answer:To solve for overrightarrow{a}⊕overrightarrow{b}, we start by analyzing the given information and applying it step by step. 1. **Given Information**: - |overrightarrow{a}|=1 since overrightarrow{a} is a unit vector. - |overrightarrow{b}|=sqrt{2}. - |{overrightarrow{a}-overrightarrow{b}}|=1. 2. **Finding cosθ**: - We use the formula for the magnitude of the difference of two vectors: |{overrightarrow{a}-overrightarrow{b}}|=sqrt{{|overrightarrow{a}|}^2 - 2overrightarrow{a}⋅overrightarrow{b} + {|overrightarrow{b}|}^2}. - Substituting the given magnitudes and setting the equation equal to 1 (from |{overrightarrow{a}-overrightarrow{b}}|=1), we get: [ 1 = sqrt{1^2 - 2 times 1 times sqrt{2} times cosθ + (sqrt{2})^2}. ] - Squaring both sides to eliminate the square root gives us: [ 1 = 1 - 2sqrt{2}cosθ + 2. ] - Solving for cosθ yields: [ cosθ = frac{sqrt{2}}{2}. ] 3. **Finding sinθ**: - Knowing cosθ, we can find sinθ using the Pythagorean identity sin^2θ + cos^2θ = 1: [ sinθ = sqrt{1 - left(frac{sqrt{2}}{2}right)^2} = frac{sqrt{2}}{2}. ] 4. **Calculating overrightarrow{a}⊕overrightarrow{b}**: - Substituting sinθ and cosθ into the definition of overrightarrow{a}⊕overrightarrow{b}: [ overrightarrow{a}⊕overrightarrow{b} = left|frac{sqrt{2}}{2}overrightarrow{a} + frac{sqrt{2}}{2}overrightarrow{b}right|. ] - Using the properties of vector magnitudes and the given magnitudes of overrightarrow{a} and overrightarrow{b}, we find: [ overrightarrow{a}⊕overrightarrow{b} = sqrt{left(frac{sqrt{2}}{2}right)^2 cdot 1^2 + frac{sqrt{2}}{2} cdot frac{sqrt{2}}{2} cdot 2 + left(frac{sqrt{2}}{2}right)^2 cdot (sqrt{2})^2}. ] - Simplifying the expression inside the square root gives us: [ overrightarrow{a}⊕overrightarrow{b} = sqrt{frac{1}{2} + 1 + 1} = sqrt{frac{5}{2}} = frac{sqrt{10}}{2}. ] Therefore, the correct answer is boxed{C}.

answer:Let x be the length of the third side of the triangle. According to the triangle inequality: 1. 40 + 50 > x implies x < 90 2. 40 + x > 50 implies x > 10 3. 50 + x > 40 implies x > -10 Since x must be an integer and must satisfy all these inequalities, the smallest possible integral value for x that satisfies x > 10 and x < 90 is x = 11. Calculate the perimeter using x = 11: [ text{Perimeter} = 40 + 50 + 11 = boxed{101} text{ units} ] Conclusion: The least possible perimeter of this triangle, with integral side lengths, is boxed{101} units.

answer:Given (x, y, z) are all positive numbers. Part (1): We need to prove: [ frac{x}{yz} + frac{y}{zx} + frac{z}{xy} geq frac{1}{x} + frac{1}{y} + frac{1}{z} ] Since (x, y, z) are positive numbers, we use the AM-GM inequality. 1. Consider the term (frac{x}{yz} + frac{y}{zx}): [ frac{x}{yz} + frac{y}{zx} = frac{1}{z} left( frac{x}{y} + frac{y}{x} right) ] Applying the AM-GM inequality: [ frac{x}{y} + frac{y}{x} geq 2 sqrt{frac{x}{y} cdot frac{y}{x}} = 2 ] Thus, [ frac{x}{yz} + frac{y}{zx} geq frac{2}{z} ] 2. Similarly, [ frac{y}{zx} + frac{z}{xy} = frac{1}{x} left( frac{y}{z} + frac{z}{y} right) geq frac{2}{x} ] 3. And, [ frac{z}{xy} + frac{x}{yz} = frac{1}{y} left( frac{z}{x} + frac{x}{z} right) geq frac{2}{y} ] 4. Adding the three inequalities: [ left( frac{x}{yz} + frac{y}{zx} right) + left( frac{y}{zx} + frac{z}{xy} right) + left( frac{z}{xy} + frac{x}{yz} right) geq frac{2}{z} + frac{2}{x} + frac{2}{y} ] 5. This simplifies to: [ 2 left( frac{x}{yz} + frac{y}{zx} + frac{z}{xy} right) geq frac{2}{x} + frac{2}{y} + frac{2}{z} ] 6. Dividing both sides by 2: [ frac{x}{yz} + frac{y}{zx} + frac{z}{xy} geq frac{1}{x} + frac{1}{y} + frac{1}{z} ] Conclusion: [ boxed{frac{x}{yz} + frac{y}{zx} + frac{z}{xy} geq frac{1}{x} + frac{1}{y} + frac{1}{z}} ] Part (2): Given (x + y + z geq xyz), find the minimum value of (u = frac{x}{yz} + frac{y}{zx} + frac{z}{xy}). Using part (1): [ u = frac{x}{yz} + frac{y}{zx} + frac{z}{xy} geq frac{1}{x} + frac{1}{y} + frac{1}{z} ] 1. Using the AM-HM inequality, we know: [ frac{1}{x} + frac{1}{y} + frac{1}{z} geq frac{9}{x + y + z} ] 2. Applying the given condition (x + y + z geq xyz): Multiplying both sides by (frac{1}{xyz}), we have: [ frac{1}{x} + frac{1}{y} + frac{1}{z} geq 1 ] 3. Then use the earlier conclusion from step (1): [ u = frac{x}{yz} + frac{y}{zx} + frac{z}{xy} geq sqrt{frac{1}{x^2} + frac{1}{y^2} + frac{1}{z^2} + 2left(frac{1}{xy} + frac{1}{yz} + frac{1}{zx}right)} ] Given: [ frac{1}{xy} + frac{1}{yz} + frac{1}{zx} geq 1 ] 4. We have: [ u geq sqrt{frac{1}{x^2} + frac{1}{y^2} + frac{1}{z^2} + 2(1)} ] 5. Using (x = y = z): [ frac{1}{x^2} + frac{1}{y^2} + frac{1}{z^2} + 2(1) = 3 ] Thus: [ u geq sqrt{3} ] Conclusion: [ u_{min} = sqrt{3} ] [ boxed{sqrt{3}} ]

answer:To find the original number, we can set up the equation: x * 0.74 = 1.9832 Now, we can solve for x by dividing both sides of the equation by 0.74: x = 1.9832 / 0.74 x = 2.68 Therefore, the original number is boxed{2.68} .