Everything seems to check out. I think that's the correct answer.

answer：First, I need to find the derivative of the curve ( y = frac{1}{3}x^3 - 2 ) to determine the slope of the tangent line at the point ((-1, -frac{7}{3})).Calculating the derivative:[y' = frac{d}{dx}left(frac{1}{3}x^3 - 2right) = x^2]Next, I substitute ( x = -1 ) into the derivative to find the slope at that specific point:[y' = (-1)^2 = 1]The slope of the tangent line is 1. To find the slope angle, I use the arctangent function:[theta = arctan(1) = 45^circ]Therefore, the slope angle of the tangent line at the given point is ( 45^circ ).

answer：Alright, so I have this problem here: "If the equation ( Ax + By + C = 0 ) represents a line that is the ( x )-axis, then ( A ), ( B ), and ( C ) must satisfy which of the following?" The options are A, B, C, and D, each with different conditions. Hmm, okay, let me try to figure this out step by step.First, I remember that the ( x )-axis is a horizontal line on the coordinate plane. I also recall that the equation of the ( x )-axis is ( y = 0 ). So, if the given equation ( Ax + By + C = 0 ) is supposed to represent the ( x )-axis, it must be equivalent to ( y = 0 ).Let me write that down:( Ax + By + C = 0 ) should be the same as ( y = 0 ).Now, I need to see what this means for the coefficients ( A ), ( B ), and ( C ). Let me rearrange the given equation to see if I can make it look like ( y = 0 ).Starting with ( Ax + By + C = 0 ), if I solve for ( y ), I get:( By = -Ax - C )Then,( y = (-A/B)x - C/B )Hmm, okay. So, the equation is in the slope-intercept form ( y = mx + b ), where ( m ) is the slope and ( b ) is the y-intercept.But since the line is the ( x )-axis, which is a horizontal line, its slope should be zero. So, ( m = 0 ). That means:( -A/B = 0 )Which implies that ( A = 0 ), because if ( -A/B = 0 ), then ( A ) must be zero (since a fraction is zero only when the numerator is zero, provided the denominator is not zero).Also, the y-intercept ( b ) should be zero because the ( x )-axis crosses the y-axis at zero. So, ( -C/B = 0 ). Again, this implies that ( C = 0 ) (since ( B ) can't be zero because we're dividing by it).Wait, hold on. If ( B ) is in the denominator, then ( B ) can't be zero. So, ( B neq 0 ). That's another condition.So, summarizing:1. ( A = 0 ) (because the slope is zero)2. ( C = 0 ) (because the y-intercept is zero)3. ( B neq 0 ) (because we can't divide by zero when solving for ( y ))Let me check the options given:A: ( A cdot C = 0 )Well, if both ( A ) and ( C ) are zero, their product is zero. But does this capture all the necessary conditions? I think not, because it doesn't specify that ( B ) is not zero. So, this might not be sufficient.B: ( B neq 0 )This is part of what we need, but it doesn't say anything about ( A ) and ( C ). So, this alone isn't enough.C: ( B neq 0 ) and ( A = C = 0 )This seems to cover all the necessary conditions: ( A ) and ( C ) are zero, and ( B ) is not zero. That matches exactly what I derived earlier.D: ( A cdot C = 0 ) and ( B neq 0 )This says that the product of ( A ) and ( C ) is zero, which could mean either ( A = 0 ) or ( C = 0 ), but not necessarily both. However, we know from the equation that both ( A ) and ( C ) must be zero. So, this condition is too broad because it allows for cases where only one of them is zero, which wouldn't necessarily give us the ( x )-axis.Therefore, option C is the correct one because it specifies both ( A ) and ( C ) must be zero, and ( B ) must not be zero. This ensures that the equation simplifies to ( y = 0 ), which is the ( x )-axis.Just to make sure I didn't miss anything, let me plug these conditions back into the original equation.If ( A = 0 ), ( B neq 0 ), and ( C = 0 ), then the equation becomes:( 0 cdot x + B cdot y + 0 = 0 )Simplifying, that's:( By = 0 )Since ( B neq 0 ), we can divide both sides by ( B ):( y = 0 )Which is indeed the equation of the ( x )-axis. So, everything checks out.I think I've covered all the bases here. I considered the slope, the intercept, and the conditions on the coefficients. I also tested the conditions in the original equation to confirm they result in the ( x )-axis. So, I'm confident that option C is the right answer.**Final Answer**boxed{C}

answer：Okay, so I have this problem here: I need to find the value of 'a' such that the derivative of the function f(x) = (x² + a)/(x + 1) at x = 1 is zero. Hmm, that sounds like calculus, specifically involving derivatives. I remember that to find the derivative of a function like this, which is a quotient of two functions, I should use the quotient rule. Let me recall the quotient rule. If I have a function f(x) = g(x)/h(x), then the derivative f'(x) is [g'(x)h(x) - g(x)h'(x)] / [h(x)]². Yeah, that seems right. So, in this case, g(x) is x² + a and h(x) is x + 1. First, I need to find the derivatives of g(x) and h(x). The derivative of g(x) = x² + a should be g'(x) = 2x, since the derivative of x² is 2x and the derivative of a constant 'a' is zero. Then, the derivative of h(x) = x + 1 is h'(x) = 1, because the derivative of x is 1 and the derivative of 1 is zero. So, plugging these into the quotient rule formula, f'(x) should be [2x*(x + 1) - (x² + a)*1] divided by (x + 1) squared. Let me write that out:f'(x) = [2x(x + 1) - (x² + a)] / (x + 1)²Now, I need to simplify the numerator. Let's expand 2x(x + 1):2x(x + 1) = 2x² + 2xSo, substituting back into the numerator:2x² + 2x - x² - a = (2x² - x²) + 2x - a = x² + 2x - aTherefore, f'(x) simplifies to:f'(x) = (x² + 2x - a) / (x + 1)²Alright, now the problem states that the derivative at x = 1 is zero. So, I need to evaluate f'(1) and set it equal to zero, then solve for 'a'. Let's substitute x = 1 into the derivative:f'(1) = (1² + 2*1 - a) / (1 + 1)²Calculating the numerator:1² = 12*1 = 2So, numerator = 1 + 2 - a = 3 - aDenominator:(1 + 1)² = 2² = 4So, f'(1) = (3 - a)/4We are told that f'(1) = 0, so:(3 - a)/4 = 0To solve for 'a', I can multiply both sides by 4:3 - a = 0Then, solving for 'a':a = 3Wait, that seems straightforward. Let me double-check my steps to make sure I didn't make any mistakes.1. Identified the function as a quotient, so I used the quotient rule. That seems correct.2. Calculated g'(x) = 2x and h'(x) = 1. That looks right.3. Applied the quotient rule: [2x(x + 1) - (x² + a)] / (x + 1)². Yes, that's correct.4. Expanded and simplified the numerator: 2x² + 2x - x² - a = x² + 2x - a. That's accurate.5. Substituted x = 1 into the derivative: (1 + 2 - a)/4 = (3 - a)/4. Correct.6. Set (3 - a)/4 = 0, leading to a = 3. That makes sense.I don't see any errors in my calculations. So, the value of 'a' that makes the derivative of f(x) at x = 1 equal to zero is 3.

answer：Alright, so I have this problem here: Given the curve ( y = x^{n+1} ) where ( n ) is a positive integer, I need to find the horizontal coordinate ( x_n ) where the tangent line at the point ( (1, 1) ) intersects the x-axis. Then, I have to determine the product ( x_1 cdot x_2 cdot ldots cdot x_n ).Hmm, okay. Let me break this down step by step. First, I need to find the equation of the tangent line to the curve at the point ( (1, 1) ). To do that, I remember that the tangent line at a point on a curve has the slope equal to the derivative of the function at that point.So, let's find the derivative of ( y = x^{n+1} ). The derivative ( y' ) with respect to ( x ) is ( (n+1)x^n ). That makes sense because when you take the derivative of ( x^k ), it's ( kx^{k-1} ). So here, ( k = n+1 ), so the derivative is ( (n+1)x^n ).Now, I need to evaluate this derivative at the point ( x = 1 ) to get the slope of the tangent line at ( (1, 1) ). Plugging in ( x = 1 ), we get ( y' = (n+1)(1)^n = n+1 ). So, the slope of the tangent line is ( n+1 ).With the slope known and a point on the line ( (1, 1) ), I can write the equation of the tangent line using the point-slope form. The point-slope form is ( y - y_1 = m(x - x_1) ), where ( m ) is the slope and ( (x_1, y_1) ) is the point.Substituting the known values, the equation becomes:[y - 1 = (n+1)(x - 1)]Simplifying this, we get:[y = (n+1)(x - 1) + 1]Expanding the right-hand side:[y = (n+1)x - (n+1) + 1][y = (n+1)x - n]So, the equation of the tangent line is ( y = (n+1)x - n ).Now, I need to find where this tangent line intersects the x-axis. The x-axis is where ( y = 0 ), so I'll set ( y = 0 ) in the equation and solve for ( x ):[0 = (n+1)x - n]Adding ( n ) to both sides:[n = (n+1)x]Dividing both sides by ( n+1 ):[x = frac{n}{n+1}]So, the x-coordinate of the intersection point is ( frac{n}{n+1} ). Therefore, ( x_n = frac{n}{n+1} ).Now, the problem asks for the product ( x_1 cdot x_2 cdot ldots cdot x_n ). Let's write out what each ( x_k ) is for ( k = 1 ) to ( n ):[x_1 = frac{1}{2}, quad x_2 = frac{2}{3}, quad x_3 = frac{3}{4}, quad ldots, quad x_n = frac{n}{n+1}]So, the product becomes:[prod_{k=1}^{n} x_k = frac{1}{2} cdot frac{2}{3} cdot frac{3}{4} cdot ldots cdot frac{n}{n+1}]Looking at this product, I notice that it's a telescoping product. Each numerator cancels with the denominator of the next fraction:[frac{1}{2} cdot frac{2}{3} = frac{1}{3}, quad frac{1}{3} cdot frac{3}{4} = frac{1}{4}, quad ldots, quad frac{1}{n} cdot frac{n}{n+1} = frac{1}{n+1}]So, after all the cancellations, the entire product simplifies to ( frac{1}{n+1} ).Therefore, the value of ( x_1 cdot x_2 cdot ldots cdot x_n ) is ( frac{1}{n+1} ).**Final Answer**The value of ( x_1 cdot x_2 cdot ldots cdot x_n ) is boxed{dfrac{1}{n+1}}.

answer：Okay, so I have this problem where two strips of width 1 overlap at an angle α, and I need to find the area of their overlapping region. The answer choices are given, so that might help if I get stuck.First, I should visualize the problem. The Asymptote code draws two strips overlapping at an angle α, and the overlapping area is shaded. From the code, it looks like the strips are rectangles of width 1, and they intersect at some angle. The overlapping region is a quadrilateral, probably a parallelogram or a rhombus.I remember that when two rectangles intersect at an angle, the overlapping area can sometimes be a rhombus, especially if both rectangles have the same width. So, maybe the overlapping area here is a rhombus.To find the area of a rhombus, I know the formula is (d1 * d2) / 2, where d1 and d2 are the lengths of the diagonals. Alternatively, if I know the side length and the angle, the area can also be calculated as side^2 * sin(α). But I'm not sure which one applies here.Wait, let me think. The strips have a width of 1, so maybe the sides of the rhombus are related to that width. If the strips are at an angle α, then the overlapping area's shape depends on how they cross each other.Let me try to break it down. Each strip is a rectangle of width 1, so their intersection would be a region where both rectangles cover the same space. Since they're at an angle α, the overlapping area is likely a parallelogram.To find the area of the parallelogram, I can use the formula: base * height. But I need to figure out what the base and height are in this case.If I consider one strip as the base, the width is 1. The other strip is at an angle α, so the height relative to the base would be related to the width of the other strip and the angle α.Wait, maybe it's better to think in terms of projection. The width of one strip is 1, and when it's projected onto the direction perpendicular to the other strip, the effective width would be 1 / sin(α). Because the projection of the width onto the perpendicular direction would be 1 divided by the sine of the angle between them.So, if the base is 1 and the height is 1 / sin(α), then the area would be base * height = 1 * (1 / sin(α)) = 1 / sin(α). That seems to make sense.But let me double-check. If α is 90 degrees, then sin(α) is 1, so the area would be 1, which is correct because the overlapping area would just be a square of 1x1. If α is smaller, say 30 degrees, then sin(30) is 0.5, so the area would be 2, which seems reasonable because the overlapping region would be longer.Alternatively, if I think about the rhombus, the diagonals would be related to the widths of the strips and the angle α. The length of the diagonals can be found using trigonometry.If the strips are at an angle α, then the diagonals of the rhombus would be 1 / sin(α) and 1. So, using the formula for the area of a rhombus: (d1 * d2) / 2 = (1 / sin(α) * 1) / 2 = 1 / (2 sin(α)). Wait, that's different from what I got earlier.Hmm, now I'm confused. Which one is correct? Is the area 1 / sin(α) or 1 / (2 sin(α))?Let me think again. If the overlapping region is a parallelogram, then the area is base * height. The base is 1, and the height is the width of the other strip projected onto the direction perpendicular to the base. Since the strips are at an angle α, the height would be 1 / sin(α). So, the area should be 1 * (1 / sin(α)) = 1 / sin(α).But when I thought of it as a rhombus, I got 1 / (2 sin(α)). Maybe I made a mistake there. Let me clarify.In the case of the rhombus, the diagonals are perpendicular bisectors of each other. If one diagonal is 1, then the other diagonal would be 2 * (1 / sin(α)), because the height is 1 / sin(α) and the diagonal would be twice that. So, the area would be (1 * (2 / sin(α))) / 2 = (2 / sin(α)) / 2 = 1 / sin(α). Okay, that matches the earlier result.So, both methods give me the same answer, 1 / sin(α). That makes me more confident that the area is indeed 1 / sin(α).Let me check the answer choices. Option B is 1 / sin(α), which matches my conclusion. So, I think that's the correct answer.Just to be thorough, let me consider edge cases. If α approaches 0, the overlapping area should become very large, which makes sense because 1 / sin(α) approaches infinity as α approaches 0. If α is 90 degrees, the area is 1, which is correct. If α is 180 degrees, the strips are overlapping completely, but that's not really applicable here since they're crossing at an angle.Wait, actually, when α is 180 degrees, the strips are parallel, so the overlapping area would be a rectangle of width 1 and infinite length, but since the strips are of finite length in the Asymptote code, maybe it's not considered here. But in the problem statement, it's just two strips crossing at an angle α, so I think the formula still holds.Another way to think about it is using coordinate geometry. Let me set up a coordinate system where one strip is along the x-axis with width 1 in the y-direction, and the other strip is rotated by α. The equations of the boundaries of the strips can be written, and then I can find the intersection points to determine the vertices of the overlapping region.For the first strip, the boundaries are y = 0 and y = 1. For the second strip, rotated by α, the boundaries would be lines at angle α and α + 180 degrees, each offset by 1 unit in the direction perpendicular to the strip.The distance between the parallel lines of the second strip is 1, so the equations of these lines can be written using the angle α. The overlapping region would be a parallelogram bounded by these four lines.To find the area, I can calculate the distance between the two lines of the second strip, which is 1, and the distance between the two lines of the first strip, which is also 1. The area of the parallelogram would then be the product of these distances divided by sin(α), which is 1 * 1 / sin(α) = 1 / sin(α).This method also gives me the same result, reinforcing that the area is indeed 1 / sin(α).I think I've considered multiple approaches and they all lead to the same conclusion. So, I'm pretty confident that the area of the overlapping region is 1 / sin(α), which corresponds to option B.**Final Answer**boxed{B}