💡First, I need to determine the commission rate per sale. The salesperson earns 180 for making 15 sales. To find the rate, I'll divide the total earnings by the number of sales.Next, with the commission rate per sale identified, I can calculate the earnings for 20 sales by multiplying the rate by the new number of sales.Finally, I'll present the calculated earnings as the final answer.

💡Okay, so I've got this problem here about the ratio of the interior angles of a triangle being 1:2:3, and I need to figure out what type of triangle it is. The options are acute, equilateral, obtuse, or right triangle. Hmm, let's see how to approach this.First, I remember that in any triangle, the sum of the interior angles is always 180 degrees. That's a basic property of triangles. So, if the angles are in the ratio 1:2:3, I can think of them as parts of 180 degrees. Maybe I can represent each angle as a multiple of a variable to make it easier.Let's say the smallest angle is 'x'. Then, according to the ratio, the other angles would be '2x' and '3x'. So, the three angles are x, 2x, and 3x. Now, if I add them all up, they should equal 180 degrees. Let me write that down:x + 2x + 3x = 180Combining like terms, that's:6x = 180Okay, so to find the value of x, I can divide both sides by 6:x = 180 / 6x = 30So, the smallest angle is 30 degrees. Then, the next angle is 2x, which is 2 * 30 = 60 degrees, and the largest angle is 3x, which is 3 * 30 = 90 degrees.Wait a minute, one of the angles is 90 degrees. That means the triangle has a right angle. So, it must be a right triangle. Let me double-check to make sure I didn't make any mistakes.Adding up the angles: 30 + 60 + 90 = 180 degrees. Yep, that checks out. So, the triangle has angles of 30, 60, and 90 degrees, which confirms it's a right triangle.I don't think it's an acute triangle because an acute triangle has all angles less than 90 degrees, and here we have a 90-degree angle. It's definitely not equilateral because all angles in an equilateral triangle are 60 degrees, and we have angles of different measures here. And it's not obtuse because an obtuse triangle has one angle greater than 90 degrees, but here the largest angle is exactly 90 degrees.So, after going through this step by step, I'm pretty confident that the triangle is a right triangle.

💡Alright, so I have this problem about an ellipse. Let me try to figure it out step by step. First, part (Ⅰ) asks for the standard equation of the ellipse C. I remember that the standard form of an ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a) is the semi-major axis and (b) is the semi-minor axis. The problem gives me the foci of the ellipse: (F_1(-2sqrt{2}, 0)) and (F_2(2sqrt{2}, 0)). So, the distance between each focus and the center is (c). Since the foci are on the x-axis, the major axis is along the x-axis. It also says that for any point P on the ellipse, the sum of the distances from P to each focus is (4sqrt{3}). I remember that this sum is equal to (2a), so (2a = 4sqrt{3}), which means (a = 2sqrt{3}). Now, I need to find (b). I recall the relationship between (a), (b), and (c) for an ellipse: (c^2 = a^2 - b^2). Given that the foci are at ((-2sqrt{2}, 0)) and ((2sqrt{2}, 0)), the distance from the center to each focus is (c = 2sqrt{2}). So, plugging in the values I have:(c^2 = (2sqrt{2})^2 = 8)(a^2 = (2sqrt{3})^2 = 12)Now, using (c^2 = a^2 - b^2):(8 = 12 - b^2)So, (b^2 = 12 - 8 = 4)Therefore, (b = 2).Now, plugging (a^2) and (b^2) into the standard equation:(frac{x^2}{12} + frac{y^2}{4} = 1)Okay, that seems straightforward. So, part (Ⅰ) is done.Moving on to part (Ⅱ). This one is a bit more involved. It says that a line (l_2) passes through the point ((0, 1)) with slope (k) and intersects the ellipse at points M and N. We need to investigate whether the dot product (overrightarrow{RM} cdot overrightarrow{RN}) is always a constant, regardless of the value of (k). If it is, find the constant; if not, explain why.First, let me understand the setup. The ellipse is given by (frac{x^2}{12} + frac{y^2}{4} = 1). The point R is the lower vertex of the ellipse. Since the major axis is along the x-axis, the vertices along the minor axis (y-axis) are at ((0, b)) and ((0, -b)). Since (b = 2), the lower vertex R is at ((0, -2)).So, R is at (0, -2). Points M and N are intersections of the line (l_2) with the ellipse. The line (l_2) has a slope (k) and passes through (0, 1), so its equation is (y = kx + 1).I need to find the points M and N where this line intersects the ellipse. Then, compute the vectors (overrightarrow{RM}) and (overrightarrow{RN}), take their dot product, and see if it's a constant.Let me start by finding the points of intersection between the line and the ellipse.Substitute (y = kx + 1) into the ellipse equation:(frac{x^2}{12} + frac{(kx + 1)^2}{4} = 1)Let me expand this:(frac{x^2}{12} + frac{k^2x^2 + 2kx + 1}{4} = 1)Multiply both sides by 12 to eliminate denominators:(x^2 + 3(k^2x^2 + 2kx + 1) = 12)Expand the terms:(x^2 + 3k^2x^2 + 6kx + 3 = 12)Combine like terms:((1 + 3k^2)x^2 + 6kx + (3 - 12) = 0)Simplify:((1 + 3k^2)x^2 + 6kx - 9 = 0)So, this is a quadratic in x. Let me denote it as:(A x^2 + B x + C = 0), where:- (A = 1 + 3k^2)- (B = 6k)- (C = -9)Let me denote the roots as (x_1) and (x_2), which correspond to the x-coordinates of points M and N.From quadratic theory, I know that:- (x_1 + x_2 = -B/A = -6k / (1 + 3k^2))- (x_1 x_2 = C/A = -9 / (1 + 3k^2))Now, the coordinates of M and N can be written as:- (M(x_1, kx_1 + 1))- (N(x_2, kx_2 + 1))Now, I need to find vectors (overrightarrow{RM}) and (overrightarrow{RN}). Since R is at (0, -2), these vectors are:- (overrightarrow{RM} = (x_1 - 0, (kx_1 + 1) - (-2)) = (x_1, kx_1 + 3))- (overrightarrow{RN} = (x_2 - 0, (kx_2 + 1) - (-2)) = (x_2, kx_2 + 3))Now, the dot product (overrightarrow{RM} cdot overrightarrow{RN}) is:(x_1 x_2 + (kx_1 + 3)(kx_2 + 3))Let me expand this:(x_1 x_2 + [k^2 x_1 x_2 + 3k x_1 + 3k x_2 + 9])Combine like terms:(x_1 x_2 + k^2 x_1 x_2 + 3k(x_1 + x_2) + 9)Factor out (x_1 x_2):((1 + k^2) x_1 x_2 + 3k(x_1 + x_2) + 9)Now, substitute the known values from the quadratic equation:(x_1 + x_2 = -6k / (1 + 3k^2))(x_1 x_2 = -9 / (1 + 3k^2))Plugging these into the expression:((1 + k^2)(-9 / (1 + 3k^2)) + 3k(-6k / (1 + 3k^2)) + 9)Let me compute each term step by step.First term: ((1 + k^2)(-9 / (1 + 3k^2)))Second term: (3k(-6k / (1 + 3k^2)))Third term: 9Compute the first term:(-9(1 + k^2) / (1 + 3k^2))Compute the second term:(-18k^2 / (1 + 3k^2))So, putting it all together:(-9(1 + k^2)/(1 + 3k^2) - 18k^2/(1 + 3k^2) + 9)Combine the first two terms:([-9(1 + k^2) - 18k^2] / (1 + 3k^2) + 9)Expand the numerator:(-9 - 9k^2 - 18k^2 = -9 - 27k^2)So, the expression becomes:((-9 - 27k^2)/(1 + 3k^2) + 9)Let me write 9 as (9(1 + 3k^2)/(1 + 3k^2)) to have a common denominator:((-9 - 27k^2 + 9(1 + 3k^2)) / (1 + 3k^2))Expand the numerator:(-9 - 27k^2 + 9 + 27k^2 = 0)So, the entire expression simplifies to 0.Wow, that's interesting. So, regardless of the value of (k), the dot product (overrightarrow{RM} cdot overrightarrow{RN}) is always 0.Wait, does that mean that vectors RM and RN are always perpendicular? Because their dot product is zero. That seems like a special property of this ellipse and the line passing through (0,1). Let me just double-check my calculations to make sure I didn't make a mistake.Starting from the dot product:(overrightarrow{RM} cdot overrightarrow{RN} = x_1 x_2 + (kx_1 + 3)(kx_2 + 3))Expanding:(x_1 x_2 + k^2 x_1 x_2 + 3k x_1 + 3k x_2 + 9)Which is:((1 + k^2) x_1 x_2 + 3k(x_1 + x_2) + 9)Substituting:(x_1 + x_2 = -6k/(1 + 3k^2))(x_1 x_2 = -9/(1 + 3k^2))So,((1 + k^2)(-9/(1 + 3k^2)) + 3k(-6k/(1 + 3k^2)) + 9)Compute each term:First term: (-9(1 + k^2)/(1 + 3k^2))Second term: (-18k^2/(1 + 3k^2))Third term: 9Combine first and second terms:([-9(1 + k^2) - 18k^2]/(1 + 3k^2))Which is:(-9 - 9k^2 - 18k^2 = -9 - 27k^2)So,((-9 - 27k^2)/(1 + 3k^2) + 9)Express 9 as (9(1 + 3k^2)/(1 + 3k^2)):((-9 - 27k^2 + 9 + 27k^2)/(1 + 3k^2) = 0/(1 + 3k^2) = 0)Yes, that seems correct. So, regardless of the slope (k), the dot product is always zero. Therefore, (overrightarrow{RM} cdot overrightarrow{RN}) is always a constant, specifically zero.I wonder why this happens. Maybe because the point (0,1) has some special relation to the ellipse or the point R. It might be related to the properties of ellipses and their foci, but I'm not entirely sure. However, the algebra checks out, so I think the conclusion is correct.**Final Answer**Ⅰ. The standard equation of ellipse (C) is (boxed{dfrac{x^2}{12} + dfrac{y^2}{4} = 1}).Ⅱ. The dot product (overrightarrow{RM} cdot overrightarrow{RN}) is always a constant value of (boxed{0}).

💡Okay, so I have this problem where there's a region R in the plane with an area of 15 units squared. Then, they apply a matrix transformation to R, resulting in a new region R'. I need to find the area of R'. Hmm, okay, let me think about how to approach this.First, I remember that when you apply a linear transformation to a region in the plane, the area of the transformed region is scaled by the absolute value of the determinant of the transformation matrix. So, the key here is to find the determinant of the given matrix and then multiply it by the original area to get the new area.The matrix given is:[begin{pmatrix} 3 & 4 8 & -2 end{pmatrix}]Alright, so I need to calculate the determinant of this matrix. The formula for the determinant of a 2x2 matrix (begin{pmatrix} a & b c & d end{pmatrix}) is (ad - bc). Let me apply that here.So, for our matrix, a is 3, b is 4, c is 8, and d is -2. Plugging these into the formula:Determinant = (3)(-2) - (4)(8)Let me compute that step by step. First, 3 times -2 is -6. Then, 4 times 8 is 32. So, the determinant is -6 - 32, which is -38.Wait a second, the determinant is negative. But area can't be negative, right? So, I think I need to take the absolute value of the determinant to get the scaling factor for the area. That makes sense because the determinant's sign tells us about the orientation of the transformation, but the area scaling is just the magnitude.So, the absolute value of -38 is 38. That means the area of the transformed region R' is 38 times the area of the original region R.The original area is 15 units squared, so multiplying that by 38 should give me the new area. Let me do that calculation:38 * 15Hmm, 38 times 10 is 380, and 38 times 5 is 190. So, adding those together, 380 + 190 equals 570. So, the area of R' should be 570 units squared.Wait, let me double-check my calculations to make sure I didn't make any mistakes. First, the determinant: (3)(-2) is indeed -6, and (4)(8) is 32. So, -6 - 32 is -38. Absolute value is 38. Then, 38 times 15: 38*10=380, 38*5=190, so 380+190=570. Yep, that seems correct.I think I got it. The area scales by the absolute value of the determinant, which is 38, so the new area is 38*15=570. That should be the answer.**Final Answer**The area of region ( R' ) is boxed{570} units squared.

💡Okay, so I have this problem here: Given that x is a real number, the statement "x squared equals 1" is what kind of condition for "x equals 1"? The options are A: Sufficient but not necessary, B: Necessary but not sufficient, C: Necessary and sufficient, D: Neither sufficient nor necessary.Hmm, let me think about this. I remember that in logic, a condition can be sufficient, necessary, both, or neither. A sufficient condition means that if it's met, then the statement must be true. A necessary condition means that for the statement to be true, the condition must be met, but meeting the condition doesn't necessarily make the statement true.So, in this case, the statement is "x equals 1," and the condition is "x squared equals 1." Let me break it down.First, if x equals 1, then x squared is definitely 1. So, "x squared equals 1" is a necessary condition for "x equals 1." That means that in order for x to be 1, it's necessary that x squared is 1. So, the condition is necessary.But is it sufficient? Well, if x squared equals 1, does that necessarily mean that x equals 1? Hmm, no, because x could also be -1. So, x squared equals 1 doesn't guarantee that x is 1; it could be either 1 or -1. Therefore, the condition is not sufficient on its own.So putting it together, "x squared equals 1" is necessary for "x equals 1," but it's not sufficient because it doesn't uniquely determine x as 1. Therefore, the correct answer should be B: Necessary but not sufficient.Wait, let me double-check. If I have x equals 1, then x squared is 1. So, the condition is necessary. But if x squared is 1, x could be 1 or -1, so it's not enough to conclude x is 1. Yeah, that makes sense. So, it's necessary but not sufficient.I don't think it's C because it's not both necessary and sufficient. It's only necessary. It's also not D because it is necessary. So, A is wrong because it's not sufficient. So, yeah, B is the right choice.**Final Answer**boxed{B}

💡Okay, so I have this problem where I need to figure out which set of vectors can represent the vector a = (3,2). There are four options: A, B, C, and D. Each option gives two vectors, e1 and e2, and I need to check if I can express a as a linear combination of these e1 and e2 vectors.First, I remember that to represent a vector as a linear combination of two other vectors, those two vectors need to span the space, which in this case is 2D. That means the vectors e1 and e2 should not be parallel; they should form a basis for R². If they are parallel, they can't span the entire plane, so I can't represent every vector in R², including a = (3,2).Let me go through each option one by one.**Option A: e1 = (0,0), e2 = (1,2)**Hmm, e1 is the zero vector. That's interesting. So, if I try to write a = (3,2) as a combination of e1 and e2, it would look like:(3,2) = λ*(0,0) + μ*(1,2)Which simplifies to:3 = 0*λ + 1*μ => 3 = μ2 = 0*λ + 2*μ => 2 = 2*μBut from the first equation, μ is 3, and from the second equation, μ would have to be 1. That's a contradiction. So, there's no solution here. Therefore, Option A can't represent vector a.**Option B: e1 = (-1,2), e2 = (5,-2)**Alright, let's set up the equation:(3,2) = λ*(-1,2) + μ*(5,-2)Breaking this down into components:For the x-component: 3 = -λ + 5μFor the y-component: 2 = 2λ - 2μNow, I have a system of two equations:1. -λ + 5μ = 32. 2λ - 2μ = 2I can solve this system using substitution or elimination. Let's try elimination. Maybe multiply the second equation by something to make the coefficients of λ or μ the same.If I multiply the second equation by 5, I get:10λ - 10μ = 10Now, let's look at the first equation:-λ + 5μ = 3If I multiply the first equation by 10, I get:-10λ + 50μ = 30Now, add the two equations together:(-10λ + 50μ) + (10λ - 10μ) = 30 + 10This simplifies to:40μ = 40So, μ = 1.Now, plug μ = 1 back into one of the original equations, say the second one:2λ - 2*1 = 22λ - 2 = 22λ = 4λ = 2So, λ = 2 and μ = 1. Let me check if these values satisfy the first equation:-2 + 5*1 = -2 + 5 = 3Yes, that works. So, both equations are satisfied. Therefore, vector a can be expressed as a combination of e1 and e2 in Option B. So, Option B is valid.But just to be thorough, I should check the other options too.**Option C: e1 = (3,5), e2 = (6,10)**Let's set up the equation:(3,2) = λ*(3,5) + μ*(6,10)Breaking it down:3 = 3λ + 6μ2 = 5λ + 10μHmm, looking at these equations, I notice that the second equation is just (5/3) times the first equation if we ignore the constants. Let me check:First equation multiplied by (5/3):(3)*(5/3) = 5 = 5λ + 10μBut the second equation is 2 = 5λ + 10μSo, 5 = 5λ + 10μ and 2 = 5λ + 10μThat's a contradiction because 5 ≠ 2. Therefore, there's no solution here. So, Option C can't represent vector a.**Option D: e1 = (2,-3), e2 = (-2,3)**Set up the equation:(3,2) = λ*(2,-3) + μ*(-2,3)Breaking it down:3 = 2λ - 2μ2 = -3λ + 3μLet me write these equations:1. 2λ - 2μ = 32. -3λ + 3μ = 2I can simplify both equations by dividing by 2 and 3 respectively:1. λ - μ = 1.52. -λ + μ = 2/3Now, let's add these two equations:(λ - μ) + (-λ + μ) = 1.5 + 2/3Simplify:0 = 1.5 + 0.666...0 = 2.166...That's not possible. So, there's no solution here either. Therefore, Option D can't represent vector a.So, after checking all options, only Option B works. Therefore, the answer is B.