answer:Let's denote the number of matches won in the initial M matches as W. Since the team won P% of these matches, we can write: W = P% of M W = (P/100) * M After the winning streak of N matches, the total number of matches played by the team becomes M + N, and the new winning percentage is Q%. Therefore, the total number of matches won after the winning streak is: Total wins after streak = Q% of (M + N) Total wins after streak = (Q/100) * (M + N) Since the team won all N matches during the winning streak, the number of wins after the streak is the initial number of wins W plus the N wins from the streak: Total wins after streak = W + N Now we can set up the equation: (Q/100) * (M + N) = W + N Substitute W with (P/100) * M: (Q/100) * (M + N) = (P/100) * M + N Now, we need to solve for N, the number of matches won during the winning streak. To do this, we can multiply both sides of the equation by 100 to get rid of the denominators: Q * (M + N) = P * M + 100N Now, we can distribute Q on the left side: QM + QN = PM + 100N Next, we can move all terms involving N to one side and the rest to the other side: QN - 100N = PM - QM Factor out N on the left side: N(Q - 100) = PM - QM Now, we can solve for N: N = (PM - QM) / (Q - 100) This gives us the number of matches won during the winning streak. However, we need to ensure that the values of P, Q, M, and N are such that the equation makes sense (e.g., Q should be greater than P to have a positive number of matches won during the streak, and Q should be less than boxed{100} because it's a percentage).

answer:1. **Define the Sequences**: - Let ( n_k = 5k^2 ) and ( m_k = 2k ), where ( k = 0, 1, 2, ldots ) 2. **Calculate ( sqrt{n_k + m_k^2} )**: [ sqrt{n_k + m_k^2} = sqrt{5k^2 + (2k)^2} = sqrt{5k^2 + 4k^2} = sqrt{9k^2} = 3k ] - This shows that ( sqrt{n_k + m_k^2} ) simplifies to an integer, ( 3k ), for all ( k ). 3. **Calculate ( sqrt{n_k - m_k^2} )**: [ sqrt{n_k - m_k^2} = sqrt{5k^2 - (2k)^2} = sqrt{5k^2 - 4k^2} = sqrt{k^2} = k ] - This shows that ( sqrt{n_k - m_k^2} ) simplifies to an integer, ( k ), for all ( k ). 4. **Verify the Conditions**: - Both sequences ( {m_k} ) and ( {n_k} ) consist of distinct integers for each ( k ). - The sequences ( left{ sqrt{n_k + m_k^2} right} ) and ( left{ sqrt{n_k - m_k^2} right} ) are entirely composed of integers. **Conclusion**: We have shown that the sequences ( {m_k} ) and ( {n_k} ), defined as ( m_k = 2k ) and ( n_k = 5k^2 ) respectively, satisfy the required conditions for all integers ( k geq 0 ). Therefore, the solution is complete. [ blacksquare ]

answer:First, let's find out what x is when it is 20 percent greater than 52. 20 percent of 52 is calculated as: 0.20 * 52 = 10.4 So, x is 10.4 greater than 52: x = 52 + 10.4 = 62.4 Next, x is decreased by 10 percent. To find 10 percent of x: 0.10 * 62.4 = 6.24 Now, we decrease x by this amount: x = 62.4 - 6.24 = 56.16 Finally, x is increased by 15 percent. To find 15 percent of x: 0.15 * 56.16 = 8.424 We increase x by this amount: x = 56.16 + 8.424 = 64.584 So, the final value of x is boxed{64.584} .

answer:Let's denote the amount of solution Y that needs to be added as ( y ) milliliters. The amount of alcohol in 300 milliliters of solution X is 15% of 300 milliliters: ( 0.15 times 300 = 45 ) milliliters of alcohol. The amount of alcohol in ( y ) milliliters of solution Y is 45% of ( y ): ( 0.45 times y ) milliliters of alcohol. When we mix the two solutions, we want to have a solution that is 25% alcohol by volume. The total volume of the new solution will be ( 300 + y ) milliliters, and the amount of alcohol in this new solution should be 25% of ( 300 + y ): ( 0.25 times (300 + y) ) milliliters of alcohol. The equation representing the total amount of alcohol in the new solution is the sum of the alcohol from solution X and solution Y: ( 45 + 0.45y = 0.25 times (300 + y) ). Now we can solve for ( y ): ( 45 + 0.45y = 75 + 0.25y ) ( 0.45y - 0.25y = 75 - 45 ) ( 0.20y = 30 ) ( y = frac{30}{0.20} ) ( y = 150 ) milliliters. So, 150 milliliters of solution Y must be added to 300 milliliters of solution X to create a solution that is boxed{25} percent alcohol by volume.