answer:Solution. a) To determine the velocity of the motion, we find the derivative of the path with respect to time:v=frac{d x}{d t}=6 t^{2}-8 t+2=2left(3 t^{2}-4 t+1right)and for determining the acceleration of the motion - the derivative of the velocity with respect to time:a=frac{d v}{d t}=2(5 t-4)=4(3 t-2)b) If t=0, then v=2 (m/s) (initial velocity); if t=3, then v=32(mathrm{~m} / mathrm{s}).c) The condition v=0 means that 3 t^{2}-4 t+1=0. Solving this equation, we get t=frac{2}{3} pm sqrt{frac{4}{9}-frac{1}{3}}=frac{2}{3} pm frac{1}{3}. Therefore, the value v=0 is achieved twice: first at the moment t=1 / 3 s, and then at the moment t=1.Let's find the abscissas of the moving point at these moments in time:left.xright|_{t=1 / 3}=left(2 t^{3}-4 t^{2}+2 t+3right)_{t=1 / 3}=3 frac{8}{27} mathrm{~m} ;left.xright|_{t=1}=3(mathrm{~m})

answer:triangle For a natural n, the inequality holds:(n+1)^{2} leqslant n^{2}+3 n<(n+2)^{2}(check it!). But since the numbers (n+1)^{2} and (n+2)^{2} are the two closest perfect squares to each other, the number n^{2}+3 n will be a perfect square only when the left inequality turns into an equality:n^{2}+3 n=(n+1)^{2}, quad n^{2}+3 n=n^{2}+2 n+1, quad n=1Answer: 1.The method of solution we used in problem 663 is called the "pinching" method.

answer:# Answer: 3## Exact match of the answer -1 point## Solution.Notice that the person living on the 5th floor is definitely lying, as there is no one living above them, including liars. Therefore, there are 0 liars above them, and 0 or more knights below. Now consider the resident on the 1st floor. They definitely told the truth, as there are 0 knights below them, and at least 1 liar living above, specifically the one on the 5th floor.Now let's focus on the resident on the 4th floor. Above them lives exactly 1 liar, and below - at least 1 knight on the 1st floor. Therefore, their statement is false in any case. This means they are a liar. Then, the resident on the 2nd floor, similar to the reasoning about the one on the 1st floor, is a knight. The only one left to determine is the resident on the 3rd floor. Above them live 2 liars, and below - 2 knights. Therefore, their statement is false. Hence, they are a liar.#

answer:# Solution:Let s be the distance between points A and B, and v_{1}, v_{2} be the speeds of the cyclists. Then frac{s}{2 v_{1}}=frac{s-24}{v_{2}} quad and quad frac{s-15}{v_{1}}=frac{s}{2 v_{2}} . quad From this, quad frac{s}{2(s-24)}=frac{(s-15) cdot 2}{s} ; quad s^{2}=4 s^{2}-4 cdot 39 s+60 cdot 24 ; s^{2}-52 s+480=0 ; s_{1,2}=26 pm 14 . s_{1}=40, quad s_{2}=12 does not satisfy the conditions s>15, s>24.Answer: 40 km.

answer:(1) Given f(x)= frac {x^{2}}{1+x^{2}}+ log _{2}x,f(2)= frac {4}{1+4}+1= frac {9}{5},f( frac {1}{2})= frac { frac {1}{4}}{1+ frac {1}{4}}-1=- frac {4}{5}, (2 points)f(4)= frac {16}{1+16}+2= frac {50}{17},f( frac {1}{4})= frac { frac {1}{16}}{1+ frac {1}{16}}-2=- frac {33}{17}, (4 points)f(2)+f( frac {1}{2})=1, f(4)+f( frac {1}{4})=1. (6 points)(2) Given f(x)+f( frac {1}{x})= frac {x^{2}}{1+x^{2}}+ log _{2}x+ frac { frac {1}{x^{2}}}{1+ frac {1}{x^{2}}}+ log _{2} frac {1}{x}=1, (9 points)f(1)+f(2)+f(3)+…+f(2016)+f( frac {1}{2})+f( frac {1}{3})+…f( frac {1}{2016})=f(1)+[f(2)+f( frac {1}{2})]+[f(3)+f( frac {1}{3})]+…+[f(2016)+f( frac {1}{2016})]= frac {1}{2}+2015×1= boxed{frac {4031}{2}}. (12 points)

answer:## Solution.Let A B=3 (dm), A D=6 (dm), A A_{1}=4 (dm) (Fig. 11.16); A_{1} H- height. Then A_{1} H=A A_{1} sin 30^{circ}=2 (dm). The area of the base S_{text {base }}=A B cdot A D sin 45^{circ}=9 sqrt{2}left(right. dm left.{ }^{2}right). VolumeV=S_{text {base }} cdot A_{1} H=9 sqrt{2} cdot 2=18 sqrt{2}left(right. dm left.^{3}right).Answer: 18 sqrt{2} dm ^{3}.