answer：Okay, so I have this problem to evaluate the sum:frac{1}{log _{2} 12 sqrt{5}}+frac{1}{log _{3} 12 sqrt{5}}+frac{1}{log _{4} 12 sqrt{5}}+frac{1}{log _{5} 12 sqrt{5}}+frac{1}{log _{6} 12 sqrt{5}} .Hmm, that's a sum of five terms, each of which is the reciprocal of a logarithm with different bases. The argument of each logarithm is the same: 12 times the square root of 5. So, all the logs are of the same number, but with different bases from 2 to 6.I remember that there's a change of base formula for logarithms. Maybe that can help me simplify each term. The change of base formula says that:log_b a = frac{ln a}{ln b}So, if I take the reciprocal of that, it would be:frac{1}{log_b a} = frac{ln b}{ln a}Which is the same as:log_a bWait, that's interesting. So, each term in the sum can be rewritten using the change of base formula. Specifically, each term is the reciprocal of a logarithm, which is equivalent to the logarithm with the base and the argument swapped. So, for each term:frac{1}{log_b (12 sqrt{5})} = log_{12 sqrt{5}} bSo, substituting that into the original sum, the expression becomes:log_{12 sqrt{5}} 2 + log_{12 sqrt{5}} 3 + log_{12 sqrt{5}} 4 + log_{12 sqrt{5}} 5 + log_{12 sqrt{5}} 6Now, since all the logarithms have the same base, I can combine them using the logarithm addition rule, which states that:log_b a + log_b c = log_b (a times c)So, applying this rule to all five terms, the sum becomes:log_{12 sqrt{5}} (2 times 3 times 4 times 5 times 6)Let me compute the product inside the logarithm:First, 2 × 3 is 6.6 × 4 is 24.24 × 5 is 120.120 × 6 is 720.So, the product is 720.Therefore, the sum simplifies to:log_{12 sqrt{5}} 720Now, I need to evaluate this logarithm. Let me denote the base as ( 12 sqrt{5} ) and the argument as 720.I can express both the base and the argument in terms of their prime factors to see if they can be related.First, let's factorize 12 and 5:12 is 2² × 3.So, ( 12 sqrt{5} = 2^2 times 3 times 5^{1/2} ).Similarly, 720 can be factorized as:720 ÷ 2 = 360360 ÷ 2 = 180180 ÷ 2 = 9090 ÷ 2 = 4545 ÷ 3 = 1515 ÷ 3 = 55 ÷ 5 = 1So, 720 = 2⁴ × 3² × 5¹.So, 720 is 2⁴ × 3² × 5.Now, let me write both the base and the argument in terms of exponents:Base: ( 12 sqrt{5} = 2^2 times 3 times 5^{1/2} )Argument: 720 = 2⁴ × 3² × 5So, I can write 720 as (12 √5) raised to some power. Let me denote:720 = (12 √5)^kI need to find k such that:(2² × 3 × 5^{1/2})^k = 2⁴ × 3² × 5Let me compute the left side:(2² × 3 × 5^{1/2})^k = 2^{2k} × 3^k × 5^{k/2}Set this equal to the right side:2^{2k} × 3^k × 5^{k/2} = 2⁴ × 3² × 5¹Therefore, equate the exponents:For 2: 2k = 4 ⇒ k = 2For 3: k = 2 ⇒ k = 2For 5: k/2 = 1 ⇒ k = 2All exponents give k = 2, so k = 2.Therefore, 720 = (12 √5)^2Thus, the logarithm becomes:log_{12 sqrt{5}} 720 = log_{12 sqrt{5}} (12 sqrt{5})^2 = 2Because logarithm of a number to its own base is 1, so (12√5)^2 would be 2.Therefore, the entire sum simplifies to 2.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I converted each reciprocal logarithm to a logarithm with swapped base and argument. That seems correct because of the change of base formula.2. Then, I combined all the logarithms into a single logarithm by multiplying their arguments. That also seems correct, as log_b a + log_b c = log_b (a*c).3. Calculated the product 2×3×4×5×6 as 720. Let me verify that:2×3=6, 6×4=24, 24×5=120, 120×6=720. Yes, that's correct.4. Then, I expressed both the base (12√5) and the argument (720) in terms of their prime factors.12√5 is 2² × 3 × 5^{1/2}, and 720 is 2⁴ × 3² × 5. Correct.5. Then, I set up the equation (12√5)^k = 720 and solved for k by equating the exponents.For 2: 2k = 4 ⇒ k=2For 3: k=2For 5: k/2=1 ⇒ k=2All give k=2, so 720 is indeed (12√5)^2.Therefore, log_{12√5} 720 = 2.So, the sum is 2.Therefore, the final answer is 2.**Final Answer**boxed{2}

answer：To determine the number that must be added to 8 to result in -5, I start by setting up the equation:8 + x = -5Next, I isolate x by subtracting 8 from both sides of the equation:x = -5 - 8Finally, I perform the subtraction to find the value of x:x = -13Therefore, the number that must be added to 8 to give -5 is -13.

answer：First, I need to determine how many complete laps Joneal makes around the track. Since the circumference of the track is 50 feet, I divide the total distance he ran, which is 5280 feet, by 50 to find the number of laps.Next, I'll calculate the remainder after dividing 5280 by 50. This remainder will tell me how far Joneal ran beyond the complete laps. This distance will determine his stopping point on the track.Once I have the remaining distance, I'll figure out which quarter of the circle corresponds to that distance. Each quarter of the track is 12.5 feet long. By dividing the remaining distance by 12.5, I can identify the specific quarter where Joneal stopped.Finally, based on the calculation, I'll determine whether Joneal stopped in quarter A, B, C, or D.

answer：First, I need to determine how many red and blue cubes Grady gives to Gage. Grady has 20 red cubes and gives away 2/5 of them. Calculating 2/5 of 20 gives 8 red cubes. Grady also has 15 blue cubes and gives away 1/3 of them, which is 5 blue cubes.Next, I'll add the cubes Gage already has to the ones he receives from Grady. Gage initially has 10 red cubes and receives 8 more, totaling 18 red cubes. For the blue cubes, Gage has 12 initially and receives 5 more, making a total of 17 blue cubes.Finally, to find the total number of cubes Gage has, I'll add the red and blue cubes together: 18 red cubes plus 17 blue cubes equals 35 cubes in total.

answer：To determine the coefficient of ( x^8 ) in the expansion of ( (x - 1)^9 ), I can use the Binomial Theorem. The Binomial Theorem states that ( (a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n - k} b^k ).In this case, ( a = x ), ( b = -1 ), and ( n = 9 ). I'm interested in the term where the power of ( x ) is 8, which corresponds to ( k = 1 ) in the expansion.First, I calculate the binomial coefficient ( binom{9}{1} ), which is 9. Then, I raise ( x ) to the power of ( 9 - 1 = 8 ) and ( -1 ) to the power of 1, which is -1.Multiplying these together, the term is ( 9 times x^8 times (-1) = -9x^8 ).Therefore, the coefficient of ( x^8 ) in the expansion is -9.

answer：Okay, so I have this problem here where I need to find positive integers m and n such that x = m + sqrt(n) is a solution to the equation x² - 10x + 1 = sqrt(x)(x + 1). Then, I have to find m + n. Hmm, let me try to figure this out step by step.First, let me write down the equation again to make sure I have it correctly:x² - 10x + 1 = sqrt(x)(x + 1)Alright, so x is given as m + sqrt(n), where m and n are positive integers. That means x is a number that's a sum of an integer and a square root of another integer. So, x is probably irrational unless sqrt(n) is an integer, but since n is a positive integer, sqrt(n) is either integer or irrational. But since m is also an integer, x is either integer or irrational. But the equation involves sqrt(x), so x must be a perfect square? Wait, not necessarily. Because sqrt(x) is on the right side, so x must be non-negative, which it is since m and n are positive integers.But maybe I can square both sides to eliminate the square root? Let me try that.But before that, let me see if I can manipulate the equation a bit. Let's bring everything to one side:x² - 10x + 1 - sqrt(x)(x + 1) = 0Hmm, that might not be very helpful. Maybe I can rearrange the equation:x² - 10x + 1 = sqrt(x)(x + 1)Let me denote sqrt(x) as t. Then, x = t². Let me substitute that into the equation:(t²)² - 10(t²) + 1 = t(t² + 1)Simplify that:t⁴ - 10t² + 1 = t³ + tBring all terms to one side:t⁴ - t³ - 10t² - t + 1 = 0Hmm, so now I have a quartic equation in terms of t. That might be a bit complicated, but maybe I can factor it or find rational roots.Wait, since t is sqrt(x), and x is m + sqrt(n), then t is sqrt(m + sqrt(n)). Hmm, that seems a bit messy. Maybe I can think of t as a sum of square roots? Like t = a + b, where a and b are positive real numbers? Maybe, but I'm not sure.Alternatively, maybe I can factor the quartic equation. Let me try to factor it.Looking at t⁴ - t³ - 10t² - t + 1. Let me try rational roots. The possible rational roots are ±1.Testing t = 1: 1 - 1 - 10 - 1 + 1 = -10 ≠ 0Testing t = -1: 1 + 1 - 10 + 1 + 1 = -6 ≠ 0So no rational roots. Hmm, maybe it factors into quadratics?Let me assume it factors as (t² + at + b)(t² + ct + d). Let's expand this:t⁴ + (a + c)t³ + (ac + b + d)t² + (ad + bc)t + bdSet this equal to t⁴ - t³ - 10t² - t + 1.So, equate coefficients:1. a + c = -1 (from t³ term)2. ac + b + d = -10 (from t² term)3. ad + bc = -1 (from t term)4. bd = 1 (constant term)Since bd = 1, possible integer solutions are b = 1, d = 1 or b = -1, d = -1.Let me try b = 1 and d = 1 first.Then, equation 1: a + c = -1Equation 2: ac + 1 + 1 = ac + 2 = -10 ⇒ ac = -12Equation 3: a*1 + b*c = a + c = -1 (Wait, ad + bc = a*1 + 1*c = a + c = -1, which is consistent with equation 1.)So, we have a + c = -1 and a*c = -12.So, we can solve for a and c.Let me set up the quadratic equation: Let a and c be roots of x² + x -12 = 0.Solutions: x = [-1 ± sqrt(1 + 48)] / 2 = [-1 ± 7]/2.So, x = (6)/2 = 3 or x = (-8)/2 = -4.So, a = 3, c = -4 or a = -4, c = 3.So, let's test both possibilities.First, a = 3, c = -4.Then, the factors would be (t² + 3t + 1)(t² -4t +1). Let me multiply them:First, t² * t² = t⁴t² * (-4t) = -4t³t² * 1 = t²3t * t² = 3t³3t * (-4t) = -12t²3t * 1 = 3t1 * t² = t²1 * (-4t) = -4t1 * 1 = 1Now, adding all these up:t⁴ + (-4t³ + 3t³) + (t² -12t² + t²) + (3t -4t) +1Simplify:t⁴ - t³ -10t² - t +1Yes! That's exactly the quartic we had. So, the quartic factors as (t² + 3t +1)(t² -4t +1).So, t⁴ - t³ -10t² - t +1 = (t² + 3t +1)(t² -4t +1) = 0Therefore, the solutions are the roots of t² + 3t +1 = 0 and t² -4t +1 = 0.But t is sqrt(x), which is a real number, so t must be positive. So, let's solve each quadratic:First quadratic: t² + 3t +1 = 0Discriminant: 9 - 4 = 5Solutions: t = [-3 ± sqrt(5)] / 2But t must be positive, so t = [-3 + sqrt(5)] / 2 ≈ (-3 + 2.236)/2 ≈ negative, which is invalid.So, no positive solutions from this quadratic.Second quadratic: t² -4t +1 = 0Discriminant: 16 - 4 = 12Solutions: t = [4 ± sqrt(12)] / 2 = [4 ± 2*sqrt(3)] / 2 = 2 ± sqrt(3)Since t must be positive, both solutions are positive: 2 + sqrt(3) ≈ 3.732 and 2 - sqrt(3) ≈ 0.2679.So, t = 2 + sqrt(3) or t = 2 - sqrt(3).But t = sqrt(x). So, sqrt(x) = 2 + sqrt(3) or sqrt(x) = 2 - sqrt(3).Therefore, x = (2 + sqrt(3))² or x = (2 - sqrt(3))².Let me compute both:First, x = (2 + sqrt(3))² = 4 + 4*sqrt(3) + 3 = 7 + 4*sqrt(3)Second, x = (2 - sqrt(3))² = 4 - 4*sqrt(3) + 3 = 7 - 4*sqrt(3)So, the solutions for x are 7 + 4*sqrt(3) and 7 - 4*sqrt(3).But in the problem, x is given as m + sqrt(n), where m and n are positive integers.Looking at 7 + 4*sqrt(3), that can be written as 7 + sqrt(16*3) = 7 + sqrt(48). So, m =7, n=48.Similarly, 7 - 4*sqrt(3) can be written as 7 + sqrt(48) but with a negative sign, but since x must be positive, both 7 + 4*sqrt(3) and 7 - 4*sqrt(3) are positive because 4*sqrt(3) ≈6.928, so 7 - 6.928 ≈0.072, which is positive.But the problem says x = m + sqrt(n), where m and n are positive integers. So, 7 - 4*sqrt(3) is 7 + sqrt(48) with a negative sign? Wait, no. Wait, 7 - 4*sqrt(3) is equal to 7 + sqrt(48) only if 4*sqrt(3) is sqrt(48). But 4*sqrt(3) is sqrt(16*3)=sqrt(48). So, 7 - sqrt(48). So, that would be m=7, n=48, but with a negative sign. However, the problem says x = m + sqrt(n), so it's positive. So, 7 - sqrt(48) is positive, but it's written as 7 + sqrt(n). Wait, but 7 - sqrt(48) is not of the form m + sqrt(n) unless sqrt(n) is negative, which it isn't. So, maybe only 7 + sqrt(48) is acceptable.Wait, but 7 - 4*sqrt(3) is equal to 7 - sqrt(48). So, if x is equal to m + sqrt(n), then 7 - sqrt(48) is not of the form m + sqrt(n) because sqrt(n) is positive, so m must be 7 and sqrt(n) must be -sqrt(48), which is not possible because sqrt(n) is positive. So, only x =7 + sqrt(48) is acceptable.Therefore, m=7 and n=48, so m + n = 7 + 48 = 55.Wait, but before I conclude, let me verify that x =7 + sqrt(48) satisfies the original equation.Compute x =7 + sqrt(48). Let's compute x² -10x +1 and sqrt(x)(x +1) and see if they are equal.First, compute x²:x =7 + sqrt(48)x² = (7)^2 + 2*7*sqrt(48) + (sqrt(48))^2 = 49 + 14*sqrt(48) + 48 = 97 + 14*sqrt(48)Then, compute -10x:-10x = -10*(7 + sqrt(48)) = -70 -10*sqrt(48)Add 1:x² -10x +1 = (97 +14*sqrt(48)) + (-70 -10*sqrt(48)) +1 = (97 -70 +1) + (14*sqrt(48) -10*sqrt(48)) = 28 + 4*sqrt(48)Now, compute sqrt(x)(x +1):sqrt(x) = sqrt(7 + sqrt(48)). Hmm, that seems complicated, but maybe we can express it in terms of t.Wait, earlier, we had t = sqrt(x) = 2 + sqrt(3). So, sqrt(x) = 2 + sqrt(3). So, sqrt(x) = 2 + sqrt(3).Then, x +1 = (7 + sqrt(48)) +1 =8 + sqrt(48). But sqrt(48)=4*sqrt(3), so x +1 =8 +4*sqrt(3).Therefore, sqrt(x)(x +1) = (2 + sqrt(3))(8 +4*sqrt(3)).Let me compute that:Multiply (2 + sqrt(3))(8 +4*sqrt(3)):= 2*8 + 2*4*sqrt(3) + sqrt(3)*8 + sqrt(3)*4*sqrt(3)=16 + 8*sqrt(3) +8*sqrt(3) +4*3=16 +16*sqrt(3) +12=28 +16*sqrt(3)Wait, but earlier, x² -10x +1 was 28 +4*sqrt(48). Let me compute sqrt(48):sqrt(48)=4*sqrt(3). So, 4*sqrt(48)=4*4*sqrt(3)=16*sqrt(3). So, x² -10x +1=28 +16*sqrt(3), which is equal to sqrt(x)(x +1)=28 +16*sqrt(3). So, yes, it checks out.Therefore, x=7 + sqrt(48) is indeed a solution.So, m=7, n=48, so m +n=55.But just to be thorough, let me check the other solution x=7 - sqrt(48). Let me see if that also satisfies the equation.Compute x=7 - sqrt(48). Let's compute x² -10x +1 and sqrt(x)(x +1).First, x²:x=7 - sqrt(48)x²=(7)^2 - 2*7*sqrt(48) + (sqrt(48))^2=49 -14*sqrt(48) +48=97 -14*sqrt(48)Compute -10x:-10x= -10*(7 - sqrt(48))= -70 +10*sqrt(48)Add 1:x² -10x +1= (97 -14*sqrt(48)) + (-70 +10*sqrt(48)) +1= (97 -70 +1) + (-14*sqrt(48)+10*sqrt(48))=28 -4*sqrt(48)Now, compute sqrt(x)(x +1):sqrt(x)=sqrt(7 - sqrt(48)). Hmm, earlier, we had t = sqrt(x) =2 - sqrt(3). Because when t² -4t +1=0, t=2 - sqrt(3). So, sqrt(x)=2 - sqrt(3).Then, x +1=(7 - sqrt(48)) +1=8 - sqrt(48)=8 -4*sqrt(3).So, sqrt(x)(x +1)=(2 - sqrt(3))(8 -4*sqrt(3)).Compute that:=2*8 +2*(-4*sqrt(3)) + (-sqrt(3))*8 + (-sqrt(3))*(-4*sqrt(3))=16 -8*sqrt(3) -8*sqrt(3) +4*3=16 -16*sqrt(3) +12=28 -16*sqrt(3)But x² -10x +1=28 -4*sqrt(48)=28 -16*sqrt(3). So, yes, it's equal to sqrt(x)(x +1)=28 -16*sqrt(3). So, it also satisfies the equation.But in the problem, x is given as m + sqrt(n), where m and n are positive integers. So, x=7 - sqrt(48) is equal to 7 + sqrt(48) with a negative sign on the sqrt(48). But sqrt(n) is positive, so 7 - sqrt(48) is not of the form m + sqrt(n) unless m=7 and sqrt(n)= -sqrt(48), which is not possible because sqrt(n) is positive. Therefore, only x=7 + sqrt(48) is acceptable.Therefore, m=7, n=48, so m +n=55.Wait, but just to make sure, let me check if there are any other possible factorizations or solutions. The quartic equation factored into two quadratics, each giving two roots, but only two of them were positive. So, we have two positive solutions for t, which gave us two positive solutions for x. But only one of them is in the form m + sqrt(n). So, that must be the solution.Therefore, I think the answer is 55.**Final Answer**boxed{55}