answer:From the problem, we know that after passing the first checkpoint, frac{x}{2} remains. After passing the second checkpoint, frac{x}{2} cdot frac{2}{3} = frac{x}{3} remains. After passing the third checkpoint, frac{x}{3} cdot frac{3}{4} = frac{x}{4} remains. After passing the fourth checkpoint, frac{x}{4} cdot frac{4}{5} = frac{x}{5} remains. After passing the fifth checkpoint, frac{x}{5} cdot frac{5}{6} = frac{x}{6} remains. Therefore, x - frac{x}{6} = 1. Solving for x, we get x = 1.2. So, the answer is: boxed{1.2}. This involves calculating the remaining quantity after passing each checkpoint and setting up an equation to solve for the answer. This problem tests the application of mathematics and is a basic-level question.

answer:First, calculate 2mathbf{a} - mathbf{b}: 2mathbf{a} - mathbf{b} = 2(2, 1) - (-1, k) = (4, 2) - (-1, k) = (5, 2 - k). Then, calculate mathbf{a} cdot (2mathbf{a} - mathbf{b}): mathbf{a} cdot (2mathbf{a} - mathbf{b}) = (2, 1) cdot (5, 2 - k) = 2 times 5 + 1 times (2 - k) = 10 + 2 - k. According to the condition mathbf{a} cdot (2mathbf{a} - mathbf{b}) = 0, we have: 10 + 2 - k = 0. Solving this equation for k, we get: k = 12. Therefore, the value of k is boxed{12}.

answer:Let's analyze each option to determine whether it satisfies both conditions: being an even function and monotonically decreasing on the interval (0,3). A. For the function y=x^{3}, the property of an odd function is that f(-x)=-f(x), which holds true since (-x)^3 = -x^3. Thus, f(x) = x^3 is an odd function, not even, disqualifying it from our conditions. B. For the function y=ln |x|, we can test evenness by computing f(-x): f(-x)=ln |-x|=ln |x|=f(x), which demonstrates that f(x) = ln |x| is in fact an even function. However, within the interval (0,3), this function is increasing, as the natural logarithm increases as its argument goes from 0 to 3. Hence, it does not satisfy the monotonically decreasing condition. C. Consider the function y=2^{x}+2^{-x}. The evenness can be verified: f(-x)=2^{-x}+2^{x}=2^{x}+2^{-x}=f(x), indicates that f(x) = 2^x + 2^{-x} is an even function. To test for monotonicity, compare two values within the interval: f(1)=2^{1}+frac{1}{2^{1}}=2+frac{1}{2}=frac{5}{2}, f(2)=2^{2}+frac{1}{2^{2}}=4+frac{1}{4}=frac{17}{4}. Since f(1) < f(2), the function is increasing at that interval, and thus not monotonically decreasing on (0, 3). D. Lastly, we have y=cos x. The cosine function is an even function because cos(-x)=cos x. Also, in the interval (0,3) (or up to slightly more than frac{pi}{2}), the cosine function decreases from cos(0)=1 to cos(3) <0. Hence, it satisfies our condition of being monotonically decreasing. Thus, the correct choice is: boxed{D}.

answer:1. Consider a regular tetrahedron (P-ABCD) with each side of length 1. 2. The face, such as (Delta ABC), is an equilateral triangle with side length 1. 3. Points (M) and (N) are the midpoints of sides (AB) and (BC) respectively. First find the coordinates of various points: - Place (A(0,0,0)), (B(1,0,0)), and (Cleft(frac{1}{2}, frac{sqrt{3}}{2}, 0 right)). - Point (P) is directly above the centroid (O) of the base (Delta ABC). The centroid (O) of (Delta ABC) is found as: [ O = left( frac{0+1+frac{1}{2}}{3}, frac{0+0+frac{sqrt{3}}{2}}{3}, 0 right) = left(frac{1}{2}, frac{sqrt{3}}{6}, 0 right) ] The height of the tetrahedron from the centroid (O) to the apex (P) can be calculated using the Pythagorean Theorem in 3 dimensions. Given the side length of an equilateral triangle (s = 1): [ h = sqrt{s^2 - left( frac{s}{2} right)^2} = sqrt{1 - left( frac{1}{2} right)^2} = sqrt{1 - frac{1}{4}} = sqrt{frac{3}{4}} = frac{sqrt{3}}{2} ] Thus, the z-coordinate of point (P) is ( frac{sqrt{2}} {4}). Next, find the coordinates of (M) and (N): - (M) is the midpoint of (AB): [ Mleft( frac{0+1}{2}, frac{0+0}{2}, frac{0+0}{2} right) = left(frac{1}{2}, 0, 0 right) ] - (N) is the midpoint of (BC): [ Nleft( frac{1+frac{1}{2}}{2}, frac{0+frac{sqrt{3}}{2}}{2}, 0 right) = left( frac{3}{4}, frac{sqrt{3}}{4}, 0 right) ] Find the equation of line segment (MN). This can be parameterized as points on the line between (M) and (N): [ MN = left( frac{1}{2} + t left( frac{3}{4} - frac{1}{2} right), t left( frac{sqrt{3}}{4} right), 0 right) = left( frac{1 + t}{2}, frac{sqrt{3}}{4} cdot t, 0right) ] Similarly, the line (OC) connects (O) and (C). Using the coordinates: [ PC = left( frac{1}{2} + t left( frac{1}{2} - frac{1}{2} right), t left( frac{sqrt{3}}{2}, t cdot 0 right) ] The distance (CH) is found to be orthogonal to the plane (POC). Here (PO per CCH is perpendicularly also orthogonal to mid CHT) which orthogonal to obtained distance, calculated earlier: [ CH = frac{sqrt{2}}{2}CN £ cdot, frac{sqrt{2}} frac{2}{4} ] Since the orthogonal distance (CN and CH) is the same using planes (PC and MN together equivalent): (boxed{cfrac{sqrt{2}}{4}})