answer：Okay, so I have this problem here: In the expansion of (1 + 3x)^n, where n is a positive integer and n is at least 6, the coefficients of x^5 and x^6 are equal. I need to find the value of n. Hmm, let me think about how to approach this.First, I remember that the binomial expansion of (a + b)^n is given by the sum from k=0 to n of (n choose k) * a^(n - k) * b^k. In this case, a is 1 and b is 3x. So, each term in the expansion will be (n choose k) * 1^(n - k) * (3x)^k, which simplifies to (n choose k) * 3^k * x^k.Therefore, the coefficient of x^k in the expansion is (n choose k) * 3^k. So, for x^5, the coefficient is (n choose 5) * 3^5, and for x^6, it's (n choose 6) * 3^6.The problem states that these two coefficients are equal. So, I can set up the equation:(n choose 5) * 3^5 = (n choose 6) * 3^6.Alright, let me write that down more clearly:C(n, 5) * 3^5 = C(n, 6) * 3^6.Where C(n, k) is the combination of n things taken k at a time.Now, I can simplify this equation. Let me first divide both sides by 3^5 to make it simpler:C(n, 5) = C(n, 6) * 3.So, C(n, 5) = 3 * C(n, 6).Hmm, okay. Now, I know that C(n, k) = n! / (k! (n - k)!). So, let's write both sides using factorials.Left side: C(n, 5) = n! / (5! (n - 5)!).Right side: 3 * C(n, 6) = 3 * [n! / (6! (n - 6)!)].So, putting it together:n! / (5! (n - 5)!) = 3 * [n! / (6! (n - 6)!)].Hmm, okay. Let me see if I can simplify this equation. First, I notice that n! appears on both sides, so I can divide both sides by n! to cancel that out:1 / (5! (n - 5)!) = 3 / (6! (n - 6)!).Alright, that's better. Now, let's compute the factorials:5! is 120, and 6! is 720. So, substituting those in:1 / (120 * (n - 5)!) = 3 / (720 * (n - 6)!).Wait, but (n - 5)! is equal to (n - 5) * (n - 6)! So, I can write (n - 5)! as (n - 5)(n - 6)!.So, substituting that into the left side:1 / [120 * (n - 5)(n - 6)!] = 3 / [720 * (n - 6)!].Hmm, now I can see that (n - 6)! appears in both denominators, so I can multiply both sides by (n - 6)! to cancel that out:1 / [120 * (n - 5)] = 3 / 720.Alright, that simplifies things. Now, let's compute 3 / 720. 3 divided by 720 is 1 / 240.So, now the equation is:1 / [120 * (n - 5)] = 1 / 240.So, 1 divided by [120(n - 5)] equals 1 divided by 240. Therefore, the denominators must be equal:120(n - 5) = 240.So, solving for n:120(n - 5) = 240.Divide both sides by 120:n - 5 = 240 / 120.240 divided by 120 is 2, so:n - 5 = 2.Therefore, n = 2 + 5 = 7.Wait, so n is 7? Let me check if that makes sense.Let me verify by plugging n = 7 into the coefficients.First, the coefficient of x^5 is C(7, 5) * 3^5.C(7, 5) is 21, and 3^5 is 243, so 21 * 243. Let me compute that: 21 * 243.21 * 240 is 5040, and 21 * 3 is 63, so total is 5040 + 63 = 5103.Now, the coefficient of x^6 is C(7, 6) * 3^6.C(7, 6) is 7, and 3^6 is 729, so 7 * 729. Let me compute that: 7 * 700 is 4900, and 7 * 29 is 203, so total is 4900 + 203 = 5103.Yes, both coefficients are 5103 when n = 7. So, that checks out.Therefore, the value of n is 7.**Final Answer**The value of ( n ) is boxed{7}.

answer：To determine the area of the triangle formed by the line ( x + y - 2 = 0 ) and the coordinate axes, I first need to find the points where the line intersects the axes.**Finding the x-intercept:**I set ( y = 0 ) in the equation of the line:[x + 0 - 2 = 0 implies x = 2]So, the line intersects the x-axis at the point ( (2, 0) ).**Finding the y-intercept:**Next, I set ( x = 0 ) in the equation of the line:[0 + y - 2 = 0 implies y = 2]Thus, the line intersects the y-axis at the point ( (0, 2) ).**Calculating the Area:**The triangle formed by these intercepts and the origin is a right-angled triangle with legs of length 2 units each. The area ( A ) of a right-angled triangle is given by:[A = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 2 times 2 = 2]Therefore, the area of the triangle is 2 square units.

answer：Okay, so I have this problem about Mario walking on a moving sidewalk. Let me try to understand it step by step. First, the problem says that there's a moving sidewalk from Point A to Point B. When the sidewalk is turned off, it takes Mario 90 seconds to walk from A to B. That means Mario's walking speed is such that he covers the distance between A and B in 90 seconds when he's walking on a stationary sidewalk.Then, when the sidewalk is moving, but Mario isn't walking, it takes him 45 seconds to be carried from A to B. So, the moving sidewalk alone can cover the same distance in half the time it takes Mario to walk it. That suggests that the moving sidewalk is faster than Mario's walking speed.Now, the question is asking how long it will take Mario to walk from A to B when he is both walking and the sidewalk is moving. So, in this case, both his walking speed and the sidewalk's speed are contributing to his movement from A to B. I need to find the time it takes for him to cover the distance when both speeds are acting together.Let me think about how to model this. I remember that when two objects are moving in the same direction, their speeds add up. So, Mario's effective speed when both walking and on the moving sidewalk should be the sum of his walking speed and the sidewalk's speed.To solve this, I think I need to find the distance between A and B first. Since I have two different times and two different speeds, I can use the formula:Distance = Speed × TimeLet me denote the distance between A and B as D.When the sidewalk is off, Mario walks the distance D in 90 seconds. So, his walking speed, let's call it Vm, is:Vm = D / 90Similarly, when the sidewalk is moving and Mario isn't walking, the sidewalk carries him the same distance D in 45 seconds. So, the speed of the sidewalk, let's call it Vs, is:Vs = D / 45Now, when Mario is both walking and the sidewalk is moving, his effective speed will be Vm + Vs. So, the combined speed is:V_total = Vm + Vs = (D / 90) + (D / 45)Let me compute that:First, note that D / 45 is the same as 2D / 90, because 45 is half of 90. So, D / 45 = 2D / 90.Therefore, V_total = (D / 90) + (2D / 90) = 3D / 90 = D / 30.So, Mario's effective speed when both walking and on the moving sidewalk is D / 30 per second.Now, the time it takes to cover distance D at this effective speed is:Time = Distance / Speed = D / (D / 30) = 30 seconds.Wait, that seems straightforward, but let me double-check my steps.1. Calculated Mario's walking speed as D / 90.2. Calculated the sidewalk's speed as D / 45.3. Added the two speeds: D / 90 + D / 45 = D / 90 + 2D / 90 = 3D / 90 = D / 30.4. Then, time = D / (D / 30) = 30 seconds.Yes, that seems correct. So, the time it takes Mario to walk from A to B when both walking and the sidewalk is moving is 30 seconds.Alternatively, I can think of it in terms of rates. Let me denote the distance as 1 unit for simplicity. So, if the distance is 1, then:- Mario's walking rate is 1/90 per second.- The sidewalk's rate is 1/45 per second.Combined, their rates add up: 1/90 + 1/45 = 1/90 + 2/90 = 3/90 = 1/30 per second.Therefore, the time taken is 1 / (1/30) = 30 seconds.Yes, that confirms my previous result.I think that's solid. I don't see any mistakes in my reasoning. So, the answer should be 30 seconds.**Final Answer**boxed{30}

answer：First, I observe that the sequence alternates between ★ and ▲ symbols, starting with ★.Given that there are 15 ★ symbols, I can determine the total number of symbols in the sequence. Since each pair of symbols (★▲) contains one ★, the total number of symbols is 15 multiplied by 2, which equals 30.In this alternating sequence, the number of ▲ symbols is equal to the number of ★ symbols. Therefore, there are also 15 ▲ symbols in the sequence.

answer：First, I'll define the total number of books as ( N ), and it's given that ( 500 < N < 650 ).When all books are packed in small boxes that hold 12 books each, there are 7 books remaining. This means ( N ) divided by 12 leaves a remainder of 7. Mathematically, this can be expressed as:[ N = 12k + 7 ]where ( k ) is an integer representing the number of small boxes used.When all books are packed in large boxes that hold 25 books each, there is space for 5 more books. This implies that ( N ) divided by 25 leaves a remainder of 20 (since 25 - 5 = 20). This can be written as:[ N = 25m + 20 ]where ( m ) is an integer representing the number of large boxes used.Now, I'll set the two expressions for ( N ) equal to each other:[ 12k + 7 = 25m + 20 ]Simplifying this equation:[ 12k - 25m = 13 ]To solve for integers ( k ) and ( m ), I'll look for values that satisfy this equation within the given range of ( N ). By testing possible values, I find that when ( k = 18 ) and ( m = 7 ), the equation holds:[ 12(18) - 25(7) = 216 - 175 = 13 ]Substituting ( k = 18 ) back into the equation for ( N ):[ N = 12(18) + 7 = 216 + 7 = 223 ]However, 223 is below the minimum range of 500. To find a value within the specified range, I'll consider the periodic nature of the problem. The least common multiple of 12 and 25 is 300, so adding 300 to 223 gives:[ N = 223 + 300 = 523 ]Finally, I'll verify that 523 satisfies both conditions:- ( 523 div 12 = 43 ) with a remainder of 7.- ( 523 div 25 = 20 ) with a remainder of 23, which means there's space for 2 more books (since 25 - 23 = 2). However, this doesn't align with the original condition of space for 5 more books.Upon re-evaluating, I realize that the correct remainder when dividing by 25 should be 20. Therefore, the correct value within the range is:[ N = 523 + 300 = 823 ]But 823 exceeds the maximum range of 650. Thus, there is no solution within the given range that satisfies both conditions.

answer：To solve the equation ( log_{10}(a^2 - 15a) = 2 ), I first convert the logarithmic equation to its exponential form. This gives me ( a^2 - 15a = 10^2 ), which simplifies to ( a^2 - 15a = 100 ).Next, I rearrange the equation to form a quadratic equation: ( a^2 - 15a - 100 = 0 ). To solve this quadratic equation, I use the quadratic formula ( a = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -15 ), and ( c = -100 ).Calculating the discriminant, ( b^2 - 4ac = (-15)^2 - 4(1)(-100) = 225 + 400 = 625 ). Since the discriminant is positive, there are two real solutions.Substituting the values into the quadratic formula, I find the solutions: ( a = frac{15 pm 25}{2} ). This results in two real numbers: ( a = 20 ) and ( a = -5 ).Finally, I verify the solutions by substituting them back into the original logarithmic equation to ensure they are valid. Both solutions satisfy the equation, confirming that the set of solutions consists of two integers.