answer:1. **Understanding the discount structure**: - One free window for every five purchased. - Additional 5% discount if more than ten windows are acquired in total. 2. **Calculate individual purchases**: - **Greg's purchase**: Greg requires 9 windows. He gets one free window for buying five, and pays for the remaining four: [ text{Cost for Greg} = 8 times 120 = 960 text{ dollars} ] - **Susan's purchase**: Susan requires 13 windows. She will get three free windows (two for the first ten and one more for the extra five): [ text{Cost for Susan} = (13 - 2) times 120 = 1320 text{ dollars} ] - Applying 5% discount since Susan is buying more than ten windows: [ text{Discounted cost for Susan} = 0.95 times 1320 = 1254 text{ dollars} ] - **Total cost if purchased separately**: [ text{Total Separate Cost} = 960 + 1254 = 2214 text{ dollars} ] 3. **Calculating joint purchase**: - **Joint purchase**: Together, Greg and Susan need 22 windows. They get four free windows: [ text{Cost for Joint Purchase} = (22 - 4) times 120 = 2160 text{ dollars} ] - Applying 5% discount since they are buying more than ten windows: [ text{Total Joint Discounted Cost} = 0.95 times 2160 = 2052 text{ dollars} ] 4. **Calculating savings**: - **Savings**: [ text{Savings} = text{Total separate cost} - text{Total joint discounted cost} = 2214 - 2052 = 162 text{ dollars} ] 5. **Conclusion**: - Greg and Susan will save 162 dollars if they purchase the windows together rather than separately. The final answer is boxed{(C) 162}

answer:Given i(z-1) = 1 + i, we have z-1 = frac{1+i}{i} = frac{(1+i)(-i)}{-i^2} = 1-i, Therefore, z = 2-i. Hence, the correct choice is: boxed{A}. This problem involves transforming the given equation and then simplifying it using the algebraic operations of complex numbers. It tests the basic computational skills related to the algebraic operations of complex numbers.

answer:Let us first assume that for some integers ( n ) and ( r ) (where ( n geq r + 3 )), the binomial coefficients ( binom{n}{r}, binom{n}{r+1}, binom{n}{r+2}, binom{n}{r+3} ) form an arithmetic sequence. This would imply the following relationships: [ 2 binom{n}{r+1} = binom{n}{r} + binom{n}{r+2} 2 binom{n}{r+2} = binom{n}{r+1} + binom{n}{r+3} ] Recall the definition of binomial coefficients: [ binom{n}{r} = frac{n!}{r!(n-r)!} ] We can rewrite the above relationships using the binomial coefficient formula: **From the first relationship:** [ 2 binom{n}{r+1} = binom{n}{r} + binom{n}{r+2} ] Substitute the binomial coefficient formulas: [ 2 frac{n!}{(r+1)!(n-r-1)!} = frac{n!}{r!(n-r)!} + frac{n!}{(r+2)!(n-r-2)!} ] Divide both sides by ( n! ): [ 2 frac{1}{(r+1)!(n-r-1)!} = frac{1}{r!(n-r)!} + frac{1}{(r+2)!(n-r-2)!} ] Rewrite using the factorial properties: [ 2 frac{1}{(r+1)(n-r-1)} = frac{1}{(n-r)(n-r-1)} + frac{1}{(r+1)(r+2)} ] Similarly, for the second relationship: [ 2 binom{n}{r+2} = binom{n}{r+1} + binom{n}{r+3} ] Substitute the binomial coefficient formulas again: [ 2 frac{n!}{(r+2)!(n-r-2)!} = frac{n!}{(r+1)!(n-r-1)!} + frac{n!}{(r+3)!(n-r-3)!} ] Divide both sides by ( n! ): [ 2 frac{1}{(r+2)!(n-r-2)!} = frac{1}{(r+1)!(n-r-1)!} + frac{1}{(r+3)!(n-r-3)!} ] Rewrite using the factorial properties: [ 2 frac{1}{(r+2)(n-r-2)} = frac{1}{(n-r-1)(n-r-2)} + frac{1}{(r+2)(r+3)} ] Upon simplifying both equations, we obtain the following system of equations: 1. [ frac{2}{(r+1)(n-r-1)} = frac{1}{(n-r-1)(n-r)} + frac{1}{(r+1)(r+2)} ] 2. [ frac{2}{(r+2)(n-r-2)} = frac{1}{(n-r-2)(n-r-1)} + frac{1}{(r+2)(r+3)} ] Next, let's clean up these equations by clearing the denominators and simplifying: For equation 1, multiply through by ( (r+1)(r+2)(n-r-1)(n-r) ): [ 2 (r+2) (n-r) = (r+1)(r+2) + (r+2) (n-r)(n-r-1) ] For equation 2, multiply through by ( (r+2)(r+3)(n-r-2)(n-r-1) ): [ 2 (r+3) (n-r-2) = (r+2)(r+3) + (r+2)(n-r-1)(n-r-2) ] These simplify to quadratic forms: 1. [ n^2 - (4r+5)n + 4r(r+2) + 2 = 0 ] 2. [ n^2 - (4r+9)n + 4(r+1)(r+3) + 2 = 0 ] Subtract the second equation from the first: [ left( n^2 - (4r+5)n + 4r(r+2) + 2 right) - left( n^2 - (4r+9)n + 4(r+1)(r+3) + 2 right) = 0 ] Simplify: [ (4r+9) - (4r+5) = 4(r+1)(r+3) - 4r(r+2) ] This simplifies to: [ 4 = 4((r+1)(r+3) - r(r+2)) ] [ 4 = 4(r^2 + 4r + 3 - r^2 - 2r) ] Simplify further: [ 4 = 8r + 12 - 4r ] [ 4 = 4r + 12 ] Subtract 12 from both sides: [ -8 = 4r ] Divide both sides by 4: [ r = -2 ] However, this contradicts the assumption that ( r ) and ( n ) are positive integers, hence: Given that the purported value of ( n = 2r + 3 ), a contradiction occurs. Using the nature of binomial coefficients: [ binom{2r+3}{r} = binom{2r+3}{r+3} < binom{2r+3}{r+1} = binom{2r+3}{r+2} ] This is inconsistent with the properties of an arithmetic sequence, leading to the conclusion: (boxed{text{The binomial coefficients cannot form an arithmetic sequence}} ). (blacksquare)

answer:1. **Triangle and Circle Properties:** triangle ABC is isosceles with angle B = 80^circ, so angle A = angle C = 50^circ. Angles angle BAO and angle BCO are 90^circ. 2. **Angle Analysis:** angle OAC = angle OCA = 65^circ (as angle A = 50^circ and angle OAC = 90^circ - frac{50^circ}{2}). Then, angle AOC = 180^circ - 65^circ - 65^circ = 50^circ. 3. **Central and Inscribed Angles:** angle AOB = angle COB = 90^circ (half of 360^circ - angle AOC). 4. **Length Calculations:** DO = AO = BO cdot sin(45^circ) = frac{BO sqrt{2}}{2}. BD = BO - DO = BO - frac{BO sqrt{2}}{2}. 5. **Ratio Calculation:** frac{BD}{BO} = frac{BO - frac{BO sqrt{2}}{2}}{BO} = 1 - frac{sqrt{2}}{2}. Conclusion: [ frac{BD{BO} = 1 - frac{sqrt{2}}{2}} ] The final answer is boxed{B}.