answer:1. **Define the terms and expressions for the sums:** Let the first term of the arithmetic sequence be ( a ) and the common difference be ( d ). The ( n )-th term of an arithmetic sequence is ( a + (n-1)d ). 2. **Expression for the sum of the first 30 terms:** The sum of the first ( n ) terms is: [ S_{30} = frac{30}{2} times (a + (a + 29d)) = 15(2a + 29d) ] Given ( S_{30} = 150 ), we have: [ 15(2a + 29d) = 150 ] Simplifying: [ 2a + 29d = 10 quad text{(Equation 1)} ] 3. **Expression for the sum of the next 70 terms (terms 31 to 100):** [ S_{31-100} = frac{70}{2} times ((a + 30d) + (a + 99d)) = 35(2a + 129d) ] Given ( S_{31-100} = 4900 ), we have: [ 35(2a + 129d) = 4900 ] Simplifying: [ 2a + 129d = 140 quad text{(Equation 2)} ] 4. **Solve the system of equations:** Subtract Equation 1 from Equation 2: [ (2a + 129d) - (2a + 29d) = 140 - 10 ] [ 100d = 130 ] [ d = 1.3 ] Substitute ( d = 1.3 ) back into Equation 1: [ 2a + 29 times 1.3 = 10 ] [ 2a + 37.7 = 10 ] [ 2a = -27.7 ] [ a = -13.85 ] 5. **Conclusion:** The first term of the sequence is ( -13.85 ), which gives us our final answer: ( -13.85 ). The final answer is boxed{textbf{B}}.

answer:Given the system of equations: [ begin{aligned} &cos x + cos y = cos (x+y), &sin x + sin y = sin (x+y). end{aligned} ] 1. **Rewrite and Simplify the Equations Using Known Trigonometric Identities**: The sum-to-product identities provide a useful way to manipulate these equations: [ cos x + cos y = 2 cos left( frac{x+y}{2} right) cos left( frac{x-y}{2} right) ] and [ sin x + sin y = 2 sin left( frac{x+y}{2} right) cos left( frac{x-y}{2} right). ] For cos(x+y) and sin(x+y), we know: [ cos (x+y) = 2 cos^2 left( frac{x+y}{2} right) - 1 ] and [ sin (x+y) = 2 sin left( frac{x+y}{2} right) cos left( frac{x+y}{2} right). ] 2. **Reduce the Problem to a Simplified Equation**: Substituting the identities into the original equations: [ 2 cos left( frac{x+y}{2} right) cos left( frac{x-y}{2} right) = 2 cos^2 left( frac{x+y}{2} right) - 1, ] [ 2 sin left( frac{x+y}{2} right) cos left ( frac{x-y}{2} right) = 2 sin left( frac{x+y}{2} right) cos left( frac{x+y}{2} right). ] 3. **Solve the Simplified Equations**: For the first equation: [ 2 cos left( frac{x+y}{2} right) cos left( frac{x-y}{2} right) - 2 cos^2 left( frac{x+y}{2} right) + 1 = 0, ] [ 2 cos left( frac{x+y}{2} right) left( cos left( frac{x-y}{2} right) - cos left( frac{x+y}{2} right) right) + 1 = 0. ] For the second equation: [ 2 sin left( frac{x+y}{2} right) left( cos left( frac{x-y}{2} right) - cos left( frac{x+y}{2} right) right) = 0. ] The product of two terms equals zero if any of the terms is zero. We analyze these possibilities: 4. **Consider the Case Where (cos left( frac{x-y}{2} right) - cos left( frac{x+y}{2} right) = 0)**: If cos left( frac{x-y}{2} right) - cos left( frac{x+y}{2} right) = 0: [ cos left( frac{x-y}{2} right) = cos left( frac{x+y}{2} right). ] This would imply identical angles or periodic differences, contradicting the trigonometry. Therefore: Ensure (sin left( frac{x+y}{2} right) = 0): [ sin left( frac{x+y}{2} right) = 0 quad Rightarrow quad frac{x+y}{2} = kpi quad Rightarrow quad x+y = 2kpi. ] 5. **Substitute (x+y = 2kpi) Back into the Simplified Equations**: Substitute into the original equation: [ cos y = cos(2kpi - x) = cos x. ] Thus, [ cos x = frac{1}{2}, ] leading to: [ x = pm frac{pi}{3} + 2mpi quad (m in mathbb{Z}). ] For (y): [ y = 2kpi - x = 2npi mp frac{pi}{3} quad (n in mathbb{Z}). ] 6. **Conclusion**: The solutions to the system are: [ begin{aligned} x &= pm frac{pi}{3} + 2mpi quad (m in mathbb{Z}), y &= mp frac{pi}{3} + 2npi quad (n in mathbb{Z}). end{aligned} ] Thus, the solution to the system is: [ boxed{left{ left( pm frac{pi}{3} + 2mpi, mp frac{pi}{3} + 2npi right) mid m, n in mathbb{Z} right}} ]

answer:First, let's determine how many weeks are in two months. Typically, a month has about 4 weeks, so two months would have: 2 months x 4 weeks/month = 8 weeks Lyka plans to save 15 per week for 8 weeks, so the total amount she will save is: 15/week x 8 weeks = 120 Lyka already has 40, so the total amount she will have after saving for two months is: 40 (current savings) + 120 (amount saved over 8 weeks) = 160 Therefore, the smartphone is worth boxed{160} .

answer:1. Calculate the maximum weight possible using these weights: [ 1 + 4 + 9 = 14 text{ lb} ] Thus, the maximum weight that can be measured is 14 lb. 2. Determine the number of different weights that can be combined: - Directly, weights of 1 lb, 4 lb, and 9 lb can be combined in various ways (considering that each weight can be used at most once): - Using 1 lb alone. - Using 4 lb alone. - Using 9 lb alone. - Using 1 lb and 4 lb together: 1 + 4 = 5 text{ lb}. - Using 1 lb and 9 lb together: 1 + 9 = 10 text{ lb}. - Using 4 lb and 9 lb together: 4 + 9 = 13 text{ lb}. - Using all three weights: 1 + 4 + 9 = 14 text{ lb}. - This lists weights of 1, 4, 9, 5, 10, 13, and 14 lb. - Therefore, the number of different possible weights is 7. - Conclusion: The maximum weight that can be measured is 14 lb, and the number of different weights that can be achieved is 7. Thus, the answer is 14 text{ lb text{ and } 7 text{ weights}}. The final answer is boxed{C) 14 lb and 7 weights}