answer:Since there are villagers who tell the truth and those who lie, The first person said: "All 4 of us are liars." Therefore, the first person must be a liar, The second person said: "Among us 4, only one is a liar." If the second person is not lying, then what he said is correct, meaning only the first person is a liar, and the other three are truth-tellers, This contradicts what the third person said: "Among us 4, there are two liars." Therefore, the second person is a liar, The statement made by the second person is incorrect, Therefore, among the four, there is not just one liar; the second person also lies, If the third person is lying, meaning the statement that there are two liars among the four is incorrect, then there are three liars among the four, Only the fourth person is not lying, Hence, the correct choice is boxed{text{D}}.

answer:To calculate Sandy's monthly phone bill expense, we follow these steps: 1. Determine Kim's age in two years: Kim is currently 10 years old, so in two years, she will be 10 + 2 = 12 years old. 2. Calculate Sandy's age in two years: Since Sandy will be three times as old as Kim in two years, Sandy's age at that time will be 3 times 12 = 36 years old. 3. Find Sandy's current age: To find Sandy's current age, we subtract 2 years from her age in two years, giving us 36 - 2 = 34 years old. 4. Calculate Sandy's monthly phone bill expense: Sandy's monthly phone bill expense is ten times her current age. Therefore, her monthly expense is 10 times 34 = 340. So, Sandy's monthly phone bill expense is boxed{340}.

answer:1. **Find the largest power of 7 less than 3500**: - 7^3 = 343 - 7^4 = 2401 - 7^5 = 16807 (too large) 2. **Determine the largest multiple of 7^4 less than 3500**: - 2401 times 1 = 2401 (as 2401 times 2 = 4802 is too large) 3. **Subtract and proceed with the remainder**: - 3500 - 2401 = 1099 4. **Find the largest power of 7 that fits in 1099**: - 7^3 = 343 - 343 times 3 = 1029 (as 343 times 4 = 1372 is too large) 5. **Subtract and move to the next**: - 1099 - 1029 = 70 6. **Find the largest power of 7 that fits in 70**: - 7^2 = 49 - 49 times 1 = 49 (as 49 times 2 = 98 is too large) 7. **Subtract and move to the next**: - 70 - 49 = 21 8. **Find the largest power of 7 that fits in 21**: - 7^1 = 7 - 7 times 3 = 21 9. **Final subtraction (if any)**: - 21 - 21 = 0 This process can now be summarized in base 7: - 3500_{10} = 1 cdot 7^4 + 3 cdot 7^3 + 1 cdot 7^2 + 3 cdot 7^1 + 0 cdot 7^0 - Therefore, 3500_{10} = boxed{13130_7}.

answer:Given a convex polygon, we need to demonstrate that a homothety of this polygon with a coefficient of (frac{1}{2}) can fit inside of itself. 1. **Identify the Largest Triangle:** Consider the triangle (ABC) with the largest area among all triangles whose vertices are vertices of the original convex polygon. 2. **Construct Homothetic Image:** Draw lines through the vertices (A), (B), and (C) of the triangle (ABC) that are parallel to the opposite sides. The lines drawn parallel to side (BC) through (A), to side (CA) through (B), and to side (AB) through (C) will form a larger triangle (A'B'C'), which we call the "doubled" triangle. 3. **Hypothesis:** The entire original polygon is contained within this "doubled" triangle (A'B'C'). Let's assume, for the sake of contradiction, that there is at least one vertex (X) of the polygon that lies outside the triangle (A'B'C'). 4. **Analyzing the Contradiction:** Consider the position of the point (X) relative to (B'C'): - If (X) lies on the opposite side of (BC) relative to (X) and outside (B'C'), then the area of the triangle (XBC) is: [ S_{XBC} > S_{ABC}, ] because (BC) is a common base and the height from (X) to (BC) is larger than the height from (A) to (BC). This contradicts our assumption that (ABC) is the largest-area triangle. - Similarly, if (X) were outside (C'A') or (A'B'), we would face analogous contradictions, proving in each case: [ S_{XCA} > S_{ABC} quad text{or} quad S_{XAB} > S_{ABC}, ] respectively. 5. **Verification via Homothety:** Now, let’s consider a homothety centered at the centroid of triangle (ABC) with a coefficient of (-frac{1}{2}). This transformation would map the larger triangle (A'B'C') back to the original triangle (ABC), and thus compress the entire original polygon into a smaller congruent polygon that fully lies inside (ABC). 6. **Conclusion:** Therefore, by demonstrating through contradiction and geometric homothety, it has been proven that the homothetic image of the convex polygon with a scaling factor of (frac{1}{2}) can indeed be contained within the original polygon. (blacksquare)