answer:1. **Identify the current situation**: Chelsea leads by 75 points after 100 shots. Let k represent the points Chelsea has scored so far. 2. **Determine the maximum possible score for the opponent**: The opponent can score up to 10 points per shot for the remaining 50 shots. Hence, if the opponent scores all bullseyes, their final score would be (k - 75) + 50 times 10 = k + 425. 3. **Calculate Chelsea's score if she hits n bullseyes**: For Chelsea's next 50 shots, if n shots are bullseyes (10 points each), she scores 10n. For the remaining 50 - n shots, assuming the minimum score of 5 points each, she scores 5(50 - n). Therefore, Chelsea's total score after 150 shots would be k + 10n + 5(50 - n) = k + 10n + 250 - 5n = k + 5n + 250. 4. **Set up the inequality for Chelsea to guarantee a win**: Chelsea's total needs to be greater than the opponent's potential maximum. [ k + 5n + 250 > k + 425 ] Simplifying, we get: [ 5n + 250 > 425 ] [ 5n > 175 ] [ n > 35 ] 5. **Find the minimum integer value of n**: The smallest integer greater than 35 is 36. Thus, Chelsea needs at least 36 bullseyes to guarantee her victory. The final answer is boxed{36 textbf{(C)}}

answer:To find the first four terms of the series expansion of the particular solution to the differential equation [ y' + x y^2 = 2 cos x, ] subject to the initial condition ( y(0) = 1 ), we proceed as follows: 1. **Series Representation:** Assume the solution ( y(x) ) can be represented as a power series: [ y(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + cdots ] 2. **Initial Condition:** From the initial condition ( y(0) = 1 ), it follows that: [ a_0 = 1 ] 3. **Compute ( y'(x) ):** Differentiating the assumed series term by term, we get: [ y'(x) = a_1 + 2 a_2 x + 3 a_3 x^2 + cdots ] 4. **Compute ( y^2 ):** Squaring the series representation for ( y(x) ): [ y^2(x) = (a_0 + a_1 x + a_2 x^2 + a_3 x^3 + cdots)^2 ] The first few terms of ( y^2(x) ) expanded are: [ y^2(x) = a_0^2 + 2a_0 a_1 x + (2 a_0 a_2 + a_1^2) x^2 + (2 a_0 a_3 + 2 a_1 a_2) x^3 + cdots ] Substituting ( a_0 = 1 ): [ y^2(x) = 1 + 2a_1 x + (2a_2 + a_1^2) x^2 + (2a_3 + 2a_1 a_2) x^3 + cdots ] 5. **Series for ( cos x ):** The Taylor series expansion of ( cos x ) around ( x = 0 ) is: [ cos x = 1 - frac{x^2}{2!} + frac{x^4}{4!} - cdots ] Hence, [ 2 cos x = 2 - x^2 + frac{x^4}{12} - cdots ] 6. **Substitution into the Differential Equation:** Substitute the power series expressions for ( y', y^2 ), and ( cos x ) into the original differential equation: [ y' + x y^2 = 2 cos x ] Hence, [ (a_1 + 2a_2 x + 3a_3 x^2 + cdots) + x (1 + 2a_1 x + (2a_2 + a_1^2) x^2 + (2a_3 + 2a_1 a_2) x^3 + cdots) = 2 - x^2 + cdots ] 7. **Equating Coefficients:** Compare coefficients of like powers of ( x ) on both sides of the equation. - For the constant term ( x^0 ): [ a_1 = 2 ] - For the ( x^1 ) term: [ 2a_2 + 2a_1 = 0 implies 2a_2 + 2(2) = 0 implies a_2 = -1 ] - For the ( x^2 ) term: [ 3a_3 + 1 + 2a_1 a_2 = -1 implies 3a_3 + 1 + 2(2)(-1) = -1 implies 3a_3 + 1 - 4 = -1 implies 3a_3 = 2 implies a_3 = frac{2}{3} ] 8. **Constructing the Series:** Using the coefficients ( a_0, a_1, a_2, ) and ( a_3 ) found, the first four terms of the series expansion are: [ y(x) = 1 + 2x - x^2 + frac{2}{3} x^3 + cdots ] Hence, the particular solution expanded to the first four terms is: [ boxed{1 + 2x - x^2 + frac{2}{3} x^3} ]

answer:: **1. Claim:** There exists an integer ( k ) such that ( S_{2023} < k^2 < S_{2024} ). **2. Proof:** Let ( k ) be the smallest integer such that ( S_{2023} < k^2 ). Note that: [ k^2 = 1+3+5+cdots+O_k, ] where ( O_k = 2k-1 ), the ( k^{text {th }} ) odd number. Furthermore, observe that ( O_k leq p_{2023} ), the ( 2023^{text{rd}} ) prime. This follows from the fact that ( S_{2023} = 2+3+5+7+11+cdots+p_{2023} ) is a sum of 2023 primes, and if we compare this with the sum of all the odd integers up to ( p_{2023} ), this second sum is surely larger since it includes every number in the first sum, plus many more odd numbers (since there are gaps in the primes). Note that this is a very crude estimate! By the minimality of ( k ), we know that: [ (k-1)^2 = 1+3+5+cdots+(O_k-2) leq S_{2023}. ] **3. Adding ( O_k ) to both sides yields:** [ begin{aligned} 1+3+5+cdots+(O_k-2)+O_k &= k^2 &leq S_{2023} + O_k &leq S_{2023} + p_{2023} & < S_{2023} + p_{2024} & = S_{2024}. end{aligned} ] **Conclusion:** Therefore, we have shown that there exists an integer ( k ) such that ( S_{2023} < k^2 < S_{2024} ), thus proving the claim. (blacksquare) **4. Alternative Version:** We argue by contradiction. Suppose there is no perfect square between ( S_{2023} ) and ( S_{2024} ). Then there is some positive integer ( r ) such that: [ r^2 leq S_{2023} < S_{2024} leq (r+1)^2. ] Consequently, ( S_{2024} - S_{2023} ) (which is the ( 2024^{text{th}} ) prime) is at most: [ (r+1)^2 - r^2 = 2r + 1, ] and ( S_{2023} ) is at most the sum of all primes no larger than ( 2r ). This sum is made up of 2 and some, but not all, of the odd numbers up to ( 2r ) (in particular, 1 and 9 are missing). Thus, [ begin{aligned} S_{2023} &leq (1+3+5+7+9+cdots+(2r-1)) - (1+9) + 2 &= r^2 - 8, end{aligned} ] which contradicts the premise that ( r^2 leq S_{2023} ). Therefore, there must be a perfect square between ( S_{2023} ) and ( S_{2024} ).

answer:Let's denote the number of years that have passed since Jessica's mother's death as ( x ). When Jessica's mother died, Jessica was half her mother's age. This means that if we add ( x ) years to Jessica's current age (40), we will get twice the age she was when her mother died. So, if we subtract ( x ) from 40, we will get the age Jessica was when her mother died. Let's call Jessica's age when her mother died ( J ), and her mother's age at that time ( M ). We have: ( J = 40 - x ) (Jessica's age when her mother died) Since Jessica was half her mother's age at that time, we can write: ( M = 2J ) ( M = 2(40 - x) ) ( M = 80 - 2x ) We are also told that if her mother were alive ( x ) years later, she would have been 70. This means that her mother's age at death (M) plus ( x ) years would equal 70: ( M + x = 70 ) ( 80 - 2x + x = 70 ) ( 80 - x = 70 ) ( x = 80 - 70 ) ( x = 10 ) So, boxed{10} years have passed since Jessica's mother's death.