answer:1. **Compute the Total Number of Passwords without the Digit 7**: - A password consists of 4 digits, each digit can be any of the 10 digits (0 through 9) except for 7. This means each digit has 9 possible choices. - Therefore, the total number of such passwords is: [ 9^4 = 6561 ] 2. **Compute the Number of Passwords with All Different Digits**: - Given that the password must not contain the digit 7, the remaining digits are 0, 1, 2, 3, 4, 5, 6, 8, and 9 (9 digits in total). - We need to count the number of 4-digit passwords where all four digits are different. - We select 4 digits out of the 9 available digits. The number of ways to choose 4 digits out of 9 is given by the combination: [ binom{9}{4} = frac{9!}{4!(9-4)!} = frac{9 cdot 8 cdot 7 cdot 6}{4 cdot 3 cdot 2 cdot 1} = 126 ] - Each of these 4 chosen digits can be arranged in (4!) (factorial of 4) ways: [ 4! = 24 ] - Therefore, the number of 4-digit passwords with all different digits is: [ binom{9}{4} cdot 4! = 126 cdot 24 = 3024 ] 3. **Compute the Number of Passwords with At Least Two Digits the Same**: - The total number of passwords without the digit 7 is 6561 (as computed in step 1). - The number of passwords with all different digits is 3024 (as computed in step 2). - Therefore, the number of passwords containing at least two identical digits is: [ 6561 - 3024 = 3537 ] # Conclusion: [ boxed{3537} ]

answer:Let a = e^{ix}, b = e^{iy}, and c = e^{iz}. Then, [begin{align*} a + b + c &= e^{ix} + e^{iy} + e^{iz} &= (cos x + cos y + cos z) + i (sin x + sin y + sin z) &= 1 + 0i = 1. end{align*}] The complex conjugates are given by, [begin{align*} frac{1}{a} + frac{1}{b} + frac{1}{c} &= e^{-ix} + e^{-iy} + e^{-iz} &= (cos x + cos y + cos z) - i (sin x + sin y + sin z) &= 1. end{align*}] Multiplying the two results, we have: [(a + b + c) left(frac{1}{a} + frac{1}{b} + frac{1}{c}right) = 1 times 1 = 1.] Squaring a + b + c = 1 yields: [(a + b + c)^2 = 1^2 = 1 = a^2 + b^2 + c^2 + 2(ab + ac + bc).] Thus, we find ab + ac + bc = 0. Now consider, [begin{align*} a^2 + b^2 + c^2 &= e^{2ix} + e^{2iy} + e^{2iz} &= (cos 2x + cos 2y + cos 2z) + i (sin 2x + sin 2y + sin 2z). end{align*}] Using the earlier derived a^2 + b^2 + c^2 = 1, the real part gives us the desired sum: [cos 2x + cos 2y + cos 2z = boxed{1}.]

answer:1. **Data Points**: The percentages of households in Maple Town using solar energy are: - 2000: 6% - 2010: 12% - 2015: 24% - 2020: 48% 2. **Trend Analysis**: Observing the given data points provides insight into the growth trend of solar energy usage among households: - 2000 to 2010: (12% - 6% = 6%) - 2010 to 2015: (24% - 12% = 12%) - 2015 to 2020: (48% - 24% = 24%) The increase is doubling every subsequent period, indicating an exponential growth pattern. 3. **Graph Selection**: The ideal graphical representation would begin with a lower steepness that becomes much steeper over each subsequent five-year period, illustrating the doubling effect observed. **Conclusion**: The right graph for depicting this trend should initiate at 6% in 2000 and end at 48% in 2020, with an increasingly steep curve illustrating the rising pattern in solar energy adaptation. Hence, the solution for the graphical depiction of this progression is modeled by an exponential growth graph. Thus the correct graph must depict this escalating growth pattern with text{Graph depicting exponential growth}. The final answer is boxed{B) Exponential growth}

answer:Given that quadrilateral ABCD is a rhombus with a perimeter of 20cm and diagonal AC=6cm, we can deduce several things step by step: 1. Since ABCD is a rhombus, all sides are equal. Therefore, each side AB = frac{20cm}{4} = 5cm. 2. The diagonals of a rhombus are perpendicular bisectors of each other. Therefore, AC perp BD and AC is bisected into two equal parts at O, the intersection point. Thus, OA = frac{1}{2}AC = 3cm. 3. In the right triangle OAB, we can use the Pythagorean theorem to find OB: [OB = sqrt{AB^2 - OA^2} = sqrt{5^2 - 3^2} = sqrt{25 - 9} = sqrt{16} = 4cm.] 4. Since OB is half of BD (because the diagonals bisect each other), we find that BD = 2 times OB = 2 times 4cm = 8cm. 5. The area of rhombus ABCD can be calculated using the formula involving the product of its diagonals: [S_{rhombusABCD} = frac{1}{2} times AC times BD = frac{1}{2} times 6cm times 8cm = 24cm^2.] Therefore, the area of the rhombus is boxed{24cm^2}.