answer:To solve the game probability, we observe that A, B, and C each have 2 possible gestures to make, resulting in 2 times 2 times 2 = 2^3 = 8 total possible outcomes for the three players' gestures in a single game round. For A to win, A's gesture must differ from both B and C. Therefore, the winning outcomes for A are as follows: 1. A shows black (B), B shows white (W), C shows white (W): BW-W-W 2. A shows white (W), B shows black (B), C shows black (B): W-B-B There are 2 winning outcomes for A out of the 8 possible outcomes. Therefore, the probability of A winning is the ratio of winning outcomes to total possible outcomes: P(text{A wins}) = frac{text{Number of winning outcomes for A}}{text{Total possible outcomes}} = frac{2}{8} = frac{1}{4} Thus, the probability of A winning the game in one round is boxed{frac{1}{4}}.

answer:1. We start by considering the fractional part of ( n sqrt{2} ), denoted as ( {n sqrt{2}} ). By definition, this is given by: [ {n sqrt{2}} = n sqrt{2} - lfloor n sqrt{2} rfloor ] where ( lfloor x rfloor ) is the integer part of ( x ). 2. We need to find the maximum value of ( c ) such that: [ {n sqrt{2}} geq frac{c}{n} quad forall n in mathbb{N} ] 3. Let ( lfloor n sqrt{2} rfloor = m ). Then: [ n sqrt{2} = m + {n sqrt{2}} ] Squaring both sides, we get: [ 2n^2 = m^2 + 2m {n sqrt{2}} + {n sqrt{2}}^2 ] 4. Rearranging, we have: [ 2n^2 - m^2 = 2m {n sqrt{2}} + {n sqrt{2}}^2 ] Since ( 0 leq {n sqrt{2}} < 1 ), we can approximate: [ 2n^2 - m^2 approx 2m {n sqrt{2}} ] 5. For large ( n ), the term ( {n sqrt{2}} ) is small, so we can approximate further: [ 2n^2 - m^2 approx 2m {n sqrt{2}} ] This implies: [ {n sqrt{2}} approx frac{2n^2 - m^2}{2m} ] 6. We know that ( m approx n sqrt{2} ), so: [ {n sqrt{2}} approx frac{2n^2 - (n sqrt{2})^2}{2n sqrt{2}} = frac{2n^2 - 2n^2}{2n sqrt{2}} = 0 ] This approximation is too crude, so we need a more precise approach. 7. Consider the continued fraction expansion of ( sqrt{2} ). The convergents of ( sqrt{2} ) are of the form ( frac{p_k}{q_k} ) where ( p_k ) and ( q_k ) are integers. For these convergents, we have: [ left| sqrt{2} - frac{p_k}{q_k} right| < frac{1}{q_k^2} ] This implies: [ left| n sqrt{2} - m right| < frac{1}{n} ] where ( m = lfloor n sqrt{2} rfloor ). 8. Therefore, we have: [ {n sqrt{2}} = n sqrt{2} - m approx frac{1}{n} ] 9. To find the best possible ( c ), we need to maximize ( c ) such that: [ {n sqrt{2}} geq frac{c}{n} ] From the above approximation, we see that the best possible ( c ) is: [ c = frac{sqrt{2}}{4} ] 10. Finally, we need to determine for which ( n in mathbb{N} ) the equality holds: [ {n sqrt{2}} = frac{c}{n} ] This occurs when: [ n sqrt{2} - lfloor n sqrt{2} rfloor = frac{sqrt{2}}{4n} ] The final answer is ( boxed{ c = frac{sqrt{2}}{4} } )

answer:Let's calculate the number of pages Mack writes each day: Monday: He writes for 60 minutes at a rate of 1 page every 30 minutes. 60 minutes / 30 minutes per page = 2 pages Tuesday: He writes for 45 minutes at a rate of 1 page every 15 minutes. 45 minutes / 15 minutes per page = 3 pages Wednesday: He writes 5 pages (as given). Thursday: During the first 30 minutes, he writes at a rate of 1 page every 10 minutes. 30 minutes / 10 minutes per page = 3 pages During the remaining 60 minutes, he writes at a rate of 1 page every 20 minutes. 60 minutes / 20 minutes per page = 3 pages Now, let's add up the pages from each day: Monday: 2 pages Tuesday: 3 pages Wednesday: 5 pages Thursday: 3 pages (first 30 minutes) + 3 pages (remaining 60 minutes) = 6 pages Total pages from Monday to Thursday: 2 + 3 + 5 + 6 = 16 pages Mack writes a total of boxed{16} pages in his journal from Monday to Thursday.

answer:The original expression equals sin^2 alpha + (-cos alpha) cdot (-cos alpha) + 1 =sin^2 alpha + cos^2 alpha + 1 =1+1=2. Therefore, the correct choice is boxed{text{B}}. **Analysis:** This can be derived using trigonometric identities and the Pythagorean identity.