answer：Okay, so I have this expression to simplify: (3y - 2) multiplied by (5y¹² + 4y¹¹ + 6y⁹ + 3y⁸). Hmm, I remember from algebra that when you have two polynomials multiplied together, you need to distribute each term in the first polynomial to each term in the second polynomial. This is often called the distributive property or the FOIL method when dealing with binomials, but since the second polynomial has four terms, it might take a bit longer.Let me write this out step by step. First, I'll distribute the 3y across each term in the second polynomial. So, 3y times 5y¹². When you multiply variables with exponents, you add the exponents, right? So y times y¹² is y¹³. Then, 3 times 5 is 15. So that term becomes 15y¹³.Next, 3y times 4y¹¹. Again, multiplying the coefficients: 3 times 4 is 12. And y times y¹¹ is y¹². So that term is 12y¹².Then, 3y times 6y⁹. 3 times 6 is 18, and y times y⁹ is y¹⁰. So that gives me 18y¹⁰.Lastly, 3y times 3y⁸. 3 times 3 is 9, and y times y⁸ is y⁹. So that term is 9y⁹.Okay, so after distributing the 3y, I have: 15y¹³ + 12y¹² + 18y¹⁰ + 9y⁹.Now I need to distribute the -2 across each term in the second polynomial. Let's do that step by step as well. Starting with -2 times 5y¹². That's -10y¹².Next, -2 times 4y¹¹. That gives me -8y¹¹.Then, -2 times 6y⁹. That's -12y⁹.Lastly, -2 times 3y⁸. That results in -6y⁸.So after distributing the -2, I have: -10y¹² - 8y¹¹ - 12y⁹ - 6y⁸.Now, I need to combine these two results. Let me write them together:15y¹³ + 12y¹² + 18y¹⁰ + 9y⁹ - 10y¹² - 8y¹¹ - 12y⁹ - 6y⁸.Now, I should combine like terms. Like terms are terms with the same exponent. Let me rearrange them in order of descending exponents to make it easier.Starting with the highest exponent, which is y¹³. There's only one term: 15y¹³.Next, y¹². There are two terms: 12y¹² and -10y¹². Adding those together: 12y¹² - 10y¹² = 2y¹².Then, y¹¹. There's only one term: -8y¹¹.Next, y¹⁰. Only one term: 18y¹⁰.Then, y⁹. There are two terms: 9y⁹ and -12y⁹. Adding those: 9y⁹ - 12y⁹ = -3y⁹.Lastly, y⁸. Only one term: -6y⁸.Putting it all together, the simplified polynomial is:15y¹³ + 2y¹² - 8y¹¹ + 18y¹⁰ - 3y⁹ - 6y⁸.I think that's it. Let me double-check to make sure I didn't miss any terms or make any calculation errors. Starting from the distribution:3y * 5y¹² = 15y¹³3y * 4y¹¹ = 12y¹²3y * 6y⁹ = 18y¹⁰3y * 3y⁸ = 9y⁹-2 * 5y¹² = -10y¹²-2 * 4y¹¹ = -8y¹¹-2 * 6y⁹ = -12y⁹-2 * 3y⁸ = -6y⁸Combining like terms:15y¹³12y¹² -10y¹² = 2y¹²-8y¹¹18y¹⁰9y⁹ -12y⁹ = -3y⁹-6y⁸Yes, that looks correct. So the final simplified polynomial is 15y¹³ + 2y¹² - 8y¹¹ + 18y¹⁰ - 3y⁹ - 6y⁸.

answer：First, I need to determine the width of the margin on each side of the sheet. The sheet measures 20 cm by 30 cm, and there are margins on all sides. The top and bottom margins are each 3 cm, and the side margins are of unknown width, which I'll denote as ( x ) cm.Next, I'll calculate the total area of the sheet by multiplying its length and width:[text{Total Area} = 20 text{ cm} times 30 text{ cm} = 600 text{ cm}^2]The typing area is the portion of the sheet that is actually used for typing. To find the dimensions of this area, I'll subtract the margins from the total dimensions. For the width, I'll subtract twice the side margin (( 2x )) from the total width:[text{Typing Width} = 20 text{ cm} - 2x]For the length, I'll subtract twice the top and bottom margins (each 3 cm) from the total length:[text{Typing Length} = 30 text{ cm} - 2 times 3 text{ cm} = 24 text{ cm}]Now, I'll calculate the area of the typing region by multiplying the typing width and length:[text{Typing Area} = (20 - 2x) text{ cm} times 24 text{ cm}]The problem states that the percentage of the page used by the typist is 64%. Therefore, the typing area should be 64% of the total area:[text{Typing Area} = 0.64 times 600 text{ cm}^2 = 384 text{ cm}^2]Setting up the equation for the typing area:[(20 - 2x) times 24 = 384]To solve for ( x ), I'll divide both sides of the equation by 24:[20 - 2x = frac{384}{24} = 16]Next, I'll isolate ( x ) by subtracting 16 from both sides:[20 - 16 = 2x][4 = 2x]Finally, I'll divide both sides by 2 to find the value of ( x ):[x = frac{4}{2} = 2]Therefore, the width of the margin on each side is 2 cm.

answer：Okay, so I have this problem where there's a line and a circle, and they intersect at points A and B. The circle is given by the equation x² + y² = 4, which I recognize as a circle centered at the origin with a radius of 2. The line is given by kx - y - 2 = 0, and I need to find the value of k such that the vectors OA and OB are perpendicular to each other, meaning their dot product is zero.First, I should probably write down what I know. The circle has radius 2, so any point on the circle satisfies x² + y² = 4. The line is kx - y - 2 = 0, which I can rearrange to y = kx - 2. That might be helpful because it expresses y in terms of x, which I can substitute into the circle's equation.So, substituting y = kx - 2 into x² + y² = 4, I get:x² + (kx - 2)² = 4Expanding that, I have:x² + k²x² - 4kx + 4 = 4Simplify that:(1 + k²)x² - 4kx + 4 - 4 = 0Which simplifies further to:(1 + k²)x² - 4kx = 0Wait, that seems too simple. Let me check my steps again. Starting from substitution:x² + (kx - 2)² = 4Expanding (kx - 2)² gives k²x² - 4kx + 4. So, the equation becomes:x² + k²x² - 4kx + 4 = 4Subtracting 4 from both sides:x² + k²x² - 4kx = 0Factor out x:x(1 + k²)x - 4k = 0Wait, that doesn't look right. Maybe I should factor differently. Let me try again.Starting from:x² + (kx - 2)² = 4Which is:x² + k²x² - 4kx + 4 = 4Combine like terms:(1 + k²)x² - 4kx + 4 - 4 = 0So, (1 + k²)x² - 4kx = 0Factor out x:x[(1 + k²)x - 4k] = 0So, the solutions are x = 0 or (1 + k²)x - 4k = 0If x = 0, then from y = kx - 2, y = -2. So, one point is (0, -2). The other solution is when (1 + k²)x - 4k = 0, so x = 4k / (1 + k²). Then, y = k*(4k / (1 + k²)) - 2 = (4k² / (1 + k²)) - 2 = (4k² - 2(1 + k²)) / (1 + k²) = (4k² - 2 - 2k²) / (1 + k²) = (2k² - 2) / (1 + k²) = 2(k² - 1)/(1 + k²)So, the two points of intersection are A(0, -2) and B(4k/(1 + k²), 2(k² - 1)/(1 + k²))Now, I need to find vectors OA and OB. Since O is the origin, OA is just the vector from O to A, which is (0, -2), and OB is the vector from O to B, which is (4k/(1 + k²), 2(k² - 1)/(1 + k²))The dot product of OA and OB should be zero because they are perpendicular. So, OA · OB = 0.Calculating the dot product:(0)*(4k/(1 + k²)) + (-2)*(2(k² - 1)/(1 + k²)) = 0Simplify:0 + (-4(k² - 1))/(1 + k²) = 0So, (-4(k² - 1))/(1 + k²) = 0For this fraction to be zero, the numerator must be zero:-4(k² - 1) = 0Divide both sides by -4:k² - 1 = 0So, k² = 1Taking square roots:k = ±1Therefore, the possible values of k are 1 and -1.Wait, but let me double-check because sometimes when dealing with intersections, especially with circles, there might be cases where the line is tangent or doesn't intersect, but in this case, since we have two points A and B, the line must intersect the circle at two distinct points, so the discriminant should be positive.Looking back at the quadratic equation we had:(1 + k²)x² - 4kx = 0This is a quadratic in x, and the discriminant D is [(-4k)]² - 4*(1 + k²)*0 = 16k²Since 16k² is always non-negative, and for k ≠ 0, it's positive, so the line will intersect the circle at two points as long as k is not zero. But since we found k = ±1, which are non-zero, it's fine.So, I think my answer is correct. k can be 1 or -1, so the answer is ±1, which is option C.**Final Answer**boxed{C}

answer：Okay, so I have this problem about finding the polar equation of a circle. Let me try to understand what it's asking. First, it says that circle C passes through the point P with coordinates (√2, π/4) in polar form. Then, the center of this circle is the intersection point of the line given by the equation ρ sin(θ - π/3) = -√3/2 with the polar axis. Alright, so I need to find the polar equation of circle C. Let me break this down step by step.First, I think I should find the center of the circle. The center is where the given line intersects the polar axis. The polar axis is the horizontal axis in polar coordinates, which corresponds to θ = 0 in rectangular coordinates, right? So, to find the intersection point, I need to convert that line equation into rectangular coordinates and then find where it crosses the x-axis (since the polar axis is the x-axis in rectangular coordinates).Let me recall how to convert polar equations to rectangular coordinates. The general conversion formulas are:- x = ρ cos θ- y = ρ sin θGiven the line equation: ρ sin(θ - π/3) = -√3/2I can use the sine difference identity to expand sin(θ - π/3). The identity is:sin(A - B) = sin A cos B - cos A sin BSo, applying this to sin(θ - π/3):sin θ cos(π/3) - cos θ sin(π/3)I know that cos(π/3) is 1/2 and sin(π/3) is √3/2. So substituting these values in:sin θ * (1/2) - cos θ * (√3/2)So, the equation becomes:ρ [ (1/2) sin θ - (√3/2) cos θ ] = -√3/2Now, let's express this in terms of x and y. Remember that ρ sin θ = y and ρ cos θ = x. So substituting these in:(1/2) y - (√3/2) x = -√3/2So, the equation in rectangular coordinates is:(1/2) y - (√3/2) x = -√3/2Let me write this more neatly:(-√3/2) x + (1/2) y = -√3/2I can multiply both sides by 2 to eliminate the denominators:-√3 x + y = -√3So, the equation simplifies to:y = √3 x - √3Wait, is that right? Let me check my steps again. Original equation after substitution:(1/2) y - (√3/2) x = -√3/2Multiply both sides by 2:y - √3 x = -√3So, rearranged:y = √3 x - √3Yes, that seems correct.Now, I need to find where this line intersects the polar axis. The polar axis is the x-axis, where y = 0. So, I can set y = 0 in the equation and solve for x.Setting y = 0:0 = √3 x - √3Adding √3 to both sides:√3 x = √3Divide both sides by √3:x = 1So, the intersection point is (1, 0) in rectangular coordinates. Therefore, the center of the circle is at (1, 0) in rectangular coordinates, which is (1, 0) in polar coordinates as well since it's on the polar axis.Wait, actually, in polar coordinates, (1, 0) is the same as (1, 0) because it's on the polar axis. So, the center is at (1, 0) in polar coordinates.Now, the circle passes through the point P(√2, π/4). Let me convert this point to rectangular coordinates to find its distance from the center. The distance will be the radius of the circle.To convert P(√2, π/4) to rectangular coordinates:x = ρ cos θ = √2 cos(π/4)y = ρ sin θ = √2 sin(π/4)I know that cos(π/4) and sin(π/4) are both √2/2. So:x = √2 * (√2/2) = (2)/2 = 1y = √2 * (√2/2) = 1So, point P is at (1, 1) in rectangular coordinates.Now, the center of the circle is at (1, 0), and the point P is at (1, 1). So, the distance between these two points is the radius.Calculating the distance between (1, 0) and (1, 1):Distance = √[(1 - 1)^2 + (1 - 0)^2] = √[0 + 1] = √1 = 1So, the radius of the circle is 1.Now, knowing the center and the radius, I can write the equation of the circle in rectangular coordinates. The general equation is:(x - h)^2 + (y - k)^2 = r^2Where (h, k) is the center and r is the radius. Substituting the values:(x - 1)^2 + (y - 0)^2 = 1^2Simplifying:(x - 1)^2 + y^2 = 1Now, I need to convert this equation back into polar coordinates. Let me recall that:x = ρ cos θy = ρ sin θSubstituting these into the equation:(ρ cos θ - 1)^2 + (ρ sin θ)^2 = 1Let me expand this:(ρ cos θ - 1)^2 + (ρ sin θ)^2 = 1Expanding the first term:(ρ^2 cos^2 θ - 2ρ cos θ + 1) + ρ^2 sin^2 θ = 1Combine like terms:ρ^2 cos^2 θ + ρ^2 sin^2 θ - 2ρ cos θ + 1 = 1Factor out ρ^2 from the first two terms:ρ^2 (cos^2 θ + sin^2 θ) - 2ρ cos θ + 1 = 1I know that cos^2 θ + sin^2 θ = 1, so:ρ^2 (1) - 2ρ cos θ + 1 = 1Simplify:ρ^2 - 2ρ cos θ + 1 = 1Subtract 1 from both sides:ρ^2 - 2ρ cos θ = 0Factor out ρ:ρ (ρ - 2 cos θ) = 0So, either ρ = 0 or ρ = 2 cos θBut ρ = 0 is just the origin, which is a single point, not the entire circle. So, the equation of the circle is ρ = 2 cos θ.Let me double-check this result. If I plug in θ = 0, then ρ = 2 cos 0 = 2, which would be the point (2, 0) in polar coordinates. But our center is at (1, 0) with radius 1, so the point (2, 0) is on the circle, which makes sense because it's one radius away from the center. Similarly, if θ = π/2, ρ = 2 cos(π/2) = 0, which is the origin, but that's not on our circle. Wait, that seems conflicting.Wait, no. Actually, when θ = π/2, ρ = 0, which is the origin, but our circle is centered at (1, 0) with radius 1, so the origin is inside the circle, not on it. So, maybe I made a mistake in the conversion.Wait, let me think again. The equation in rectangular coordinates is (x - 1)^2 + y^2 = 1. So, when θ = π/2, which is the positive y-axis, plugging into polar coordinates, x = 0, y = ρ. So, substituting into the rectangular equation:(0 - 1)^2 + (ρ)^2 = 11 + ρ^2 = 1ρ^2 = 0ρ = 0So, that's correct, the origin is the only point on the circle along the y-axis. But in polar coordinates, θ = π/2, ρ = 0 is just the origin, so that's consistent.Another check: when θ = π, ρ = 2 cos π = -2. But in polar coordinates, negative ρ means we go in the opposite direction, so θ = π, ρ = -2 would be the same as θ = 0, ρ = 2, which is the point (2, 0). But our circle is centered at (1, 0) with radius 1, so (2, 0) is on the circle, which is correct.Wait, but θ = π, ρ = -2 is equivalent to θ = 0, ρ = 2, which is on the circle. So, that seems okay.Another point: θ = π/4. Let's see what ρ is. ρ = 2 cos(π/4) = 2*(√2/2) = √2. So, the point (√2, π/4) is on the circle, which matches the given point P. So, that's correct.Therefore, the polar equation is ρ = 2 cos θ.I think that makes sense. Let me recap:1. Convert the given line equation to rectangular coordinates.2. Find its intersection with the polar axis (x-axis) to get the center.3. Convert the given point P to rectangular coordinates.4. Calculate the distance between the center and P to find the radius.5. Write the equation of the circle in rectangular coordinates.6. Convert that equation back to polar coordinates.7. Verify the result with specific points.Everything seems to check out. So, the polar equation of the circle is ρ = 2 cos θ.

answer：First, identify the smallest and largest three-digit numbers, which are 100 and 999, respectively.Next, determine the smallest multiple of 13 that is greater than or equal to 100. This is found by dividing 100 by 13 and rounding up to the nearest whole number, resulting in 8. Therefore, the smallest multiple is 13 multiplied by 8, which equals 104.Then, find the largest multiple of 13 that is less than or equal to 999. This is done by dividing 999 by 13 and rounding down to the nearest whole number, giving 76. Thus, the largest multiple is 13 multiplied by 76, which equals 988.Finally, calculate the total number of three-digit multiples of 13 by subtracting the smallest multiple index from the largest multiple index and adding 1. This results in 76 minus 8 plus 1, which equals 69.

answer：Okay, so I have this problem here about finding the equation of a tangent line to the curve ( C ): ( f(x) = x^3 + 1 ) that's perpendicular to the line ( y = -frac{1}{3}x - 4 ). The options are given as A, B, C, D, and I need to figure out which one is correct.First, I remember that if two lines are perpendicular, their slopes multiply to -1. So, the given line has a slope of ( -frac{1}{3} ). Therefore, the slope of the tangent line we're looking for should be the negative reciprocal of ( -frac{1}{3} ), which is 3. That makes sense because ( -frac{1}{3} times 3 = -1 ), confirming they're perpendicular.Next, I need to find the point on the curve ( f(x) = x^3 + 1 ) where the tangent line has a slope of 3. To do this, I should find the derivative of ( f(x) ) because the derivative gives the slope of the tangent line at any point ( x ).Calculating the derivative: ( f'(x) = 3x^2 ). So, the slope of the tangent at any point ( x ) is ( 3x^2 ). We want this slope to be 3, so I set up the equation:( 3x^2 = 3 )Dividing both sides by 3 gives:( x^2 = 1 )Taking the square root of both sides, I find that ( x = 1 ) or ( x = -1 ). So, there are two points on the curve where the tangent has a slope of 3.Now, I need to find the corresponding ( y )-coordinates for these ( x )-values on the curve ( f(x) = x^3 + 1 ).For ( x = 1 ):( f(1) = (1)^3 + 1 = 1 + 1 = 2 )So, the point is ( (1, 2) ).For ( x = -1 ):( f(-1) = (-1)^3 + 1 = -1 + 1 = 0 )So, the point is ( (-1, 0) ).Now, I can write the equations of the tangent lines at these two points using the point-slope form of a line, which is ( y - y_1 = m(x - x_1) ), where ( m ) is the slope and ( (x_1, y_1) ) is the point.First tangent line at ( (1, 2) ):( y - 2 = 3(x - 1) )Simplifying:( y - 2 = 3x - 3 )( y = 3x - 1 )To write it in standard form ( Ax + By + C = 0 ):( 3x - y - 1 = 0 )Second tangent line at ( (-1, 0) ):( y - 0 = 3(x - (-1)) )Simplifying:( y = 3(x + 1) )( y = 3x + 3 )In standard form:( 3x - y + 3 = 0 )So, the two possible equations for the tangent lines are ( 3x - y - 1 = 0 ) and ( 3x - y + 3 = 0 ). Looking back at the options, option C is ( 3x - y - 1 = 0 ) or ( 3x - y + 3 = 0 ), which matches my results.Wait, but let me double-check to make sure I didn't make any mistakes. For the point ( (-1, 0) ), plugging back into the tangent line equation ( y = 3x + 3 ), when ( x = -1 ), ( y = 3(-1) + 3 = -3 + 3 = 0 ), which is correct. And for ( x = 1 ), ( y = 3(1) - 1 = 2 ), which also checks out.I think I did everything correctly. So, the answer should be option C.**Final Answer**boxed{C}