answer:- We start by establishing the ratio of the height to the shadow length of the tree, which is (frac{60 text{feet}}{18 text{feet}} = frac{10}{3}). - Applying the same ratio to Raul, whose shadow is 3 feet long, we multiply 3 feet by (frac{10}{3}). - Performing the multiplication, we get (3 times frac{10}{3} = 10) feet. Conclusion: boxed{10 text{feet}}

answer:1. **Setting Variables:** - ( t_B ) is the time for the motorboat to travel from A to C. - ( p ) is the motorboat speed in still water, ( r ) is the river current speed, and ( s ) is the additional kayak speed relative to the river. 2. **Formulating Distances:** - The distance from A to C is ( (p + r) t_B ). - The kayak travels ( (r + s) t_K ) in ( t_K ) hours, where ( t_K = 12 ) since it never stops. - When coming back, the motorboat travels ( (p - r) (12 - t_B) ) upstream. 3. **Equating Total Distances for Point of Meeting:** - Set up the meeting point equation: [ (p + r) t_B + (p - r) (12 - t_B) = (r + s) cdot 12 ] - Simplify and solve for ( t_B ): [ p t_B + r t_B + p cdot 12 - p t_B - r cdot 12 + r t_B = r cdot 12 + s cdot 12 ] [ 12p = r cdot 12 + s cdot 12 ] [ p = r + s ] [ (2r + s) t_B = r cdot 12 + s cdot 6 ] [ t_B = frac{r cdot 12 + s cdot 6}{2r + s} ] - The motorboat took ( t_B = frac{r cdot 12 + s cdot 6}{2r + s} ) hours to travel from dock A to dock C. Conclusion: The final answer for the time ( t_B ) it took for the motorboat to reach dock C before turning back is (frac{r cdot 12 + s cdot 6{2r + s}}) hours. The final answer is boxed{D) (frac{r cdot 12 + s cdot 6}{2r + s})}

answer:The negation of a universal proposition is an existential proposition. The negation of "all" is "there exists at least one", and the negation of "is odd" is "is not odd". Therefore, the correct negation of the given statement "Every integer divisible by 5 is odd" would be "There exists at least one integer divisible by 5 that is not odd". Step by step, we can conclude the following: - The original statement asserts that the property of being divisible by 5 implies being odd for every integer. - To negate this universal statement, we need to find just one counter-example that violates the specified implication. - The counter-example would be an integer that is divisible by 5 but does not have the property of being odd, which means it is even. Hence we get the correct answer: boxed{C. There exists an integer divisible by 5 that is not odd}

answer:(1) Proof: Since nge2, we have that a_n = S_n - S_{n-1}. Thus, S_n - S_{n-1} = frac { 2S_{ n }^{ 2 }}{2S_{n}-1}. Multiplying both sides by (2S_n - 1) gives us: (S_n - S_{n-1})(2S_n - 1) = 2S_n^2. Simplify the equation to obtain: frac {1}{S_{n}}- frac {1}{S_{n-1}} = 2 quad (nge2). From this, we see that the sequence {frac {1}{S_{n}}} has first term frac {1}{S_{1}}=1 and common difference 2, hence it is an arithmetic sequence. (2) From (1), we know that frac {1}{S_{n}} = 2n - 1, which means S_n = frac{1}{2n-1}, so S_{n+1} = frac{1}{2n+1}. Define F(n) = frac{(1+S_1)(1+S_2)ldots(1+S_n)}{sqrt{2n+1}}, and consider the ratio: frac{F(n+1)}{F(n)} = frac{(1+S_{n+1})sqrt{2n+1}}{sqrt{2n+3}} = sqrt{ frac{(2n+2)^2}{(2n+1)(2n+3)}} > 1. Therefore, F(n) is increasing for ninmathbb{N}^*. To ensure F(n)geq k for all n, it suffices to have F(1)geq k, since F(1) is the minimum of F(n). As F(1) = frac{2}{sqrt{3}}, we must have 0<kleqfrac{2}{sqrt{3}}. The maximum value of k is thus boxed{frac{2}{sqrt{3}}}.