answer：Okay, so I have this system of inequalities to solve. It looks like there are two inequalities here:1. ( x - 2 geq -5 ) (Let's call this Inequality ①)2. ( 3x < x + 2 ) (Let's call this Inequality ②)I need to find the values of ( x ) that satisfy both of these inequalities at the same time. Hmm, okay, let's take it step by step.Starting with Inequality ①: ( x - 2 geq -5 ). I think I need to solve for ( x ) here. To isolate ( x ), I should get rid of that "-2" on the left side. How do I do that? Well, if I add 2 to both sides of the inequality, that should work.So, adding 2 to both sides:( x - 2 + 2 geq -5 + 2 )Simplifying that:( x geq -3 )Okay, so from the first inequality, ( x ) is greater than or equal to -3. Got it.Now, moving on to Inequality ②: ( 3x < x + 2 ). Again, I need to solve for ( x ). Let me see. I have ( 3x ) on the left and ( x + 2 ) on the right. Maybe I can subtract ( x ) from both sides to get all the ( x ) terms on one side.Subtracting ( x ) from both sides:( 3x - x < x + 2 - x )Simplifying that:( 2x < 2 )Alright, now I have ( 2x < 2 ). To solve for ( x ), I should divide both sides by 2.Dividing both sides by 2:( frac{2x}{2} < frac{2}{2} )Which simplifies to:( x < 1 )So, from the second inequality, ( x ) is less than 1.Now, I have two conditions:1. ( x geq -3 )2. ( x < 1 )I need to find the values of ( x ) that satisfy both of these conditions simultaneously. So, ( x ) has to be greater than or equal to -3 and less than 1. Let me visualize this on a number line. Starting from -3, which is included (since it's "greater than or equal to"), and going up to, but not including, 1. So, the solution set is all real numbers from -3 to 1, including -3 but not including 1.To write this in interval notation, it would be ([-3, 1)). But since the question asks for the solution set, I think writing it as an inequality is also acceptable.So, combining both inequalities, the solution is:( -3 leq x < 1 )Let me just double-check my steps to make sure I didn't make any mistakes.For Inequality ①:( x - 2 geq -5 )Adding 2 to both sides:( x geq -3 )That seems correct.For Inequality ②:( 3x < x + 2 )Subtracting ( x ) from both sides:( 2x < 2 )Dividing by 2:( x < 1 )That also looks correct.Combining both, ( x ) has to be between -3 and 1, including -3 but not 1. Yeah, that makes sense.I don't think I missed anything here. So, the solution set is all real numbers from -3 up to, but not including, 1.

answer：Okay, so I have this problem here about ear muffs sales. Let me try to understand what it's asking. It says that before December, customers bought 1,346 ear muffs from the mall. Then, during December, they bought some more, and after that, there were none left. In total, customers bought 7,790 ear muffs. The question is asking how many ear muffs were bought during December.Alright, so I need to find out the number of ear muffs bought in December. Let's break this down. Before December, 1,346 were sold. Then, in December, some more were sold, and the total became 7,790. So, the total is the sum of what was sold before December and what was sold during December.Let me write this out as an equation to make it clearer. Let’s call the number of ear muffs bought during December "D." So, the total ear muffs bought would be:1,346 (before December) + D (during December) = 7,790 (total)So, to find D, I need to subtract the number sold before December from the total. That makes sense because if I know the total and I know part of it, subtracting gives me the other part.So, D = 7,790 - 1,346Let me do that subtraction. 7,790 minus 1,346.Starting from the rightmost digit:0 - 6: Hmm, I can't subtract 6 from 0, so I need to borrow from the next digit to the left. The 9 becomes 8, and the 0 becomes 10.Now, 10 - 6 = 4Next digit: 8 - 4 = 4Next digit: 7 - 3 = 4Next digit: 7 - 1 = 6So, putting it all together, 7,790 - 1,346 equals 6,444.Wait, let me double-check that to make sure I didn't make a mistake. Maybe I should add 1,346 and 6,444 to see if it equals 7,790.1,346 + 6,444Adding the units place: 6 + 4 = 10, write down 0 and carry over 1.Tens place: 4 + 4 = 8, plus the carried over 1 makes 9.Hundreds place: 3 + 4 = 7.Thousands place: 1 + 6 = 7.So, 1,346 + 6,444 = 7,790, which matches the total given. That means my subtraction was correct.Therefore, the number of ear muffs bought during December is 6,444.I think that makes sense. Before December, they sold 1,346, then in December they sold 6,444, and altogether that's 7,790. And since there are none left, that means all the ear muffs were sold by the end of December.Just to visualize it, imagine the mall had a certain number of ear muffs in stock. Before December, they sold 1,346, and then in December, they sold the rest, which was 6,444, leaving nothing. So, the total stock must have been 7,790 ear muffs.Another way to think about it is if you have a total amount and you know part of it, subtracting gives you the other part. It's like having a whole pizza and eating a slice; the remaining slices are what's left after you subtract the slice you ate from the whole pizza.In this case, the whole pizza is the total ear muffs sold, which is 7,790. The slice eaten before December is 1,346, so the remaining slices, or the ear muffs sold in December, are 6,444.I think I've got it. The key steps were identifying what was given (sales before December and total sales), setting up the equation to find the unknown (December sales), performing the subtraction, and then verifying the answer by adding back to ensure it matches the total.It's always good to check your work, especially with arithmetic, to avoid simple mistakes. I did that by adding the two numbers to see if they sum up to the total, and they did, so I'm confident in the answer.So, in summary, to find out how many ear muffs were bought during December, subtract the number bought before December from the total number bought, and that gives the number bought in December.

answer：First, I recognize that the sum of the interior angles of a triangle is always (180^{circ}).Given the ratio of the angles is (1:3:4), I'll let the smallest angle be (alpha). This means the other two angles are (3alpha) and (4alpha).Adding these together: (alpha + 3alpha + 4alpha = 8alpha).Since the total is (180^{circ}), I set up the equation (8alpha = 180^{circ}).Solving for (alpha), I find (alpha = frac{180^{circ}}{8} = 22.5^{circ}).Finally, the largest angle is (4alpha = 4 times 22.5^{circ} = 90^{circ}).

answer：Okay, so I need to find the equation of the tangent line to the curve f(x) = x³ - 3x² + 6 at the point (1, f(1)). Hmm, I remember that to find the tangent line, I need two things: the slope of the tangent at that point and a point through which the tangent passes. I already have the point, which is (1, f(1)), so I just need to find the slope.First, let me find f(1). Plugging x = 1 into the function:f(1) = (1)³ - 3*(1)² + 6 = 1 - 3 + 6 = 4.So the point is (1, 4). Got that down.Now, to find the slope of the tangent line at x = 1, I need to compute the derivative of f(x). The derivative f'(x) gives the slope of the tangent at any point x on the curve.Let me find f'(x). The function is f(x) = x³ - 3x² + 6. Taking the derivative term by term:- The derivative of x³ is 3x².- The derivative of -3x² is -6x.- The derivative of 6 is 0, since the derivative of a constant is zero.So, putting it all together, f'(x) = 3x² - 6x.Now, I need to evaluate this derivative at x = 1 to get the slope at that specific point.f'(1) = 3*(1)² - 6*(1) = 3 - 6 = -3.Okay, so the slope of the tangent line at x = 1 is -3.Now that I have the slope and a point on the line, I can use the point-slope form of a line to write the equation of the tangent. The point-slope form is:y - y₁ = m(x - x₁)Where m is the slope and (x₁, y₁) is the point. Substituting the values I have:y - 4 = -3(x - 1)Let me simplify this equation to get it into a more standard form.First, distribute the -3 on the right side:y - 4 = -3x + 3Now, add 4 to both sides to solve for y:y = -3x + 3 + 4y = -3x + 7Alternatively, I can write this in standard form, which is Ax + By + C = 0. Starting from y = -3x + 7, I can add 3x to both sides:3x + y = 7Then, subtract 7 from both sides:3x + y - 7 = 0So, the equation of the tangent line is 3x + y - 7 = 0.Let me double-check my work to make sure I didn't make any mistakes.First, f(1) = 1 - 3 + 6 = 4. That seems correct.Then, f'(x) = 3x² - 6x. Plugging in x = 1 gives 3 - 6 = -3. That also seems correct.Using point-slope form: y - 4 = -3(x - 1). Distribute the -3: y - 4 = -3x + 3. Add 4: y = -3x + 7. Convert to standard form: 3x + y - 7 = 0. Yep, that all checks out.I think I did everything right. The tangent line at (1, 4) has a slope of -3, and the equation in standard form is 3x + y - 7 = 0.

answer：Okay, so I have these two problems to solve. Let me start with the first one.**Problem 1:** Let ( p(x) ) be any real coefficient polynomial with a positive leading coefficient ( p_0 ) and no real zeros. I need to prove that there exist real coefficient polynomials ( f(x) ) and ( g(x) ) such that ( p(x) = (f(x))^2 + (g(x))^2 ).Hmm, okay. So ( p(x) ) is a real polynomial with no real roots and a positive leading coefficient. Since it has no real roots, all its roots must be complex. And because the coefficients are real, the complex roots must come in conjugate pairs. So if ( a + bi ) is a root, then ( a - bi ) is also a root. Let me think about the degree of ( p(x) ). Since all roots are complex and they come in pairs, the degree must be even. Let's say the degree is ( 2n ). Then, ( p(x) ) can be factored into quadratic factors, each corresponding to a pair of complex conjugate roots. Each quadratic factor would look like ( (x - (a + bi))(x - (a - bi)) = (x - a)^2 + b^2 ). So, ( p(x) ) can be written as the product of these quadratic terms. But how does that help me express ( p(x) ) as a sum of squares of two polynomials? Maybe I can use the fact that each quadratic factor is a sum of squares. For example, ( (x - a)^2 + b^2 ) is already a sum of squares. Wait, but ( p(x) ) is a product of such quadratic terms. So, maybe I can represent ( p(x) ) as a product of sums of squares. But I need it as a single sum of squares, not a product. I recall that the product of sums of squares can be expressed as another sum of squares using the identity ( (a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2 ). Maybe I can apply this recursively to factor ( p(x) ) into a sum of squares.Let me try with a simple case. Suppose ( p(x) ) is a quadratic polynomial with no real roots, so it can be written as ( (x - a)^2 + b^2 ). Then, ( f(x) = x - a ) and ( g(x) = b ) would work, since ( p(x) = f(x)^2 + g(x)^2 ).Now, if ( p(x) ) is of degree 4, it can be written as the product of two quadratics, each of which is a sum of squares. Let’s say ( p(x) = (f_1(x)^2 + g_1(x)^2)(f_2(x)^2 + g_2(x)^2) ). Using the identity I mentioned earlier, this product can be expressed as ( (f_1(x)f_2(x) - g_1(x)g_2(x))^2 + (f_1(x)g_2(x) + f_2(x)g_1(x))^2 ). So, if I define ( f(x) = f_1(x)f_2(x) - g_1(x)g_2(x) ) and ( g(x) = f_1(x)g_2(x) + f_2(x)g_1(x) ), then ( p(x) = f(x)^2 + g(x)^2 ).This seems to work for degree 4. Maybe I can generalize this for any even degree. Since ( p(x) ) is of degree ( 2n ), I can factor it into ( n ) quadratic terms, each of which is a sum of squares. Then, by repeatedly applying the product identity, I can combine them into a single sum of squares.So, inductively, if it works for degree 2 and 4, it should work for any even degree. Therefore, there exist real polynomials ( f(x) ) and ( g(x) ) such that ( p(x) = f(x)^2 + g(x)^2 ).Wait, but I need to make sure that ( f(x) ) and ( g(x) ) are real polynomials. Since each quadratic factor is a sum of squares of real polynomials, and the product identity preserves real coefficients, the resulting ( f(x) ) and ( g(x) ) should indeed be real polynomials.Okay, that makes sense. I think I have a rough idea for Problem 1.**Problem 2:** Prove that if ( Q(x) ) is a real coefficient polynomial with a positive leading coefficient and there exists a real number ( a ) such that ( Q(a) < 0 ), then ( Q(x) ) must have a real zero.Hmm, this seems related to the Intermediate Value Theorem. Since ( Q(x) ) is a polynomial, it's continuous everywhere. If ( Q(a) < 0 ) and the leading coefficient is positive, then as ( x ) approaches infinity, ( Q(x) ) tends to positive infinity. So, there must be some point where ( Q(x) ) crosses from negative to positive, implying a real root.Let me formalize this. Since ( Q(x) ) has a positive leading coefficient, as ( x to infty ), ( Q(x) to +infty ). Similarly, as ( x to -infty ), the behavior depends on the degree. If the degree is even, ( Q(x) to +infty ) as ( x to -infty ) as well. If the degree is odd, ( Q(x) to -infty ) as ( x to -infty ).But regardless, since ( Q(a) < 0 ) for some real ( a ), and ( Q(x) to +infty ) as ( x to infty ), there must be some ( b > a ) such that ( Q(b) > 0 ). By the Intermediate Value Theorem, since ( Q(x) ) is continuous, there exists some ( c ) between ( a ) and ( b ) where ( Q(c) = 0 ). Hence, ( Q(x) ) has a real zero.Alternatively, if I use the result from Problem 1, suppose ( Q(x) ) has no real zeros. Then, by Problem 1, ( Q(x) ) can be written as ( f(x)^2 + g(x)^2 ). But then ( Q(a) = f(a)^2 + g(a)^2 geq 0 ) for all real ( a ), which contradicts the given that ( Q(a) < 0 ). Therefore, ( Q(x) ) must have a real zero.Wait, that's another approach using Problem 1. So, if ( Q(x) ) had no real zeros, it could be expressed as a sum of squares, which would make it non-negative everywhere. But since ( Q(a) < 0 ), it must have a real zero.I think both approaches are valid. The first one using the Intermediate Value Theorem is more elementary, while the second one ties it back to Problem 1, which is also neat.So, summarizing:For Problem 1, since ( p(x) ) has no real roots, it can be factored into quadratic terms, each of which is a sum of squares. Using the product identity for sums of squares, we can express ( p(x) ) as a single sum of squares of two polynomials.For Problem 2, either by the Intermediate Value Theorem or by contradiction using Problem 1, we conclude that ( Q(x) ) must have a real zero.I think I've got a good grasp on both problems now.**Final Answer**(1) boxed{p(x) = f(x)^2 + g(x)^2}(2) boxed{text{Q(x) must have a real root.}}

answer：Alright, so I've got this problem here: Let ( S ) be the set of all positive integer divisors of ( 72,000 ). I need to find how many numbers are the product of two distinct elements of ( S ). The options are A) 445, B) 451, C) 460, D) 470.Hmm, okay. First, I think I need to understand what ( S ) is. Since ( S ) is the set of all positive integer divisors of ( 72,000 ), I should probably start by factoring ( 72,000 ) into its prime components. That should help me figure out how many divisors there are and what they look like.So, let's factor ( 72,000 ). I know that ( 72,000 ) is ( 72 times 1000 ). Breaking that down further, ( 72 ) is ( 8 times 9 ), which is ( 2^3 times 3^2 ), and ( 1000 ) is ( 10^3 ), which is ( (2 times 5)^3 = 2^3 times 5^3 ). So putting it all together, ( 72,000 = 2^6 times 3^2 times 5^3 ). Got that.Now, the number of positive divisors of a number is found by adding one to each of the exponents in its prime factorization and then multiplying those together. So for ( 72,000 ), the exponents are 6, 2, and 3. Adding one to each gives 7, 3, and 4. Multiplying those together: ( 7 times 3 times 4 = 84 ). So there are 84 positive divisors of ( 72,000 ). That means ( S ) has 84 elements.Each element of ( S ) can be written in the form ( 2^a times 3^b times 5^c ) where ( 0 leq a leq 6 ), ( 0 leq b leq 2 ), and ( 0 leq c leq 3 ). So any divisor is determined by choosing exponents ( a ), ( b ), and ( c ) within those ranges.Now, the problem asks for the number of numbers that are the product of two distinct elements of ( S ). So, if I take two distinct divisors ( d_1 ) and ( d_2 ) from ( S ), their product ( d_1 times d_2 ) will be another number, and I need to count how many distinct such numbers there are.Wait, but if I just take all possible products ( d_1 times d_2 ) where ( d_1 ) and ( d_2 ) are distinct, that could potentially result in duplicate numbers, right? So I need to count the distinct products.Hmm, maybe there's a smarter way to approach this rather than enumerating all possible products, which would be 84 choose 2, which is a lot. Instead, perhaps I can think about the prime factorization of the product.Since each ( d_1 ) and ( d_2 ) are of the form ( 2^{a_1} times 3^{b_1} times 5^{c_1} ) and ( 2^{a_2} times 3^{b_2} times 5^{c_2} ) respectively, their product will be ( 2^{a_1 + a_2} times 3^{b_1 + b_2} times 5^{c_1 + c_2} ).So, the exponents in the product will be the sums of the exponents from ( d_1 ) and ( d_2 ). Therefore, the product will be another number whose prime exponents are within certain ranges.Specifically, since ( a_1 ) and ( a_2 ) can each be between 0 and 6, their sum ( a_1 + a_2 ) can be between 0 and 12. Similarly, ( b_1 + b_2 ) can be between 0 and 4, and ( c_1 + c_2 ) can be between 0 and 6.Therefore, the product ( d_1 times d_2 ) can be any number of the form ( 2^p times 3^q times 5^r ) where ( 0 leq p leq 12 ), ( 0 leq q leq 4 ), and ( 0 leq r leq 6 ).Wait a second, that sounds like the prime factorization of ( 72,000^2 ). Let me check: ( 72,000^2 = (2^6 times 3^2 times 5^3)^2 = 2^{12} times 3^4 times 5^6 ). Yep, exactly. So the product ( d_1 times d_2 ) is a divisor of ( 72,000^2 ).Therefore, the number of possible products is equal to the number of divisors of ( 72,000^2 ). But wait, not exactly, because ( d_1 ) and ( d_2 ) are distinct, so we have to exclude cases where ( d_1 = d_2 ). But actually, in the count of divisors of ( 72,000^2 ), each divisor can be expressed as ( d_1 times d_2 ) where ( d_1 ) and ( d_2 ) are divisors of ( 72,000 ), but we need to ensure that ( d_1 ) and ( d_2 ) are distinct.But hold on, the total number of divisors of ( 72,000^2 ) is ( (12 + 1)(4 + 1)(6 + 1) = 13 times 5 times 7 = 455 ). So there are 455 divisors of ( 72,000^2 ). But not all of these can be expressed as the product of two distinct divisors of ( 72,000 ).Specifically, some of these divisors can only be expressed as the product of a divisor with itself, i.e., ( d times d ). So, for example, the divisor 1 can only be expressed as ( 1 times 1 ), and the divisor ( 2^{12} ) can only be expressed as ( 2^6 times 2^6 ), and similarly for ( 3^4 ) and ( 5^6 ). So these four cases cannot be expressed as the product of two distinct divisors.Therefore, the number of products that are the product of two distinct divisors is ( 455 - 4 = 451 ).Wait, but let me make sure. Are there only four such cases where the product can only be expressed as the product of a divisor with itself? Let me think.The number 1 is ( 1 times 1 ), which is the only way to get 1 as a product of two divisors. Similarly, ( 2^{12} ) is ( (2^6) times (2^6) ), which is the only way to get ( 2^{12} ) since ( 2^6 ) is the highest power of 2 in ( 72,000 ). Similarly, ( 3^4 ) is ( (3^2) times (3^2) ), and ( 5^6 ) is ( (5^3) times (5^3) ). So yes, these four divisors can only be expressed as the product of a divisor with itself.Therefore, subtracting these four cases from the total 455 gives us 451, which should be the number of products that are the product of two distinct divisors.Let me double-check my reasoning. The total number of possible products ( d_1 times d_2 ) where ( d_1 ) and ( d_2 ) are divisors of ( 72,000 ) is equal to the number of divisors of ( 72,000^2 ), which is 455. However, among these, the four cases where ( d_1 = d_2 ) (i.e., 1, ( 2^{12} ), ( 3^4 ), and ( 5^6 )) cannot be expressed as the product of two distinct divisors. Therefore, subtracting these four cases gives us 451, which is the number of products that are the product of two distinct elements of ( S ).Yes, that seems correct. So the answer should be 451, which is option B.