💡Okay, so I have this statistics problem here. Let me try to figure it out step by step. The problem says:"If the mean of the data ( X_{1}, X_{2}, ldots, X_{n} ) is ( overline{x} ), and the variance is ( S^{2} ), then the mean and variance of ( 2X_{1}+3 ), ( 2X_{2}+3 ), ..., ( 2X_{n}+3 ) are respectively ( )"And the options are:A: ( overline{x} ) and ( S )B: ( 2overline{x}+3 ) and ( 4S^{2} )C: ( 2overline{x}+3 ) and ( S^{2} )D: ( 2overline{x}+3 ) and ( 4S^{2}+12S+9 )Alright, so I need to find the new mean and variance after each data point is transformed by multiplying by 2 and then adding 3. Let me recall how transformations affect mean and variance.First, the mean. If I have a set of data points and I apply a linear transformation to each, like ( Y_i = aX_i + b ), then the new mean ( overline{Y} ) is ( aoverline{X} + b ). So in this case, ( a = 2 ) and ( b = 3 ). That should make the new mean ( 2overline{x} + 3 ). So that part seems straightforward.Now, for the variance. Variance measures how spread out the data is. If I multiply each data point by a constant, the variance gets multiplied by the square of that constant. So, if ( Y_i = aX_i + b ), then the variance ( S_Y^2 = a^2 S_X^2 ). Adding a constant ( b ) doesn't affect the variance because it just shifts all the data points by the same amount, which doesn't change their spread.So in this case, since we're multiplying each ( X_i ) by 2, the variance should be multiplied by ( 2^2 = 4 ). Adding 3 doesn't change the variance, so the new variance should be ( 4S^2 ).Let me double-check that. The original variance is ( S^2 ), which is calculated as the average of the squared deviations from the mean. So, ( S^2 = frac{1}{n} sum_{i=1}^{n} (X_i - overline{x})^2 ).If we transform each ( X_i ) to ( Y_i = 2X_i + 3 ), then the new mean ( overline{Y} = 2overline{x} + 3 ). The deviations from the new mean are ( Y_i - overline{Y} = (2X_i + 3) - (2overline{x} + 3) = 2(X_i - overline{x}) ).So, the squared deviations are ( [2(X_i - overline{x})]^2 = 4(X_i - overline{x})^2 ). Therefore, the new variance ( S_Y^2 = frac{1}{n} sum_{i=1}^{n} 4(X_i - overline{x})^2 = 4 times frac{1}{n} sum_{i=1}^{n} (X_i - overline{x})^2 = 4S^2 ).Okay, that confirms it. So, the new mean is ( 2overline{x} + 3 ) and the new variance is ( 4S^2 ). Looking back at the options, that's option B.Wait, let me just make sure I didn't make any mistakes. Sometimes when dealing with variance, especially with transformations, it's easy to mix up the constants or the operations. Let me think again.If I have ( Y_i = aX_i + b ), then:- The new mean ( overline{Y} = aoverline{X} + b ). That makes sense because you're scaling and shifting each data point, so the average also scales and shifts accordingly.- The new variance ( S_Y^2 = a^2 S_X^2 ). The reason is that variance is about the spread, and scaling each data point by ( a ) scales the spread by ( a ), hence the variance (which is the square of the spread) by ( a^2 ). Adding ( b ) doesn't affect the spread, so it doesn't change the variance.So, in this case, ( a = 2 ), ( b = 3 ). Therefore, ( overline{Y} = 2overline{x} + 3 ), and ( S_Y^2 = 4S^2 ). That still points me to option B.Is there any chance I could have confused this with something else? Maybe if the transformation was different, like ( Y_i = 2(X_i + 3) ), but that would still be the same as ( 2X_i + 6 ), which is a different transformation. But in this case, it's ( 2X_i + 3 ), so the mean is shifted by 3, not scaled.Wait, hold on, actually, in the transformation ( Y_i = 2X_i + 3 ), the 3 is added after scaling, so it's a vertical shift. So, the mean is scaled by 2 and then shifted by 3, which is consistent with ( 2overline{x} + 3 ). If it were ( 2(X_i + 3) ), then it would be scaling after shifting, which would be ( 2X_i + 6 ), but that's not our case here.So, no, I think I'm correct. The mean is ( 2overline{x} + 3 ), and the variance is ( 4S^2 ). So, the answer is option B.Just to be thorough, let me consider what the other options are saying.Option A: ( overline{x} ) and ( S ). That would be the case if there was no transformation, or if the transformation was just adding 3 without scaling. But since we have scaling, the mean and variance should change, so A is incorrect.Option C: ( 2overline{x} + 3 ) and ( S^2 ). This correctly changes the mean but incorrectly keeps the variance the same. Since we scaled the data by 2, the variance should increase by a factor of 4, so C is incorrect.Option D: ( 2overline{x} + 3 ) and ( 4S^2 + 12S + 9 ). Hmm, this seems like someone might have tried to compute the variance incorrectly by expanding ( (2X_i + 3 - 2overline{x} - 3)^2 ) as ( (2(X_i - overline{x}) + 3 - 3)^2 ), which is actually ( 4(X_i - overline{x})^2 ). So, why is there an extra ( 12S + 9 )?Wait, maybe they thought of expanding ( (2X_i + 3 - 2overline{x} - 3)^2 ) as ( (2(X_i - overline{x}))^2 ), which is ( 4(X_i - overline{x})^2 ), so the variance is ( 4S^2 ). So, where does ( 12S + 9 ) come from? Maybe they incorrectly included the mean in the variance calculation?Alternatively, perhaps they thought of the variance as ( E[(Y - mu_Y)^2] ) and tried to compute it as ( E[(2X + 3 - 2mu_X - 3)^2] = E[(2(X - mu_X))^2] = 4E[(X - mu_X)^2] = 4S^2 ). So, I don't see how ( 12S + 9 ) comes into play. Maybe if someone incorrectly added the constants when expanding the square?Wait, let's see. If someone incorrectly expands ( (2X_i + 3 - 2overline{x} - 3)^2 ) as ( (2X_i - 2overline{x})^2 + 3^2 - 2 times 2X_i times 3 + ldots ), but that doesn't make sense because the 3 and -3 cancel out. So, no, that shouldn't be the case.Alternatively, maybe they thought of the variance of ( 2X + 3 ) as ( Var(2X) + Var(3) ), but ( Var(3) = 0 ) because variance of a constant is zero. So, that would still give ( 4S^2 ).Alternatively, perhaps they confused variance with standard deviation? If ( S ) is the standard deviation, then ( Var(Y) = (2S)^2 = 4S^2 ). But in the problem, ( S^2 ) is already the variance, so ( S ) would be the standard deviation. So, if someone thought ( S ) was the variance, they might have miscalculated. But in the problem, it's clearly stated that the variance is ( S^2 ), so ( S ) is the standard deviation.Alternatively, maybe they tried to compute the variance incorrectly by including the mean in the squared term. For example, if they thought ( Var(Y) = E[Y^2] - (E[Y])^2 ), then ( E[Y^2] = E[(2X + 3)^2] = 4E[X^2] + 12E[X] + 9 ), and ( (E[Y])^2 = (2E[X] + 3)^2 = 4(E[X])^2 + 12E[X] + 9 ). Then, ( Var(Y) = E[Y^2] - (E[Y])^2 = 4E[X^2] + 12E[X] + 9 - [4(E[X])^2 + 12E[X] + 9] = 4(E[X^2] - (E[X])^2) = 4Var(X) = 4S^2 ). So, even if you compute it that way, you still get ( 4S^2 ). So, where does ( 4S^2 + 12S + 9 ) come from?Wait, maybe they thought of ( Var(Y) = Var(2X + 3) = Var(2X) + Var(3) + 2Cov(2X, 3) ), but covariance between a variable and a constant is zero, so that still gives ( Var(Y) = 4Var(X) + 0 + 0 = 4S^2 ).Alternatively, maybe they thought of ( Var(Y) = (2)^2 Var(X) + (3)^2 Var(1) ), but Var(1) is zero because it's a constant. So, that still gives ( 4S^2 ).Alternatively, perhaps they thought that adding 3 affects the variance, which it doesn't. So, maybe they incorrectly added ( 3^2 = 9 ) to the variance, thinking that shifting affects variance, which it doesn't. So, they might have thought ( Var(Y) = 4S^2 + 9 ). But that's still not matching option D, which is ( 4S^2 + 12S + 9 ).Wait, 12S is 12 times the standard deviation, but in the variance formula, we only have squared terms. So, 12S would be linear in S, which doesn't make sense in the variance context. So, perhaps someone incorrectly expanded ( (2X + 3 - 2overline{x} - 3)^2 ) as ( (2X - 2overline{x} + 3 - 3)^2 ), which is still ( (2(X - overline{x}))^2 = 4(X - overline{x})^2 ). So, I don't see how 12S + 9 comes into play.Alternatively, maybe they thought of the standard deviation instead of variance? If ( S ) is the standard deviation, then ( Var(Y) = (2S)^2 = 4S^2 ). But in the problem, ( S^2 ) is the variance, so ( S ) is the standard deviation. So, if someone thought ( S ) was the variance, they might have miscalculated. But that still doesn't lead to ( 4S^2 + 12S + 9 ).Alternatively, maybe they thought of the mean of the squared terms? Let me see. If someone incorrectly computes ( E[(2X + 3)^2] ) as ( 4E[X^2] + 12E[X] + 9 ), and then subtracts ( (2E[X] + 3)^2 = 4(E[X])^2 + 12E[X] + 9 ), so the variance is ( 4(E[X^2] - (E[X])^2) = 4S^2 ). So, again, no 12S + 9.Alternatively, maybe they thought of the standard deviation as ( S ), and then thought of ( Var(Y) = (2S + 3)^2 = 4S^2 + 12S + 9 ). But that's incorrect because variance doesn't work that way. Variance is affected by scaling and shifting differently. Scaling affects variance quadratically, shifting doesn't affect variance at all.So, in summary, option D seems to be a result of incorrectly applying the transformation to the standard deviation instead of the variance, or perhaps incorrectly expanding the squared term in the variance formula. Either way, it's not the correct answer.So, going back, the correct answer is option B: ( 2overline{x} + 3 ) and ( 4S^2 ).Just to make sure, let me think of a simple example. Suppose I have a dataset: 1, 2, 3. The mean is 2, and the variance is ( frac{(1-2)^2 + (2-2)^2 + (3-2)^2}{3} = frac{1 + 0 + 1}{3} = frac{2}{3} ).Now, apply the transformation ( Y_i = 2X_i + 3 ). So, the new dataset is 5, 7, 9. The new mean is ( frac{5 + 7 + 9}{3} = frac{21}{3} = 7 ). Which is ( 2 times 2 + 3 = 7 ), so that checks out.The new variance is ( frac{(5-7)^2 + (7-7)^2 + (9-7)^2}{3} = frac{4 + 0 + 4}{3} = frac{8}{3} ). The original variance was ( frac{2}{3} ), so ( 4 times frac{2}{3} = frac{8}{3} ). So, that also checks out.Therefore, the transformation correctly scales the variance by 4 and shifts the mean by 3 after scaling. So, the answer is definitely B.

💡Okay, so I need to figure out where the graph of ( y = g(x) ) intersects with the line ( y = x + 2 ). The graph of ( g(x) ) is made up of five line segments connecting the points ( (-5, -3) ), ( (-3, 0) ), ( (-1, -3) ), ( (2, 3) ), ( (3, 2) ), and ( (5, 4) ). First, I think I should write down the equations of each of these line segments. Then, for each segment, I can set ( g(x) = x + 2 ) and solve for ( x ). If the solution for ( x ) lies within the segment's domain, that means there's an intersection point there. I'll need to do this for each segment and collect all valid ( x )-coordinates, then sum them up.Let me list the segments with their endpoints:1. From ( (-5, -3) ) to ( (-3, 0) )2. From ( (-3, 0) ) to ( (-1, -3) )3. From ( (-1, -3) ) to ( (2, 3) )4. From ( (2, 3) ) to ( (3, 2) )5. From ( (3, 2) ) to ( (5, 4) )Starting with the first segment: ( (-5, -3) ) to ( (-3, 0) ). I need the equation of this line. Let me calculate the slope first.Slope ( m = frac{0 - (-3)}{-3 - (-5)} = frac{3}{2} ). So the slope is ( frac{3}{2} ).Using point-slope form with point ( (-5, -3) ):( y - (-3) = frac{3}{2}(x - (-5)) )Simplify: ( y + 3 = frac{3}{2}(x + 5) )So, ( y = frac{3}{2}x + frac{15}{2} - 3 )Which simplifies to ( y = frac{3}{2}x + frac{9}{2} )Now, set this equal to ( x + 2 ):( frac{3}{2}x + frac{9}{2} = x + 2 )Subtract ( x ) from both sides:( frac{1}{2}x + frac{9}{2} = 2 )Subtract ( frac{9}{2} ):( frac{1}{2}x = 2 - frac{9}{2} = -frac{5}{2} )Multiply both sides by 2:( x = -5 )Wait, but the segment is from ( x = -5 ) to ( x = -3 ). So ( x = -5 ) is the starting point. Is this an intersection? Let me check the ( y )-value at ( x = -5 ) for both functions.For ( g(x) ), it's ( -3 ). For ( y = x + 2 ), it's ( -5 + 2 = -3 ). So yes, they intersect at ( (-5, -3) ). So ( x = -5 ) is a valid solution.Moving on to the second segment: ( (-3, 0) ) to ( (-1, -3) ). Let's find its equation.Slope ( m = frac{-3 - 0}{-1 - (-3)} = frac{-3}{2} )Using point-slope with ( (-3, 0) ):( y - 0 = -frac{3}{2}(x + 3) )Simplify: ( y = -frac{3}{2}x - frac{9}{2} )Set equal to ( x + 2 ):( -frac{3}{2}x - frac{9}{2} = x + 2 )Multiply both sides by 2 to eliminate fractions:( -3x - 9 = 2x + 4 )Bring variables to one side:( -3x - 2x = 4 + 9 )( -5x = 13 )( x = -frac{13}{5} ) or ( -2.6 )Now, check if this ( x ) is within the segment's domain, which is from ( x = -3 ) to ( x = -1 ). ( -2.6 ) is between ( -3 ) and ( -1 ), so it's valid.So, another intersection at ( x = -frac{13}{5} ).Third segment: ( (-1, -3) ) to ( (2, 3) ). Let's find its equation.Slope ( m = frac{3 - (-3)}{2 - (-1)} = frac{6}{3} = 2 )Using point-slope with ( (-1, -3) ):( y - (-3) = 2(x - (-1)) )Simplify: ( y + 3 = 2(x + 1) )So, ( y = 2x + 2 - 3 = 2x - 1 )Set equal to ( x + 2 ):( 2x - 1 = x + 2 )Subtract ( x ):( x - 1 = 2 )Add 1:( x = 3 )Check if ( x = 3 ) is within the segment's domain, which is from ( x = -1 ) to ( x = 2 ). ( 3 ) is outside this range, so no intersection here.Fourth segment: ( (2, 3) ) to ( (3, 2) ). Let's find its equation.Slope ( m = frac{2 - 3}{3 - 2} = frac{-1}{1} = -1 )Using point-slope with ( (2, 3) ):( y - 3 = -1(x - 2) )Simplify: ( y - 3 = -x + 2 )So, ( y = -x + 5 )Set equal to ( x + 2 ):( -x + 5 = x + 2 )Add ( x ) to both sides:( 5 = 2x + 2 )Subtract 2:( 3 = 2x )( x = frac{3}{2} ) or ( 1.5 )Check if ( x = 1.5 ) is within the segment's domain, which is from ( x = 2 ) to ( x = 3 ). ( 1.5 ) is less than 2, so it's outside the domain. No intersection here.Fifth segment: ( (3, 2) ) to ( (5, 4) ). Let's find its equation.Slope ( m = frac{4 - 2}{5 - 3} = frac{2}{2} = 1 )Using point-slope with ( (3, 2) ):( y - 2 = 1(x - 3) )Simplify: ( y - 2 = x - 3 )So, ( y = x - 1 )Set equal to ( x + 2 ):( x - 1 = x + 2 )Subtract ( x ):( -1 = 2 )Wait, that's not possible. So, no solution here. Therefore, no intersection on this segment.So, summarizing the intersections:1. First segment: ( x = -5 )2. Second segment: ( x = -frac{13}{5} )3. Third segment: No intersection4. Fourth segment: No intersection5. Fifth segment: No intersectionSo, the valid ( x )-coordinates are ( -5 ) and ( -frac{13}{5} ).Now, I need to sum these ( x )-coordinates:( -5 + (-frac{13}{5}) = -5 - frac{13}{5} = -frac{25}{5} - frac{13}{5} = -frac{38}{5} )Wait, that doesn't seem right. Let me double-check my calculations.Wait, in the first segment, I found ( x = -5 ), which is correct because both functions pass through ( (-5, -3) ). Then, in the second segment, I got ( x = -frac{13}{5} ), which is ( -2.6 ). That seems correct because it lies between ( -3 ) and ( -1 ).But when I added them, I think I made a mistake.( -5 + (-frac{13}{5}) ) is indeed ( -5 - 2.6 = -7.6 ), which is ( -frac{38}{5} ). But wait, the initial thought process in the problem statement had a different answer. Let me check again.Wait, in the initial problem statement, the user had a different thought process where they found intersections at ( x = -3 ) and ( x = 1 ), summing to ( -2 ). But in my detailed calculation, I found intersections at ( x = -5 ) and ( x = -2.6 ), summing to ( -7.6 ).This discrepancy suggests I might have made a mistake somewhere.Let me go back to the first segment. The equation was ( y = frac{3}{2}x + frac{9}{2} ). Setting equal to ( x + 2 ):( frac{3}{2}x + frac{9}{2} = x + 2 )Subtract ( x ):( frac{1}{2}x + frac{9}{2} = 2 )Subtract ( frac{9}{2} ):( frac{1}{2}x = 2 - frac{9}{2} = -frac{5}{2} )Multiply by 2:( x = -5 ). That seems correct.But in the initial problem statement, they found ( x = -3 ). Maybe they considered the intersection at ( (-3, 0) ). Let me check if ( y = x + 2 ) passes through ( (-3, 0) ).( y = -3 + 2 = -1 ). But ( g(-3) = 0 ). So, no, they don't intersect at ( (-3, 0) ). So my calculation is correct, the intersection is at ( x = -5 ).Wait, but in the initial problem statement, they considered the first segment and found ( x = -3 ). Maybe they made a mistake there.Let me re-examine the second segment. The equation was ( y = -frac{3}{2}x - frac{9}{2} ). Setting equal to ( x + 2 ):( -frac{3}{2}x - frac{9}{2} = x + 2 )Multiply both sides by 2:( -3x - 9 = 2x + 4 )Bring variables to one side:( -3x - 2x = 4 + 9 )( -5x = 13 )( x = -frac{13}{5} ). That's correct.So, my conclusion is that the intersections are at ( x = -5 ) and ( x = -frac{13}{5} ), summing to ( -frac{38}{5} ). But the initial problem statement had a different answer. Maybe I missed something.Wait, perhaps I made a mistake in the third segment. Let me check again.Third segment: ( (-1, -3) ) to ( (2, 3) ). Equation was ( y = 2x - 1 ). Setting equal to ( x + 2 ):( 2x - 1 = x + 2 )Subtract ( x ):( x - 1 = 2 )( x = 3 ). But the segment is from ( x = -1 ) to ( x = 2 ). So ( x = 3 ) is outside. So no intersection.Fourth segment: ( (2, 3) ) to ( (3, 2) ). Equation ( y = -x + 5 ). Setting equal to ( x + 2 ):( -x + 5 = x + 2 )( -2x = -3 )( x = frac{3}{2} ). But the segment is from ( x = 2 ) to ( x = 3 ). ( frac{3}{2} = 1.5 ) is less than 2, so no intersection.Fifth segment: ( (3, 2) ) to ( (5, 4) ). Equation ( y = x - 1 ). Setting equal to ( x + 2 ):( x - 1 = x + 2 )No solution.So, only two intersections: ( x = -5 ) and ( x = -frac{13}{5} ). Sum is ( -5 - frac{13}{5} = -frac{25}{5} - frac{13}{5} = -frac{38}{5} ).But in the initial problem statement, the answer was ( -2 ). So, I must have made a mistake.Wait, perhaps I misread the problem. Let me check the points again.The graph of ( g(x) ) consists of five line segments connecting:1. ( (-5, -3) ) to ( (-3, 0) )2. ( (-3, 0) ) to ( (-1, -3) )3. ( (-1, -3) ) to ( (2, 3) )4. ( (2, 3) ) to ( (3, 2) )5. ( (3, 2) ) to ( (5, 4) )Wait, in the initial problem statement, the user had a different approach. They considered the first segment and found ( x = -3 ), but that's not correct because ( y = x + 2 ) at ( x = -3 ) is ( -1 ), not ( 0 ). So, the intersection is at ( x = -5 ).Then, for the second segment, they found ( x = frac{7}{4} ), which is ( 1.75 ), but that's outside the segment's domain of ( x = -3 ) to ( x = -1 ). So, their calculation was wrong.In my calculation, I found ( x = -frac{13}{5} ) which is ( -2.6 ), which is within ( -3 ) to ( -1 ). So, that's correct.Then, for the third segment, they found ( x = 1 ), but in my calculation, I found ( x = 3 ), which is outside. Wait, let me check their equation.Wait, in the initial problem statement, they set ( x + 2 = frac{6}{3}x + 3 ). Wait, that's incorrect. The slope from ( (-1, -3) ) to ( (2, 3) ) is ( frac{6}{3} = 2 ), so the equation is ( y = 2x + b ). Plugging in ( (-1, -3) ):( -3 = 2(-1) + b )( -3 = -2 + b )( b = -1 )So, equation is ( y = 2x - 1 ). So, setting equal to ( x + 2 ):( 2x - 1 = x + 2 )( x = 3 ). So, their equation was correct, but they found ( x = 1 ), which is wrong. So, they made a mistake there.So, in conclusion, the correct intersections are at ( x = -5 ) and ( x = -frac{13}{5} ), summing to ( -frac{38}{5} ).But wait, the initial problem statement had a different answer. Maybe I'm missing an intersection.Wait, let me check the fifth segment again. The equation is ( y = x - 1 ). Setting equal to ( x + 2 ):( x - 1 = x + 2 )( -1 = 2 ). No solution.Fourth segment: ( y = -x + 5 ). Setting equal to ( x + 2 ):( -x + 5 = x + 2 )( -2x = -3 )( x = frac{3}{2} ). But the segment is from ( x = 2 ) to ( x = 3 ). So, ( x = 1.5 ) is outside.Third segment: ( x = 3 ) is outside.Second segment: ( x = -2.6 ).First segment: ( x = -5 ).So, only two intersections.Wait, but in the initial problem statement, they found two intersections: ( x = -3 ) and ( x = 1 ), summing to ( -2 ). But that's incorrect because ( x = -3 ) is not an intersection, and ( x = 1 ) is not on the third segment.So, my conclusion is that the correct sum is ( -frac{38}{5} ), which is ( -7.6 ).But wait, maybe I made a mistake in the second segment. Let me recheck.Second segment: from ( (-3, 0) ) to ( (-1, -3) ). Equation is ( y = -frac{3}{2}x - frac{9}{2} ). Setting equal to ( x + 2 ):( -frac{3}{2}x - frac{9}{2} = x + 2 )Multiply both sides by 2:( -3x - 9 = 2x + 4 )Bring variables to one side:( -3x - 2x = 4 + 9 )( -5x = 13 )( x = -frac{13}{5} ). That's correct.So, I think my answer is correct, and the initial problem statement had errors in their calculations.Therefore, the sum of all ( x )-coordinates is ( -5 + (-frac{13}{5}) = -frac{38}{5} ).But wait, the problem asks for the sum of all ( x )-coordinates where ( g(x) = x + 2 ). So, I think my answer is correct.However, to be thorough, let me plot these points mentally.First segment: from ( (-5, -3) ) to ( (-3, 0) ). The line ( y = x + 2 ) passes through ( (-5, -3) ) because ( -5 + 2 = -3 ). So, that's one intersection.Second segment: from ( (-3, 0) ) to ( (-1, -3) ). The line ( y = x + 2 ) at ( x = -3 ) is ( -1 ), but the segment at ( x = -3 ) is ( 0 ). So, they don't intersect at ( x = -3 ). But somewhere between ( x = -3 ) and ( x = -1 ), they intersect at ( x = -2.6 ).Third segment: from ( (-1, -3) ) to ( (2, 3) ). The line ( y = x + 2 ) at ( x = 2 ) is ( 4 ), but the segment at ( x = 2 ) is ( 3 ). So, they don't intersect at ( x = 2 ). The intersection was at ( x = 3 ), which is outside.Fourth segment: from ( (2, 3) ) to ( (3, 2) ). The line ( y = x + 2 ) at ( x = 2 ) is ( 4 ), but the segment at ( x = 2 ) is ( 3 ). So, no intersection.Fifth segment: from ( (3, 2) ) to ( (5, 4) ). The line ( y = x + 2 ) at ( x = 3 ) is ( 5 ), but the segment at ( x = 3 ) is ( 2 ). So, no intersection.Therefore, only two intersections: ( x = -5 ) and ( x = -2.6 ).Sum: ( -5 + (-2.6) = -7.6 ), which is ( -frac{38}{5} ).But the initial problem statement had a different answer. Maybe I should check if ( x = -5 ) is considered an intersection. Since both functions pass through ( (-5, -3) ), it is indeed an intersection point.So, I think my answer is correct, and the initial problem statement had errors.

💡Okay, so I have this problem where there's a point P on the terminal side of an angle α, and the coordinates of P are (√3, -1). I need to find the smallest positive value of α from the given options: A is 5π/6, B is 2π/3, C is 5π/3, and D is 11π/6.First, I remember that in the Cartesian plane, the coordinates of a point on the terminal side of an angle can help determine the angle itself. The x-coordinate is √3, which is positive, and the y-coordinate is -1, which is negative. So, this point is in the fourth quadrant because x is positive and y is negative.Now, I think about how to find the angle α. I recall that the tangent of the angle is equal to the y-coordinate divided by the x-coordinate. So, tan(α) = y/x = (-1)/√3. That simplifies to tan(α) = -1/√3.Hmm, I know that tan(π/6) is 1/√3, so tan(-π/6) would be -1/√3. That means the reference angle here is π/6, but since the point is in the fourth quadrant, the angle α should be 2π - π/6. Let me calculate that: 2π is the full circle, so subtracting π/6 gives me 11π/6.Wait, but let me make sure I'm not making a mistake here. Sometimes, I get confused between the reference angle and the actual angle in standard position. The reference angle is always positive and acute, which in this case is π/6. Since the point is in the fourth quadrant, the angle α should be measured clockwise from the positive x-axis, which would be 2π - π/6, which is indeed 11π/6.Let me check the options again. Option D is 11π/6, so that must be the answer. But just to be thorough, let me think if there's another way to approach this.Alternatively, I could use the arctangent function. If I take arctan(y/x), that would give me arctan(-1/√3). The arctangent of 1/√3 is π/6, so arctan(-1/√3) would be -π/6. But since angles are periodic with period 2π, adding 2π to -π/6 gives me 11π/6, which is the positive angle we're looking for.Another way to think about it is to consider the unit circle. On the unit circle, the point (√3/2, -1/2) corresponds to the angle 11π/6. But in our case, the coordinates are (√3, -1), which is just a scaled version of that point. So, the angle remains the same because scaling doesn't change the angle, only the distance from the origin.Wait, let me confirm that scaling doesn't affect the angle. If I have a point (x, y), and I scale it by a factor r, the new point is (rx, ry). The angle α is determined by the ratio y/x, which remains the same after scaling because (ry)/(rx) = y/x. So yes, scaling doesn't change the angle, only the radius. Therefore, the angle α is indeed 11π/6.I think I've covered all the bases here. I considered the quadrant, calculated the reference angle, used the arctangent function, and even thought about the unit circle. All methods point to 11π/6 as the smallest positive angle that corresponds to the point (√3, -1). So, I'm confident that the correct answer is D: 11π/6.

💡Okay, so I have this problem about a circle M with the equation x² + y² + 4x - 2y + 3 = 0, and a line l that passes through the point P(-3, 0). I need to find two things: first, the coordinates of the center of circle M, and second, the y-intercept of the tangent line l if it's tangent to circle M.Alright, let's start with the circle equation. I remember that the general form of a circle is x² + y² + Dx + Ey + F = 0, and the center is at (-D/2, -E/2). So, in this case, D is 4 and E is -2. That means the center should be at (-4/2, -(-2)/2), which simplifies to (-2, 1). Hmm, that seems straightforward. Let me double-check by rewriting the equation in standard form.To rewrite the equation, I need to complete the squares for both x and y terms. Starting with the x terms: x² + 4x. To complete the square, I take half of 4, which is 2, and square it to get 4. So, x² + 4x becomes (x + 2)² - 4. Similarly, for the y terms: y² - 2y. Half of -2 is -1, and squaring that gives 1. So, y² - 2y becomes (y - 1)² - 1.Putting it all together, the equation becomes:(x + 2)² - 4 + (y - 1)² - 1 + 3 = 0Simplify the constants: -4 -1 +3 = -2So, (x + 2)² + (y - 1)² - 2 = 0Which means (x + 2)² + (y - 1)² = 2Yep, that confirms the center is at (-2, 1) and the radius is √2. Cool, that part seems solid.Now, moving on to the tangent line. The line passes through P(-3, 0) and is tangent to the circle. I need to find the y-intercept of this tangent line.First, let's recall that a tangent line to a circle is perpendicular to the radius at the point of tangency. So, if I can find the slope of the radius from the center to the point of tangency, the slope of the tangent line will be the negative reciprocal.But wait, I don't know the point of tangency yet. Maybe there's another approach. I remember that the distance from the center of the circle to the tangent line is equal to the radius. So, if I can write the equation of the line in a general form and then use the distance formula, I can solve for the slope or the y-intercept.Let's denote the equation of the tangent line as y = kx + b. Since it passes through P(-3, 0), substituting x = -3 and y = 0 gives 0 = -3k + b, so b = 3k. Therefore, the equation of the line can be written as y = kx + 3k or y = k(x + 3).Alternatively, in standard form, that would be kx - y + 3k = 0.Now, the distance from the center (-2, 1) to this line should be equal to the radius √2. The distance formula from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / √(A² + B²).In our case, A = k, B = -1, and C = 3k. Plugging in the center coordinates:Distance = |k*(-2) + (-1)*1 + 3k| / √(k² + 1) = |(-2k -1 + 3k)| / √(k² + 1) = |k -1| / √(k² + 1)This distance should equal the radius √2:|k -1| / √(k² + 1) = √2To eliminate the square roots, let's square both sides:(k -1)² / (k² + 1) = 2Multiply both sides by (k² + 1):(k -1)² = 2(k² + 1)Expand the left side:k² - 2k + 1 = 2k² + 2Bring all terms to one side:k² - 2k + 1 - 2k² - 2 = 0Simplify:- k² - 2k -1 = 0Multiply both sides by -1:k² + 2k + 1 = 0This is a quadratic equation: k² + 2k + 1 = 0Factor it:(k + 1)² = 0So, k = -1 is the only solution.Now, since the slope k is -1, the equation of the tangent line is y = -1(x + 3) = -x -3.Therefore, the y-intercept b is -3.Wait, let me verify this. If the line is y = -x -3, does it pass through (-3, 0)? Plugging in x = -3: y = -(-3) -3 = 3 -3 = 0. Yes, that works.Also, let's check the distance from the center (-2,1) to this line. Using the distance formula:Distance = |(-1)*(-2) + (-1)*1 + (-3)| / √((-1)² + (-1)²) = |2 -1 -3| / √(1 +1) = |-2| / √2 = 2 / √2 = √2Which matches the radius. Perfect, that checks out.So, to recap:1. The center of circle M is (-2, 1).2. The y-intercept of the tangent line passing through P(-3, 0) is -3.I think that's all. It makes sense, and all the checks worked out.**Final Answer**The coordinates of the center of circle ( M ) are (boxed{(-2, 1)}); the y-intercept of the tangent line is (boxed{-3}).

💡Okay, so I have this problem here about a set M defined as M equals the set of all x such that x squared minus 1 equals zero. Then I have four options to choose from: A, B, C, and D. I need to figure out which one is correct.First, let me understand what the set M actually is. The definition says M is {x | x² - 1 = 0}. That means M contains all real numbers x that satisfy the equation x squared minus one equals zero. So, I should solve this equation to find out what elements are in M.Alright, solving x² - 1 = 0. Let me do that step by step. If I add 1 to both sides, I get x² = 1. Then, taking the square root of both sides, I get x equals plus or minus 1. So, the solutions are x = 1 and x = -1. Therefore, the set M is {-1, 1}. That makes sense because when you square both 1 and -1, you get 1, and subtracting 1 gives zero, which satisfies the equation.Now, looking at the options:A: M = (-1, 1)B: M = (-1, 1]C: -1 ∈ MD: 1 ⊆ MLet me go through each option one by one.Option A says M is equal to the interval (-1, 1). Hmm, wait a second. The interval (-1, 1) includes all real numbers between -1 and 1, not including -1 and 1 themselves. But from my earlier calculation, M only contains two elements: -1 and 1. So, M is not an interval; it's just a set with two points. Therefore, option A is incorrect because it's describing an interval instead of the specific set containing -1 and 1.Option B says M is equal to (-1, 1]. This is similar to option A but includes 1. So, this interval includes all numbers from -1 up to and including 1. Again, this is not correct because M is not an interval; it's just the two numbers -1 and 1. So, option B is also incorrect.Option C states that -1 is an element of M. Well, from my earlier solution, M is {-1, 1}, so yes, -1 is indeed in M. That seems correct.Option D says 1 is a subset of M. Wait, that doesn't sound right. The notation here is important. The symbol ⊆ means "is a subset of." So, for 1 to be a subset of M, 1 would have to be a set itself, and every element of 1 would have to be in M. But 1 is just a number, not a set. So, saying 1 is a subset of M is incorrect. Instead, we should be using the element symbol ∈, like in option C. So, option D is wrong because it's using the wrong notation and incorrectly applying the subset concept.So, after evaluating all options, only option C is correct because -1 is indeed an element of M.I think I got this. Just to recap:- Solved the equation x² - 1 = 0 to find M = {-1, 1}.- Checked each option: - A and B are incorrect because they describe intervals, not the specific set. - C is correct because -1 is in M. - D is incorrect because 1 is an element, not a subset.Therefore, the correct answer is option C.**Final Answer**boxed{C}

💡Okay, so I have this problem about a hyperbola. The asymptotes are given as ( y = 2x + 3 ) and ( y = 1 - 2x ). The hyperbola also passes through the point ( (4, 5) ). I need to find the distance between the foci of the hyperbola. Hmm, okay, let's break this down step by step.First, I remember that the asymptotes of a hyperbola intersect at the center of the hyperbola. So, I need to find the point where these two lines intersect. To do that, I can set the two equations equal to each other and solve for ( x ) and ( y ).So, setting ( 2x + 3 = 1 - 2x ). Let me solve for ( x ):( 2x + 3 = 1 - 2x )Adding ( 2x ) to both sides:( 4x + 3 = 1 )Subtracting 3 from both sides:( 4x = -2 )Dividing both sides by 4:( x = -frac{1}{2} )Okay, so ( x = -frac{1}{2} ). Now, plug this back into one of the asymptote equations to find ( y ). Let's use ( y = 2x + 3 ):( y = 2(-frac{1}{2}) + 3 = -1 + 3 = 2 )So, the center of the hyperbola is at ( (-frac{1}{2}, 2) ). Got that.Next, I need to figure out the equation of the hyperbola. Since the asymptotes are ( y = 2x + 3 ) and ( y = -2x + 1 ), I can see that the slopes are ( 2 ) and ( -2 ). This tells me that the hyperbola is oriented vertically because the slopes are positive and negative, and their absolute values are equal. Wait, actually, no. Let me think. If the slopes are positive and negative, it could be either. But the standard form of a hyperbola with vertical transverse axis has asymptotes with slopes ( pm frac{a}{b} ), and for horizontal transverse axis, it's ( pm frac{b}{a} ). Hmm, maybe I need to clarify.Looking at the asymptotes, they have slopes of ( 2 ) and ( -2 ). So, the slopes are ( pm 2 ). That suggests that ( frac{a}{b} = 2 ) if it's a vertical hyperbola or ( frac{b}{a} = 2 ) if it's a horizontal hyperbola. Wait, actually, no. Let me recall: for a hyperbola centered at ( (h, k) ), the asymptotes are ( y - k = pm frac{a}{b}(x - h) ) for a vertical hyperbola, and ( y - k = pm frac{b}{a}(x - h) ) for a horizontal hyperbola.Given that the slopes are ( pm 2 ), so ( frac{a}{b} = 2 ) or ( frac{b}{a} = 2 ). Hmm, so if it's a vertical hyperbola, ( frac{a}{b} = 2 ), so ( a = 2b ). If it's a horizontal hyperbola, ( frac{b}{a} = 2 ), so ( b = 2a ).But how do I know which one it is? Well, looking at the asymptotes, they have a steeper slope, which might suggest a vertical hyperbola, but I'm not sure. Maybe I can figure it out by the point it passes through.Alternatively, maybe I can write the general equation of the hyperbola given the center and the slopes of the asymptotes.So, since the center is ( (-frac{1}{2}, 2) ), and the slopes are ( pm 2 ), the equation of the hyperbola can be written in the form:( frac{(y - k)^2}{a^2} - frac{(x - h)^2}{b^2} = 1 ) for a vertical hyperbola, or( frac{(x - h)^2}{a^2} - frac{(y - k)^2}{b^2} = 1 ) for a horizontal hyperbola.Given the slopes, if it's vertical, then ( frac{a}{b} = 2 ), so ( a = 2b ). If it's horizontal, ( frac{b}{a} = 2 ), so ( b = 2a ).But I don't know which one it is yet. Maybe I can assume one and see if it works with the point ( (4, 5) ).Let me try assuming it's a vertical hyperbola first.So, equation: ( frac{(y - 2)^2}{a^2} - frac{(x + frac{1}{2})^2}{b^2} = 1 ), with ( a = 2b ).So, substituting ( a = 2b ), the equation becomes:( frac{(y - 2)^2}{(2b)^2} - frac{(x + frac{1}{2})^2}{b^2} = 1 )Simplify:( frac{(y - 2)^2}{4b^2} - frac{(x + frac{1}{2})^2}{b^2} = 1 )Now, plug in the point ( (4, 5) ):( frac{(5 - 2)^2}{4b^2} - frac{(4 + frac{1}{2})^2}{b^2} = 1 )Calculate each term:( frac{3^2}{4b^2} - frac{(4.5)^2}{b^2} = 1 )( frac{9}{4b^2} - frac{20.25}{b^2} = 1 )Combine the terms:( frac{9}{4b^2} - frac{81}{4b^2} = 1 ) (since 20.25 is 81/4)( frac{9 - 81}{4b^2} = 1 )( frac{-72}{4b^2} = 1 )Simplify:( frac{-18}{b^2} = 1 )Wait, that gives ( -18 = b^2 ), which is impossible because ( b^2 ) can't be negative. So, that means my assumption that it's a vertical hyperbola is wrong. Therefore, it must be a horizontal hyperbola.Okay, so let's try that. For a horizontal hyperbola, the equation is:( frac{(x - h)^2}{a^2} - frac{(y - k)^2}{b^2} = 1 )With the center at ( (-frac{1}{2}, 2) ), so:( frac{(x + frac{1}{2})^2}{a^2} - frac{(y - 2)^2}{b^2} = 1 )And since the slopes are ( pm 2 ), which is ( frac{b}{a} = 2 ), so ( b = 2a ).Substituting ( b = 2a ) into the equation:( frac{(x + frac{1}{2})^2}{a^2} - frac{(y - 2)^2}{(2a)^2} = 1 )Simplify:( frac{(x + frac{1}{2})^2}{a^2} - frac{(y - 2)^2}{4a^2} = 1 )Now, plug in the point ( (4, 5) ):( frac{(4 + frac{1}{2})^2}{a^2} - frac{(5 - 2)^2}{4a^2} = 1 )Calculate each term:( frac{(4.5)^2}{a^2} - frac{3^2}{4a^2} = 1 )( frac{20.25}{a^2} - frac{9}{4a^2} = 1 )Convert 9/4 to decimal to make it easier:( frac{20.25}{a^2} - frac{2.25}{a^2} = 1 )Combine the terms:( frac{20.25 - 2.25}{a^2} = 1 )( frac{18}{a^2} = 1 )So, ( a^2 = 18 ), which means ( a = sqrt{18} = 3sqrt{2} ).Since ( b = 2a ), then ( b = 2 times 3sqrt{2} = 6sqrt{2} ), so ( b^2 = (6sqrt{2})^2 = 36 times 2 = 72 ).Now, for a hyperbola, the distance between the foci is ( 2c ), where ( c ) is the distance from the center to each focus. The relationship between ( a ), ( b ), and ( c ) is ( c^2 = a^2 + b^2 ).So, let's calculate ( c ):( c^2 = a^2 + b^2 = 18 + 72 = 90 )Therefore, ( c = sqrt{90} = 3sqrt{10} ).Hence, the distance between the foci is ( 2c = 2 times 3sqrt{10} = 6sqrt{10} ).Wait, let me double-check my steps to make sure I didn't make a mistake.1. Found the center by solving the asymptotes: correct, got ( (-frac{1}{2}, 2) ).2. Determined the slopes are ( pm 2 ), considered both vertical and horizontal hyperbola.3. Tried vertical first, got a negative ( b^2 ), which is impossible, so switched to horizontal.4. For horizontal, used ( b = 2a ), substituted into the equation, plugged in the point ( (4,5) ), solved for ( a^2 = 18 ), then ( b^2 = 72 ).5. Calculated ( c = sqrt{90} = 3sqrt{10} ), so distance between foci is ( 6sqrt{10} ).Everything seems to check out. I think that's the correct answer.