answer:1. **Identify the pattern of movement**: The movement pattern and rules remain unchanged - east, north, west, south in a spiral pattern with increasing steps. 2. **Determine the number of steps in each direction**: - The step pattern starts as 1 east, 1 north, then 2 west, 2 south, and so on, always incrementing the step count in each full cycle by one in each direction. 3. **Calculate the total number of steps for complete cycles**: - The sum of steps for k complete cycles is calculated by: [ S = 2 + 4 + 6 + ldots + 4k = 2(1 + 2 + 3 + ldots + 2k) ] - The sum of the first 2k natural numbers is: [ 1 + 2 + 3 + ldots + 2k = k(2k+1) ] - Thus the total steps after k cycles is: [ S = 2k(2k+1) ] 4. **Find the cycle containing the 2023rd step**: - Compute k such that 2k(2k+1) < 2023 and 2(k+1)(2(k+1)+1) > 2023. - For k = 22, 2k(2k+1) = 44 times 45 = 1980 steps. 5. **Determine the position after 1980 steps**: - After 22 cycles, Aaron is at (-22, -22). 6. **Calculate the remaining steps**: - There are 2023 - 1980 = 43 steps left. - Moving east from (-22, -22), Aaron moves 43 steps east. 7. **Final position**: [ x = -22 + 43 = 21, quad y = -22 ] - Hence, the final position is (21, -22). Thus, the finalized position for Aaron at step p_{2023} is (21, -22). The final answer is boxed{C) (21, -22)}

answer:Given a right triangle with one angle of 30^{circ} and the hypotenuse of 12 inches, this sets up a classic 30-60-90 triangle scenario. In a 30-60-90 triangle: - The side opposite 30^{circ} (short leg) is half of the hypotenuse. - The side opposite 60^{circ} (long leg) is sqrt{3} times the short leg. Thus, the side opposite 30^{circ} (short leg) is [ frac{12}{2} = 6 text{ inches}. ] The side opposite 60^{circ} (long leg) is [ 6 sqrt{3} text{ inches}. ] The area A of a triangle is given by [ A = frac{1}{2} times text{base} times text{height}. ] Here, using the sides mentioned: [ A = frac{1}{2} times 6 times 6sqrt{3} = 18sqrt{3} text{ square inches}. ] Conclusion with the boxed answer: The area of the triangle is boxed{18sqrt{3}} square inches.

answer:Let's denote the amount of fuel A added to the tank as x gallons. Since fuel A contains 12% ethanol by volume, the amount of ethanol in fuel A is 0.12x gallons. The remaining capacity of the tank that was filled with fuel B is (200 - x) gallons. Since fuel B contains 16% ethanol by volume, the amount of ethanol in fuel B is 0.16(200 - x) gallons. The total amount of ethanol in the full fuel tank is the sum of the ethanol from fuel A and fuel B, which is given as 28 gallons. Therefore, we can set up the following equation: 0.12x + 0.16(200 - x) = 28 Now, let's solve for x: 0.12x + 32 - 0.16x = 28 -0.04x + 32 = 28 -0.04x = 28 - 32 -0.04x = -4 Now, divide both sides by -0.04 to find x: x = -4 / -0.04 x = 100 So, boxed{100} gallons of fuel A were added to the tank.

answer:(1) Since point A is outside circle O, x_{0}^{2}+y_{0}^{2} > 13. The distance d from the circle center (0,0) to line l: x_{0}x+y_{0}y=13 is d= frac{13}{sqrt{x_{0}^{2}+y_{0}^{2}}} < sqrt{13}=r. Therefore, the line and the circle intersect. (2) Since point A is on circle O, and x_{0}=2, y_{0} > 0, we have y_{0}=3, so point A(2,3). ① If line AM passes through point O, the slope of AM is K_{AM}= frac{3}{2}, hence K_{AN}=- frac{3}{2}, tan ∠MAN=| frac{K_{AM}-K_{AN}}{1+K_{AM}cdot K_{AN}}|=| frac{frac{3}{2}+frac{3}{2}}{1+frac{3}{2}(-frac{3}{2})}|= frac{12}{5}. ② Denote the slope of line AM as k, then the equation of line AM is: y=kx+3-2k. Substituting y=kx+3-2k into the equation of circle O yields: x^{2}+(kx+3-2k)^{2}=13, Simplifying gives: (k^{2}+1)x^{2}+2k(3-2k)x+(3-2k)^{2}-13=0, Since 2 is a root of the equation, 2x_{M}= frac{(3-2k)^{2}-13}{k^{2}+1}, hence x_{M}= frac{2k^{2}-6k-2}{k^{2}+1}, According to the problem, we have: k_{AN}=-k, similarly, we can obtain: x_{N}= frac{2k^{2}+6k-2}{k^{2}+1}, Therefore, the slope of line MN is k_{MN}= frac{y_{M}-y_{N}}{x_{M}-x_{N}}=kcdot frac{x_{M}+x_{N}-4}{x_{M}-x_{N}}=kcdot frac{frac{-8}{k^{2}+1}}{frac{-12k}{k^{2}+1}}= frac{2}{3}, Thus, no matter how the slope of line AM changes, the slope of line MN is always constant, which is boxed{frac{2}{3}}.