answer:Dave had to wash a total of 20 + 16 = 36 shirts before school. Since he had to wash 9 short sleeve shirts, the number of long sleeve shirts he had to wash is 36 - 9 = boxed{27} long sleeve shirts.

answer:To factorize the given quadratic expression 2x^{2}-4x+2, we follow these steps: 1. First, we factor out the greatest common factor (GCF) from all the terms. The GCF here is 2: 2x^{2}-4x+2 = 2(x^{2}-2x+1) 2. Next, we observe the expression inside the parentheses, x^{2}-2x+1. This is a perfect square trinomial, which can be written as the square of a binomial. The square root of x^{2} is x, and the square root of 1 is 1. Since the middle term is -2x, it indicates that the binomial is (x-1): x^{2}-2x+1 = (x-1)^{2} 3. Substituting back into the expression, we get: 2(x^{2}-2x+1) = 2left(x-1right)^{2} Therefore, the factorized form of 2x^{2}-4x+2 is boxed{2left(x-1right)^{2}}.

answer:From the given conditions, we have 2y + z = x + 2u geq 2sqrt{2xu} = 4sqrt{yz}, Thus, 2 + frac{z}{y} geq 4sqrt{frac{z}{y}}, Let t = sqrt{frac{z}{y}}, then t geq 1, t^2 geq 4t, So, t^2 - 4t + 2 geq 0, Since t geq 1, we have t geq 2 + sqrt{2}, So, frac{z}{y} = t^2 geq 6 + 4sqrt{2}, Since there exists a positive real number M such that M leq frac{z}{y}, We have M leq 6 + 4sqrt{2}, Thus, the maximum value of M is boxed{6 + 4sqrt{2}}. To solve this problem, we first transform the first equation of the given system to get 2y + z = x + 2u geq 2sqrt{2xu}, then combine it with the second equation to get 2y + z geq 4sqrt{yz}. Dividing both sides by y, we obtain an inequality in terms of sqrt{frac{z}{y}}. After substitution, we solve the inequality to find the range of sqrt{frac{z}{y}}, and thus the maximum value of M. This problem tests the application of basic inequalities and the ability to analyze and solve problems. Correctly using basic inequalities is the key, and it is a moderate-difficulty problem.

answer:1. **Definition and Base Cases**: - Let ( F_n ) denote the number of ways to climb a staircase of ( n ) steps. - Clearly, ( F_0 = 1 ) since there is only one way to stay at the ground step (doing nothing). - Additionally, ( F_1 = 1 ) because to reach the first step, one must take exactly one step. 2. **Recurrence Relation**: - To reach the ( (n+2) )-th step, one can either: - Come from the ( (n+1) )-th step, in which case one must have taken one step, or - Come from the ( n )-th step, in which case one must have taken two steps. - Therefore, the recurrence relation for ( F_n ) is given by: [ F_{n+2} = F_{n+1} + F_n ] 3. **Establishing the Closed Form Solution**: We need to show that for any natural number ( n ): [ F_n = frac{1}{sqrt{5}} left( varphi^{n+1} - bar{varphi}^{n+1} right) ] where ( varphi = frac{1+sqrt{5}}{2} ) and ( bar{varphi} = frac{1-sqrt{5}}{2} ). 4. **Base Case Verification for Recurrence Proof**: - For ( n = 0 ): [ frac{1}{sqrt{5}} (varphi^1 - bar{varphi}^1) = frac{1}{sqrt{5}} (varphi - bar{varphi}) = frac{1}{sqrt{5}} (sqrt{5}) = 1 = F_0 ] - For ( n = 1 ): [ frac{1}{sqrt{5}} (varphi^2 - bar{varphi}^2) = frac{1}{sqrt{5}} ((varphi + 1) - (bar{varphi} + 1)) = frac{1}{sqrt{5}} (varphi - bar{varphi}) = 1 = F_1 ] 5. **Inductive Step**: Suppose the formula holds for ( n ) and ( n+1 ). We need to show that it holds for ( n+2 ). Assume: [ F_n = frac{1}{sqrt{5}} left( varphi^{n+1} - bar{varphi}^{n+1} right) ] [ F_{n+1} = frac{1}{sqrt{5}} left( varphi^{n+2} - bar{varphi}^{n+2} right) ] Since: [ F_{n+2} = F_{n+1} + F_n ] Substitute the assumed formulas for ( F_n ) and ( F_{n+1} ): [ F_{n+2} = frac{1}{sqrt{5}} left( varphi^{n+2} - bar{varphi}^{n+2} right) + frac{1}{sqrt{5}} left( varphi^{n+1} - bar{varphi}^{n+1} right) ] Factor out the common term: [ F_{n+2} = frac{1}{sqrt{5}} left( varphi^{n+2} + varphi^{n+1} - bar{varphi}^{n+2} - bar{varphi}^{n+1} right) ] Note that: ( varphi + 1 = varphi^2 ) and ( bar{varphi} + 1 = bar{varphi}^2 ), thus: [ F_{n+2} = frac{1}{sqrt{5}} left( varphi^{n+1} (varphi + 1) - bar{varphi}^{n+1} (bar{varphi} + 1) right) = frac{1}{sqrt{5}} left( varphi^{n+1} varphi^2 - bar{varphi}^{n+1} bar{varphi}^2 right) = frac{1}{sqrt{5}} left( varphi^{n+3} - bar{varphi}^{n+3} right) ] This completes the inductive step. 6. **Conclusion**: By mathematical induction, we have shown that for all natural numbers ( n ): [ F_n = frac{1}{sqrt{5}} left( varphi^{n+1} - bar{varphi}^{n+1} right) ] (blacksquare)