answer:To determine the distance between the two parallel lines, we follow these steps closely aligned with the given solution:1. Given Lines Equations: - Line l_{1} is given by x+(m+1)y+m-2=0 - Line l_{2} is given by mx+2y+8=02. Condition for Parallelism: For two lines to be parallel, their coefficients must satisfy frac{{A}_{1}}{{A}_{2}}=frac{{B}_{1}}{{B}_{2}}neqfrac{{C}_{1}}{{C}_{2}}. Applying this to our lines, we get frac{1}{m}=frac{m+1}{2}neqfrac{m-2}{8}.3. Finding m: Solving frac{1}{m}=frac{m+1}{2} leads to m=1 (we discard m=-2 since it does not satisfy all conditions).4. Substituting m: With m=1, we rewrite both lines: - Line l_{1} becomes x+2y-1=0 - Line l_{2} becomes x+2y+8=05. Calculating Distance Between Parallel Lines: The formula for the distance d between two parallel lines Ax+By+C_{1}=0 and Ax+By+C_{2}=0 is d=frac{|C_{1}-C_{2}|}{sqrt{A^2+B^2}}. Applying this to our lines, we have: [ d = frac{|-1-8|}{sqrt{1^2+2^2}} = frac{|-9|}{sqrt{1+4}} = frac{9}{sqrt{5}} = frac{9sqrt{5}}{5sqrt{5}} = frac{9sqrt{5}}{5} ]Therefore, the distance between these two parallel lines is boxed{frac{9sqrt{5}}{5}}.

answer:## Solutionint_{0}^{2} frac{d x}{sqrt{left(16-x^{2}right)^{3}}}=Substitution:begin{aligned}& x=4 sin t ; d x=4 cos t d t & x=0 Rightarrow t=arcsin frac{0}{4}=0 & x=2 Rightarrow t=arcsin frac{2}{4}=arcsin frac{1}{2}=frac{pi}{6}end{aligned}We get:begin{aligned}& =int_{0}^{pi / 6} frac{4 cos t d t}{sqrt{left(16-16 sin ^{2} tright)^{3}}}=int_{0}^{pi / 6} frac{4 cos t d t}{4^{3} sqrt{left(1-sin ^{2} tright)^{3}}}=frac{1}{16} int_{0}^{pi / 6} frac{cos t d t}{cos ^{3} t}=frac{1}{16} int_{0}^{pi / 6} frac{d t}{cos ^{2} t}= & =left.frac{1}{16} tan tright|_{0} ^{pi / 6}=frac{1}{16} tan frac{pi}{6}-frac{1}{16} tan 0=frac{1}{16} cdot frac{1}{sqrt{3}}=frac{sqrt{3}}{48}end{aligned}Source — «http://pluspi.org/wiki/index.php/������������ %D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82 % mathrm{D} 0 % mathrm{~B} 5 % mathrm{D} 0 % mathrm{~B} 3 % mathrm{D} 1 % 80 % mathrm{D} 0 % mathrm{~B} 0 % mathrm{D} 0 % mathrm{BB} % mathrm{D} 1 % 8 mathrm{~B}+12-22 » Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals- Last edited: 15:55, 5 July 2009.- Content is available under CC-BY-SA 3.0.Created by Geeteatoo## Problem Kuznetsov Integrals 12-23## Material from PlusPi

answer:6. A.If the equation has a positive integer solution (x, y), note that a perfect square is congruent to 0 or 1 modulo 4, thus, x is odd, y is even.Let x=2 m+1, y=2 n. Substituting in, we getm(m+1)-2 n^{2}=1 text {. }Since m(m+1) is even, 2 n^{2} is even, this leads to a contradiction.Therefore, the original equation has no positive integer solutions.

answer:Solution: Let's try to solve it from the end - how to get the number 1 from 50 with the least amount of rubles, if you can only divide by 2 and subtract 1. We get: 50>25->24->12->6->3->2->1. That is, it will require 4 two-ruble and 3 one-ruble coins. It is obvious that if you use 3 two-ruble coins and fewer than 5 one-ruble coins (which corresponds to multiplying by 8), you cannot get a number greater than 40. Fewer one-ruble coins would also not suffice - this can be shown by enumeration.

answer:## Solution.The only solution to the equation is (m, n, p)=(3,1,3).By adding 1 to both sides of the equation and applying the difference of squares, we getbegin{aligned}left(5^{n}+1right)^{2} & =p^{m}+9 & & 1 text{ point} left(5^{n}-2right)left(5^{n}+4right) & =p^{m} . & & 1 text{ point}end{aligned}Both factors on the left side of the equation are natural numbers greater than 1, so for their product to equal a power of a prime number, both must be powers of that prime.Thus, there exist natural numbers a and b such that 5^{n}-2=p^{a} and 5^{n}+4=p^{b}, and a+b=m. Since the second factor is greater than the first, we conclude that b>a.By subtracting the obtained equations, we get6=left(5^{n}+4right)-left(5^{n}-2right)=p^{b}-p^{a}=p^{a}left(p^{b-a}-1right)Since the prime factors of 6 are 2 and 3, we have two cases: p=2 and p=3.If p=2, the initial equation has no solution because the expression 25^{n} is odd, while all other expressions are even.If p=3, then p^{a}=3 and p^{b-a}-1=2, so a=1 and b=2. We conclude that m=3, and from 5^{n}-2=3 we get n=1.

answer:Given 2 < a < 3, then sqrt[3]{(2-a)^{3}}+ sqrt[4]{(3-a)^{4}} = 2-a + |3-a| = 2-a + 3-a = 5-2a,Therefore, the correct choice is: boxed{A}. This problem tests the simplification of radical expressions, which is a basic question.