💡First, I recognize that the line ( y = 2x + 1 ) is tangent to the curve ( y = x^3 + ax + b ) at the point ( (1, 3) ). This means that both the function and its derivative must satisfy the given point and the slope of the tangent line.I start by finding the derivative of the curve to determine the slope at any point ( x ). The derivative of ( y = x^3 + ax + b ) is ( y' = 3x^2 + a ).Since the slope of the tangent line at ( x = 1 ) is 2, I set up the equation:[3(1)^2 + a = 2 implies 3 + a = 2 implies a = -1]Next, I use the point ( (1, 3) ) to find the value of ( b ). Substituting ( x = 1 ), ( y = 3 ), and ( a = -1 ) into the original equation:[3 = (1)^3 + (-1) + b implies 3 = 1 - 1 + b implies 3 = b]Therefore, the value of ( b ) is 3.

💡First, I need to simplify the expression (dfrac{i^{3}}{1-i}).I know that (i^{3}) is equal to (-i), so the expression becomes (dfrac{-i}{1-i}).To simplify this complex fraction, I'll multiply both the numerator and the denominator by the conjugate of the denominator, which is (1+i).This gives me (dfrac{-i(1+i)}{(1-i)(1+i)}).Expanding the numerator, I get (-i - i^{2}). Since (i^{2} = -1), this simplifies to (-i + 1).In the denominator, multiplying (1-i) by (1+i) results in (1 - i^{2}), which is (1 - (-1) = 2).So, the expression simplifies to (dfrac{1 - i}{2}), which is (dfrac{1}{2} - dfrac{1}{2}i).Therefore, the correct answer is option B.

💡First, I recognize that the universal set ( U ) consists of the elements (-1, 0, 1, 2).The set ( A ) is given as (-1, 2).To find the complement of ( A ) in ( U ), denoted as ( complement_U A ), I need to identify all elements in ( U ) that are not in ( A ).By examining the elements of ( U ) and excluding those that are in ( A ), I determine that the complement set ( complement_U A ) contains the elements ( 0 ) and ( 1 ).Therefore, the complement of ( A ) in ( U ) is ({0, 1}).

💡Alright, so I have this geometry problem here. It says that on the side AC of triangle ABC, a point X is chosen. I need to prove that if the inscribed circles of triangles ABX and BCX are tangent to each other, then point X lies on the incircle of triangle ABC. Hmm, okay, let me try to unpack this step by step.First, let me visualize the triangle ABC. Let me draw it out in my mind: triangle ABC with vertices A, B, and C. Point X is somewhere on side AC. So, we have two smaller triangles now: ABX and BCX. Each of these triangles has an incircle, which is the largest circle that fits inside the triangle and touches all three sides.The problem states that these two incircles are tangent to each other. That means they touch at exactly one point. I need to figure out what this implies about the position of point X on AC. Specifically, I need to show that X lies on the incircle of the original triangle ABC.Okay, let's recall some properties of incircles. The incircle of a triangle touches each side at one point, and these points are called the points of tangency. The distances from the vertices to these points of tangency are related to the triangle's semiperimeter.Let me denote the semiperimeter of triangle ABC as p. So, p = (AB + BC + AC)/2. Similarly, for triangles ABX and BCX, their semiperimeters would be p1 = (AB + BX + AX)/2 and p2 = (BC + CX + BX)/2, respectively.Now, the incircle of triangle ABX touches side AB at some point, say L, and side AX at point K. Similarly, the incircle of triangle BCX touches side BC at point M and side CX at point N. Since both incircles are tangent to each other, they must touch at a single point, say Y.Given that the incircles are tangent to each other, the distance between their centers must be equal to the sum of their radii. But maybe that's getting ahead of myself. Let me think about the lengths involved.In triangle ABX, the lengths from A to the point of tangency on AX would be equal to the semiperimeter minus AB. Wait, no, actually, the length from A to the point of tangency on AX is equal to (AX + AB - BX)/2. Similarly, in triangle BCX, the length from C to the point of tangency on CX is (CX + BC - BX)/2.Since the incircles are tangent to each other, the point of tangency Y must lie on the common tangent line, which in this case is side AC. Therefore, the distances from X to the points of tangency on AX and CX should be equal. That is, the length from X to the point of tangency on AX (let's call it K) should be equal to the length from X to the point of tangency on CX (let's call it N).So, XK = XN. But XK is also equal to the semiperimeter of ABX minus AB, and XN is equal to the semiperimeter of BCX minus BC. Wait, let me write that down more clearly.For triangle ABX:- The length from A to K is (AX + AB - BX)/2.- Therefore, the length from X to K is AX - (AX + AB - BX)/2 = (AX - AB + BX)/2.Similarly, for triangle BCX:- The length from C to N is (CX + BC - BX)/2.- Therefore, the length from X to N is CX - (CX + BC - BX)/2 = (CX - BC + BX)/2.Since XK = XN, we have:(AX - AB + BX)/2 = (CX - BC + BX)/2Multiplying both sides by 2:AX - AB + BX = CX - BC + BXSimplify by subtracting BX from both sides:AX - AB = CX - BCBut AX + CX = AC, so we can write CX = AC - AX. Substituting this into the equation:AX - AB = (AC - AX) - BCSimplify the right side:AX - AB = AC - AX - BCBring all terms to one side:AX - AB - AC + AX + BC = 02AX - AB - AC + BC = 0Let me rearrange:2AX = AB + AC - BCTherefore:AX = (AB + AC - BC)/2Hmm, interesting. So AX is equal to (AB + AC - BC)/2. Let me see what that represents.In triangle ABC, the length from A to the point where the incircle touches AC is given by (AB + AC - BC)/2. That's a standard formula for the point of tangency on side AC.So, if AX is equal to (AB + AC - BC)/2, that means X is exactly the point where the incircle of triangle ABC touches side AC. Therefore, X lies on the incircle of triangle ABC.Wait, let me double-check that. The point where the incircle touches side AC is indeed at a distance of (AB + AC - BC)/2 from A. Since we've shown that AX is equal to this value, X must be that point of tangency. Therefore, X lies on the incircle of ABC.So, putting it all together, if the incircles of triangles ABX and BCX are tangent to each other, then the distances from X to the points of tangency on AX and CX must be equal, leading us to the conclusion that X is the point where the incircle of ABC touches AC. Hence, X lies on the incircle of ABC.I think that makes sense. I should probably write this out more formally to make sure all the steps are clear and that I haven't missed anything.

💡Alright, so I'm trying to figure out how many cups of popcorn can be made from 2 tablespoons of popcorn kernels. Okay, let's start by understanding the problem. There are several people wanting different amounts of popcorn for movie night:- Joanie wants 3 cups.- Mitchell wants 4 cups.- Miles and Davis together want 6 cups.- Cliff wants 3 cups.So, if I add all that up, that's 3 + 4 + 6 + 3, which equals 16 cups of popcorn in total. Now, they say they will need 8 tablespoons of popcorn kernels to make all that popcorn. Hmm, so 8 tablespoons make 16 cups. That seems like a lot, but okay.Wait, so if 8 tablespoons make 16 cups, then how much does 1 tablespoon make? Let me see. If I divide 16 cups by 8 tablespoons, that's 2 cups per tablespoon. So, 1 tablespoon makes 2 cups of popcorn. That makes sense because the more kernels you have, the more popcorn you can make.But the question is asking about 2 tablespoons. So, if 1 tablespoon makes 2 cups, then 2 tablespoons should make double that, right? So, 2 tablespoons times 2 cups per tablespoon equals 4 cups. That seems straightforward.But let me double-check to make sure I'm not missing anything. The total amount of popcorn needed is 16 cups, and they're using 8 tablespoons. So, 16 divided by 8 is indeed 2 cups per tablespoon. Therefore, 2 tablespoons would make 4 cups. Yeah, that checks out.I think I've got it. So, from 2 tablespoons of popcorn kernels, you can make 4 cups of popcorn. That makes sense based on the given information.

💡First, I need to determine the amount of paint Joe has by calculating the surface area of the cube. The cube has a side length of 4, so each face has an area of (4^2 = 16). Since a cube has 6 faces, the total surface area is (6 times 16 = 96).Next, I know that the same amount of paint is used to coat the cylinder. The cylinder has a height equal to its diameter, which means the height (h) is twice the radius (r), so (h = 2r).The surface area of the cylinder includes the areas of the two circular bases and the side (lateral) surface. The formula for the surface area of a cylinder is:[SA = 2pi r^2 + 2pi r h]Substituting (h = 2r) into the formula:[SA = 2pi r^2 + 2pi r (2r) = 2pi r^2 + 4pi r^2 = 6pi r^2]Since the surface area of the cylinder is equal to the surface area of the cube:[6pi r^2 = 96]Solving for (r^2):[r^2 = frac{96}{6pi} = frac{16}{pi}]Taking the square root of both sides to find (r):[r = frac{4}{sqrt{pi}}]Now, I can calculate the volume of the cylinder using the formula:[V = pi r^2 h]Substituting (r = frac{4}{sqrt{pi}}) and (h = 2r = frac{8}{sqrt{pi}}):[V = pi left(frac{16}{pi}right) times frac{8}{sqrt{pi}} = frac{128}{sqrt{pi}}]To express this volume in the form (frac{P sqrt{2}}{pi}), I rationalize the denominator:[frac{128}{sqrt{pi}} = frac{128sqrt{2}}{pi}]Therefore, (P = 128).