💡Okay, so I have this problem here: A line that goes through the points (2, 16) and (-8, -4) intersects the y-axis at what point? Hmm, I need to find where this line crosses the y-axis. I remember that the y-axis is where x equals zero, so I need to find the point (0, y) that lies on this line.First, I think I need to find the equation of the line that passes through these two points. To do that, I should probably start by finding the slope of the line. The slope formula is (y2 - y1) divided by (x2 - x1). Let me label the points to make it clearer. Let's say (x1, y1) is (2, 16) and (x2, y2) is (-8, -4).So, plugging into the slope formula: slope m = (-4 - 16) / (-8 - 2). Let me calculate the numerator first: -4 minus 16 is -20. Then the denominator: -8 minus 2 is -10. So, the slope m is (-20)/(-10), which simplifies to 2. Okay, so the slope of the line is 2. That means for every unit I move to the right on the x-axis, the y-value increases by 2.Now that I have the slope, I can use the point-slope form of a line to find the equation. The point-slope formula is y - y1 = m(x - x1). I can use either of the two points given. Maybe I'll use (2, 16) because the numbers seem smaller and easier to work with.Plugging into the formula: y - 16 = 2(x - 2). Let me simplify this. Distribute the 2 on the right side: y - 16 = 2x - 4. Now, add 16 to both sides to solve for y: y = 2x - 4 + 16, which simplifies to y = 2x + 12.So, the equation of the line is y = 2x + 12. Now, to find where this line intersects the y-axis, I need to find the value of y when x is 0. That's because any point on the y-axis has an x-coordinate of 0.Substituting x = 0 into the equation: y = 2(0) + 12, which simplifies to y = 0 + 12, so y = 12. Therefore, the line intersects the y-axis at the point (0, 12).Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting with the slope: (-4 - 16) is indeed -20, and (-8 - 2) is -10, so -20 divided by -10 is 2. That seems correct.Then, using point-slope: y - 16 = 2(x - 2). Distribute the 2: y - 16 = 2x - 4. Adding 16 to both sides: y = 2x + 12. Yep, that looks right.Plugging in x = 0: y = 2(0) + 12 = 12. So, the y-intercept is at (0, 12). I think that's correct. Just to be thorough, maybe I can check using the other point (-8, -4). If I plug x = -8 into the equation, y should be -4.Calculating y: y = 2(-8) + 12 = -16 + 12 = -4. Perfect, that matches the given point. So, my equation is correct, and the y-intercept is indeed at (0, 12).I guess another way to think about it is using the concept of slope. The slope is 2, which means for every increase of 1 in x, y increases by 2. Starting from (2, 16), if I move left 2 units to reach x = 0, y should decrease by 4 (since 2 units left times a slope of 2 equals a decrease of 4). So, 16 - 4 = 12, which confirms the y-intercept is at (0, 12).Alternatively, I could have used the other point (-8, -4) and moved right 8 units to x = 0. Since the slope is 2, moving right 8 units would increase y by 16 (8 times 2). So, starting from y = -4 and adding 16 gives y = 12. Again, that leads me to the same conclusion.So, after checking it multiple ways, I'm confident that the line intersects the y-axis at (0, 12).

💡Okay, so I have this problem where I need to find the distance between two points A and B where the curve ρ = -2 sinθ and the line ρ sinθ = -1 intersect. Hmm, polar coordinates can sometimes be tricky, but maybe converting them to Cartesian coordinates will make it easier for me to visualize and solve.First, let me recall how to convert polar equations to Cartesian. I remember that ρ is the radius, θ is the angle, and the conversion formulas are x = ρ cosθ and y = ρ sinθ. Also, ρ² = x² + y². So, maybe I can rewrite both equations in terms of x and y.Starting with the curve ρ = -2 sinθ. If I multiply both sides by ρ, I get ρ² = -2ρ sinθ. Since ρ² is x² + y² and ρ sinθ is y, substituting those in, the equation becomes x² + y² = -2y. Hmm, that looks like a circle equation. Let me rearrange it to standard form. If I add 2y to both sides, I get x² + y² + 2y = 0. To complete the square for the y terms, I can write it as x² + (y² + 2y + 1) = 1, which simplifies to x² + (y + 1)² = 1. So, this is a circle with center at (0, -1) and radius 1. Okay, that makes sense.Now, the line is given by ρ sinθ = -1. Using the conversion formula again, ρ sinθ is just y, so this equation becomes y = -1. That's a horizontal line in Cartesian coordinates, which is straightforward.So, now I have a circle centered at (0, -1) with radius 1 and a horizontal line y = -1. I need to find where they intersect. Since the circle is centered at (0, -1) and has a radius of 1, the line y = -1 passes through the center of the circle. Therefore, the line is tangent to the circle? Wait, no, because the radius is 1, and the line is exactly at the center, so actually, the line should intersect the circle at two points.Wait, let me think again. If the circle is centered at (0, -1) and has a radius of 1, then the distance from the center to the line y = -1 is zero because the center is on the line. So, the line passes through the center, meaning it should intersect the circle at two points, diametrically opposite each other. So, the points A and B should be on the circle, lying on the line y = -1, which is the horizontal line passing through the center.But wait, if the line passes through the center, then the points of intersection should be at the maximum x-values on the circle. Since the circle has a radius of 1, the x-values can range from -1 to 1. So, substituting y = -1 into the circle equation, let's see:The circle equation is x² + (y + 1)² = 1. If y = -1, then substituting gives x² + ( -1 + 1 )² = x² + 0 = 1. So, x² = 1, which means x = 1 or x = -1. Therefore, the points of intersection are (1, -1) and (-1, -1). So, points A and B are (1, -1) and (-1, -1).Now, to find the distance between A and B, I can use the distance formula. The distance between two points (x1, y1) and (x2, y2) is sqrt[(x2 - x1)² + (y2 - y1)²]. Plugging in the points, we have:Distance AB = sqrt[(-1 - 1)² + (-1 - (-1))²] = sqrt[(-2)² + (0)²] = sqrt[4 + 0] = sqrt[4] = 2.So, the distance between points A and B is 2 units.Wait, let me double-check my steps. I converted both equations to Cartesian coordinates correctly. The circle equation became x² + (y + 1)² = 1, which is correct. The line equation became y = -1, which is straightforward. Substituting y = -1 into the circle equation gives x² = 1, so x = ±1, which gives the points (1, -1) and (-1, -1). Calculating the distance between these two points, since their y-coordinates are the same, the distance is just the difference in x-coordinates, which is 2. So, yes, that seems correct.I think I've got it right. The distance between points A and B is 2.**Final Answer**The distance between points A and B is boxed{2}.

💡First, since the circle is inscribed in the rectangle, the diameter of the circle is equal to both the width of the rectangle and the height of the rectangle. Given that the radius of the circle is 3, the diameter is (2 times 3 = 6). Therefore, the width of the rectangle is 6.Next, we use the given ratio of the length to the width of the rectangle, which is 3:1. If the width is 6, then the length is (3 times 6 = 18).Finally, to find the area of the rectangle, we multiply the length by the width:[text{Area} = text{length} times text{width} = 18 times 6 = 108]

💡Alright, so I have this problem about a circle and a line intersecting it. Let me try to figure it out step by step. First, the circle is given by the equation ( x^2 + y^2 - 8x + 6 = 0 ). I remember that to find the center and radius of a circle from its general equation, I need to complete the square. Let me do that for the x-terms.Starting with the equation:[ x^2 - 8x + y^2 + 6 = 0 ]I'll group the x terms:[ (x^2 - 8x) + y^2 + 6 = 0 ]To complete the square for ( x^2 - 8x ), I take half of -8, which is -4, square it to get 16, and add and subtract it inside the equation:[ (x^2 - 8x + 16 - 16) + y^2 + 6 = 0 ][ (x - 4)^2 - 16 + y^2 + 6 = 0 ][ (x - 4)^2 + y^2 - 10 = 0 ][ (x - 4)^2 + y^2 = 10 ]So, the circle has center at (4, 0) and radius ( sqrt{10} ). Got that.Now, the line passes through point P(0, 2) with slope k. So, the equation of the line should be ( y = kx + 2 ). That makes sense because when x=0, y=2, which is point P.This line intersects the circle at two points A and B, and N is the midpoint of AB. **Part 1: Determine the range of values for k.**Hmm, so the line intersects the circle at two distinct points, which means the distance from the center of the circle to the line must be less than the radius. If the distance is equal to the radius, the line is tangent, and if it's greater, there are no intersections.The formula for the distance from a point (x0, y0) to the line ( ax + by + c = 0 ) is:[ d = frac{|ax0 + by0 + c|}{sqrt{a^2 + b^2}} ]First, let me write the equation of the line in standard form. The line is ( y = kx + 2 ), so rearranged:[ kx - y + 2 = 0 ]So, a = k, b = -1, c = 2.The center of the circle is (4, 0). Plugging into the distance formula:[ d = frac{|k*4 + (-1)*0 + 2|}{sqrt{k^2 + (-1)^2}} ][ d = frac{|4k + 2|}{sqrt{k^2 + 1}} ]This distance must be less than the radius ( sqrt{10} ):[ frac{|4k + 2|}{sqrt{k^2 + 1}} < sqrt{10} ]To eliminate the square roots, I'll square both sides:[ frac{(4k + 2)^2}{k^2 + 1} < 10 ][ (4k + 2)^2 < 10(k^2 + 1) ]Expanding the numerator:[ 16k^2 + 16k + 4 < 10k^2 + 10 ]Subtracting ( 10k^2 + 10 ) from both sides:[ 6k^2 + 16k - 6 < 0 ]This is a quadratic inequality. Let me find the roots to determine the intervals where the inequality holds.Quadratic equation: ( 6k^2 + 16k - 6 = 0 )Using the quadratic formula:[ k = frac{-16 pm sqrt{16^2 - 4*6*(-6)}}{2*6} ][ k = frac{-16 pm sqrt{256 + 144}}{12} ][ k = frac{-16 pm sqrt{400}}{12} ][ k = frac{-16 pm 20}{12} ]So, two solutions:1. ( k = frac{-16 + 20}{12} = frac{4}{12} = frac{1}{3} )2. ( k = frac{-16 - 20}{12} = frac{-36}{12} = -3 )So, the quadratic crosses zero at k = -3 and k = 1/3. Since the coefficient of ( k^2 ) is positive, the parabola opens upwards. Therefore, the inequality ( 6k^2 + 16k - 6 < 0 ) holds between the roots.Thus, the range of k is:[ -3 < k < frac{1}{3} ]**Part 2: If ON is parallel to MP, calculate the value of k.**Hmm, ON is the line from the origin O(0,0) to N, the midpoint of AB. MP is the line from M(4,0) to P(0,2). So, if ON is parallel to MP, their slopes must be equal.First, let's find the slope of MP.Points M(4,0) and P(0,2). The slope is:[ m_{MP} = frac{2 - 0}{0 - 4} = frac{2}{-4} = -frac{1}{2} ]So, the slope of ON must also be -1/2.Let me denote point N as (h, l). Since N is the midpoint of AB, and it lies on the line y = kx + 2, so:[ l = kh + 2 ]Also, since ON has slope -1/2:[ frac{l - 0}{h - 0} = -frac{1}{2} ][ frac{l}{h} = -frac{1}{2} ][ l = -frac{1}{2}h ]So, we have two expressions for l:1. ( l = kh + 2 )2. ( l = -frac{1}{2}h )Setting them equal:[ kh + 2 = -frac{1}{2}h ][ kh + frac{1}{2}h = -2 ][ h(k + frac{1}{2}) = -2 ][ h = frac{-2}{k + frac{1}{2}} ][ h = frac{-4}{2k + 1} ]So, h is expressed in terms of k. Then, l is:[ l = -frac{1}{2}h = -frac{1}{2} left( frac{-4}{2k + 1} right) = frac{2}{2k + 1} ]So, point N is ( left( frac{-4}{2k + 1}, frac{2}{2k + 1} right) ).Now, since N is the midpoint of AB, and AB is the chord of the circle intersected by the line y = kx + 2, there's a property that the line from the center M to the midpoint N is perpendicular to the chord AB.So, the line MN is perpendicular to AB. The slope of AB is k, so the slope of MN should be -1/k.Let me compute the slope of MN.Points M(4,0) and N( ( frac{-4}{2k + 1}, frac{2}{2k + 1} ) ).Slope of MN:[ m_{MN} = frac{frac{2}{2k + 1} - 0}{frac{-4}{2k + 1} - 4} ]Simplify denominator:[ frac{-4}{2k + 1} - 4 = frac{-4 - 4(2k + 1)}{2k + 1} = frac{-4 - 8k -4}{2k + 1} = frac{-8k -8}{2k + 1} = frac{-8(k + 1)}{2k + 1} ]So, slope:[ m_{MN} = frac{frac{2}{2k + 1}}{frac{-8(k + 1)}{2k + 1}} = frac{2}{-8(k + 1)} = frac{-1}{4(k + 1)} ]But we know that slope of MN should be -1/k because it's perpendicular to AB (slope k). Therefore:[ frac{-1}{4(k + 1)} = frac{-1}{k} ]Simplify:[ frac{1}{4(k + 1)} = frac{1}{k} ]Cross-multiplying:[ k = 4(k + 1) ][ k = 4k + 4 ][ -3k = 4 ][ k = -frac{4}{3} ]Wait, let me check my steps to make sure I didn't make a mistake.First, the slope of MN was calculated as ( frac{-1}{4(k + 1)} ). Since MN is perpendicular to AB, which has slope k, the product of their slopes should be -1. So:[ m_{MN} * k = -1 ][ left( frac{-1}{4(k + 1)} right) * k = -1 ][ frac{-k}{4(k + 1)} = -1 ]Multiply both sides by 4(k + 1):[ -k = -4(k + 1) ][ -k = -4k -4 ][ 3k = -4 ][ k = -frac{4}{3} ]Yes, that's consistent. So, k is -4/3.But wait, in part 1, the range of k was between -3 and 1/3. But -4/3 is approximately -1.333, which is within -3 and 1/3. So, it's valid.Therefore, the value of k is -4/3.**Final Answer**(1) The range of ( k ) is ( boxed{(-3, dfrac{1}{3})} ).(2) The value of ( k ) is ( boxed{-dfrac{4}{3}} ).

💡First, I need to find the points where the lines intersect each other to determine the vertices of the triangle.1. **Intersection of ( y = 2x ) and ( y = 4 ):** [ 2x = 4 implies x = 2 implies text{Point } A(2,4) ]2. **Intersection of ( y = -2x ) and ( y = 4 ):** [ -2x = 4 implies x = -2 implies text{Point } B(-2,4) ]3. **Intersection of ( y = 2x ) and ( y = -2x ):** [ 2x = -2x implies 4x = 0 implies x = 0 implies y = 0 implies text{Point } O(0,0) ]Now, I have the three vertices of the triangle: ( A(2,4) ), ( B(-2,4) ), and ( O(0,0) ).Next, I'll calculate the length of the base of the triangle. The base can be the distance between points ( A ) and ( B ):[AB = 2 - (-2) = 4]The height of the triangle is the vertical distance from the base ( AB ) to the origin ( O ), which is 4 units.Finally, I'll calculate the area of the triangle using the formula:[text{Area} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 4 times 4 = 8]

💡Okay, so I need to figure out how many positive divisors of 50! are prime. Hmm, let me start by recalling what 50! means. 50! is 50 factorial, which is the product of all positive integers from 1 to 50. So, 50! = 1 × 2 × 3 × ... × 50. Now, the question is about the prime divisors of 50!. I remember that a prime divisor is a prime number that can divide another number without leaving a remainder. So, I need to find all the prime numbers that are factors of 50!.Wait, but 50! is a huge number, so directly factoring it isn't practical. Maybe there's a smarter way. I think that any prime number less than or equal to 50 will be a factor of 50! because 50! includes all numbers up to 50. So, if I can list all the prime numbers up to 50, those will be the prime divisors of 50!.Let me try to list the prime numbers less than or equal to 50. Starting from the smallest prime:2 is prime.3 is prime.4 is not prime (divisible by 2).5 is prime.6 is not prime (divisible by 2 and 3).7 is prime.8 is not prime.9 is not prime (divisible by 3).10 is not prime.11 is prime.12 is not prime.13 is prime.14 is not prime.15 is not prime.16 is not prime.17 is prime.18 is not prime.19 is prime.20 is not prime.21 is not prime.22 is not prime.23 is prime.24 is not prime.25 is not prime (divisible by 5).26 is not prime.27 is not prime.28 is not prime.29 is prime.30 is not prime.31 is prime.32 is not prime.33 is not prime.34 is not prime.35 is not prime.36 is not prime.37 is prime.38 is not prime.39 is not prime.40 is not prime.41 is prime.42 is not prime.43 is prime.44 is not prime.45 is not prime.46 is not prime.47 is prime.48 is not prime.49 is not prime (divisible by 7).50 is not prime.Let me count how many primes I listed:2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.That looks like 15 primes in total. So, there are 15 prime numbers less than or equal to 50, which means 50! has 15 prime divisors.Wait, let me double-check to make sure I didn't miss any or count incorrectly. I'll list them again:1. 22. 33. 54. 75. 116. 137. 178. 199. 2310. 2911. 3112. 3713. 4114. 4315. 47Yes, that's 15 primes. So, each of these primes divides 50! exactly, and there are no other primes beyond 50 that can divide 50! because 50! doesn't include any numbers larger than 50, and primes larger than 50 wouldn't be factors of 50!.Therefore, the number of positive divisors of 50! that are prime is 15.