answer：First, I need to identify the numbers that Alice is adding together. The first number is (3 + 2i), the second number is (-1 + 5i), and the third number is the conjugate of (4 - 3i). To find the conjugate of (4 - 3i), I change the sign of the imaginary part, resulting in (4 + 3i).Next, I'll add the real parts of the three numbers:[3 + (-1) + 4 = 6]Then, I'll add the imaginary parts:[2i + 5i + 3i = 10i]Finally, I combine the real and imaginary parts to get the total sum:[6 + 10i]

answer：Okay, so I have this matrix B, and I need to compute B raised to the 101st power. Hmm, that seems a bit intimidating at first because 101 is a pretty large exponent. But maybe there's a pattern or something I can use to simplify this.Let me write down the matrix B to get a better look:B = [0 1 0][0 0 1][1 0 0]Alright, so it's a 3x3 matrix. The first thing I notice is that it's a permutation matrix. Each row and each column has exactly one 1 and the rest are 0s. That usually means it's related to some kind of cyclic permutation. Maybe if I can figure out the cycle, I can find a pattern in the powers of B.Let me try computing the first few powers of B to see if there's a repeating pattern.Starting with B^1, which is just B itself:B^1 = [0 1 0][0 0 1][1 0 0]Now, let's compute B^2. To do that, I'll multiply B by itself:B^2 = B * BLet me perform the multiplication step by step.First row of B times first column of B:0*0 + 1*0 + 0*1 = 0First row of B times second column of B:0*1 + 1*0 + 0*0 = 0First row of B times third column of B:0*0 + 1*1 + 0*0 = 1So the first row of B^2 is [0 0 1].Second row of B times first column of B:0*0 + 0*0 + 1*1 = 1Second row of B times second column of B:0*1 + 0*0 + 1*0 = 0Second row of B times third column of B:0*0 + 0*1 + 1*0 = 0So the second row of B^2 is [1 0 0].Third row of B times first column of B:1*0 + 0*0 + 0*1 = 0Third row of B times second column of B:1*1 + 0*0 + 0*0 = 1Third row of B times third column of B:1*0 + 0*1 + 0*0 = 0So the third row of B^2 is [0 1 0].Putting it all together, B^2 is:[0 0 1][1 0 0][0 1 0]Okay, interesting. Now let's compute B^3, which is B^2 multiplied by B.B^3 = B^2 * BAgain, let's do the multiplication step by step.First row of B^2 times first column of B:0*0 + 0*0 + 1*1 = 1First row of B^2 times second column of B:0*1 + 0*0 + 1*0 = 0First row of B^2 times third column of B:0*0 + 0*1 + 1*0 = 0So the first row of B^3 is [1 0 0].Second row of B^2 times first column of B:1*0 + 0*0 + 0*1 = 0Second row of B^2 times second column of B:1*1 + 0*0 + 0*0 = 1Second row of B^2 times third column of B:1*0 + 0*1 + 0*0 = 0So the second row of B^3 is [0 1 0].Third row of B^2 times first column of B:0*0 + 1*0 + 0*1 = 0Third row of B^2 times second column of B:0*1 + 1*0 + 0*0 = 0Third row of B^2 times third column of B:0*0 + 1*1 + 0*0 = 1So the third row of B^3 is [0 0 1].Putting it all together, B^3 is:[1 0 0][0 1 0][0 0 1]Wait a minute, that's the identity matrix! So B^3 equals the identity matrix. That's a crucial observation. If B cubed is the identity, then higher powers of B can be simplified using this.So, if B^3 = I, then B^4 would be B^3 * B = I * B = B. Similarly, B^5 = B^4 * B = B * B = B^2, and B^6 = B^5 * B = B^2 * B = B^3 = I. So the pattern cycles every 3 powers.Therefore, the powers of B cycle every 3 exponents: B^1 = B, B^2 = B^2, B^3 = I, B^4 = B, B^5 = B^2, B^6 = I, and so on.So, to find B^101, I can figure out where 101 falls in this cycle. Since the cycle length is 3, I can divide 101 by 3 and find the remainder.Let me compute 101 divided by 3. 3 goes into 101 how many times? 3*33 = 99, which leaves a remainder of 2. So 101 = 3*33 + 2. Therefore, B^101 = B^(3*33 + 2) = (B^3)^33 * B^2 = I^33 * B^2 = I * B^2 = B^2.So, B^101 is equal to B^2.Earlier, I computed B^2 as:[0 0 1][1 0 0][0 1 0]Therefore, B^101 is this matrix.Let me double-check my calculations to make sure I didn't make any mistakes.First, B^1 is correct. Then, B^2: when I multiplied B by itself, I got the matrix with first row [0 0 1], second row [1 0 0], third row [0 1 0]. That seems right.Then, B^3: multiplying B^2 by B, I ended up with the identity matrix. That makes sense because each element is shifting back to its original position after three permutations.So, since the cycle is every three exponents, and 101 mod 3 is 2, B^101 is indeed B^2.I think that's solid. So, the final answer is B^2, which is:[0 0 1][1 0 0][0 1 0]**Final Answer**boxed{begin{pmatrix} 0 & 0 & 1 1 & 0 & 0 0 & 1 & 0 end{pmatrix}}

answer：First, I recognize that the decimal ( 0.3overline{45} ) consists of a non-repeating part, ( 0.3 ), and a repeating part, ( overline{45} ).To convert the repeating decimal ( 0.overline{45} ) into a fraction, I set ( x = 0.overline{45} ).By multiplying both sides by 100, I get ( 100x = 45.overline{45} ).Subtracting the original equation from this new equation eliminates the repeating part:[100x - x = 45.overline{45} - 0.overline{45}][99x = 45]Solving for ( x ), I find:[x = frac{45}{99} = frac{5}{11}]Next, I add the non-repeating part ( 0.3 ), which is ( frac{3}{10} ), to the fraction ( frac{5}{11} ):[frac{3}{10} + frac{5}{11} = frac{3 cdot 11 + 5 cdot 10}{110} = frac{33 + 50}{110} = frac{83}{110}]Therefore, the decimal ( 0.3overline{45} ) is equivalent to the fraction ( frac{83}{110} ).

answer：Let's solve the problem step by step.**Step 1: Define Variables**Let ( p ) be the total number of students in the group initially. Since 50% of the group are girls, the number of girls initially is ( 0.5p ).**Step 2: Change in Group Composition**After three girls leave and three boys join, the total number of students remains ( p ). The new number of girls is ( 0.5p - 3 ).**Step 3: Setting Up the Equation**After the change, 40% of the group are girls. Therefore, we can set up the equation:[frac{0.5p - 3}{p} = 0.4]**Step 4: Solving the Equation**To solve for ( p ), first, multiply both sides by ( p ) to clear the fraction:[0.5p - 3 = 0.4p]Simplify and solve for ( p ):[0.5p - 0.4p = 3 implies 0.1p = 3 implies p = frac{3}{0.1} = 30]**Step 5: Finding the Initial Number of Girls**Substitute ( p = 30 ) back into the expression for the number of girls:[0.5p = 0.5 times 30 = 15]**Conclusion:**The initial number of girls in the group was ( 15 ).

answer：Okay, so I have this problem where there are three unit vectors: a, b, and c. They satisfy the equation a + b + 2c = 0. I need to find the angle between vectors a and b. Hmm, let me think about how to approach this.First, since all three vectors are unit vectors, their magnitudes are all 1. That might be useful later on. The equation given is a vector equation, so maybe I can manipulate it to find some relationship between a and b.Let me rewrite the equation: a + b + 2c = 0. If I move the 2c to the other side, I get a + b = -2c. So, the sum of vectors a and b is equal to -2 times vector c. Interesting. Since c is a unit vector, -2c is a vector in the opposite direction of c with magnitude 2.Now, I need to find the angle between a and b. I remember that the dot product formula relates the angle between two vectors. The formula is a · b = |a||b|cosθ, where θ is the angle between them. Since a and b are unit vectors, their magnitudes are 1, so this simplifies to a · b = cosθ. So, if I can find the dot product of a and b, I can find the angle θ.But how do I find a · b from the given equation? Well, I have a + b = -2c. Maybe I can take the dot product of both sides with themselves to get some information about a · b.Let me try that. Taking the dot product of both sides with themselves:(a + b) · (a + b) = (-2c) · (-2c)On the left side, expanding the dot product:a · a + 2a · b + b · bSince a and b are unit vectors, a · a = 1 and b · b = 1. So, the left side becomes:1 + 2a · b + 1 = 2 + 2a · bOn the right side, (-2c) · (-2c) is equal to 4(c · c). Since c is a unit vector, c · c = 1, so the right side is 4*1 = 4.Putting it all together:2 + 2a · b = 4Subtracting 2 from both sides:2a · b = 2Dividing both sides by 2:a · b = 1Wait, so the dot product of a and b is 1. But since a and b are unit vectors, the dot product is equal to the cosine of the angle between them. So, cosθ = 1.What angle has a cosine of 1? Well, cos0° = 1, so θ must be 0°. That means the angle between vectors a and b is 0 degrees. Hmm, that seems a bit too straightforward. Let me double-check my steps.Starting from a + b = -2c. Taking the dot product of both sides with themselves:(a + b) · (a + b) = (-2c) · (-2c)Left side: a·a + 2a·b + b·b = 1 + 2a·b + 1 = 2 + 2a·bRight side: 4(c·c) = 4*1 = 4So, 2 + 2a·b = 4 => 2a·b = 2 => a·b = 1Yes, that seems correct. So, cosθ = 1, which implies θ = 0°. So, vectors a and b are in the same direction. That makes sense because when you add them together, you get a vector that's twice the length of each, but in the opposite direction of c.Wait, but if a and b are in the same direction, then their sum would be 2a (or 2b), right? So, 2a = -2c, which implies a = -c. Similarly, b = -c. So, both a and b are equal to -c. That makes sense because then a + b would be -c -c = -2c, which matches the equation given.So, if a and b are both equal to -c, they are indeed in the same direction, so the angle between them is 0°. That seems consistent.I think I'm confident with this result. So, the angle between a and b is 0 degrees.

answer：Okay, so I have this problem where I need to find the quotient when (9x^3 + 18x^2 - 8x + 5) is divided by (3x^2 + 4). Hmm, I remember that polynomial division is similar to long division with numbers, but I need to apply it to polynomials. Let me try to recall the steps.First, I think I need to set it up like a long division problem. The dividend is (9x^3 + 18x^2 - 8x + 5), and the divisor is (3x^2 + 4). So, I write it as:[begin{array}{r|ll}3x^2 + 4 & 9x^3 + 18x^2 - 8x + 5 end{array}]Wait, actually, I think the divisor goes on the outside, and the dividend is inside. Maybe it's better to write it like this:[begin{array}{ll}multicolumn{2}{r}{} cline{2-4}3x^2 + 4 & 9x^3 + 18x^2 - 8x + 5 end{array}]Okay, that looks more familiar. Now, I need to divide the first term of the dividend by the first term of the divisor. The first term of the dividend is (9x^3), and the first term of the divisor is (3x^2). So, (9x^3) divided by (3x^2) is (3x). That should be the first term of the quotient.So, I write (3x) above the division line.[begin{array}{ll}multicolumn{2}{r}{3x} cline{2-4}3x^2 + 4 & 9x^3 + 18x^2 - 8x + 5 end{array}]Next, I need to multiply the entire divisor by this term (3x) and subtract it from the dividend. Let me do that:Multiply (3x) by (3x^2 + 4):(3x times 3x^2 = 9x^3)(3x times 4 = 12x)So, the result is (9x^3 + 12x).Now, subtract this from the dividend:[begin{array}{ll}multicolumn{2}{r}{3x} cline{2-4}3x^2 + 4 & 9x^3 + 18x^2 - 8x + 5 multicolumn{2}{r}{-9x^3 - 12x} cline{2-4}multicolumn{2}{r}{0 + 18x^2 - 20x + 5} end{array}]Wait, let me check that subtraction. Subtracting (9x^3 + 12x) from (9x^3 + 18x^2 - 8x + 5) should give:(9x^3 - 9x^3 = 0)(18x^2 - 0 = 18x^2)(-8x - 12x = -20x)(5 - 0 = 5)So, the new polynomial after subtraction is (18x^2 - 20x + 5).Now, I need to bring down the next term, but since all terms are already there, I just proceed to divide the first term of this new polynomial by the first term of the divisor. The first term is (18x^2), and the divisor's first term is (3x^2). So, (18x^2) divided by (3x^2) is 6. That should be the next term in the quotient.So, I add 6 to the quotient:[begin{array}{ll}multicolumn{2}{r}{3x + 6} cline{2-4}3x^2 + 4 & 9x^3 + 18x^2 - 8x + 5 multicolumn{2}{r}{-9x^3 - 12x} cline{2-4}multicolumn{2}{r}{18x^2 - 20x + 5} end{array}]Now, multiply the entire divisor by 6:(6 times 3x^2 = 18x^2)(6 times 4 = 24)So, the result is (18x^2 + 24).Subtract this from the current polynomial:[begin{array}{ll}multicolumn{2}{r}{3x + 6} cline{2-4}3x^2 + 4 & 9x^3 + 18x^2 - 8x + 5 multicolumn{2}{r}{-9x^3 - 12x} cline{2-4}multicolumn{2}{r}{18x^2 - 20x + 5} multicolumn{2}{r}{-18x^2 - 24} cline{2-4}multicolumn{2}{r}{0 - 20x - 19} end{array}]Wait, let me verify the subtraction:(18x^2 - 18x^2 = 0)(-20x - 0 = -20x)(5 - 24 = -19)So, the remainder is (-20x - 19).Now, I check the degree of the remainder. The remainder is a first-degree polynomial, and the divisor is a second-degree polynomial. Since the degree of the remainder is less than the degree of the divisor, I can't continue dividing further.Therefore, the quotient is (3x + 6) with a remainder of (-20x - 19).But the question only asks for the quotient, so the answer should be (3x + 6).Wait a minute, let me double-check my steps because I might have made a mistake somewhere. When I subtracted (9x^3 + 12x) from the dividend, I got (18x^2 - 20x + 5). Then, dividing (18x^2) by (3x^2) gives 6, which seems correct. Multiplying 6 by the divisor gives (18x^2 + 24), and subtracting that from (18x^2 - 20x + 5) gives (-20x - 19). That seems right.But I'm a bit confused because when I initially thought about it, I thought the quotient might just be (3x). Maybe I made a mistake in the initial setup. Let me try another approach.Alternatively, I can use the method of equating coefficients. Let me assume that the quotient is (ax + b) and the remainder is (cx + d). So, we have:[(3x^2 + 4)(ax + b) + (cx + d) = 9x^3 + 18x^2 - 8x + 5]Expanding the left side:[3x^2 times ax = 3a x^33x^2 times b = 3b x^24 times ax = 4a x4 times b = 4bAdding the remainder:cx + dSo, combining like terms:(3a x^3 + 3b x^2 + (4a + c) x + (4b + d))Now, set this equal to the original polynomial:(9x^3 + 18x^2 - 8x + 5)So, we can set up equations by equating coefficients:1. For (x^3): (3a = 9) => (a = 3)2. For (x^2): (3b = 18) => (b = 6)3. For (x): (4a + c = -8)We already know (a = 3), so:(4*3 + c = -8) => (12 + c = -8) => (c = -20)4. For the constant term: (4b + d = 5)We know (b = 6), so:(4*6 + d = 5) => (24 + d = 5) => (d = -19)So, the quotient is (ax + b = 3x + 6) and the remainder is (cx + d = -20x -19), which matches what I got earlier with polynomial long division.Therefore, the quotient is indeed (3x + 6).Wait, but in my initial long division, I think I might have made a mistake in the subtraction step. Let me go back and check.In the first step, I had:Dividend: (9x^3 + 18x^2 - 8x + 5)Divisor: (3x^2 + 4)First term: (9x^3 / 3x^2 = 3x)Multiply divisor by (3x): (9x^3 + 12x)Subtract from dividend:(9x^3 + 18x^2 - 8x + 5)(-9x^3 -12x)Which gives:(0 + 18x^2 -20x +5)That's correct.Then, divide (18x^2) by (3x^2) to get 6.Multiply divisor by 6: (18x^2 +24)Subtract from (18x^2 -20x +5):(18x^2 -20x +5)(-18x^2 -24)Which gives:(0 -20x -19)So, remainder is (-20x -19), quotient is (3x +6). That seems correct.But earlier, I thought the quotient was just (3x), but that was probably a mistake. So, the correct quotient is (3x +6).Wait, but in the initial problem, the user wrote:"What is the quotient when (9x^3 + 18x^2 - 8x + 5) is divided by (3x^2 + 4)?"And in the previous response, the assistant wrote:"The quotient is (boxed{3x}), and the remainder is (frac{143}{3})."But according to my calculations, the quotient is (3x +6) with a remainder of (-20x -19). There's a discrepancy here. Did I make a mistake?Wait, let me check the initial problem again. The dividend is (9x^3 + 18x^2 -8x +5), and the divisor is (3x^2 +4). So, my calculations seem correct, but the previous answer was different.Alternatively, maybe I misapplied the polynomial long division. Let me try to perform the division step by step again.Divide (9x^3 + 18x^2 -8x +5) by (3x^2 +4).First term: (9x^3 / 3x^2 = 3x). Multiply divisor by (3x): (9x^3 +12x). Subtract from dividend:(9x^3 +18x^2 -8x +5)(-9x^3 -12x)Result: (0 +18x^2 -20x +5).Next term: (18x^2 / 3x^2 =6). Multiply divisor by 6: (18x^2 +24). Subtract:(18x^2 -20x +5)(-18x^2 -24)Result: (0 -20x -19).So, quotient is (3x +6), remainder is (-20x -19).But the previous answer was quotient (3x) and remainder (frac{143}{3}). That doesn't align with my result. Maybe the previous answer was incorrect.Alternatively, perhaps the previous answer considered only the first term of the quotient, which is (3x), but that would be incomplete because the degree of the remainder is still higher than the divisor's degree, so we can continue dividing.Wait, in polynomial division, we continue until the degree of the remainder is less than the degree of the divisor. Since the divisor is degree 2, we can have a quotient term of degree 1 (which is (3x +6)) and a remainder of degree 1, which is acceptable.Therefore, I think my result is correct, and the previous answer was incorrect. The quotient is (3x +6), and the remainder is (-20x -19).But let me check using synthetic division, although synthetic division is typically for linear divisors. Since the divisor is quadratic, maybe I can use a modified approach.Alternatively, I can perform the division using algebraic manipulation.Let me write:(9x^3 +18x^2 -8x +5 = (3x^2 +4)(ax + b) + cx + d)Expanding the right side:(3x^2(ax + b) +4(ax + b) + cx + d)= (3a x^3 + 3b x^2 +4a x +4b + cx + d)Combine like terms:= (3a x^3 + 3b x^2 + (4a + c)x + (4b + d))Set equal to the original polynomial:(9x^3 +18x^2 -8x +5)So, equate coefficients:1. (3a =9) => (a=3)2. (3b=18) => (b=6)3. (4a + c = -8)Substitute (a=3):(12 + c = -8) => (c= -20)4. (4b + d =5)Substitute (b=6):(24 + d =5) => (d= -19)So, the quotient is (3x +6) and the remainder is (-20x -19). This confirms my earlier result.Therefore, the correct quotient is (3x +6), not just (3x). The previous answer was incorrect because it stopped after the first term of the quotient, which is incomplete.So, to summarize, when dividing (9x^3 +18x^2 -8x +5) by (3x^2 +4), the quotient is (3x +6) with a remainder of (-20x -19).