💡First, I need to determine the distance the man covers while walking. He walks at a speed of 5 km/h for 5 hours. By multiplying the speed by the time, I can find the total distance.Next, with the distance known, I can calculate the time it would take for the man to cover the same distance if he runs at a speed of 15 km/h. This involves dividing the distance by the new speed.Finally, I will present the calculated time in a clear and concise manner.

💡Okay, so I have this problem here where I need to find r(r(θ)) given that r(θ) is equal to 2 divided by (1 minus θ). Hmm, let me try to figure this out step by step.First, I know that r(θ) is a function that takes θ as an input and gives me 2/(1 - θ) as the output. So, if I want to find r(r(θ)), that means I need to plug the result of r(θ) back into the function r again. In other words, I need to compute r of r(θ), which is r applied to the output of r(θ).Let me write that down. So, r(r(θ)) is equal to r evaluated at r(θ). Since r(θ) is 2/(1 - θ), that means I need to substitute 2/(1 - θ) into the function r wherever there's a θ.So, r(r(θ)) = r(2/(1 - θ)). Now, let's recall what r is. It's defined as r(x) = 2/(1 - x), right? So, if I plug x = 2/(1 - θ) into this function, I get:r(2/(1 - θ)) = 2 / [1 - (2/(1 - θ))]Okay, so now I have this expression: 2 divided by [1 minus (2 divided by (1 - θ))]. Let me simplify the denominator first because that seems a bit complicated.The denominator is 1 - (2/(1 - θ)). To combine these terms, I need a common denominator. The first term is 1, which can be written as (1 - θ)/(1 - θ) to have the same denominator as the second term. So, rewriting 1 as (1 - θ)/(1 - θ), the denominator becomes:(1 - θ)/(1 - θ) - 2/(1 - θ) = [ (1 - θ) - 2 ] / (1 - θ)Let me compute the numerator of this fraction: (1 - θ) - 2. That simplifies to 1 - θ - 2, which is (-1 - θ). So, the denominator of the entire expression is (-1 - θ)/(1 - θ).Now, going back to the entire expression for r(r(θ)), which is 2 divided by [ (-1 - θ)/(1 - θ) ]. Dividing by a fraction is the same as multiplying by its reciprocal, so this becomes:2 * [ (1 - θ)/(-1 - θ) ]Let me write that out: 2 times (1 - θ) divided by (-1 - θ). I can factor out a negative sign from the denominator to make it look nicer. So, (-1 - θ) is the same as -(1 + θ). Therefore, the expression becomes:2 * (1 - θ) / [ -(1 + θ) ] = -2 * (1 - θ) / (1 + θ)Hmm, I can also factor out a negative sign from the numerator if I want. Let's see, (1 - θ) is the same as -(θ - 1). So, substituting that in, we get:-2 * [ -(θ - 1) ] / (1 + θ) = 2 * (θ - 1) / (1 + θ)So, simplifying all that, r(r(θ)) equals 2 times (θ - 1) divided by (1 + θ). Let me just double-check my steps to make sure I didn't make a mistake.Starting from r(r(θ)) = r(2/(1 - θ)) = 2 / [1 - (2/(1 - θ))]. Then, simplifying the denominator:1 - (2/(1 - θ)) = (1 - θ - 2)/(1 - θ) = (-1 - θ)/(1 - θ). So, then, 2 divided by that is 2 * (1 - θ)/(-1 - θ) = -2*(1 - θ)/(1 + θ). Then, recognizing that (1 - θ) is -(θ - 1), so it becomes 2*(θ - 1)/(1 + θ).Yes, that seems correct. So, the final expression for r(r(θ)) is 2*(θ - 1)/(1 + θ). I think that's the simplest form it can take. Let me just write it neatly:r(r(θ)) = (2(θ - 1))/(1 + θ)Yeah, that looks good. I don't think I made any calculation errors. Each step follows logically from the previous one, and I double-checked the algebra. So, I'm confident that this is the correct result.**Final Answer**The value of ( r(r(theta)) ) is boxed{dfrac{2(theta - 1)}{1 + theta}}.

💡Okay, so I have this quadratic equation: ( x^2 - (m + 3)x + m + 1 = 0 ). There are two parts to the problem. First, I need to prove that no matter what value ( m ) takes, the equation always has two distinct real roots. Second, if one of the roots is 4 and the two roots are the lengths of the two sides of an isosceles triangle, I have to find the perimeter of that triangle.Starting with part (1). I remember that for a quadratic equation ( ax^2 + bx + c = 0 ), the discriminant ( D = b^2 - 4ac ) determines the nature of the roots. If ( D > 0 ), there are two distinct real roots; if ( D = 0 ), there's exactly one real root; and if ( D < 0 ), the roots are complex.So, for the given equation, let me identify ( a ), ( b ), and ( c ). Here, ( a = 1 ), ( b = -(m + 3) ), and ( c = m + 1 ). Plugging these into the discriminant formula:( D = [-(m + 3)]^2 - 4 * 1 * (m + 1) )Simplify that:( D = (m + 3)^2 - 4(m + 1) )Expanding ( (m + 3)^2 ):( D = m^2 + 6m + 9 - 4m - 4 )Combine like terms:( D = m^2 + 2m + 5 )Hmm, so ( D = m^2 + 2m + 5 ). I need to show that this is always positive for any real ( m ). Let me see if this quadratic in ( m ) ever equals zero or becomes negative.Looking at ( m^2 + 2m + 5 ), I can check its discriminant as well to see if it has real roots. The discriminant ( D' = (2)^2 - 4 * 1 * 5 = 4 - 20 = -16 ). Since ( D' < 0 ), the quadratic ( m^2 + 2m + 5 ) never crosses the x-axis and is always positive because the coefficient of ( m^2 ) is positive.Therefore, ( D = m^2 + 2m + 5 > 0 ) for all real ( m ), which means the original equation always has two distinct real roots, regardless of the value of ( m ). That takes care of part (1).Moving on to part (2). It says that one root is 4, and the two roots are the lengths of the two sides of an isosceles triangle. I need to find the perimeter.First, since one root is 4, I can use this information to find the value of ( m ). Let me substitute ( x = 4 ) into the equation:( (4)^2 - (m + 3)(4) + m + 1 = 0 )Calculating each term:( 16 - 4(m + 3) + m + 1 = 0 )Expanding the middle term:( 16 - 4m - 12 + m + 1 = 0 )Combine like terms:( (16 - 12 + 1) + (-4m + m) = 0 )Simplify:( 5 - 3m = 0 )Solving for ( m ):( -3m = -5 )( m = frac{5}{3} )So, ( m = frac{5}{3} ). Now, I can substitute this back into the original equation to find the other root.The original equation is ( x^2 - (m + 3)x + m + 1 = 0 ). Plugging ( m = frac{5}{3} ):( x^2 - left( frac{5}{3} + 3 right)x + frac{5}{3} + 1 = 0 )Simplify the coefficients:( frac{5}{3} + 3 = frac{5}{3} + frac{9}{3} = frac{14}{3} )( frac{5}{3} + 1 = frac{5}{3} + frac{3}{3} = frac{8}{3} )So, the equation becomes:( x^2 - frac{14}{3}x + frac{8}{3} = 0 )To make it easier, I can multiply the entire equation by 3 to eliminate the denominators:( 3x^2 - 14x + 8 = 0 )Now, I need to solve this quadratic equation. Let me try factoring it.Looking for two numbers that multiply to ( 3 * 8 = 24 ) and add up to -14. Hmm, factors of 24: 1 & 24, 2 & 12, 3 & 8, 4 & 6.Looking for a pair that adds to 14. 12 and 2: 12 + 2 = 14. But since the middle term is -14x, both numbers should be negative. So, -12 and -2.So, split the middle term:( 3x^2 - 12x - 2x + 8 = 0 )Group the terms:( (3x^2 - 12x) + (-2x + 8) = 0 )Factor out common terms:( 3x(x - 4) - 2(x - 4) = 0 )Now, factor out ( (x - 4) ):( (3x - 2)(x - 4) = 0 )So, the roots are ( x = frac{2}{3} ) and ( x = 4 ).Wait, so the two roots are 4 and ( frac{2}{3} ). But the problem says that these are the lengths of the two sides of an isosceles triangle. An isosceles triangle has two sides equal. So, does that mean both roots are the lengths of the sides, implying that the triangle has sides of length 4 and ( frac{2}{3} )?But in an isosceles triangle, two sides must be equal. So, if one root is 4, the other root is ( frac{2}{3} ). So, does that mean the triangle has two sides of length 4 and one side of length ( frac{2}{3} ), or two sides of length ( frac{2}{3} ) and one side of length 4?Wait, but in a triangle, the sum of any two sides must be greater than the third side. So, if we have sides 4, 4, and ( frac{2}{3} ), let's check the triangle inequality:4 + 4 > ( frac{2}{3} ): 8 > ( frac{2}{3} ) – true.4 + ( frac{2}{3} ) > 4: ( frac{14}{3} ) > 4, which is approximately 4.666 > 4 – true.4 + ( frac{2}{3} ) > 4: same as above.So, that works.Alternatively, if we have sides ( frac{2}{3} ), ( frac{2}{3} ), and 4, let's check:( frac{2}{3} + frac{2}{3} = frac{4}{3} ), which is approximately 1.333. Is 1.333 > 4? No, 1.333 < 4. So, that doesn't satisfy the triangle inequality. Therefore, the only possibility is that the triangle has two sides of length 4 and one side of length ( frac{2}{3} ).Therefore, the sides are 4, 4, and ( frac{2}{3} ). The perimeter is the sum of all sides:Perimeter ( P = 4 + 4 + frac{2}{3} = 8 + frac{2}{3} = frac{24}{3} + frac{2}{3} = frac{26}{3} ).So, the perimeter is ( frac{26}{3} ).Wait, but let me double-check my calculations. When I substituted ( x = 4 ) into the equation, I got ( m = frac{5}{3} ). Then, plugging back into the equation, I got the quadratic ( 3x^2 - 14x + 8 = 0 ), which factors to ( (3x - 2)(x - 4) = 0 ), giving roots ( x = frac{2}{3} ) and ( x = 4 ). That seems correct.And then, considering the triangle, since two sides must be equal, and the sides are 4 and ( frac{2}{3} ), the only valid configuration is two sides of 4 and one side of ( frac{2}{3} ), because the other way doesn't satisfy the triangle inequality.Therefore, the perimeter is indeed ( frac{26}{3} ).I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**The perimeter of the isosceles triangle is boxed{dfrac{26}{3}}.

💡Alright, so I need to figure out how many 5-primable positive integers there are that are less than 1000. Let me start by understanding what a 5-primable number is. From the problem statement, a positive number is called n-primable if it's divisible by n and each of its digits is a one-digit prime number. First, I should recall what the one-digit prime numbers are. They are 2, 3, 5, and 7. So, any number that's 5-primable must be divisible by 5 and can only consist of these digits: 2, 3, 5, or 7.Now, since the number has to be divisible by 5, I remember that a number is divisible by 5 if its last digit is either 0 or 5. But wait, 0 isn't a one-digit prime number, so the last digit can't be 0. That means the last digit must be 5. So, every 5-primable number must end with a 5.Alright, so now I know that all 5-primable numbers less than 1000 must end with 5 and can only have digits 2, 3, 5, or 7 in the other places. Also, since we're looking for numbers less than 1000, we're dealing with one-digit, two-digit, and three-digit numbers.Let me break this down by the number of digits:1. **One-digit numbers**: The only one-digit number that ends with 5 is 5 itself. So, 5 is definitely a 5-primable number.2. **Two-digit numbers**: Here, the first digit can be any of the one-digit primes (2, 3, 5, 7), and the second digit must be 5. So, the possible two-digit numbers are 25, 35, 55, and 75. That's four numbers.3. **Three-digit numbers**: For three-digit numbers, the first two digits can be any combination of 2, 3, 5, or 7, and the last digit must be 5. So, for the first digit, there are 4 choices, and for the second digit, there are also 4 choices. That means there are 4 x 4 = 16 possible three-digit numbers. Examples include 225, 235, 255, 275, 325, 335, 355, 375, 525, 535, 555, 575, 725, 735, 755, and 775.Now, let me count them all up:- One-digit: 1 number (5)- Two-digit: 4 numbers (25, 35, 55, 75)- Three-digit: 16 numbersAdding them together: 1 + 4 + 16 = 21.Wait, let me double-check to make sure I didn't miss anything. For the three-digit numbers, I considered all combinations of the first two digits being 2, 3, 5, or 7, and then appending a 5 at the end. That seems correct because each combination is unique and meets the criteria of being divisible by 5 and having only one-digit primes as digits.Is there any overlap or duplication? I don't think so because each number is unique based on its digits. Also, all numbers are less than 1000, so there's no issue there.I think that's all. So, the total number of 5-primable numbers less than 1000 is 21.

💡Alright, so I have this problem with two parts, (I) and (II). Let me try to tackle part (I) first. The problem says:Given points E(-2,0) and F(2,0), and a moving point M on curve C such that the dot product of vectors EM and FM is -3. I need to find the equation of curve C.Hmm, okay. So, E is at (-2,0) and F is at (2,0). M is some point (x,y) on curve C. The vectors EM and FM are given by subtracting the coordinates of E and F from M, right?So, vector EM would be M - E, which is (x - (-2), y - 0) = (x + 2, y). Similarly, vector FM would be M - F, which is (x - 2, y - 0) = (x - 2, y).Wait, the problem says EM · FM = -3. So, the dot product of these two vectors is -3. Let me write that out:EM · FM = (x + 2)(x - 2) + (y)(y) = -3.Let me compute that:(x + 2)(x - 2) is x² - 4, and y*y is y². So, putting it together:x² - 4 + y² = -3.Simplify that:x² + y² - 4 = -3.So, x² + y² = 1.Wait, that's the equation of a circle with radius 1 centered at the origin. But hold on, E and F are at (-2,0) and (2,0), which are points on the x-axis, 2 units away from the origin. So, M is a point such that the dot product of EM and FM is -3. Hmm, but according to my calculation, the equation is x² + y² = 1.But let me double-check. Maybe I made a mistake in the vectors. The problem says EM and FM, but I interpreted them as vectors from E to M and from F to M. Is that correct?Wait, actually, in vector notation, EM would be from E to M, which is M - E, and FM would be from F to M, which is M - F. So, yes, that's correct.So, the dot product is (x + 2, y) · (x - 2, y) = (x + 2)(x - 2) + y² = x² - 4 + y².Set that equal to -3:x² + y² - 4 = -3 ⇒ x² + y² = 1.So, curve C is the circle centered at the origin with radius 1. That seems straightforward.Wait, but in the initial problem statement, it says "a moving point M on curve C such that EM · FM = -3." So, according to my calculation, that's x² + y² = 1.But let me think again. If E and F are fixed points, and M is moving such that the dot product of EM and FM is -3, then that condition defines a circle. So, yeah, that makes sense.So, part (I) answer is x² + y² = 1.Wait, but in the initial problem, it's given that M is on curve C, so curve C is x² + y² = 1. Okay, that seems right.Now, moving on to part (II). It says:Given a fixed point A(2,1), from a point P(a, b) outside the curve C, a tangent line PQ is drawn to the curve at point Q such that |PQ| = |PA|. If a circle with center at point P intersects with curve C, find the standard equation of the circle when its radius is at a minimum.Hmm, okay. So, point P is outside curve C, which is the unit circle centered at the origin. From P, we draw a tangent PQ to the curve C at point Q, and the length of PQ is equal to the length of PA, where A is (2,1). Then, we need to find the equation of the circle centered at P with the minimum radius that intersects curve C.Wait, so the circle centered at P intersects curve C, which is the unit circle. So, the circle centered at P must intersect the unit circle. The radius of this circle needs to be minimized.But also, from point P, we draw a tangent PQ to curve C, and |PQ| = |PA|. So, the length of the tangent from P to C is equal to the distance from P to A.Hmm, okay. Let me recall that the length of the tangent from a point P(a,b) to the unit circle x² + y² = 1 is given by sqrt(a² + b² - 1). Because the formula for the length of the tangent from (a,b) to the circle x² + y² = r² is sqrt(a² + b² - r²). Here, r = 1, so it's sqrt(a² + b² - 1).So, |PQ| = sqrt(a² + b² - 1). And |PA| is the distance from P(a,b) to A(2,1), which is sqrt[(a - 2)² + (b - 1)²].Given that |PQ| = |PA|, so:sqrt(a² + b² - 1) = sqrt[(a - 2)² + (b - 1)²].Let me square both sides to eliminate the square roots:a² + b² - 1 = (a - 2)² + (b - 1)².Expanding the right-hand side:(a - 2)² = a² - 4a + 4,(b - 1)² = b² - 2b + 1.So, adding them together:a² - 4a + 4 + b² - 2b + 1 = a² + b² - 4a - 2b + 5.So, the equation becomes:a² + b² - 1 = a² + b² - 4a - 2b + 5.Subtract a² + b² from both sides:-1 = -4a - 2b + 5.Bring all terms to one side:-1 - 5 = -4a - 2b,-6 = -4a - 2b,Multiply both sides by (-1):6 = 4a + 2b,Divide both sides by 2:3 = 2a + b.So, 2a + b = 3.So, point P(a,b) lies on the line 2a + b = 3.Okay, so P is on the line 2a + b = 3, and it's outside the unit circle C: x² + y² = 1.Now, we need to find the circle centered at P with the minimum radius such that it intersects curve C.So, the circle centered at P(a,b) with radius R intersects the unit circle x² + y² = 1. The condition for two circles to intersect is that the distance between their centers is less than or equal to the sum of their radii and greater than or equal to the absolute difference of their radii.In this case, the unit circle has center at (0,0) and radius 1, and the other circle has center at P(a,b) and radius R.So, the distance between centers is sqrt(a² + b²). For the circles to intersect, we must have:|R - 1| ≤ sqrt(a² + b²) ≤ R + 1.But since P is outside the unit circle, sqrt(a² + b²) > 1. So, the condition simplifies to:sqrt(a² + b²) ≤ R + 1,andsqrt(a² + b²) ≥ R - 1.But since sqrt(a² + b²) > 1, the second inequality implies that R - 1 ≤ sqrt(a² + b²). But since sqrt(a² + b²) > 1, R must be at least sqrt(a² + b²) - 1.But we need to minimize R. So, the minimal R is when R = sqrt(a² + b²) - 1.But wait, is that correct? Let me think.If we have two circles intersecting, the minimal radius R for the circle centered at P such that it intersects the unit circle would be when the circle is tangent to the unit circle. So, the minimal R would be sqrt(a² + b²) - 1, because the distance between centers is sqrt(a² + b²), and for tangency, R = distance - 1.But in our case, the circle must intersect the unit circle, not necessarily be tangent. So, the minimal R would be when the circle is tangent, because any smaller R would not intersect.Therefore, the minimal radius R is sqrt(a² + b²) - 1.But we need to express R in terms of a and b, but we also have the condition that 2a + b = 3.So, since 2a + b = 3, we can express b in terms of a: b = 3 - 2a.So, substituting into sqrt(a² + b²):sqrt(a² + (3 - 2a)²) = sqrt(a² + 9 - 12a + 4a²) = sqrt(5a² - 12a + 9).So, R = sqrt(5a² - 12a + 9) - 1.Now, to minimize R, we need to minimize sqrt(5a² - 12a + 9). Since sqrt is a monotonically increasing function, minimizing sqrt(5a² - 12a + 9) is equivalent to minimizing 5a² - 12a + 9.So, let's find the minimum of the quadratic function f(a) = 5a² - 12a + 9.The vertex of a quadratic function f(a) = pa² + qa + r is at a = -q/(2p).Here, p = 5, q = -12, so a = -(-12)/(2*5) = 12/10 = 6/5 = 1.2.So, the minimum occurs at a = 6/5.Then, b = 3 - 2a = 3 - 2*(6/5) = 3 - 12/5 = 15/5 - 12/5 = 3/5.So, point P is at (6/5, 3/5).Now, compute sqrt(5a² - 12a + 9) at a = 6/5:First, compute 5a²: 5*(36/25) = 180/25 = 36/5.Then, -12a: -12*(6/5) = -72/5.Add 9: 9 = 45/5.So, total: 36/5 - 72/5 + 45/5 = (36 - 72 + 45)/5 = (9)/5.So, sqrt(9/5) = 3/sqrt(5) = (3√5)/5.Therefore, R = (3√5)/5 - 1.So, the minimal radius is (3√5)/5 - 1.Therefore, the equation of the circle centered at P(6/5, 3/5) with radius (3√5)/5 - 1 is:(x - 6/5)² + (y - 3/5)² = [(3√5)/5 - 1]².Simplify the right-hand side:[(3√5)/5 - 1]² = (3√5/5)² - 2*(3√5/5)*1 + 1² = (9*5)/25 - (6√5)/5 + 1 = 45/25 - (6√5)/5 + 1 = 9/5 - (6√5)/5 + 1 = (9/5 + 5/5) - (6√5)/5 = 14/5 - (6√5)/5.But actually, we can just leave it as [(3√5)/5 - 1]² for the equation.So, putting it all together, the standard equation is:(x - 6/5)² + (y - 3/5)² = [(3√5)/5 - 1]².I think that's the answer for part (II).Wait, let me double-check my steps.1. Found that curve C is x² + y² = 1.2. For part (II), set up the condition |PQ| = |PA|, which led to 2a + b = 3.3. Expressed R in terms of a and b, then substituted b = 3 - 2a.4. Found that R = sqrt(5a² - 12a + 9) - 1.5. Minimized 5a² - 12a + 9 by finding vertex at a = 6/5, then b = 3/5.6. Computed sqrt(5*(6/5)^2 - 12*(6/5) + 9) = 3√5/5.7. So, R = 3√5/5 - 1.8. Therefore, the circle equation is (x - 6/5)² + (y - 3/5)² = (3√5/5 - 1)².Yes, that seems correct.**Final Answer**(I) The equation of curve ( C ) is (boxed{x^2 + y^2 = 1}).(II) The standard equation of the circle with the minimum radius is (boxed{left(x - frac{6}{5}right)^2 + left(y - frac{3}{5}right)^2 = left(frac{3sqrt{5}}{5} - 1right)^2}).

💡Alright, so I've got this quadratic equation: x squared minus (m plus 3)x plus m plus 2 equals zero. And I'm told that one of the roots is the square of 9. Hmm, okay. Let me try to figure out what that means and how to find the value of m.First off, the equation is x² - (m + 3)x + (m + 2) = 0. Quadratic equations usually have two roots, right? And these roots can be found using the quadratic formula, but maybe there's a simpler way here since I know one of the roots.Wait, the problem says one of the roots is the square of 9. So, the square of 9 is 81. That means x = 81 is one of the solutions to this equation. Okay, so if I plug x = 81 into the equation, it should satisfy it. Let me try that.Plugging in x = 81: (81)² - (m + 3)(81) + (m + 2) = 0.Calculating (81)²: 81 times 81 is 6561. So, 6561 minus (m + 3) times 81 plus (m + 2) equals zero.Now, let's expand the middle term: (m + 3) times 81 is 81m + 243. So, substituting back in, the equation becomes: 6561 - (81m + 243) + (m + 2) = 0.Let's distribute the negative sign: 6561 - 81m - 243 + m + 2 = 0.Now, combine like terms. First, the constants: 6561 - 243 is 6318, and then 6318 + 2 is 6320. So, the constants add up to 6320.Next, the m terms: -81m + m is -80m.So, putting it all together: 6320 - 80m = 0.Now, to solve for m, I can move 6320 to the other side: -80m = -6320.Dividing both sides by -80: m = (-6320)/(-80).The negatives cancel out, so m = 6320 / 80.Let's do that division: 6320 divided by 80. Well, 80 times 79 is 6320 because 80 times 70 is 5600, and 80 times 9 is 720, so 5600 + 720 is 6320.So, m equals 79.Wait, let me double-check that. If m is 79, then the original equation becomes x² - (79 + 3)x + (79 + 2) = x² - 82x + 81 = 0.Let's see if x = 81 is a root. Plugging in x = 81: (81)² - 82*81 + 81.Calculating each term: 81 squared is 6561, 82 times 81 is 6642, and then plus 81.So, 6561 - 6642 + 81.6561 - 6642 is -81, and -81 + 81 is 0. Perfect, that checks out.Just to be thorough, what's the other root? Since the product of the roots is c/a, which is 81/1 = 81. So, if one root is 81, the other root must be 81 divided by 81, which is 1.Let me verify that. If x = 1 is a root, plugging into the equation: 1² - 82*1 + 81 = 1 - 82 + 81 = 0. Yep, that works too.So, everything seems consistent. The value of m is indeed 79.