answer:To find how many days of the week could start a 30-day month such that the number of Tuesdays and Fridays are equal: 1. **Weekday Distribution for a 30-Day Month**: - Every day of the week will occur 4 times in the first 28 days, regardless of the starting day. - The distribution of the final 2 extra days depends on the weekday the month begins. 2. **Modulo Calculation**: - Since 30 equiv 2 pmod{7}, the starting day determines the weekdays of the extra 2 days. 3. **Analyzing Each Starting Day**: - **Sunday**: The extra days are Sunday and Monday. Resulting Tuesdays and Fridays each occur 4 times. - **Monday**: The extra days are Monday and Tuesday. Resulting Tuesdays occur 5 times, Fridays 4 times. - **Tuesday**: The extra days are Tuesday and Wednesday. Resulting Tuesdays occur 5 times, Fridays 4 times. - **Wednesday**: The extra days are Wednesday and Thursday. Resulting Tuesdays and Fridays each occur 4 times. - **Thursday**: The extra days are Thursday and Friday. Resulting Tuesdays and Fridays each occur 5 times. - **Friday**: The extra days are Friday and Saturday. Resulting Fridays occur 5 times, Tuesdays 4 times. - **Saturday**: The extra days are Saturday and Sunday. Resulting Tuesdays and Fridays each occur 4 times. 4. **Identifying Equal Number of Tuesdays and Fridays**: - The month starts on Sunday, Wednesday, or Saturday. 5. **Conclusion**: There are three days of the week (Sunday, Wednesday, and Saturday) when the month can start such that the number of Tuesdays and Fridays are equal. The final answer is 3. The final answer is boxed{textbf{(B)} 3}

answer:We need to calculate the pairs for each combination of different colors: white-brown, white-blue, white-red, brown-blue, brown-red, blue-red. 1. **White and Brown**: 5 text{ white socks} times 3 text{ brown socks} = 15 choices. 2. **White and Blue**: 5 text{ white socks} times 3 text{ blue socks} = 15 choices. 3. **White and Red**: 5 text{ white socks} times 1 text{ red sock} = 5 choices. 4. **Brown and Blue**: 3 text{ brown socks} times 3 text{ blue socks} = 9 choices. 5. **Brown and Red**: 3 text{ brown socks} times 1 text{ red sock} = 3 choices. 6. **Blue and Red**: 3 text{ blue socks} times 1 text{ red sock} = 3 choices. Total number of choices: 15 + 15 + 5 + 9 + 3 + 3 = boxed{50} choices. Conclusion: By considering each pairing of different colored socks and multiplying the number of socks in each pair, we ensure the solution adheres to the requirement of different colors and accounts for all possible combinations.

answer:In this new scenario, the green squares cannot share any side with red squares. We analyze each case based on the number of green squares: Case 1: No green squares - All squares are red. - **Number of ways:** 1 Case 2: One green square - The green square can be placed in any position, as it is surrounded only by red squares or edges. - **Number of ways:** 9 (each position in the grid) Case 3: More than one green square - If there are multiple green squares, they must be placed such that they do not share any side with red squares. This limits configurations heavily; essentially, only non-adjacent green squares are possible. - However, placing more than one green square becomes impossible without violating the side-sharing rule due to the connectivity of the grid. Adding the ways from each case: [ 1 + 9 = 10 ] Thus, the total number of ways to paint the squares under the given conditions is 10. The final answer is boxed{text{(C)} 10}

answer:From the given information, we have: z=1+i. Thus, frac{2}{z} = frac{2(1-i)}{(1+i)(1-i)} = 1-i. Hence, the answer is boxed{text{C}}. To solve this problem, we multiplied both the numerator and the denominator by the conjugate of the complex number z. This question primarily tests the multiplication and division operations of complex numbers in algebraic form.