answer:1. **Setup and Definitions:** - Let ( B ) be the intersection point of the two semi-infinite plane mirrors. - Let ( theta ) be the angle of incidence of the incident ray. - Let ( l ) be the distance between the point of incidence and ( B ). 2. **Claim:** - We claim that the circle ( omega ) centered at ( B ) with radius ( l cos theta ) satisfies the given conditions. 3. **Proof of the Claim:** - Let ( P_1 ) be the point of incidence of the incident ray on the first mirror. - Let ( P_2 ) be the point of incidence of the reflected ray on the second mirror. - Let ( theta_1 ) and ( theta_2 ) be the angles of incidence on the first and second mirrors, respectively. 4. **Using the Sine Rule:** - By the sine rule in the triangle formed by ( B ), ( P_1 ), and ( P_2 ), we have: [ frac{BP_1}{BP_2} = frac{cos theta_2}{cos theta_1} ] 5. **Implication of the Sine Rule:** - This implies that for every point of incidence ( P ) at which the angle of incidence is ( theta ) with length ( l ), the product ( l cos theta ) is constant. 6. **Perpendicular Distance:** - Notice that ( l cos theta ) is actually just the perpendicular distance from ( B ) to the incident ray. 7. **Conclusion:** - Since ( l cos theta ) is constant for all points of incidence, all lines of incident/reflected rays are tangential to the circle ( omega ) centered at ( B ) with radius ( l cos theta ). (blacksquare)

answer:1. **Contextual Setting:** - Robinson Crusoe **ended up on a deserted island**. Being alone, the dynamics of societal exchange that typically involve money were **entirely absent**. 2. **Utility of Money on the Island:** - In **modern society**, money serves several essential functions: - **Medium of Exchange:** Used to facilitate transactions between parties. - **Store of Value:** Can be saved and retrieved in the future, retaining its value over time. - **Unit of Account:** Provides a common base for prices. - **Standard of Deferred Payment:** Widely accepted way to settle debts. - On the island, **Crusoe found himself in a context where these functions were rendered useless**: - No one was present to engage in exchange. - Physical survival needs (food, water, shelter) couldn't be met with currency. 3. **Loss of Money's Function as a Medium of Exchange:** - **Precisely because Crusoe lacked others to trade with**, the **primary economic function of money**, that is, **serving as a medium of exchange, no longer applied**. - Money loses its essence when there is **no one to transact with**. 4. **Illustrative Example and Realization:** - The exclamation from Crusoe upon finding money: > "Useless trash, and what good is it to me now? I'd gladly exchange all this gold for any of these trifling knives. There's nothing I can use you for. So off to the bottom of the sea with you." - This demonstrates how money **lost its value** on the deserted island due to the absence of its practical economic utility. 5. **Additional Qualities for Money:** - For an object to **become money**, it must possess the following properties: - **Durability:** Physically withstand wear and tear. - **Portability:** Easily carried and transferred. - **Divisibility:** Broken down into smaller units without losing value. - **Acceptability:** Widely recognized and accepted in exchange for goods and services. - **Uniformity:** Consistent value among units of the same face value. - **Limited Supply:** Scarcity that maintains value over time. # Conclusion: Robinson Crusoe's experience on the island highlighted the crucially contextual nature of money's value as a medium of exchange. On the deserted island, money became worthless because it lacked utility in an environment devoid of exchange opportunities. blacksquare

answer:1. **Understand the Structure of the Rectangular Prism:** - A parallelepiped (2 times 1) is a rectangular prism with dimensions (2 times 1 times 1). - When unfolded into a net, it will create a 2D layout consisting of the 6 faces of the prism: two (2 times 1) rectangles and four (1 times 1) squares. 2. **Identify the Total Area of the Net:** - The net comprises - Two (2 times 1) rectangles: each area = (2) - Four (1 times 1) squares: each area = (1) - Total area of the net: [ text{Total area} = 2 times 2 + 4 times 1 = 4 + 4 = 8 ] 3. **Determine the Number of Squares in the Net:** - When viewing the layout visually, the net will arrange these faces in a way that forms a continuous shape all lying flat. - For a net of (2 times 1 times 1), there are 10 unit squares altogether (since two (2 times 1) rectangles occupy four squares and four 1x1 squares cover four more squares, leading eventually to 10 squares). 4. **Determine the Pattern of Cut Squares:** - According to the prompt, one square was cut out, resulting in nine remaining squares. 5. **Analyse and Draw Possible Nets:** - The drawing should show full net expansion of the original (2 times 1 times 1) paralellipiped. 6. **Illustrate the Possible Configurations:** - Considering that paper square can be cut should align with any possible configuration. # Draw the complete unfolded net and mark where the square could be removed. There are multiple ways to represent this, but for simplicity, marking several viable topographies reveals how cutting affects integrity. **Example of One Possible Correct Configuration:** ```markdown (Consider visual matrix with X being the marks denoting where the square is cut out.) ___________ | | | | | | |------|----| | | | | X |____| |_______ ___| ``` This confirms: The cut could be anywhere, provided it cuts from 10 square net of unfolded (2 times 1) into (1). Net checking reveals other way does meet criteria too. Note, the visual references hoped to can simplify conceptualization, understanding step-by-step. [ boxed{text{One of the possible correct configurations is drawn above.}} ]

answer:To simplify and then evaluate the given expression left(x-3yright)^{2}-left(x-yright)left(x+2yright) with x=frac{1}{2} and y=-1, we follow these steps: 1. Expand and simplify the expression: begin{align*} left(x-3yright)^{2}-left(x-yright)left(x+2yright) &= x^{2}-6xy+9y^{2}-(x^{2}+xy-2y^{2}) &= x^{2}-6xy+9y^{2}-x^{2}-xy+2y^{2} &= -7xy+11y^{2} end{align*} 2. Substitute x=frac{1}{2} and y=-1 into the simplified expression: begin{align*} -7xy+11y^{2} &= -7 times frac{1}{2} times (-1) + 11 times (-1)^{2} &= frac{7}{2} + 11 times 1 &= frac{7}{2} + 11 &= frac{7}{2} + frac{22}{2} &= frac{29}{2} end{align*} Therefore, the evaluated expression is boxed{frac{29}{2}}.