💡Alright, so I have this problem here about real numbers a and b where a + b equals 1, and it says that ab is less than or equal to 1/4. I need to figure out how many of the converse, inverse, and contrapositive of this statement are true. Hmm, okay, let's break this down step by step.First, let me make sure I understand the original proposition. It says that for any real numbers a and b, if a + b = 1, then ab ≤ 1/4. So, given that a and b add up to 1, their product can't be more than 1/4. That makes sense because I remember something about the maximum product of two numbers with a fixed sum. I think it's related to the arithmetic mean-geometric mean inequality (AM-GM). Let me recall that.The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. So, if I have two numbers a and b, then (a + b)/2 ≥ √(ab). In this case, since a + b = 1, plugging that in, we get 1/2 ≥ √(ab). Squaring both sides, we get 1/4 ≥ ab. So, that confirms that ab is indeed less than or equal to 1/4. Cool, so the original proposition is true.Now, I need to consider the converse, inverse, and contrapositive of this statement. Let me remember what each of these means.The converse of a statement "If P, then Q" is "If Q, then P." So, in this case, the converse would be: If ab ≤ 1/4, then a + b = 1. Hmm, is that necessarily true? Let me think. Suppose I have two numbers a and b such that their product is less than or equal to 1/4. Does that mean their sum has to be exactly 1? I don't think so. For example, take a = 0 and b = 0. Then ab = 0, which is less than 1/4, but a + b = 0, which is not 1. So, the converse is false.Next, the inverse of a statement "If P, then Q" is "If not P, then not Q." So, the inverse here would be: If a + b ≠ 1, then ab > 1/4. Is this true? Let me test it. Suppose a = 2 and b = 0. Then a + b = 2, which is not equal to 1, but ab = 0, which is not greater than 1/4. So, the inverse is also false.Lastly, the contrapositive of a statement "If P, then Q" is "If not Q, then not P." So, the contrapositive here would be: If ab > 1/4, then a + b ≠ 1. Is this true? Well, since the original statement is true, its contrapositive must also be true. Because in logic, a statement and its contrapositive are logically equivalent. So, if ab is greater than 1/4, then it can't be the case that a + b = 1. That makes sense because we already know that when a + b = 1, ab can't exceed 1/4. So, the contrapositive is true.To summarize:- Original proposition: True- Converse: False- Inverse: False- Contrapositive: TrueBut wait, the question specifically asks about the converse, inverse, and contrapositive. So, among these three, how many are true? The contrapositive is true, while the converse and inverse are false. So, only one of them is true.I should double-check my reasoning to make sure I didn't make any mistakes. Let's go through each one again.**Converse:** If ab ≤ 1/4, then a + b = 1. I provided a counterexample where a = 0 and b = 0, which satisfies ab ≤ 1/4 but a + b ≠ 1. So, the converse is indeed false.**Inverse:** If a + b ≠ 1, then ab > 1/4. I gave a counterexample with a = 2 and b = 0, where a + b ≠ 1 but ab = 0, which is not greater than 1/4. So, the inverse is false.**Contrapositive:** If ab > 1/4, then a + b ≠ 1. Since the original statement is true, and the contrapositive is logically equivalent, this must be true. Plus, intuitively, if the product ab is larger than 1/4, their sum can't be 1 because we know the maximum product when the sum is 1 is exactly 1/4.Therefore, my conclusion remains that only the contrapositive is true among the converse, inverse, and contrapositive. So, the number of true propositions is one.**Final Answer**The number of true propositions is boxed{1}.

💡Okay, so I have this problem about two intersecting circles. They intersect at two points: (1, 3) and (m, 1). The centers of both circles lie on the line x - y + c/2 = 0. I need to find m + c. The options are A: -1, B: 2, C: 3, D: 0.Hmm, let me think. I remember that when two circles intersect, the line connecting their centers is perpendicular to their common chord. The common chord is the line that passes through both intersection points. So, maybe I can find the equation of the common chord first.The two points of intersection are (1, 3) and (m, 1). So, the slope of the common chord can be calculated using these two points. The slope formula is (y2 - y1)/(x2 - x1). Plugging in the points, that would be (1 - 3)/(m - 1) = (-2)/(m - 1). Let me write that down: slope = -2/(m - 1).But wait, I also know that the line connecting the centers is perpendicular to the common chord. The line connecting the centers has the equation x - y + c/2 = 0. Let me find its slope. The equation can be rewritten as y = x + c/2, so the slope is 1.Since the common chord is perpendicular to this line, the product of their slopes should be -1. So, slope of common chord * slope of centers line = -1.We already have the slope of the centers line as 1, so slope of common chord must be -1. Therefore, -2/(m - 1) = -1.Let me solve for m. Multiply both sides by (m - 1): -2 = -1*(m - 1). Simplify: -2 = -m + 1. Then, add m to both sides: m - 2 = 1. Add 2 to both sides: m = 3.Okay, so m is 3. Now, I need to find c. Since the centers lie on the line x - y + c/2 = 0, and the midpoint of the common chord should also lie on this line. The midpoint of the two intersection points (1, 3) and (3, 1) can be found by averaging the coordinates.Midpoint x-coordinate: (1 + 3)/2 = 2. Midpoint y-coordinate: (3 + 1)/2 = 2. So, the midpoint is (2, 2). This point lies on the line x - y + c/2 = 0.Plugging in (2, 2): 2 - 2 + c/2 = 0. Simplify: 0 + c/2 = 0. Therefore, c/2 = 0, so c = 0.Wait, so c is 0? That seems straightforward. So, m is 3 and c is 0. Therefore, m + c = 3 + 0 = 3.Let me double-check my steps. First, I found the slope of the common chord correctly, set it to be perpendicular to the centers' line, solved for m, got 3. Then, found the midpoint, which is (2, 2), substituted into the centers' line equation, found c = 0. So, m + c is 3.Looking at the options, C is 3. So, I think that's the answer.**Final Answer**boxed{C}

💡Okay, so I have this problem here about determining which set of three line segments can form a triangle when connected end to end. There are four options: A, B, C, and D. Each option gives three lengths in centimeters. I need to figure out which one can actually form a triangle.First, I remember that for three line segments to form a triangle, they must satisfy the triangle inequality theorem. This theorem states that the sum of the lengths of any two sides must be greater than the length of the remaining side. So, for any three sides a, b, and c, the following must all be true:- a + b > c- a + c > b- b + c > aIf any of these conditions are not met, the three segments cannot form a triangle.Alright, let's go through each option one by one.**Option A: 1 cm, 2 cm, 3 cm**Let's check the triangle inequalities:1. 1 + 2 > 3? - 1 + 2 = 3, which is equal to 3, not greater. So, this condition fails. Since one of the conditions fails, these segments cannot form a triangle.**Option B: 3 cm, 4 cm, 5 cm**Checking the triangle inequalities:1. 3 + 4 > 5? - 3 + 4 = 7, which is greater than 5. Good.2. 3 + 5 > 4? - 3 + 5 = 8, which is greater than 4. Good.3. 4 + 5 > 3? - 4 + 5 = 9, which is greater than 3. Good.All three conditions are satisfied, so these segments can form a triangle.**Option C: 4 cm, 5 cm, 10 cm**Checking the triangle inequalities:1. 4 + 5 > 10? - 4 + 5 = 9, which is less than 10. This condition fails.Since one condition fails, these segments cannot form a triangle.**Option D: 6 cm, 9 cm, 2 cm**Checking the triangle inequalities:1. 6 + 9 > 2? - 6 + 9 = 15, which is greater than 2. Good.2. 6 + 2 > 9? - 6 + 2 = 8, which is less than 9. This condition fails. Since one condition fails, these segments cannot form a triangle.So, after checking all options, only Option B satisfies all the triangle inequality conditions and can form a triangle.

💡Okay, so I have this problem here where I need to find the area of a set S defined by the condition |[x+y]| + |[x−y]| ≤ 1. The floor function [x] is the greatest integer less than or equal to x. Hmm, this seems a bit tricky, but let me try to break it down step by step.First, I know that the floor function [x] takes a real number x and gives the largest integer less than or equal to x. So, for example, [2.3] is 2, and [−1.7] is −2. Got it. So, [x+y] and [x−y] will give me integers based on the values of x+y and x−y.The condition given is |[x+y]| + |[x−y]| ≤ 1. Since both |[x+y]| and |[x−y]| are non-negative, their sum being less than or equal to 1 means that each of them can be 0 or 1, but their sum can't exceed 1. So, possible cases are:1. |[x+y]| = 0 and |[x−y]| = 02. |[x+y]| = 0 and |[x−y]| = 13. |[x+y]| = 1 and |[x−y]| = 0These are the only cases because if either |[x+y]| or |[x−y]| were greater than 1, their sum would exceed 1, which isn't allowed.Let me analyze each case separately.**Case 1: |[x+y]| = 0 and |[x−y]| = 0**If |[x+y]| = 0, then [x+y] must be 0. Similarly, [x−y] must be 0. So, we have:0 ≤ x + y < 1 (since [x+y] = 0)0 ≤ x − y < 1 (since [x−y] = 0)So, this defines a region where both x + y and x − y are between 0 and 1. Let me visualize this on the coordinate plane.If I plot x + y = 0 and x + y = 1, these are straight lines. Similarly, x − y = 0 and x − y = 1 are also straight lines. The region where both inequalities hold is a diamond-shaped area bounded by these four lines.**Case 2: |[x+y]| = 0 and |[x−y]| = 1**Here, [x+y] = 0 and [x−y] can be either 1 or −1 because the absolute value is 1.First, let's consider [x−y] = 1:0 ≤ x + y < 1 (from [x+y] = 0)1 ≤ x − y < 2 (from [x−y] = 1)Similarly, for [x−y] = −1:0 ≤ x + y < 1−1 ≤ x − y < 0So, these are two additional regions adjacent to the diamond from Case 1.**Case 3: |[x+y]| = 1 and |[x−y]| = 0**This is symmetric to Case 2. Here, [x+y] can be 1 or −1, and [x−y] = 0.First, [x+y] = 1:1 ≤ x + y < 20 ≤ x − y < 1And [x+y] = −1:−1 ≤ x + y < 00 ≤ x − y < 1Again, these are two regions adjacent to the diamond from Case 1.So, putting all these cases together, the set S is composed of a central diamond and four adjacent regions, each extending from the diamond.Now, to find the area, I need to calculate the area of each of these regions and sum them up.Let's start with the central diamond from Case 1.**Central Diamond (Case 1):**The inequalities are:0 ≤ x + y < 10 ≤ x − y < 1This is a square rotated by 45 degrees. To find its area, I can find the coordinates of its vertices.The lines x + y = 0 and x + y = 1 intersect the lines x − y = 0 and x − y = 1.Solving these equations:1. Intersection of x + y = 0 and x − y = 0: Adding both equations: 2x = 0 ⇒ x = 0. Then y = 0. So, point (0, 0).2. Intersection of x + y = 0 and x − y = 1: Adding: 2x = 1 ⇒ x = 0.5. Then y = −0.5. So, point (0.5, −0.5).3. Intersection of x + y = 1 and x − y = 0: Adding: 2x = 1 ⇒ x = 0.5. Then y = 0.5. So, point (0.5, 0.5).4. Intersection of x + y = 1 and x − y = 1: Adding: 2x = 2 ⇒ x = 1. Then y = 0. So, point (1, 0).Wait a minute, that doesn't seem right. If I plot these points, they don't form a diamond. Let me check my calculations.Wait, actually, the four lines x + y = 0, x + y = 1, x − y = 0, and x − y = 1 form a diamond with vertices at (0, 0), (1, 0), (0.5, 0.5), and (0.5, −0.5). Hmm, but that seems like a square with side length √2/2, but rotated.Alternatively, maybe it's better to calculate the area using integration or by recognizing it as a square.Alternatively, since the diamond is bounded by four lines, the distance between the parallel lines x + y = 0 and x + y = 1 is √2, and similarly for x − y = 0 and x − y = 1. So, the area would be (√2 * √2)/2 = 1. Wait, that can't be right because the area of a diamond (which is a square rotated by 45 degrees) with diagonals of length √2 each would have area (d1*d2)/2 = (√2 * √2)/2 = 1.But actually, the distance between x + y = 0 and x + y = 1 is 1/√2, not √2. Because the distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is |c1 - c2| / √(a² + b²). So, for x + y = 0 and x + y = 1, the distance is |1 - 0| / √(1 + 1) = 1/√2.Similarly, the distance between x − y = 0 and x − y = 1 is also 1/√2.So, the diamond is a square with side length 1/√2, but rotated. The area of a square is side², so (1/√2)² = 1/2. But wait, that doesn't seem right because the diamond should have an area larger than that.Wait, maybe I'm confusing the side length with the distance between the lines. Let me think differently.The diamond is bounded by four lines, so it's a square with diagonals along the lines x + y = 0 and x − y = 0. The length of each diagonal is the distance between x + y = 0 and x + y = 1, which is 1/√2, and similarly for the other diagonal.But actually, the diagonals of the diamond are the distances between the parallel lines. So, the length of each diagonal is 1/√2. The area of a rhombus (which is what the diamond is) is (d1*d2)/2, where d1 and d2 are the lengths of the diagonals. So, area = (1/√2 * 1/√2)/2 = (1/2)/2 = 1/4. Hmm, that seems too small.Wait, maybe I'm overcomplicating this. Let me try to find the coordinates of the vertices again.The four lines are:1. x + y = 02. x + y = 13. x − y = 04. x − y = 1The intersection points are:- Intersection of x + y = 0 and x − y = 0: (0, 0)- Intersection of x + y = 0 and x − y = 1: (0.5, -0.5)- Intersection of x + y = 1 and x − y = 0: (0.5, 0.5)- Intersection of x + y = 1 and x − y = 1: (1, 0)Wait, so the four vertices are (0,0), (0.5, -0.5), (1, 0), and (0.5, 0.5). Connecting these points, it forms a diamond shape.To find the area, I can use the shoelace formula.Let me list the coordinates in order:(0,0), (0.5, -0.5), (1, 0), (0.5, 0.5), and back to (0,0).Using the shoelace formula:Area = ½ |(0* -0.5 + 0.5*0 + 1*0.5 + 0.5*0) - (0*0.5 + (-0.5)*1 + 0*0.5 + 0.5*0)|Calculating step by step:First part:0* -0.5 = 00.5*0 = 01*0.5 = 0.50.5*0 = 0Sum = 0 + 0 + 0.5 + 0 = 0.5Second part:0*0.5 = 0(-0.5)*1 = -0.50*0.5 = 00.5*0 = 0Sum = 0 + (-0.5) + 0 + 0 = -0.5So, Area = ½ |0.5 - (-0.5)| = ½ |1| = 0.5So, the area of the central diamond is 0.5.Wait, that seems small, but it matches the shoelace calculation. Okay, moving on.**Case 2: |[x+y]| = 0 and |[x−y]| = 1**This gives two regions:1. [x+y] = 0 and [x−y] = 12. [x+y] = 0 and [x−y] = -1Let's handle each subcase.**Subcase 2.1: [x+y] = 0 and [x−y] = 1**So, 0 ≤ x + y < 1 and 1 ≤ x − y < 2.Let me find the region defined by these inequalities.First, x + y < 1 and x − y ≥ 1.Let me solve these inequalities:From x − y ≥ 1, we get x − y = 1 + k, where k ≥ 0.But since [x−y] = 1, x−y is in [1, 2).Similarly, x + y is in [0,1).So, the region is bounded by:x + y ≥ 0x + y < 1x − y ≥ 1x − y < 2Let me find the vertices of this region.The lines are:1. x + y = 02. x + y = 13. x − y = 14. x − y = 2Find their intersection points.Intersection of x + y = 1 and x − y = 1:Adding: 2x = 2 ⇒ x = 1. Then y = 0.Intersection of x + y = 1 and x − y = 2:Adding: 2x = 3 ⇒ x = 1.5. Then y = -0.5.Intersection of x + y = 0 and x − y = 1:Adding: 2x = 1 ⇒ x = 0.5. Then y = -0.5.Intersection of x + y = 0 and x − y = 2:Adding: 2x = 2 ⇒ x = 1. Then y = -1.But wait, our region is bounded by x + y < 1 and x − y < 2, so the upper boundaries are x + y =1 and x − y=2, but we also have x + y ≥0 and x − y ≥1.So, the vertices are:- Intersection of x + y =1 and x − y=1: (1,0)- Intersection of x + y=1 and x − y=2: (1.5, -0.5)- Intersection of x + y=0 and x − y=2: (1, -1)- Intersection of x + y=0 and x − y=1: (0.5, -0.5)Wait, but x + y=0 and x − y=2 intersect at (1, -1), but x + y=0 and x − y=1 intersect at (0.5, -0.5). So, the region is a quadrilateral with vertices at (1,0), (1.5, -0.5), (1, -1), and (0.5, -0.5).Let me apply the shoelace formula to find the area.List the coordinates in order:(1,0), (1.5, -0.5), (1, -1), (0.5, -0.5), back to (1,0).Calculating:First part:1*(-0.5) = -0.51.5*(-1) = -1.51*(-0.5) = -0.50.5*0 = 0Sum = -0.5 -1.5 -0.5 + 0 = -2.5Second part:0*1.5 = 0(-0.5)*1 = -0.5(-1)*0.5 = -0.5(-0.5)*1 = -0.5Sum = 0 -0.5 -0.5 -0.5 = -1.5Area = ½ |(-2.5) - (-1.5)| = ½ |(-2.5 + 1.5)| = ½ |-1| = 0.5So, the area for this subcase is 0.5.**Subcase 2.2: [x+y] = 0 and [x−y] = -1**So, 0 ≤ x + y <1 and -1 ≤ x − y <0.Similarly, let's find the vertices.The lines are:1. x + y =02. x + y =13. x − y = -14. x − y =0Find their intersection points.Intersection of x + y =1 and x − y = -1:Adding: 2x =0 ⇒ x=0. Then y=1.Intersection of x + y =1 and x − y =0:Adding: 2x=1 ⇒ x=0.5. Then y=0.5.Intersection of x + y =0 and x − y = -1:Adding: 2x = -1 ⇒ x= -0.5. Then y=0.5.Intersection of x + y =0 and x − y =0:x=0, y=0.So, the vertices are:(0,1), (0.5,0.5), (0,0), (-0.5,0.5)Wait, let me check:Wait, x + y =1 and x − y = -1 intersect at (0,1).x + y =1 and x − y =0 intersect at (0.5,0.5).x + y =0 and x − y = -1 intersect at (-0.5,0.5).x + y =0 and x − y =0 intersect at (0,0).So, the region is a quadrilateral with vertices at (0,1), (0.5,0.5), (0,0), (-0.5,0.5).Applying the shoelace formula:List the coordinates in order:(0,1), (0.5,0.5), (0,0), (-0.5,0.5), back to (0,1).Calculating:First part:0*0.5 =00.5*0=00*0.5=0-0.5*1= -0.5Sum =0 +0 +0 -0.5= -0.5Second part:1*0.5=0.50.5*0=00*(-0.5)=00.5*0=0Sum=0.5 +0 +0 +0=0.5Area=½ |(-0.5) -0.5|=½ |-1|=0.5So, the area for this subcase is also 0.5.So, total area for Case 2 is 0.5 + 0.5 =1.**Case 3: |[x+y]| =1 and |[x−y]| =0**This is similar to Case 2, but now [x+y] can be 1 or -1, and [x−y]=0.So, two subcases:1. [x+y] =1 and [x−y] =02. [x+y] =-1 and [x−y] =0**Subcase 3.1: [x+y] =1 and [x−y] =0**So, 1 ≤ x + y <2 and 0 ≤ x − y <1.Let me find the vertices.Lines:1. x + y =12. x + y =23. x − y =04. x − y =1Intersection points:x + y =1 and x − y =0: (0.5,0.5)x + y =1 and x − y =1: (1,0)x + y =2 and x − y =0: (1,1)x + y =2 and x − y =1: (1.5,0.5)So, the vertices are (0.5,0.5), (1,0), (1.5,0.5), (1,1).Applying shoelace formula:List the coordinates:(0.5,0.5), (1,0), (1.5,0.5), (1,1), back to (0.5,0.5).Calculating:First part:0.5*0=01*0.5=0.51.5*1=1.51*0.5=0.5Sum=0 +0.5 +1.5 +0.5=2.5Second part:0.5*1=0.50*1.5=00.5*1=0.51*0.5=0.5Sum=0.5 +0 +0.5 +0.5=1.5Area=½ |2.5 -1.5|=½ |1|=0.5So, area is 0.5.**Subcase 3.2: [x+y] =-1 and [x−y] =0**So, -1 ≤ x + y <0 and 0 ≤ x − y <1.Let me find the vertices.Lines:1. x + y =-12. x + y =03. x − y =04. x − y =1Intersection points:x + y =-1 and x − y =0: (-0.5,-0.5)x + y =-1 and x − y =1: (0,-1)x + y =0 and x − y =0: (0,0)x + y =0 and x − y =1: (0.5,-0.5)So, the vertices are (-0.5,-0.5), (0,-1), (0.5,-0.5), (0,0).Applying shoelace formula:List the coordinates:(-0.5,-0.5), (0,-1), (0.5,-0.5), (0,0), back to (-0.5,-0.5).Calculating:First part:-0.5*(-1)=0.50*(-0.5)=00.5*0=00*(-0.5)=0Sum=0.5 +0 +0 +0=0.5Second part:-0.5*0=0-1*0.5=-0.5-0.5*0=00*(-0.5)=0Sum=0 -0.5 +0 +0=-0.5Area=½ |0.5 - (-0.5)|=½ |1|=0.5So, the area for this subcase is 0.5.Therefore, total area for Case 3 is 0.5 +0.5=1.**Total Area:**Now, adding up all the areas:- Central diamond (Case 1): 0.5- Case 2:1- Case 3:1Total area=0.5 +1 +1=2.5Wait, 2.5 is equal to 5/2. So, the area is 5/2.But let me double-check because sometimes I might have missed something.Wait, in Case 1, I found the area as 0.5, and each of the other cases added 1 each, so total 2.5, which is 5/2. That seems correct.But let me visualize again. The central diamond is 0.5, and each of the four regions (two from Case 2 and two from Case 3) adds 0.5 each, so total 0.5 + 4*0.5=2.5. Yes, that makes sense.So, the area of the region S is 5/2.**Final Answer**boxed{frac{5}{2}}

💡First, I recognize that the given equation ( y^2 = 8x ) represents a parabola. To find the latus rectum, I need to determine the value of ( p ) in the standard form of a parabola equation.I compare ( y^2 = 8x ) with the standard form ( y^2 = 4px ). By equating the coefficients, I find that ( 4p = 8 ), which means ( p = 2 ).Next, I recall that the equation of the latus rectum for a parabola opening to the right is ( x = -p ). Substituting ( p = 2 ) into this equation gives ( x = -2 ).Therefore, the equation of the latus rectum for the parabola ( y^2 = 8x ) is ( x = -2 ).

💡First, I need to determine the scale factor used to create the print from the original painting. I'll start by comparing the widths of both the original painting and the print.The original painting is 15 inches wide, and the print is 37.5 inches wide. To find the scale factor, I'll divide the width of the print by the width of the original painting:Scale Factor = 37.5 inches / 15 inches = 2.5This means that the print is 2.5 times larger than the original painting in both height and width.Next, I'll use this scale factor to find the height of the original painting. The print's height is given as 25 inches. Using the scale factor, I can set up the following equation:Print Height = Scale Factor × Original Height25 inches = 2.5 × Original HeightTo find the original height, I'll divide the print height by the scale factor:Original Height = 25 inches / 2.5 = 10 inchesTherefore, the original painting is 10 inches tall.