answer:To find 33 1/3% of 360, you can convert the percentage to a decimal and then multiply by the number. 33 1/3% is equal to 33.333...% (repeating decimal). To convert this to a decimal, divide by 100: 33.333...% ÷ 100 = 0.333333... Now, multiply this decimal by 360: 0.333333... × 360 = 120 So, 33 1/3% of 360 is boxed{120} .

answer:To find the opposite of a number, we simply change its sign. For the number -3, changing its sign gives us: 1. Start with the number: -3 2. Change its sign: -(-3) 3. Simplify: 3 Therefore, the opposite of -3 is 3. So, the correct answer is boxed{B}.

answer:Given that the angle angle F_{1}PF_{2} is 60^{circ}, the point P lies somewhere on the ellipse and is perpendicularly above the x-axis from the left focus F_1. The coordinates of P would either be (-c, frac {b^{2}}{a}) or (-c,- frac {b^{2}}{a}), but since P is above the x-axis, we choose the positive y-coordinate. From the property of an ellipse, the distance between the foci is 2c, and since the angle at P is 60^{circ}, we can form an equilateral triangle with the line segment F_{1}P and PF_{2} being two of its sides. Therefore, the distance PF_{2} is equal to F_{1}P, which is the perpendicular from F_1 to P on the ellipse. Hence, we have: frac {2c}{ frac {b^{2}}{a}}= sqrt {3}, which after simplifying gives: 2ac= sqrt {3}b^{2} = sqrt{3}(a^{2}-c^{2}). Then, upon rearranging the terms and substituting e = frac{c}{a} (since eccentricity e is the ratio of the distance between the foci and the length of the major axis), we get: sqrt {3}e^{2}+2e- sqrt {3}=0. Solving this quadratic equation for e yields two solutions: 1. e = frac{sqrt{3}}{3} (acceptable since e is positive) or 2. e = -sqrt{3} (rejected since eccentricity cannot be negative). Thus, the eccentricity of the ellipse is boxed{frac{sqrt{3}}{3}}.

answer:We need to find all real functions f(x) that satisfy the following conditions: a) There exists a finite interval in which f(x) is bounded, b) For any pair of values x_{1}, x_{2}, [ fleft(x_{1}+x_{2}right)=fleft(x_{1}right)+fleft(x_{2}right), ] c) (f(1)=1). Step-by-Step Solution: 1. **Applying the Functional Equation for Integers:** Let's use the functional equation given in (b). For any integer k, we write: [ f(k cdot x) = k cdot f(x) ] - **Proof by Mathematical Induction:** - **Base case:** For (k = 1), clearly (f(1 cdot x) = f(x)), which holds true. - **Inductive step:** Assume (f(n cdot x) = n cdot f(x)) holds for some integer (n). For (n + 1), apply the functional equation: [ f((n + 1) cdot x) = f(n cdot x + x) = f(n cdot x) + f(x) = n cdot f(x) + f(x) = (n + 1) cdot f(x) ] By induction, the statement holds for all natural numbers (n). For negative integers, note: [ f(-n cdot x) + f(n cdot x) = f(0) Rightarrow -n cdot f(x) + n cdot f(x) = 0 Rightarrow f(-n cdot x) = -n cdot f(x) ] Therefore, (f(k cdot x) = k cdot f(x)) holds for all integers (k). 2. **Finding (f(0))**: We apply the functional equation to x_1 = x and x_2 = 0: [ f(x + 0) = f(x) + f(0) Rightarrow f(x) = f(x) + f(0) Rightarrow f(0) = 0 ] 3. **Extension to Rational Numbers:** Suppose r is a rational number such that r = frac{k}{m}, where k and m are integers and m neq 0. Using the functional equation, we get: [ k cdot f(x) = f(k cdot x) = fleft(m cdot left(frac{k}{m} xright)right) = m cdot fleft(frac{k}{m} xright) ] Dividing both sides by m: [ fleft(frac{k}{m} cdot xright) = frac{k}{m} cdot f(x) ] Thus, for any rational number (r): [ f(r cdot x) = r cdot f(x) ] Applying this to (f(1)), we get: [ f(r) = r quad text{for all rational } r ] 4. **Extension to Irrational Numbers Using Boundedness:** Let (x_0) be an irrational number, and suppose (f(x)) is bounded in the interval ((a, b)). For any rational number (r), there exists some real number (t) such that: [ t r in (a, b) quad text{and} quad r x_0 + t r in mathbb{Q} ] Using the functional equation: [ f(r x_0 + t r) = f(r x_0) + f(t r) Rightarrow r x_0 + t r = r f(x_0) + f(t r) ] Simplifying, we have: [ t r - f(t r) = r (f(x_0) - x_0) ] Since (r) can be arbitrarily large and the left side is bounded, the expression can only hold if: [ f(x_0) = x_0 ] Therefore, the condition must hold for all real numbers (x), implying: [ f(x) = x ] **Conclusion:** The only function that satisfies all given conditions is: [ boxed{f(x) = x} ]