answer:To solve the given problem, we follow a step-by-step approach closely related to the provided solution: 1. We are given the equation {(2x+1)}^{5}={a}_{0}+{a}_{1}x+⋅⋅⋅+{a}_{5}{x}^{5}. This equation represents the expansion of {(2x+1)}^{5} in terms of powers of x. 2. To find {a}_{0}, we substitute x=0 into the equation: begin{align*} {(2(0)+1)}^{5} &= {a}_{0}+{a}_{1}(0)+⋅⋅⋅+{a}_{5}(0)^{5} {1}^{5} &= {a}_{0} 1 &= {a}_{0}. end{align*} 3. To find the sum a_{1}+a_{2}+ldots+a_{5}, we substitute x=1 into the original equation: begin{align*} {(2(1)+1)}^{5} &= {a}_{0}+{a}_{1}(1)+⋅⋅⋅+{a}_{5}(1)^{5} {(2+1)}^{5} &= {a}_{0}+{a}_{1}+⋅⋅⋅+{a}_{5} {3}^{5} &= 1 + (a_{1}+a_{2}+ldots+a_{5}). end{align*} 4. Solving for a_{1}+a_{2}+ldots+a_{5}, we subtract {a}_{0}=1 from both sides: begin{align*} {3}^{5} - 1 &= a_{1}+a_{2}+ldots+a_{5} 243 - 1 &= a_{1}+a_{2}+ldots+a_{5} 242 &= a_{1}+a_{2}+ldots+a_{5}. end{align*} Therefore, the sum a_{1}+a_{2}+ldots+a_{5}={3}^{5}-1, which corresponds to choice boxed{text{A: }3^{5}-1}.

answer:To solve for the number of positive integers in the range 14 to 40 for which D(n) = 3, we analyze the binary representations of numbers within the range and count those with exactly three transitions between adjacent digits. Case Analysis: For D(n) = 3, the binary representation of n must have exactly three transitions. Examples of general forms include 1...10...01...10...0 and 1...10...01...10...1, where the ellipses represent sequences of the same digit. **Binary Range Details:** - Lowest binary for n = 14 is 1110_2. - Highest binary for n = 40 is 101000_2. **Specific Number Analysis:** - Analyze the numbers with binary digits fitting into our specified forms and within the given range. - Examples include: 1110_2 (D(n)=3), 1010_2 (D(n)=3), 10100_2 (D(n)=3). Consider valid transitions: - Binary length 4: 1010_2. - Binary length 5: 10101_2, 11010_2. - Binary length 6: Substring the binary forms with 3 transitions, and count only those fulfilling 14 leq n leq 40. Count of valid n's in the range: - For binary length 4: 1010_2 = 10 (only this one is valid and within range). - For binary length 5: 10101_2 = 21. - For binary length 6: 11010_2 = 26. Conclusion: Thus, the valid n's are 10, 21, 26 making a total of 3 numbers. The result is therefore 3 integers. The final answer is boxed{textbf{(C)} 3}.

answer:Given that P_1, P_2, …, P_n are points on the parabola C: y^{2}=4x, with their abscissas being x_1, x_2, …, x_n, respectively, and F is the focus of parabola C, x_1+x_2+…+x_n=10, Then, |P_1F|+|P_2F|+…+|P_nF| =(x_1+1)+(x_2+1)+…+(x_n+1) =x_1+x_2+…+x_n+n =n+10. Thus, the answer is boxed{A}. By the properties of a parabola, we know that |P_nF|=x_n+ dfrac {p}{2}=x_n+1, which helps us to find the result. This problem tests the method of finding the sum of a set of line segments in a parabola and is of moderate difficulty. When solving the problem, carefully read the question and pay attention to the reasonable application of the properties of the parabola.

answer:To find the base area of the water tank, we need to understand the relationship between the volume of water being added to the tank and the rate at which the height of the water increases. We know that 1 liter of water is equivalent to 1000 cubic centimeters (cm³) of volume. Since the tank is being filled at a rate of 1 liter per minute, this means that 1000 cm³ of water is being added to the tank each minute. We are also given that the height of the water increases at a rate of 10 centimeters per minute. This means that for every minute, the height of the water goes up by 10 cm. The volume of water added each minute can be calculated using the formula for the volume of a rectangular prism (which is the shape of the water tank): Volume = Base Area × Height Since we know the volume added each minute (1000 cm³) and the rate at which the height increases (10 cm/min), we can rearrange the formula to solve for the base area: Base Area = Volume / Height Plugging in the values we have: Base Area = 1000 cm³ / 10 cm Base Area = 100 cm² Therefore, the base area of the water tank is boxed{100} square centimeters.