answer:To find the point (a,b), we identify the formula of the line perpendicular to y = -4x + 6 that passes through the chocolate at (14,13). Since the slope of y = -4x + 6 is -4, the slope of the perpendicular line is frac{1}{4}. The equation of the perpendicular through (14,13) is [ y - 13 = frac{1}{4}(x - 14) Rightarrow y = frac{1}{4}x + frac{45}{2}. ] We find the intersection of this line with y = -4x + 6: [ frac{1}{4}x + frac{45}{2} = -4x + 6. ] To solve for x, [ frac{1}{4}x + 4x = 6 - frac{45}{2} Rightarrow frac{17}{4}x = -frac{33}{2} Rightarrow x = -frac{33}{2} times frac{4}{17} = -frac{66}{17}. ] Substituting x = -frac{66}{17} back into y = -4x + 6: [ y = -4left(-frac{66}{17}right) + 6 = frac{264}{17} + 6 = frac{264}{17} + frac{102}{17} = frac{366}{17}. ] So, (a, b) = left(-frac{66}{17}, frac{366}{17}right) and a+b = -frac{66}{17} + frac{366}{17} = boxed{300/17}.

answer:(I) Solution: The derivative of the function f(x)=(1+x^{2})e^{x}-a(a > 0) is f′(x)=(1+2x+x^{2})e^{x}=(1+x)^{2}e^{x}geqslant 0, f(x) is increasing on (-∞,+∞), Given that x_{0} > 0 is a zero of f(x), we have f(0)=1-a < f(x_{0})=0, i.e., a > 1, and at this time f(a)=(1+a^{2})e^{a}-a > 2a-a=a > 0, then f(x) has only one zero in (0,a), thus a > 1 and f(x) has exactly one zero on (-∞,+∞); (II) Proof: f′(x)=e^{x}(x+1)^{2}, Let the point be P(x_{0},y_{0}), then f′(x_{0})=e^{x_{0}}(x_{0}+1)^{2}, Since the tangent line to y=f(x) at point P is parallel to the x-axis, Therefore, f′(x_{0})=0, i.e., e^{x0}(x_{0}+1)^{2}=0, Thus, x_{0}=-1, Substituting x_{0}=-1 into y=f(x) gives y_{0}= dfrac {2}{e}-a. Therefore, k_{OP}=a- dfrac {2}{e}, Therefore, f′(m)=e^{m}(m+1)^{2}=a- dfrac {2}{e}, To prove m+1leqslant 3a- dfrac {2}{e}, we need to prove (m+1)^{3}leqslant a- dfrac {2}{e}, it is required to prove (m+1)^{3}leqslant e^{m}(m+1)^{2}, i.e., to prove m+1leqslant e^{m}, Therefore, construct the function g(m)=e^{m}-(m+1), then g′(m)=e^{m}-1, from g′(m)=0 we get m=0. When m∈(0,+∞), g′(m) > 0, When m∈(-∞,0), g′(m) < 0, Therefore, the minimum value of g(m) is g(0)=0, Therefore, g(m)=e^{m}-(m+1)geqslant 0, Therefore, e^{m}geqslant m+1, Therefore, e^{m}(m+1)^{2}geqslant (m+1)^{3}, i.e., a- dfrac {2}{e}geqslant (m+1)^{3}, Therefore, m+1leqslant 3a- dfrac {2}{e}. Thus, we have boxed{m+1leqslant 3a- dfrac {2}{e}}.

answer:Let ( n ) be the number of members in the choir. According to the problem, ( n equiv 4 mod 6 ) and ( n equiv 4 mod 10 ). 1. **Finding the LCM of 6 and 10**: [ text{lcm}(6, 10) = 30 ] 2. **Formulating the general solution**: Since ( n equiv 4 ) (mod 6) and ( n equiv 4 ) (mod 10), by the Chinese remainder theorem, ( n equiv 4 mod 30 ). 3. **Finding the valid numbers**: ( n = 30k + 4 ) for some integer ( k ). The valid numbers between 50 and 150 are: [ begin{align*} k = 2 & Rightarrow n = 30 times 2 + 4 = 64 k = 3 & Rightarrow n = 30 times 3 + 4 = 94 end{align*} ] The next value, ( k = 4 ) gives ( n = 124 ), which is still within the range. Thus, the possible number of members in the choir are (boxed{64, 94, 124}).

answer:1. **Calculate the circumference of the Earth at the equator:** Using the formula for the circumference of a circle, C = 2pi r, where r is the radius: [ C = 2pi times 3960 = 7920pi text{ miles} ] 2. **Determine the time taken for the jet to fly around the Earth:** The time, T, needed to travel a distance at a constant speed is: [ T = frac{text{Distance}}{text{Speed}} = frac{7920pi}{550} ] Simplifying this: [ T = frac{7920pi}{550} approx 14.4pi text{ hours} ] Using pi approx 3.14: [ 14.4pi approx 14.4 times 3.14 = 45.216 text{ hours} ] 3. **Conclusion with boxed answer:** The calculated time of approximately 45.216 hours suggests the best estimate for the number of hours of flight is 45. The final answer is boxed{B}