answer:First, we need to find the LCM of 45 and 75. The prime factorization of 45 is 3^2 cdot 5, and for 75, it is 3 cdot 5^2. To find the LCM, we take the highest power of all prime factors present: - For 3, the highest power is 3^2. - For 5, the highest power is 5^2. Thus, the LCM of 45 and 75 is 3^2 cdot 5^2 = 9 cdot 25 = 225. Next, we check if 225 is divisible by 20. Since 20 is 2^2 cdot 5, and 225 does not contain the factor 2, it is not divisible by 20. Therefore, the smallest positive integer that is a multiple of both 45 and 75 but not a multiple of 20 is boxed{225}.

answer:Let's denote the projection of overset{→}{a} on overset{→}{a} + overset{→}{b} as text{proj}_{overset{→}{a} + overset{→}{b}}overset{→}{a}. The formula for the projection of a vector overset{→}{a} onto another vector overset{→}{b} is given by: text{proj}_{overset{→}{b}}overset{→}{a} = frac{overset{→}{a} cdot overset{→}{b}}{|overset{→}{b}|^2} overset{→}{b} In this case, overset{→}{b} = overset{→}{a} + overset{→}{b}, so we have: text{proj}_{overset{→}{a} + overset{→}{b}}overset{→}{a} = frac{overset{→}{a} cdot (overset{→}{a} + overset{→}{b})}{|overset{→}{a} + overset{→}{b}|^2} (overset{→}{a} + overset{→}{b}) Since overset{→}{a} and overset{→}{b} are unit vectors, |overset{→}{a}| = |overset{→}{b}| = 1. Also, given that |overset{→}{a} + overset{→}{b}| = sqrt{2}|overset{→}{a} - overset{→}{b}|, we can square both sides to get rid of the square root: |overset{→}{a} + overset{→}{b}|^2 = 2|overset{→}{a} - overset{→}{b}|^2 Expanding both sides, we get: overset{→}{a} cdot overset{→}{a} + 2overset{→}{a} cdot overset{→}{b} + overset{→}{b} cdot overset{→}{b} = 2(overset{→}{a} cdot overset{→}{a} - 2overset{→}{a} cdot overset{→}{b} + overset{→}{b} cdot overset{→}{b}) Simplifying this expression, we get: 3overset{→}{a} cdot overset{→}{b} = 1 So, overset{→}{a} cdot overset{→}{b} = dfrac{1}{3}. Substituting this back into our projection formula, we get: text{proj}_{overset{→}{a} + overset{→}{b}}overset{→}{a} = frac{dfrac{1}{3} + 1}{|overset{→}{a} + overset{→}{b}|^2} (overset{→}{a} + overset{→}{b}) From the given condition |overset{→}{a} + overset{→}{b}| = sqrt{2}|overset{→}{a} - overset{→}{b}|, we can find that |overset{→}{a} + overset{→}{b}| = sqrt{6}. Therefore, text{proj}_{overset{→}{a} + overset{→}{b}}overset{→}{a} = frac{dfrac{4}{3}}{sqrt{6}} (overset{→}{a} + overset{→}{b}) Finally, we can simplify this expression to get: text{proj}_{overset{→}{a} + overset{→}{b}}overset{→}{a} = boxed{frac{sqrt{6}}{3}}

answer:1. Let (x) be the portion of the cake eaten by Pechkin. According to the problem, Uncle Fyodor ate half the portion eaten by Pechkin, so Uncle Fyodor ate (frac{x}{2}). 2. The problem states that Cat Matroskin ate half the portion of the cake not eaten by Pechkin. The portion not eaten by Pechkin is (1 - x), so Cat Matroskin's portion is (frac{1 - x}{2}). 3. The sum of the portions eaten by Uncle Fyodor and Cat Matroskin, therefore, is: [ frac{x}{2} + frac{1 - x}{2} = frac{x + 1 - x}{2} = frac{1}{2} ] This means Uncle Fyodor and Cat Matroskin together ate half of the cake. 4. According to the problem, Sharik ate one-tenth of the cake, denoted as (0.1). 5. The total cake eaten by Pechkin and Sharik is half of the cake. Hence we write the equation: [ x + 0.1 = 0.5 ] 6. Solve for (x): [ x + 0.1 = 0.5 implies x = 0.5 - 0.1 implies x = 0.4 ] # Conclusion: The portion of the cake eaten by postman Pechkin is (boxed{0.4}).

answer:Since the slope of the line 2x+y+1=0 is k_{1}=-2, the slope of the line perpendicular to 2x+y+1=0 is k_{2}= frac {-1}{k_{1}}= frac {1}{2}. Comparing options A, B, C, and D, only option B has a slope equal to frac {1}{2}. Therefore, the correct answer is: boxed{text{B}} By converting the line into slope-intercept form, it is easy to find that the slope of the given line is k_{1}=-2. Thus, the slope of the line perpendicular to the given line is k_{2}= frac {-1}{k_{1}}= frac {1}{2}. By comparing each option, we can find the answer to this question. This question provides a given line and asks for a line perpendicular to it, focusing on the basic quantities and forms of lines, as well as the relationships between lines, which makes it a basic question.