answer:Let's use algebra to solve this problem. Let the number of girls be represented by G and the number of boys be represented by B. According to the problem, the number of boys is 16 less than twice the number of girls. This can be written as: B = 2G - 16 We also know that the total number of students in eighth grade is 68, which is the sum of the number of boys and girls: G + B = 68 Now we have a system of two equations: 1) B = 2G - 16 2) G + B = 68 We can substitute the expression for B from the first equation into the second equation: G + (2G - 16) = 68 Combine like terms: 3G - 16 = 68 Add 16 to both sides of the equation: 3G = 68 + 16 3G = 84 Divide both sides by 3 to solve for G: G = 84 / 3 G = 28 So, there are boxed{28} girls in the eighth grade.

answer:First, recognize that ( x - 1 = frac{sqrt{3}}{1 + frac{sqrt{3}}{1 + dots}} ), so ( frac{sqrt{3}}{x-1} = 1 + frac{sqrt{3}}{1 + frac{sqrt{3}}{1 + dots}} = x ). This gives us the equation ( sqrt{3} = x(x-1) ), or equivalently ( x^2 - x - sqrt{3} = 0 ). Next, find ( frac{1}{(x+2)(x-3)} ) in terms of ( x ). The denominator simplifies to ( x^2 - x - 6 ). Substitute ( x^2 - x ) from the quadratic equation: [ frac{1}{(x+2)(x-3)} = frac{1}{sqrt{3} - 6} ] To rationalize the denominator, multiply numerator and denominator by the conjugate of ( sqrt{3} - 6 ): [ frac{1}{sqrt{3} - 6} = frac{sqrt{3} + 6}{(sqrt{3} - 6)(sqrt{3} + 6)} = frac{sqrt{3} + 6}{3 - 36} = frac{sqrt{3} + 6}{-33} ] [ = frac{6 + sqrt{3}}{-33} ] Here, ( A = 6 ), ( B = 3 ), and ( C = -33 ). Therefore, ( |A|+|B|+|C| = 6 + 3 + 33 = boxed{42} ).

answer:Solution: (1) Let the center of the circle be M(m, 0), where m in mathbb{Z}. According to the condition that the circle is tangent to the line 4x+3y-29=0, we have frac {|4m+0-29|}{ sqrt {16+9}} = 5, which simplifies to |4m-29|=25. Considering m is an integer, we find m=1. Therefore, the equation of the circle is (x-1)^2+y^2=25. (2) Substituting the line ax-y+5=0 (a>0) into the equation of the circle, we get (a^2+1)x^2+2(5a-1)x+1=0. Since the line ax-y+5=0 intersects the circle at points A and B, we have Delta = 4(5a-1)^2-4(a^2+1) > 0, which simplifies to 12a^2-5a > 0. Solving this inequality, we find a > frac{5}{12} or a < 0. Therefore, the range of a is boxed{(-infty, 0) cup left(frac{5}{12}, +inftyright)}.

answer:To determine the order of a, b, and c, we can convert the logarithmic expressions to a more comparable form by exploiting the properties of logarithms. First, let's look at a: a = log_{4}3 It's clear that 3 is less than the base 4, so a must be less than 1. Hence, we have: a < log_{4}4 = 1 Now, let's examine b: b = log_{3}4 In this case, 4 is greater than the base 3, so b must be greater than 1. Thus, we get: b > log_{3}3 = 1 Finally, for c, we can use the change of base formula or understand that since both the base and the argument are reciprocal to each other, the log value must be negative, and since any real number is greater than a negative number, we have: c = log_{frac{3}{4}} frac{4}{3} Since frac{4}{3} is greater than the base frac{3}{4}, and considering the fact that increasing functions would normally imply that c > 1, we need to recognize that this is a logarithm with a base smaller than 1 (0 < frac{3}{4} < 1), which in turn means the function is decreasing. Therefore, c must be less than 0: c < log_{frac{3}{4}} 1 = 0 Combining all these findings, we can deduce the order: b > a > c Hence, the correct order is b > a > c. The correct choice is: boxed{text{A: }b > a > c}