answer:If John sneezes 40 times in 2 minutes, we first need to convert the minutes into seconds to have a consistent unit of time. 2 minutes = 2 * 60 seconds = 120 seconds Now, we divide the total time by the number of sneezes to find out how many seconds it takes for him to sneeze once. 120 seconds / 40 sneezes = 3 seconds per sneeze So, it takes John boxed{3} seconds to sneeze once.

answer:First, let's calculate the total weight of the contents in one box: 1 pencil: 28.3 grams 1 eraser: 15.7 grams 2 paperclips: 3.5 grams each, so 2 * 3.5 = 7 grams 1 small stapler: 42.2 grams Total weight of one box = 28.3 + 15.7 + 7 + 42.2 = 93.2 grams Now, let's calculate the total weight of the contents if we have 5 pencils, 3 erasers, 4 paperclips, and 2 small staplers: 5 pencils: 5 * 28.3 grams = 141.5 grams 3 erasers: 3 * 15.7 grams = 47.1 grams 4 paperclips: 4 * 3.5 grams = 14 grams 2 small staplers: 2 * 42.2 grams = 84.4 grams Total weight of the new contents = 141.5 + 47.1 + 14 + 84.4 = 287 grams Since the boxes are identical, we only need to calculate the weight of the contents. The combined weight of the contents in all of the boxes with the new quantities is boxed{287} grams.

answer:Given the equation: [ log_{a} sqrt{4+x} + 3 log_{a^{2}}(4-x) - log_{a^{4}}(16-x^{2})^{2} = 2 ] we need to solve for (x). First, we note that the equation only makes sense if ( -4 < x < 4 ). We can rewrite each logarithm using properties of logarithms: 1. Using ( log_b x = frac{log_c x}{log_c b} ), we rewrite the terms in the equation. [ log_{a} sqrt{4+x} = frac{1}{2} log_{a} (4+x) ] 2. Using ( log_{a^{2}} b = frac{1}{2} log_a b ): [ 3 log_{a^{2}} (4-x) = 3 cdot frac{1}{2} log_a (4-x) = frac{3}{2} log_a (4-x) ] 3. Using ( log_{a^{4}} b = frac{1}{4} log_a b ): [ log_{a^{4}} (16-x^{2})^{2} = 2 cdot frac{1}{4} log_a (16-x^{2}) = frac{1}{2} log_a (16-x^{2}) ] Substituting back, our equation becomes: [ frac{1}{2} log_a (4+x) + frac{3}{2} log_a (4-x) - frac{1}{2} log_a (16-x^{2}) = 2 ] We can multiply through by 2 to clear the fractions: [ log_a (4+x) + 3 log_a (4-x) - log_a (16-x^{2}) = 4 ] We use the properties of logarithms to combine terms: [ log_a [(4+x) (4-x)^3 (16-x^2)^{-1}] = 4 ] We note that ( (4-x)^3 (16-x^2)^{-1} = (4-x)^2 ): [ (4+x) (4-x) (4-x)^2 (16 - x^2)^{-1} = 1 ] Simplifying: [ (4+x) = a^4 ] Thus: [ 4+x = a^4 ] Solving for (x): [ x = 4 - a^4 ] For the condition (-4 < x < 4) to hold: [ -4 < 4 - a^4 < 4 ] [ -8 < -a^4 < 0 implies 0 < a^4 < 8 implies 0 < a < sqrt[4]{8} ] Therefore, if (0 < a < sqrt[4]{8}), the solution to the equation is: [ x = 4 - a^4 ] Thus, the only solution is ( x = 4 - a^4 ) when ( 0 < a < sqrt[4]{8} ). (boxed{x = 4 - a^4})

answer:Since the axis of symmetry of f(x) is x=2, we have b=-4. Therefore, f(x)=x^{2}-4x+c, and f'(x)=2x-4. Thus, the line f'(x) does not pass through the second quadrant. Hence, the correct answer is boxed{text{B}}. By determining the value of b based on the axis of symmetry, calculating the derivative of the function f(x), and judging based on the equation of the line, we can solve the problem. This question examines the problem of the equation of a line, investigates the properties of quadratic functions, and the application of derivatives, making it a basic question.