answer：Okay, so I have this problem here: "If the graph of the quadratic function ( y = (m + 1)x^{|m|} ) opens downwards, then the value of ( m ) is ____." Hmm, I need to figure out what ( m ) is. Let me think step by step.First, I remember that a quadratic function is generally in the form ( y = ax^2 + bx + c ). But in this case, the function is given as ( y = (m + 1)x^{|m|} ). So, it's a bit different because the exponent is ( |m| ) instead of 2. Wait, but for it to be a quadratic function, the exponent should be 2, right? Because quadratic means degree 2.So, maybe the first thing I need to do is ensure that the exponent ( |m| ) is equal to 2. That way, the function is quadratic. Let me write that down:( |m| = 2 )Solving for ( m ), this equation has two solutions: ( m = 2 ) or ( m = -2 ). So, ( m ) can be either 2 or -2.But wait, the problem also says that the graph opens downwards. I remember that for a quadratic function ( y = ax^2 + bx + c ), the direction it opens depends on the coefficient ( a ). If ( a ) is positive, it opens upwards, and if ( a ) is negative, it opens downwards.In our function, the coefficient is ( (m + 1) ). So, for the graph to open downwards, ( (m + 1) ) must be negative. Let me write that condition:( m + 1 < 0 )Solving this inequality:( m < -1 )Okay, so ( m ) has to be less than -1. Now, let's go back to our earlier solutions for ( m ): 2 and -2. Which of these satisfy ( m < -1 )?Looking at ( m = 2 ): 2 is not less than -1, so that doesn't work.Looking at ( m = -2 ): -2 is less than -1, so that works.Therefore, ( m ) must be -2.Wait, let me double-check to make sure I didn't make a mistake. If ( m = -2 ), then the function becomes:( y = (-2 + 1)x^{|-2|} = (-1)x^{2} )Which simplifies to ( y = -x^2 ). That's a quadratic function that opens downwards, which is exactly what the problem is asking for. So, that seems correct.Just to be thorough, let me check if ( m = 2 ) would work. If ( m = 2 ), the function becomes:( y = (2 + 1)x^{|2|} = 3x^2 )That opens upwards, which is the opposite of what we want. So, yeah, ( m = 2 ) is not a solution.Is there any other possible value for ( m )? Well, we started by setting ( |m| = 2 ), which only gives ( m = 2 ) or ( m = -2 ). There are no other real numbers that satisfy ( |m| = 2 ). So, those are the only two possibilities, and only ( m = -2 ) satisfies the condition that the graph opens downwards.Therefore, I'm confident that the value of ( m ) is -2.**Final Answer**The value of ( m ) is boxed{-2}.

answer：Okay, so I have this problem where there are two circles, C₁ and C₂. C₁ is given by the equation x² + y² = 5, and C₂ is given by x² + y² - 4x - 8y - m = 0. The problem says that these two circles are externally tangent to each other, and I need to find the real number m.First, I should recall what it means for two circles to be externally tangent. If two circles are externally tangent, it means they touch each other at exactly one point, and the distance between their centers is equal to the sum of their radii. So, I need to find the centers and radii of both circles, then set up an equation where the distance between the centers is equal to the sum of the radii, and solve for m.Starting with circle C₁: x² + y² = 5. This is a standard equation of a circle centered at the origin (0,0) with radius sqrt(5). Let me write that down:Center of C₁: (0, 0)Radius of C₁: r₁ = √5Now, moving on to circle C₂: x² + y² - 4x - 8y - m = 0. This equation isn't in the standard form, so I need to rewrite it by completing the squares for both x and y terms.Let me rearrange the equation:x² - 4x + y² - 8y = mNow, complete the square for x:Take the coefficient of x, which is -4, divide by 2 to get -2, square it to get 4. So, I add and subtract 4.Similarly, for y:Take the coefficient of y, which is -8, divide by 2 to get -4, square it to get 16. So, I add and subtract 16.So, rewriting the equation:(x² - 4x + 4) - 4 + (y² - 8y + 16) - 16 = mSimplify:(x - 2)² + (y - 4)² - 20 = mBring the constant to the right side:(x - 2)² + (y - 4)² = m + 20So, the standard form of C₂ is (x - 2)² + (y - 4)² = m + 20.Therefore, the center of C₂ is (2, 4), and the radius r₂ is sqrt(m + 20).Wait, hold on. The radius has to be a real number, so m + 20 must be positive. So, m > -20. I should keep that in mind.So, center of C₂: (2, 4)Radius of C₂: r₂ = √(m + 20)Now, since the circles are externally tangent, the distance between their centers should be equal to the sum of their radii.Let me compute the distance between centers first. The centers are (0,0) and (2,4). The distance formula is sqrt[(x2 - x1)² + (y2 - y1)²].So, distance d = sqrt[(2 - 0)² + (4 - 0)²] = sqrt[4 + 16] = sqrt[20] = 2*sqrt(5).So, d = 2√5.Now, the sum of the radii is r₁ + r₂ = √5 + √(m + 20).Since they are externally tangent, we have:d = r₁ + r₂So,2√5 = √5 + √(m + 20)Let me solve for m.Subtract √5 from both sides:2√5 - √5 = √(m + 20)Simplify left side:√5 = √(m + 20)Now, square both sides:(√5)² = (√(m + 20))²So,5 = m + 20Subtract 20 from both sides:5 - 20 = mSo,m = -15Wait, let me double-check my steps.1. Center of C₁ is (0,0), radius √5. That's correct.2. Rewrote C₂ by completing the square:x² - 4x + y² - 8y = mCompleting the square:(x - 2)² - 4 + (y - 4)² - 16 = mSo, (x - 2)² + (y - 4)² = m + 20. So, center (2,4), radius √(m + 20). That seems correct.3. Distance between centers: sqrt[(2)^2 + (4)^2] = sqrt[4 + 16] = sqrt[20] = 2√5. Correct.4. Sum of radii: √5 + √(m + 20). Since externally tangent, distance equals sum of radii. So, 2√5 = √5 + √(m + 20). Subtract √5: √5 = √(m + 20). Square both sides: 5 = m + 20. So, m = -15.Wait, but let me think again. Is m = -15 acceptable? Earlier, I noted that m + 20 must be positive for the radius to be real. So, m + 20 > 0 => m > -20. Since -15 > -20, that's fine.So, m = -15 is acceptable.Alternatively, let me check if I made any mistake in the algebra.Starting from:2√5 = √5 + √(m + 20)Subtract √5:√5 = √(m + 20)Square both sides:5 = m + 20So, m = -15. That seems correct.Alternatively, let me plug m = -15 back into the equation of C₂ and see if it's consistent.C₂: x² + y² - 4x - 8y - (-15) = 0 => x² + y² - 4x - 8y + 15 = 0Completing the square:x² - 4x + y² - 8y = -15(x - 2)^2 - 4 + (y - 4)^2 - 16 = -15(x - 2)^2 + (y - 4)^2 - 20 = -15(x - 2)^2 + (y - 4)^2 = 5So, radius is √5, same as C₁. Wait, but if both circles have the same radius and their centers are 2√5 apart, then the distance between centers is 2√5, which is equal to the sum of radii (√5 + √5 = 2√5). So, that makes sense for external tangency.Wait, but if both circles have the same radius, and the distance between centers is twice the radius, then they are externally tangent. So, that seems consistent.Alternatively, if the radii were different, the distance would be the sum of the two different radii. But in this case, both radii are √5, so the sum is 2√5, which is equal to the distance between centers. So, that's correct.Therefore, m = -15 is the correct answer.Wait, hold on. Wait, when I plugged m = -15 back into C₂, I got radius √5, same as C₁. So, both circles have the same radius and are externally tangent. That's possible, right? Because if two circles have the same radius and are externally tangent, the distance between centers is 2r, which is exactly what we have here.Yes, that makes sense. So, I think my answer is correct.But just to be thorough, let me check if m = -15 is the only solution.Suppose, hypothetically, that I made a mistake in the step where I set the distance equal to the sum of radii. What if I had considered internal tangency instead? Then, the distance between centers would be equal to the difference of radii. But in this problem, it's specified as externally tangent, so it should be the sum.But just for my own understanding, let me see what would happen if I considered internal tangency.If it were internally tangent, then the distance between centers would be |r₁ - r₂|.So, in that case:2√5 = |√5 - √(m + 20)|But since 2√5 is positive, we can write:2√5 = |√5 - √(m + 20)|Which would lead to two cases:Case 1: 2√5 = √5 - √(m + 20)But √5 - √(m + 20) must be positive, so √5 > √(m + 20). Then,2√5 = √5 - √(m + 20)Subtract √5:√5 = -√(m + 20)But square roots are non-negative, so this would imply √5 = negative number, which is impossible. So, no solution in this case.Case 2: 2√5 = - (√5 - √(m + 20)) = √(m + 20) - √5So,2√5 = √(m + 20) - √5Add √5 to both sides:3√5 = √(m + 20)Square both sides:9*5 = m + 20 => 45 = m + 20 => m = 25But wait, if m = 25, then the radius of C₂ is √(25 + 20) = √45 = 3√5.Then, the distance between centers is 2√5, and the difference of radii is |√5 - 3√5| = 2√5, which would make them internally tangent.But in the problem statement, it's specified as externally tangent, so m = 25 would be the case for internal tangency, which is not what we want.Therefore, m = -15 is the correct answer for external tangency.So, I think I've considered all possible cases and my answer is consistent.**Final Answer**The real number ( m ) is boxed{-15}.

answer：Okay, so I have this problem here: I need to show that if ( n ) is an integer and the expression ( frac{n^{5}}{120} + frac{n^{3}}{24} + frac{n}{30} ) is an integer, then ( n ) must be a multiple of 6. Hmm, interesting. Let me try to figure this out step by step.First, let me write down the expression again to make sure I have it right:[frac{n^{5}}{120} + frac{n^{3}}{24} + frac{n}{30}]So, the problem states that this sum is an integer, and we need to prove that ( n ) is a multiple of 6. That means ( n ) should be divisible by both 2 and 3, right? So, maybe I can approach this by checking the divisibility conditions separately for 2 and 3.Let me think about how to handle this. The expression has three terms with denominators 120, 24, and 30. Maybe I can combine these terms into a single fraction to simplify things. Let me try that.First, find a common denominator for 120, 24, and 30. Let's see, 120 is already a multiple of 24 and 30. Let me check:- 120 divided by 24 is 5, so yes, 120 is a multiple of 24.- 120 divided by 30 is 4, so yes, 120 is a multiple of 30.So, the least common denominator (LCD) is 120. Let me rewrite each term with denominator 120:- ( frac{n^{5}}{120} ) is already with denominator 120.- ( frac{n^{3}}{24} = frac{5n^{3}}{120} ) because 24*5=120.- ( frac{n}{30} = frac{4n}{120} ) because 30*4=120.So, combining all three terms:[frac{n^{5}}{120} + frac{5n^{3}}{120} + frac{4n}{120} = frac{n^{5} + 5n^{3} + 4n}{120}]Therefore, the original expression simplifies to:[frac{n^{5} + 5n^{3} + 4n}{120}]And this is given to be an integer. Let me denote this integer as ( k ), so:[frac{n^{5} + 5n^{3} + 4n}{120} = k]Which implies:[n^{5} + 5n^{3} + 4n = 120k]So, ( n^{5} + 5n^{3} + 4n ) is divisible by 120. Since 120 factors into ( 2^3 times 3 times 5 ), the expression must be divisible by 8, 3, and 5. So, to prove that ( n ) is a multiple of 6, I need to show that ( n ) is divisible by 2 and 3. Maybe I can check the divisibility by 2 and 3 separately.Let me first check divisibility by 2. So, if ( n ) is even, then ( n ) is divisible by 2. If ( n ) is odd, then ( n ) is not divisible by 2. Let me see what happens in both cases.Case 1: ( n ) is even.If ( n ) is even, then ( n = 2m ) for some integer ( m ). Let me substitute this into the expression:[(2m)^5 + 5(2m)^3 + 4(2m) = 32m^5 + 40m^3 + 8m]Each term here is clearly divisible by 8, since 32, 40, and 8 are all multiples of 8. So, the entire expression is divisible by 8, which is good because 120 is divisible by 8.But wait, the problem is that even if ( n ) is even, the expression is divisible by 8, but we need to ensure that ( n ) is a multiple of 6, so we also need to check divisibility by 3.Case 2: ( n ) is odd.If ( n ) is odd, let's see what happens. Let me substitute ( n = 2m + 1 ) into the expression:[(2m + 1)^5 + 5(2m + 1)^3 + 4(2m + 1)]This seems complicated, but maybe I can compute it modulo 8 to see if it's divisible by 8.Let me compute each term modulo 8:First, ( (2m + 1)^5 ). Since ( 2m ) is even, ( 2m + 1 ) is odd. Let me compute ( (2m + 1)^5 ) modulo 8.Note that for any odd integer ( a ), ( a^1 equiv 1 ) mod 8, ( a^2 equiv 1 ) mod 8, ( a^3 equiv 1 ) mod 8, and so on. Wait, is that true?Wait, no. Let me test with specific numbers. Let me take ( a = 1 ): ( 1^5 = 1 equiv 1 ) mod 8.( a = 3 ): ( 3^5 = 243 ). 243 divided by 8 is 30 with remainder 3, so 243 ≡ 3 mod 8.Similarly, ( a = 5 ): ( 5^5 = 3125 ). 3125 divided by 8: 8*390=3120, so 3125 ≡ 5 mod 8.( a = 7 ): ( 7^5 = 16807 ). 16807 divided by 8: 8*2100=16800, so 16807 ≡ 7 mod 8.So, in general, for odd ( a ), ( a^5 equiv a ) mod 8. So, ( (2m + 1)^5 equiv (2m + 1) ) mod 8.Similarly, ( (2m + 1)^3 equiv (2m + 1) ) mod 8.So, let's compute each term modulo 8:1. ( (2m + 1)^5 equiv (2m + 1) ) mod 8.2. ( 5(2m + 1)^3 equiv 5(2m + 1) ) mod 8.3. ( 4(2m + 1) equiv 4(2m + 1) ) mod 8.So, adding them all together:[(2m + 1) + 5(2m + 1) + 4(2m + 1) = (1 + 5 + 4)(2m + 1) = 10(2m + 1)]Now, 10(2m + 1) mod 8: 10 ≡ 2 mod 8, so 2*(2m + 1) = 4m + 2 mod 8.So, the entire expression modulo 8 is ( 4m + 2 ). For this to be divisible by 8, ( 4m + 2 equiv 0 ) mod 8.Which implies:( 4m equiv -2 ) mod 8, or ( 4m equiv 6 ) mod 8.Divide both sides by 2: ( 2m equiv 3 ) mod 4.But 2m is even, and 3 is odd, so this is impossible. Therefore, if ( n ) is odd, the expression ( n^5 + 5n^3 + 4n ) is congruent to 4m + 2 mod 8, which is not 0. Hence, the expression is not divisible by 8, which contradicts the fact that it's equal to 120k, which is divisible by 8.Therefore, ( n ) cannot be odd. So, ( n ) must be even. That takes care of the divisibility by 2.Now, moving on to divisibility by 3. So, we need to show that ( n ) must be divisible by 3. Let me check the expression modulo 3.So, if ( n ) is not divisible by 3, then ( n equiv 1 ) or ( 2 ) mod 3. Let's compute ( n^5 + 5n^3 + 4n ) modulo 3 in both cases.Case 1: ( n equiv 0 ) mod 3.Then, ( n^5 equiv 0 ), ( 5n^3 equiv 0 ), ( 4n equiv 0 ). So, the entire expression is 0 mod 3, which is fine.Case 2: ( n equiv 1 ) mod 3.Compute each term:- ( n^5 equiv 1^5 = 1 ) mod 3.- ( 5n^3 equiv 5*1^3 = 5 equiv 2 ) mod 3.- ( 4n equiv 4*1 = 4 equiv 1 ) mod 3.Adding them up: 1 + 2 + 1 = 4 ≡ 1 mod 3. So, the expression is congruent to 1 mod 3, which is not 0. Therefore, if ( n equiv 1 ) mod 3, the expression is not divisible by 3, which contradicts the fact that it's equal to 120k, which is divisible by 3.Case 3: ( n equiv 2 ) mod 3.Compute each term:- ( n^5 equiv 2^5 = 32 equiv 2 ) mod 3 (since 32 divided by 3 is 10 with remainder 2).- ( 5n^3 equiv 5*(2^3) = 5*8 = 40 equiv 1 ) mod 3 (since 40 divided by 3 is 13 with remainder 1).- ( 4n equiv 4*2 = 8 equiv 2 ) mod 3.Adding them up: 2 + 1 + 2 = 5 ≡ 2 mod 3. So, the expression is congruent to 2 mod 3, which is not 0. Therefore, if ( n equiv 2 ) mod 3, the expression is not divisible by 3, which again contradicts the fact that it's equal to 120k.Therefore, the only possibility is that ( n equiv 0 ) mod 3. So, ( n ) must be divisible by 3.Since we've already established that ( n ) must be even (divisible by 2) and now we know it must be divisible by 3, it follows that ( n ) is divisible by the least common multiple of 2 and 3, which is 6. Therefore, ( n ) must be a multiple of 6.Wait, hold on, let me just make sure I didn't miss anything. So, I considered both cases for divisibility by 2 and 3, and in each case, if ( n ) wasn't divisible by 2 or 3, the expression wasn't divisible by 8 or 3, respectively, which contradicts the given condition. So, that should cover all possibilities, right?Just to recap:1. If ( n ) is odd, the expression isn't divisible by 8, which is a factor of 120. So, ( n ) must be even.2. If ( n ) is not divisible by 3, the expression isn't divisible by 3, which is another factor of 120. So, ( n ) must be divisible by 3.Therefore, combining both, ( n ) must be divisible by 6.I think that's solid. I don't see any gaps in the reasoning. So, yeah, that should do it.**Final Answer**boxed{6}

answer：First, I need to determine the amount of water in the dried mushrooms. Since the dried mushrooms weigh 2.5 kg and contain 12% water, the water in them is 0.12 multiplied by 2.5 kg, which equals 0.3 kg.Next, I'll calculate the amount of dry matter in the dried mushrooms. The dry matter is the total weight of the dried mushrooms minus the water content. So, 2.5 kg minus 0.3 kg gives 2.2 kg of dry matter.This 2.2 kg of dry matter comes from the fresh mushrooms. Therefore, the fresh mushrooms also contain 2.2 kg of dry matter.To find the amount of water in the fresh mushrooms, I'll subtract the dry matter from the total weight of the fresh mushrooms. That is, 22 kg minus 2.2 kg equals 19.8 kg of water.Finally, to determine the percentage of water in the fresh mushrooms, I'll divide the water content by the total weight of the fresh mushrooms and multiply by 100. So, 19.8 kg divided by 22 kg multiplied by 100 gives 90%.Therefore, the fresh mushrooms contain 90% water.

answer：Okay, so I have this problem about a right-angled triangle. The altitude drawn to the hypotenuse is 12 units, and the perimeter is 60 units. I need to find the lengths of the legs. Hmm, let's see. I remember that in a right-angled triangle, the altitude to the hypotenuse relates the two segments it creates. Also, the perimeter is the sum of all sides, so that should help too.First, let me visualize the triangle. It's right-angled, so let's call the legs 'a' and 'b', and the hypotenuse 'c'. The altitude to the hypotenuse is given as 12. I recall that the area of the triangle can be expressed in two ways: one as half the product of the legs, and another as half the product of the hypotenuse and the altitude. So, maybe I can set up an equation using that.Let me write down the area expressions:Area = (1/2) * a * b Area = (1/2) * c * 12Since both equal the area, I can set them equal to each other:(1/2) * a * b = (1/2) * c * 12 Simplifying, that becomes:a * b = 12 * c So, a * b = 12c. Got that.Also, the perimeter is 60, so:a + b + c = 60Okay, so I have two equations:1. a * b = 12c 2. a + b + c = 60I need a third equation because I have three variables: a, b, c. Since it's a right-angled triangle, Pythagoras' theorem applies:a² + b² = c²So now I have three equations:1. a * b = 12c 2. a + b + c = 60 3. a² + b² = c²Hmm, so I need to solve this system of equations. Let me see how I can approach this.Maybe I can express c from the second equation in terms of a and b, then substitute into the other equations.From equation 2: c = 60 - a - bSo, substitute c into equation 1:a * b = 12 * (60 - a - b) Let me compute that:a * b = 720 - 12a - 12b Bring all terms to one side:a * b + 12a + 12b - 720 = 0Hmm, that's a quadratic in terms of a and b. Maybe I can factor this or find another relation.Alternatively, let's think about equation 3. Since c = 60 - a - b, substitute that into equation 3:a² + b² = (60 - a - b)² Let me expand the right-hand side:(60 - a - b)² = 60² - 2*60*(a + b) + (a + b)² = 3600 - 120(a + b) + a² + 2ab + b²So, equation 3 becomes:a² + b² = 3600 - 120(a + b) + a² + 2ab + b²Subtract a² + b² from both sides:0 = 3600 - 120(a + b) + 2abSo,2ab - 120(a + b) + 3600 = 0 Divide both sides by 2:ab - 60(a + b) + 1800 = 0Hmm, interesting. Now, from equation 1, we have ab = 12c, and c = 60 - a - b, so ab = 12*(60 - a - b) = 720 - 12a - 12b.So, ab = 720 - 12a - 12bLet me plug this into the equation I just got from equation 3:(ab) - 60(a + b) + 1800 = 0 Substitute ab:(720 - 12a - 12b) - 60(a + b) + 1800 = 0Let me expand this:720 - 12a - 12b - 60a - 60b + 1800 = 0Combine like terms:720 + 1800 = 2520 -12a - 60a = -72a -12b - 60b = -72bSo, equation becomes:2520 - 72a - 72b = 0Let me factor out -72:2520 - 72(a + b) = 0 So,72(a + b) = 2520 Divide both sides by 72:a + b = 2520 / 72 Let me compute that:2520 ÷ 72. Let's see, 72*35 = 2520, because 72*30=2160, 72*5=360, so 2160+360=2520. So, a + b = 35.Wait, so a + b = 35. But from equation 2, a + b + c = 60, so c = 60 - (a + b) = 60 - 35 = 25.So, c = 25.Now, from equation 1: ab = 12c = 12*25 = 300.So, ab = 300, and a + b = 35.So, now I have a system:a + b = 35 a * b = 300This is a quadratic equation. Let me write it as:x² - (a + b)x + ab = 0 Which is:x² - 35x + 300 = 0Let me solve this quadratic equation.The discriminant D = 35² - 4*1*300 = 1225 - 1200 = 25So, sqrt(D) = 5Thus, solutions are:x = [35 ± 5]/2So,x = (35 + 5)/2 = 40/2 = 20 x = (35 - 5)/2 = 30/2 = 15So, the legs are 15 and 20.Let me verify this.So, legs are 15 and 20, hypotenuse is 25.Check if the perimeter is 15 + 20 + 25 = 60. Yes, that's correct.Check if the area is (15*20)/2 = 150. Also, the area should be (25*12)/2 = 150. Correct.Also, check Pythagoras: 15² + 20² = 225 + 400 = 625 = 25². Correct.So, everything checks out.**Final Answer**The legs are boxed{15} and boxed{20}.

answer：First, I need to determine the total number of boxes Kaylee has already sold. She sold 12 boxes of lemon biscuits, 5 boxes of chocolate biscuits, and 4 boxes of oatmeal biscuits. Adding these together gives 12 + 5 + 4, which equals 21 boxes sold so far.Next, I'll subtract the number of boxes she has already sold from her total goal. Kaylee needs to sell 33 boxes in total. Subtracting the 21 boxes she has already sold from 33 gives 12 boxes remaining.Therefore, Kaylee needs to sell 12 more boxes of biscuits to meet her goal.