💡Okay, so I need to find the inverse of the matrix A, which is given as:[ A = begin{bmatrix} 3 & 4 1 & 2 end{bmatrix} ]Hmm, I remember that to find the inverse of a 2x2 matrix, there's a specific formula. Let me try to recall it. I think it involves the determinant of the matrix and swapping some elements while changing the signs of others. First, I should calculate the determinant of matrix A. The determinant for a 2x2 matrix [ begin{bmatrix} a & b c & d end{bmatrix} ] is calculated as ( ad - bc ). So, applying that to matrix A:The elements are:- a = 3- b = 4- c = 1- d = 2So, determinant ( |A| = (3)(2) - (4)(1) = 6 - 4 = 2 ).Okay, the determinant is 2. Since the determinant is not zero, the matrix is invertible, which is good. If the determinant was zero, the matrix wouldn't have an inverse, and I'd be stuck.Now, the formula for the inverse of a 2x2 matrix is:[ A^{-1} = frac{1}{|A|} begin{bmatrix} d & -b -c & a end{bmatrix} ]So, plugging in the values from matrix A:[ A^{-1} = frac{1}{2} begin{bmatrix} 2 & -4 -1 & 3 end{bmatrix} ]Now, I need to multiply each element of this matrix by ( frac{1}{2} ):- The element in the first row, first column: ( frac{1}{2} times 2 = 1 )- The element in the first row, second column: ( frac{1}{2} times (-4) = -2 )- The element in the second row, first column: ( frac{1}{2} times (-1) = -frac{1}{2} )- The element in the second row, second column: ( frac{1}{2} times 3 = frac{3}{2} )Putting it all together, the inverse matrix ( A^{-1} ) is:[ A^{-1} = begin{bmatrix} 1 & -2 -frac{1}{2} & frac{3}{2} end{bmatrix} ]Let me double-check my calculations to make sure I didn't make any mistakes. Calculating the determinant again: ( 3 times 2 = 6 ) and ( 4 times 1 = 4 ), so ( 6 - 4 = 2 ). That's correct.Then, swapping a and d, and changing the signs of b and c:Original matrix:[ begin{bmatrix} 3 & 4 1 & 2 end{bmatrix} ]After swapping a and d and changing signs:[ begin{bmatrix} 2 & -4 -1 & 3 end{bmatrix} ]Dividing each element by the determinant, which is 2:- 2/2 = 1- -4/2 = -2- -1/2 = -0.5 or -1/2- 3/2 = 1.5 or 3/2So, the inverse matrix is indeed:[ begin{bmatrix} 1 & -2 -frac{1}{2} & frac{3}{2} end{bmatrix} ]Just to be thorough, I can verify this by multiplying the original matrix A by its inverse and checking if the result is the identity matrix.Let's compute ( A times A^{-1} ):[ begin{bmatrix} 3 & 4 1 & 2 end{bmatrix} times begin{bmatrix} 1 & -2 -frac{1}{2} & frac{3}{2} end{bmatrix} ]Calculating element-wise:First row, first column:( 3 times 1 + 4 times (-frac{1}{2}) = 3 - 2 = 1 )First row, second column:( 3 times (-2) + 4 times frac{3}{2} = -6 + 6 = 0 )Second row, first column:( 1 times 1 + 2 times (-frac{1}{2}) = 1 - 1 = 0 )Second row, second column:( 1 times (-2) + 2 times frac{3}{2} = -2 + 3 = 1 )So, the product is:[ begin{bmatrix} 1 & 0 0 & 1 end{bmatrix} ]Which is indeed the identity matrix. That confirms that my inverse matrix is correct.I think I've got it right. The inverse of matrix A is:[ begin{bmatrix} 1 & -2 -frac{1}{2} & frac{3}{2} end{bmatrix} ]**Final Answer**The inverse matrix of ( A ) is boxed{begin{bmatrix} 1 & -2 -frac{1}{2} & frac{3}{2} end{bmatrix}}.

💡Okay, so I have this equation to solve: |y - 8| + 3y = 11. Hmm, absolute value equations can sometimes be tricky because of the modulus, but I think I remember that I need to consider different cases based on the value inside the absolute value. First, let me recall that the absolute value function |x| equals x if x is non-negative and -x if x is negative. So, in this case, |y - 8| will be equal to y - 8 if y - 8 is non-negative, which means y ≥ 8, and it will be equal to -(y - 8) if y - 8 is negative, which means y < 8. So, I guess I need to split this problem into two separate cases: one where y is greater than or equal to 8, and another where y is less than 8. Let me write that down.**Case 1: y ≥ 8**In this case, |y - 8| = y - 8. So, substituting that into the original equation, I get:y - 8 + 3y = 11Let me simplify this equation step by step. Combining like terms:y + 3y = 4ySo, 4y - 8 = 11Now, I'll solve for y:4y = 11 + 84y = 19y = 19/4y = 4.75Wait a minute, I assumed that y is greater than or equal to 8 in this case, but 4.75 is less than 8. That doesn't make sense. So, this solution doesn't satisfy the condition y ≥ 8. Therefore, y = 4.75 is not a valid solution in this case.Hmm, so maybe there's no solution in this case. Let me check my calculations again to make sure I didn't make a mistake.Starting again:|y - 8| + 3y = 11Case 1: y ≥ 8So, |y - 8| = y - 8Substituting:y - 8 + 3y = 11Combine like terms:4y - 8 = 11Add 8 to both sides:4y = 19Divide by 4:y = 19/4 = 4.75Yes, that's correct. But since 4.75 < 8, it doesn't satisfy the condition for this case. So, no solution in this case.**Case 2: y < 8**In this case, |y - 8| = -(y - 8) = -y + 8Substituting into the original equation:-y + 8 + 3y = 11Simplify:(-y + 3y) + 8 = 112y + 8 = 11Now, solve for y:2y = 11 - 82y = 3y = 3/2y = 1.5Okay, so y = 1.5. Let me check if this satisfies the condition y < 8. Yes, 1.5 is less than 8, so that's good.Now, let me verify this solution by plugging it back into the original equation:|1.5 - 8| + 3*(1.5) = | -6.5 | + 4.5 = 6.5 + 4.5 = 11Yes, that works out. So, y = 1.5 is a valid solution.Wait, but in the first case, I got y = 4.75, which didn't satisfy the condition, so it's not a solution. In the second case, I got y = 1.5, which does satisfy the condition and checks out when plugged back into the original equation.So, the only solution is y = 1.5.I think that's it. Let me just recap to make sure I didn't miss anything.I considered both cases for the absolute value expression. In the first case, assuming y ≥ 8 led to a solution that didn't satisfy the condition, so it was discarded. In the second case, assuming y < 8 gave a valid solution that satisfied both the condition and the original equation.Therefore, the solution is y = 1.5.**Final Answer**The solution is boxed{dfrac{3}{2}}.

💡Okay, so I have this function f(x) = x² - 6x + 5, and I need to plot the points P(x, y) where f(x) - f(y) is greater than or equal to zero. Hmm, let me try to break this down step by step.First, I should probably write out what f(x) - f(y) actually is. Since f(x) is x² - 6x + 5, then f(y) would be y² - 6y + 5. So, subtracting f(y) from f(x), I get:f(x) - f(y) = (x² - 6x + 5) - (y² - 6y + 5)Let me simplify that. The +5 and -5 cancel out, so it becomes:f(x) - f(y) = x² - 6x - y² + 6yHmm, okay. Maybe I can rearrange this to make it easier to analyze. Let's group the x terms and the y terms:f(x) - f(y) = (x² - y²) - 6x + 6yNow, I remember that x² - y² can be factored into (x - y)(x + y). Let me try that:f(x) - f(y) = (x - y)(x + y) - 6(x - y)Oh, wait, I see a common factor of (x - y) here. Let me factor that out:f(x) - f(y) = (x - y)(x + y - 6)Alright, so now the expression is factored into two parts: (x - y) and (x + y - 6). The inequality we need to satisfy is f(x) - f(y) ≥ 0, which translates to:(x - y)(x + y - 6) ≥ 0Okay, so this product needs to be greater than or equal to zero. That means either both factors are positive, or both are negative.Let me consider the two cases separately.**Case 1: Both factors are non-negative**- (x - y) ≥ 0- (x + y - 6) ≥ 0So, from the first inequality, x ≥ y. From the second inequality, x + y ≥ 6.**Case 2: Both factors are non-positive**- (x - y) ≤ 0- (x + y - 6) ≤ 0From the first inequality, x ≤ y. From the second inequality, x + y ≤ 6.So, the solution set consists of points where either:1. x ≥ y and x + y ≥ 6, or2. x ≤ y and x + y ≤ 6Now, I need to visualize this on the Cartesian plane.First, let's plot the lines x = y and x + y = 6.- The line x = y is a straight line at 45 degrees passing through the origin.- The line x + y = 6 is a straight line with a slope of -1, intersecting the x-axis at (6, 0) and the y-axis at (0, 6).These two lines intersect at the point where x = y and x + y = 6. Solving these equations:From x = y, substitute into x + y = 6:x + x = 6 ⇒ 2x = 6 ⇒ x = 3 ⇒ y = 3So, they intersect at (3, 3).Now, let's analyze the regions defined by these lines.**For Case 1: x ≥ y and x + y ≥ 6**- x ≥ y is the region above the line x = y.- x + y ≥ 6 is the region above the line x + y = 6.The intersection of these two regions is the area above both lines. This would be the region where both conditions are satisfied, which is the area above the line x + y = 6 and above the line x = y.**For Case 2: x ≤ y and x + y ≤ 6**- x ≤ y is the region below the line x = y.- x + y ≤ 6 is the region below the line x + y = 6.The intersection of these two regions is the area below both lines. This would be the area below the line x + y = 6 and below the line x = y.So, combining both cases, the solution set consists of two regions:1. The region above both lines x = y and x + y = 6.2. The region below both lines x = y and x + y = 6.These regions are separated by the lines x = y and x + y = 6, which intersect at (3, 3).To ensure I'm correct, let me test a point in each region.**Testing Case 1: Above both lines**Let's choose the point (4, 2).- Check x ≥ y: 4 ≥ 2, which is true.- Check x + y ≥ 6: 4 + 2 = 6, which is equal, so it's on the boundary.Now, plug into f(x) - f(y):f(4) = 16 - 24 + 5 = -3f(2) = 4 - 12 + 5 = -3f(4) - f(2) = -3 - (-3) = 0, which satisfies the inequality.Another point, say (5, 1):- x ≥ y: 5 ≥ 1, true.- x + y = 6: 5 + 1 = 6, on the boundary.f(5) = 25 - 30 + 5 = 0f(1) = 1 - 6 + 5 = 0f(5) - f(1) = 0 - 0 = 0, which is okay.Now, a point strictly above both lines, say (4, 3):- x ≥ y: 4 ≥ 3, true.- x + y = 7 ≥ 6, true.f(4) = -3f(3) = 9 - 18 + 5 = -4f(4) - f(3) = -3 - (-4) = 1 ≥ 0, which works.**Testing Case 2: Below both lines**Let's choose the point (2, 4).- x ≤ y: 2 ≤ 4, true.- x + y = 6, on the boundary.f(2) = -3f(4) = -3f(2) - f(4) = -3 - (-3) = 0, which is okay.Another point, say (1, 5):- x ≤ y: 1 ≤ 5, true.- x + y = 6, on the boundary.f(1) = 0f(5) = 0f(1) - f(5) = 0 - 0 = 0, which is fine.A point strictly below both lines, say (2, 2):- x ≤ y: 2 ≤ 2, true.- x + y = 4 ≤ 6, true.f(2) = -3f(2) = -3f(2) - f(2) = 0, which satisfies the inequality.Wait, but (2, 2) is on the line x = y, so it's included.Another point, say (1, 1):- x ≤ y: 1 ≤ 1, true.- x + y = 2 ≤ 6, true.f(1) = 0f(1) = 0f(1) - f(1) = 0, which is okay.And a point like (0, 0):- x ≤ y: 0 ≤ 0, true.- x + y = 0 ≤ 6, true.f(0) = 5f(0) = 5f(0) - f(0) = 0, which is fine.Wait, but f(0) is 5, so f(0) - f(0) is 0, which is okay.But what about a point like (0, 1):- x ≤ y: 0 ≤ 1, true.- x + y = 1 ≤ 6, true.f(0) = 5f(1) = 0f(0) - f(1) = 5 - 0 = 5 ≥ 0, which works.Similarly, (1, 0):- x ≤ y: 1 ≤ 0, false.- So, it's not in the region.But wait, (1, 0) is below x + y = 1, which is less than 6, but x > y here, so it's not in Case 2.So, seems like the regions are correctly identified.I think I've got a good handle on this. The solution set is the union of the regions above both lines and below both lines, including the boundary lines themselves.To summarize:- The inequality f(x) - f(y) ≥ 0 is satisfied by points (x, y) where either: - x ≥ y and x + y ≥ 6, or - x ≤ y and x + y ≤ 6These regions are separated by the lines x = y and x + y = 6, which intersect at (3, 3).So, plotting these regions on the Cartesian plane would show two distinct areas: one above both lines and one below both lines, with the lines themselves included as boundaries.

💡Okay, so I have this problem where a line passes through two vectors, p and q. And I need to find a certain value m such that the vector m times p plus five-eighths times q lies on that line. Hmm, let me try to visualize this.First, I remember that a line in vector form can be parameterized. That usually looks something like p plus some scalar multiple of the direction vector. The direction vector between p and q would be q minus p, right? So, the parametric equation of the line should be p plus s times (q minus p), where s is a scalar parameter.Let me write that down:Line equation: p + s(q - p)Now, I need this line to pass through the vector m*p + (5/8)*q. So, I can set up an equation where m*p + (5/8)*q equals p + s(q - p). That should help me solve for m.So, setting them equal:m*p + (5/8)*q = p + s(q - p)Let me simplify the right side:p + s(q - p) = p + s*q - s*p = (1 - s)*p + s*qSo now, the equation is:m*p + (5/8)*q = (1 - s)*p + s*qSince p and q are vectors, their coefficients must be equal on both sides for the equation to hold. So, I can set up two equations:1. For the coefficient of p: m = 1 - s2. For the coefficient of q: 5/8 = sSo, from the second equation, s is 5/8. Plugging that into the first equation:m = 1 - (5/8) = 3/8Wait, so m is 3/8? Let me check that again.If s is 5/8, then the point on the line is p + (5/8)(q - p) = p + (5/8)q - (5/8)p = (1 - 5/8)p + (5/8)q = (3/8)p + (5/8)q. Yeah, that matches m*p + (5/8)q, so m must be 3/8.I think that makes sense. The parameter s is 5/8, which means we're moving 5/8 of the way from p to q along the line. So, the coefficient of p would be 1 - 5/8, which is 3/8. Therefore, m is 3/8.I don't see any mistakes in my reasoning. The steps seem logical: parameterize the line, set the given vector equal to the parametric form, equate coefficients, solve for m. Yeah, I'm confident that m is 3/8.**Final Answer**The value of ( m ) is boxed{dfrac{3}{8}}.

💡Alright, so I have this trigonometry problem here: I need to find the value of sin 600 degrees plus tan 240 degrees, and then choose the correct option from A to D. Let me think about how to approach this step by step.First, I remember that trigonometric functions are periodic, which means their values repeat after certain intervals. For sine and cosine, the period is 360 degrees, while for tangent, the period is 180 degrees. So, maybe I can reduce these angles to something within the first rotation (0 to 360 degrees) to make it easier to compute.Starting with sin 600 degrees. Since the sine function has a period of 360 degrees, I can subtract 360 degrees from 600 degrees to find an equivalent angle. Let's do that: 600 minus 360 is 240 degrees. So, sin 600 degrees is the same as sin 240 degrees.Now, sin 240 degrees. Hmm, 240 degrees is in the third quadrant where sine is negative. I also know that 240 degrees is 180 degrees plus 60 degrees. So, sin 240 degrees is the same as sin (180 + 60) degrees. Using the sine addition formula, sin (180 + θ) is equal to -sin θ. Therefore, sin 240 degrees is -sin 60 degrees.I remember that sin 60 degrees is √3/2, so sin 240 degrees is -√3/2. Therefore, sin 600 degrees is also -√3/2.Okay, that takes care of the sine part. Now, moving on to tan 240 degrees. Let's see, 240 degrees is also in the third quadrant where tangent is positive because both sine and cosine are negative there, and tangent is sine over cosine. So, tan 240 degrees is positive.Again, 240 degrees can be expressed as 180 degrees plus 60 degrees. So, tan 240 degrees is tan (180 + 60) degrees. Using the tangent addition formula, tan (180 + θ) is equal to tan θ. Therefore, tan 240 degrees is tan 60 degrees.I recall that tan 60 degrees is √3. So, tan 240 degrees is √3.Now, putting it all together: sin 600 degrees plus tan 240 degrees is equal to (-√3/2) + √3.Hmm, let me compute that. To add these two terms, I need to have a common denominator. The first term is -√3/2, and the second term is √3, which can be written as 2√3/2. So, adding them together:-√3/2 + 2√3/2 = ( -√3 + 2√3 ) / 2 = (√3)/2.So, the value of sin 600 degrees plus tan 240 degrees is √3/2.Looking back at the options provided:A: -√3/2B: √3/2C: -1/2 + √3D: 1/2 + √3The correct answer is B: √3/2.Wait a second, let me double-check my steps to make sure I didn't make a mistake. Starting with sin 600 degrees: 600 minus 360 is 240, correct. Sin 240 is in the third quadrant, so negative. 240 is 180 + 60, so sin 240 is -sin 60, which is -√3/2. That seems right.Tan 240 degrees: 240 is 180 + 60, so tan 240 is tan 60, which is √3. That also seems correct.Adding them: -√3/2 + √3. Converting √3 to 2√3/2, so -√3/2 + 2√3/2 is indeed √3/2. So, yes, the answer is B.I think that's solid. I don't see any errors in my reasoning.**Final Answer**boxed{B}

💡First, I'll denote the unknown divisor as ( d ).From the problem, when ( x ) is divided by ( d ), the remainder is 5. This can be expressed as:[x = kd + 5]where ( k ) is an integer.Next, when ( 7x ) is divided by ( d ), the remainder is 8. This can be written as:[7x = jd + 8]where ( j ) is another integer.Substituting the expression for ( x ) from the first equation into the second equation:[7(kd + 5) = jd + 8]Expanding this, we get:[7kd + 35 = jd + 8]Rearranging the equation to isolate terms involving ( d ):[7kd - jd = 8 - 35][(7k - j)d = -27]Since ( d ) must be a positive integer, we can rewrite this as:[(7k - j)d = 27]Now, ( d ) must be a divisor of 27. The possible values for ( d ) are 1, 3, 9, and 27.However, the remainder when dividing by ( d ) must be less than ( d ). Given that the remainder is 5, ( d ) must be greater than 5. Therefore, the possible values for ( d ) are 9 and 27.Testing these values:- If ( d = 9 ): [ x = 9k + 5 ] [ 7x = 7(9k + 5) = 63k + 35 ] [ 63k + 35 = 9j + 8 ] [ 35 = 9j + 8 - 63k ] This simplifies to: [ 35 equiv 8 pmod{9} ] [ 35 mod 9 = 8 ] This holds true.- If ( d = 27 ): [ x = 27k + 5 ] [ 7x = 7(27k + 5) = 189k + 35 ] [ 189k + 35 = 27j + 8 ] [ 35 = 27j + 8 - 189k ] This simplifies to: [ 35 equiv 8 pmod{27} ] [ 35 mod 27 = 8 ] This also holds true.Since both 9 and 27 satisfy the conditions, the smallest possible divisor is 9.