💡Okay, so I have this problem about finding the number of polynomials of the form ( x^4 + ax^3 + bx^2 + cx + 2400 ), where ( a ), ( b ), and ( c ) are real numbers. The condition is that whenever ( r ) is a root, then ( frac{-1 + isqrt{3}}{2} cdot r ) is also a root. Hmm, that seems a bit tricky, but let me try to break it down.First, I recognize that ( frac{-1 + isqrt{3}}{2} ) is a complex number. In fact, it's one of the cube roots of unity, often denoted as ( omega ). The cube roots of unity are ( 1 ), ( omega ), and ( omega^2 ), where ( omega = frac{-1 + isqrt{3}}{2} ) and ( omega^2 = frac{-1 - isqrt{3}}{2} ). These satisfy the equation ( omega^3 = 1 ) and ( 1 + omega + omega^2 = 0 ).Given that whenever ( r ) is a root, so is ( omega cdot r ), this suggests that the roots of the polynomial come in cycles of three. That is, if ( r ) is a root, then ( omega r ) and ( omega^2 r ) must also be roots. However, the polynomial is of degree 4, which complicates things because 4 isn't a multiple of 3. So, how does this work?Let me think. If the polynomial has real coefficients, then complex roots must come in conjugate pairs. So, if ( r ) is a complex root, then its complex conjugate ( overline{r} ) must also be a root. But in this case, the transformation ( r mapsto omega r ) is given, which is a rotation in the complex plane by 120 degrees. So, if ( r ) is a root, then ( omega r ) and ( omega^2 r ) must also be roots. This forms a set of three roots: ( r ), ( omega r ), and ( omega^2 r ).But since the polynomial is of degree 4, we can't have just three roots; we need a fourth root. Let me consider the possibilities for the fourth root. It could either be another root that's transformed by ( omega ), but that would require another set of three roots, which would make the polynomial degree 6, which is not the case here. Alternatively, the fourth root could be a real number because real roots don't need to be accompanied by their complex conjugates—they are their own conjugates.Wait, but if the fourth root is real, say ( s ), then does ( omega s ) also have to be a root? Because the condition says that whenever ( r ) is a root, so is ( omega r ). So, if ( s ) is a real root, then ( omega s ) must also be a root. But ( omega s ) is complex unless ( s = 0 ). If ( s = 0 ), then ( omega cdot 0 = 0 ), so 0 would just be a repeated root. But in that case, the polynomial would have a root at 0 with multiplicity 1 or more.Wait, but if 0 is a root, then the constant term of the polynomial would be 0. However, in our polynomial, the constant term is 2400, which is not zero. So, 0 cannot be a root. Therefore, the fourth root cannot be a real number because that would require ( omega cdot s ) to be a root, which would have to be complex, but then we would have more than four roots, which is impossible.Hmm, so maybe the fourth root has to coincide with one of the existing roots? That is, perhaps one of the roots is repeated. So, if ( r ) is a root, then ( omega r ) and ( omega^2 r ) are also roots, and the fourth root is ( r ) again. But then, the polynomial would have a root ( r ) with multiplicity 2, and ( omega r ) and ( omega^2 r ) each with multiplicity 1. But then, the polynomial would have degree 4, which fits.Wait, but if ( r ) is a root with multiplicity 2, then ( omega r ) and ( omega^2 r ) would each have multiplicity 1, but then the polynomial would have roots ( r, r, omega r, omega^2 r ). But then, the polynomial would be ( (x - r)^2 (x - omega r)(x - omega^2 r) ). Let me compute this.First, ( (x - omega r)(x - omega^2 r) = x^2 - (omega + omega^2) r x + (omega cdot omega^2) r^2 ). Since ( omega + omega^2 = -1 ) and ( omega cdot omega^2 = 1 ), this simplifies to ( x^2 + r x + r^2 ).Then, the polynomial becomes ( (x - r)^2 (x^2 + r x + r^2) ). Let me expand this:First, ( (x - r)^2 = x^2 - 2 r x + r^2 ).Multiplying this by ( x^2 + r x + r^2 ):( (x^2 - 2 r x + r^2)(x^2 + r x + r^2) ).Let me compute this step by step:1. Multiply ( x^2 ) by each term in the second polynomial: - ( x^2 cdot x^2 = x^4 ) - ( x^2 cdot r x = r x^3 ) - ( x^2 cdot r^2 = r^2 x^2 )2. Multiply ( -2 r x ) by each term in the second polynomial: - ( -2 r x cdot x^2 = -2 r x^3 ) - ( -2 r x cdot r x = -2 r^2 x^2 ) - ( -2 r x cdot r^2 = -2 r^3 x )3. Multiply ( r^2 ) by each term in the second polynomial: - ( r^2 cdot x^2 = r^2 x^2 ) - ( r^2 cdot r x = r^3 x ) - ( r^2 cdot r^2 = r^4 )Now, let's combine all these terms:- ( x^4 )- ( r x^3 - 2 r x^3 = -r x^3 )- ( r^2 x^2 - 2 r^2 x^2 + r^2 x^2 = 0 )- ( -2 r^3 x + r^3 x = -r^3 x )- ( r^4 )So, the polynomial simplifies to ( x^4 - r x^3 - r^3 x + r^4 ).But our original polynomial is ( x^4 + a x^3 + b x^2 + c x + 2400 ). Comparing coefficients, we have:- Coefficient of ( x^4 ): 1 in both, which matches.- Coefficient of ( x^3 ): ( -r ) vs. ( a ). So, ( a = -r ).- Coefficient of ( x^2 ): 0 vs. ( b ). So, ( b = 0 ).- Coefficient of ( x ): ( -r^3 ) vs. ( c ). So, ( c = -r^3 ).- Constant term: ( r^4 ) vs. 2400. So, ( r^4 = 2400 ).Therefore, ( r ) must satisfy ( r^4 = 2400 ). Since ( r ) is a real number (because the coefficients are real and the roots come in complex conjugate pairs, but in this case, we've already accounted for the complex roots via the ( omega ) transformation), ( r ) can be either the positive or negative real fourth root of 2400.So, ( r = sqrt[4]{2400} ) or ( r = -sqrt[4]{2400} ). Let me compute ( sqrt[4]{2400} ). First, ( 2400 = 16 times 150 = 16 times 25 times 6 = 16 times 25 times 6 ). So, ( sqrt[4]{2400} = sqrt{sqrt{2400}} ). Let's compute ( sqrt{2400} ):( sqrt{2400} = sqrt{16 times 150} = 4 sqrt{150} ). Then, ( sqrt{150} = sqrt{25 times 6} = 5 sqrt{6} ). So, ( sqrt{2400} = 4 times 5 sqrt{6} = 20 sqrt{6} ). Therefore, ( sqrt[4]{2400} = sqrt{20 sqrt{6}} ).Hmm, that's a bit messy, but it's a real number. So, ( r ) can be ( sqrt[4]{2400} ) or ( -sqrt[4]{2400} ). Therefore, there are two possible values for ( r ), which would give two different polynomials.Wait, but hold on. Let me check if both ( r ) and ( -r ) would give distinct polynomials. If ( r ) is positive, then ( a = -r ), and if ( r ) is negative, then ( a = -r ) would be positive. Similarly, ( c = -r^3 ) would be negative if ( r ) is positive, and positive if ( r ) is negative. So, yes, these would result in different polynomials.But wait, earlier I thought that the fourth root had to coincide with one of the existing roots, but now I see that by having a repeated root, we can satisfy the condition. So, in this case, the polynomial would have roots ( r, r, omega r, omega^2 r ), which are four roots in total, and the polynomial would have real coefficients because the complex roots come in conjugate pairs.But let me double-check if this is the only possibility. Suppose instead that the fourth root is another root, say ( s ), which is not related to ( r ) by the ( omega ) transformation. But then, according to the given condition, ( omega s ) must also be a root, which would require another root, and so on, leading to more than four roots, which is impossible. Therefore, the only way to satisfy the condition with a degree 4 polynomial is to have one root repeated, and the other two roots being ( omega r ) and ( omega^2 r ).Therefore, there are two possible polynomials, corresponding to ( r = sqrt[4]{2400} ) and ( r = -sqrt[4]{2400} ). Hence, the number of such polynomials is 2.Wait, but hold on again. Let me think about the constant term. The constant term is ( r^4 ), which is equal to 2400. So, ( r^4 = 2400 ). Therefore, ( r ) can be any real fourth root of 2400, which are ( sqrt[4]{2400} ) and ( -sqrt[4]{2400} ). So, that's two real roots. Therefore, there are two possible polynomials.But wait, in the polynomial ( (x - r)^2 (x^2 + r x + r^2) ), the coefficient ( b ) is zero. But in the original polynomial, the coefficient ( b ) is arbitrary. So, does this mean that ( b ) must be zero? Yes, according to the expansion, the coefficient of ( x^2 ) is zero. Therefore, in the given polynomial, ( b ) must be zero. So, the polynomial must have ( b = 0 ).Therefore, the only polynomials that satisfy the given condition are those where ( b = 0 ), and ( r ) is either ( sqrt[4]{2400} ) or ( -sqrt[4]{2400} ). Therefore, there are two such polynomials.Wait, but let me check if ( r = 0 ) is a possibility. If ( r = 0 ), then the polynomial would have a root at 0 with multiplicity 4, but the constant term would be zero, which contradicts the given constant term of 2400. Therefore, ( r ) cannot be zero.Therefore, the only possibilities are ( r = sqrt[4]{2400} ) and ( r = -sqrt[4]{2400} ), leading to two distinct polynomials. Hence, the answer should be 2, which is option C.But wait, let me think again. The problem says "whenever ( r ) is a root, so is ( omega r )". So, if ( r ) is a root, then ( omega r ) is also a root. But if ( r ) is zero, then ( omega r = 0 ), which is still a root. However, as we saw earlier, the constant term would be zero, which is not the case here. Therefore, ( r ) cannot be zero.Alternatively, suppose that the polynomial has a root ( r ), then ( omega r ) and ( omega^2 r ) must also be roots. But since the polynomial is of degree 4, we can't have all three roots unless one of them is repeated. So, the only way is to have one root repeated, as I considered earlier.Therefore, there are two possible polynomials, corresponding to ( r ) being positive or negative. Hence, the answer is 2, which is option C.Wait, but in the initial expansion, I got that the polynomial is ( x^4 - r x^3 - r^3 x + r^4 ). Comparing to the given polynomial ( x^4 + a x^3 + b x^2 + c x + 2400 ), we have:- ( a = -r )- ( b = 0 )- ( c = -r^3 )- ( r^4 = 2400 )So, ( r ) must satisfy ( r^4 = 2400 ), which has two real solutions: ( r = sqrt[4]{2400} ) and ( r = -sqrt[4]{2400} ). Therefore, there are two polynomials, one with ( r = sqrt[4]{2400} ) and another with ( r = -sqrt[4]{2400} ). Hence, the number of such polynomials is 2.But wait, let me check if these two polynomials are distinct. If ( r ) is positive, then ( a = -r ) is negative, and ( c = -r^3 ) is negative. If ( r ) is negative, then ( a = -r ) is positive, and ( c = -r^3 ) is positive. Therefore, the coefficients ( a ) and ( c ) would have opposite signs in the two cases, making the polynomials distinct. Therefore, there are indeed two distinct polynomials.Therefore, the answer is 2, which corresponds to option C.

💡Okay, so I have this problem here with two vectors, a and b. Vector a is (-3, 4) and vector b is (1, m). The question says that vector a is perpendicular to (a - b), and I need to find the value of m. The options are A: 11/2, B: 7, C: -7, D: -11/2.Hmm, perpendicular vectors. I remember that if two vectors are perpendicular, their dot product is zero. So, if vector a is perpendicular to (a - b), then the dot product of a and (a - b) should be zero. That makes sense.Let me write that down. So, a · (a - b) = 0.First, I need to find the vector (a - b). Since a is (-3, 4) and b is (1, m), subtracting b from a would be:a - b = (-3 - 1, 4 - m) = (-4, 4 - m).Okay, so now I have vector a as (-3, 4) and vector (a - b) as (-4, 4 - m). Now, I need to compute their dot product and set it equal to zero.The dot product of two vectors (x1, y1) and (x2, y2) is x1*x2 + y1*y2. So, applying that here:a · (a - b) = (-3)*(-4) + 4*(4 - m).Let me compute each part step by step. First, (-3)*(-4) is 12. Then, 4*(4 - m) is 16 - 4m. So, adding those together:12 + 16 - 4m = 0.Wait, 12 + 16 is 28, so 28 - 4m = 0.Now, I need to solve for m. Let me subtract 28 from both sides:-4m = -28.Then, divide both sides by -4:m = (-28)/(-4) = 7.So, m is 7. Let me check the options. Option B is 7, so that should be the answer.But just to make sure I didn't make any mistakes, let me go through the steps again.1. Vector a is (-3, 4), vector b is (1, m).2. Vector (a - b) is (-3 - 1, 4 - m) = (-4, 4 - m).3. Dot product of a and (a - b) is (-3)*(-4) + 4*(4 - m).4. Calculating each term: (-3)*(-4) = 12, 4*(4 - m) = 16 - 4m.5. Adding them: 12 + 16 - 4m = 28 - 4m.6. Setting equal to zero: 28 - 4m = 0.7. Solving: -4m = -28 => m = 7.Yep, that seems correct. I don't think I made any calculation errors. So, m is 7, which is option B.Just another thought: sometimes when dealing with perpendicular vectors, it's good to visualize. If vector a is (-3, 4), that's in the second quadrant. Vector (a - b) is (-4, 4 - m). For them to be perpendicular, their slopes should be negative reciprocals. Maybe I can check that as another method.The slope of vector a is (4)/(-3) = -4/3. The slope of vector (a - b) is (4 - m)/(-4). For them to be perpendicular, the product of their slopes should be -1.So, (-4/3) * [(4 - m)/(-4)] = -1.Simplify this:(-4/3) * [(4 - m)/(-4)] = (-4/3) * [-(4 - m)/4] = (-4/3) * [(-4 + m)/4] = (-4/3) * [(-4 + m)/4].Simplify numerator and denominator:The 4 in the numerator and denominator cancels out, so we have (-4/3) * [(-4 + m)/4] = (-1/3) * (-4 + m).So, (-1/3)*(-4 + m) = (4 - m)/3.Set this equal to -1:(4 - m)/3 = -1.Multiply both sides by 3:4 - m = -3.Subtract 4 from both sides:-m = -7.Multiply both sides by -1:m = 7.Same result! So, that confirms it. Using the slope method also gives m = 7. So, I'm confident that the answer is 7, which is option B.I think I've covered all the steps and even double-checked using a different method. I don't see any mistakes in my reasoning, so I'm pretty sure m is 7.**Final Answer**boxed{B}

💡First, I need to determine the width of the metallic sheet. The sheet has a length of 48 meters and an unknown width, which I'll denote as ( w ) meters.When a square of side 3 meters is cut off from each corner of the metallic sheet, the dimensions of the resulting open box are affected. Specifically:- The length of the box becomes ( 48 - 2 times 3 = 42 ) meters.- The width of the box becomes ( w - 2 times 3 = w - 6 ) meters.- The height of the box is equal to the side of the square cut off, which is 3 meters.The volume ( V ) of the box is given by the formula:[V = text{length} times text{width} times text{height}]Substituting the known values:[3780 = 42 times (w - 6) times 3]Simplifying the equation:[3780 = 126 times (w - 6)][frac{3780}{126} = w - 6][30 = w - 6][w = 30 + 6][w = 36]Therefore, the width of the metallic sheet is 36 meters.

💡First, I need to determine how many questions my friend missed. I know that I missed 5 times as many questions as my friend, and I missed 36 questions.To find out how many questions my friend missed, I can set up the equation:Friend's missed questions = My missed questions / 5Plugging in the numbers:Friend's missed questions = 36 / 5 = 7.2Since it's not possible to miss a fraction of a question, I'll round 7.2 to the nearest whole number, which is 7.Now, to find the total number of questions missed together, I'll add my missed questions to my friend's missed questions:Total missed questions = 36 + 7 = 43Therefore, together, we missed a total of 43 questions.

💡First, I need to calculate the square root of 0.000001. Recognizing that 0.000001 is equivalent to (10^{-6}), taking the square root gives me (10^{-3}).Next, I need to find the cube root of (10^{-3}). The cube root of (10^{-3}) is (10^{-1}), which is equal to 0.1.Finally, I express the result as a decimal to the nearest thousandth, which is 0.100.

💡Okay, so I have this problem with four propositions about lines and planes, and I need to figure out which ones are true. Let me go through each one step by step.**Proposition 1:** If line ( l ) is perpendicular to plane ( alpha ) and line ( m ) is also perpendicular to plane ( alpha ), then ( l ) is parallel to ( m ).Hmm, I remember that if two lines are both perpendicular to the same plane, they must be parallel. Because they're both going in the same direction, perpendicular to the plane. So, yeah, this seems true. I think I can mark this as correct.**Proposition 2:** If line ( m ) is contained in plane ( beta ), line ( n ) is the projection of line ( l ) onto plane ( beta ), and ( m ) is perpendicular to ( l ), then ( m ) is perpendicular to ( n ).Alright, projections can be tricky. So, if ( n ) is the projection of ( l ) onto ( beta ), that means ( n ) lies in ( beta ) and is the shadow of ( l ) when light is shone perpendicular to ( beta ). Now, if ( m ) is in ( beta ) and perpendicular to ( l ), does that mean it's also perpendicular to ( n )?I think about the three perpendiculars theorem. It says that if a line is perpendicular to a plane, then it's perpendicular to every line in that plane. But here, ( m ) is in ( beta ) and perpendicular to ( l ). Since ( n ) is the projection, which is like the component of ( l ) in ( beta ), maybe ( m ) is also perpendicular to ( n ). I'm not entirely sure, but I think this might be true. I'll tentatively say it's correct.**Proposition 3:** If line ( m ) is contained in plane ( alpha ) and line ( n ) is parallel to ( m ), then ( n ) is parallel to ( alpha ).Wait, if ( n ) is parallel to ( m ), which is in ( alpha ), does that mean ( n ) is parallel to the entire plane? I think not necessarily. Because ( n ) could be parallel to ( m ) but still lie within another plane that intersects ( alpha ). So, ( n ) might not be parallel to ( alpha ); it could be skew or intersecting. So, this proposition doesn't seem correct. I think it's false.**Proposition 4:** If plane ( alpha ) is perpendicular to plane ( gamma ) and plane ( beta ) is also perpendicular to plane ( gamma ), then ( alpha ) is parallel to ( beta ).Hmm, if two planes are both perpendicular to a third plane, does that make them parallel? I don't think so. For example, think of three mutually perpendicular planes, like the walls and floor in a room. The two walls are both perpendicular to the floor, but they intersect each other, so they aren't parallel. Therefore, this proposition is false.So, summarizing:1. True2. True3. False4. FalseTherefore, the correct propositions are ① and ②, which corresponds to option A.**Final Answer**boxed{A}