answer:If Josh lost 7 marbles and now has 9 marbles left, we can find out how many marbles he had initially by adding the marbles he lost to the marbles he has left. So, the initial number of marbles Josh had = number of marbles he lost + number of marbles he has left = 7 marbles + 9 marbles = 16 marbles Josh initially had boxed{16} marbles in his collection.

answer:Given that ABCD is a rectangle, AB = 4 and AD = 2: 1. Midpoint calculations: - M on overline{AD} implies AM = 1. - N on overline{BC} implies BN = 2. - Since D and B are at the same y-coordinate, and A and C are at the same x-coordinate, points M and N lie on a horizontal line. 2. Coordinate placements (let D = (0,0), C = (4,0), A = (0,2), B = (4,2)): - M = (0,1), N = (4,1). 3. Using coordinates to find lengths: - MN = 4. - Diagonal AC = sqrt{AB^2 + AD^2} = sqrt{16 + 4} = sqrt{20} = 2sqrt{5}. 4. Calculating sin theta: - theta is the angle between overline{AC} (diagonal) and overline{MN} (horizontal line). Hence, theta corresponds to the angle in a right triangle where opposite side is vertical distance from A to line MN which is 1 (from y=2 to y=1). - Using sin theta, for triangle side calculations: [ sin theta = frac{text{opposite side}}{text{hypotenuse}} = frac{1}{2sqrt{5}} = frac{1}{2sqrt{5}} times frac{sqrt{5}}{sqrt{5}} = frac{sqrt{5}}{10} ] - sin theta = boxed{frac{sqrt{5}}{10}}.

answer:1. Let the angles ( angle BAC) and ( angle BCA ) of the triangle (ABC) be (2alpha) and (2gamma) respectively. 2. Note that the points (M) and (N) are the points of tangency of the incircle with sides (AB) and (BC) respectively. 3. The angles (API) and (AMI) are right angles, hence the points (A, P, M,) and (I) lie on a circle with (overline{AI}) as the diameter. This is because of the Inscribed Angle Theorem which states that an angle inscribed in a semicircle is a right angle. 4. Therefore, ( angle AMP = angle AIP ) due to the inscribed angle subtending the same arc (AP). 5. Because ( ell ) is parallel to (AC), we have ( angle AIP = angle IAC = alpha ) (Corresponding Angles Postulate). [ angle AMP = alpha ] 6. Similarly, considering the points (C, Q, N, I) with (CQ) perpendicular to ( ell ), we find that these points lie on another circle for the same reason: ( angle CQI = 90^circ ) means the points are concyclic. [ angle QNC = gamma ] 7. Now, turning our attention to the triangle (MBN), since it is isosceles with (MB = NM) (tangents from common external point B: (BM = BN)), we find the base angles: [ angle BMN = frac{180^circ - angle MBN}{2} = alpha + gamma ] 8. Next, use the previously identified angles to consider ( angle PMN): [ angle PMN = angle PMA + left(180^circ - angle BMNright) = 180^circ - gamma ] 9. On the other side: [ angle PQN = angle ICN = gamma ] 10. Since (angle PMN) and (angle PQN) sums to (180^circ): [ angle PMN + angle PQN = 180^circ ] This implies that the points (M, N, P,) and (Q) must lie on a single circle by the Converse of the Inscribed Angle Theorem. # Conclusion: [ boxed{text{The points } M, N, P, text{ and } Q text{ lie on one circle.}} ]

answer:Mrs. Hilt has already placed 125.0 rocks around the border. She plans to have a total of 189 rocks in the completed border. To find out how many more rocks she has, we subtract the number of rocks she has already placed from the total number of rocks she plans to have: 189 (total planned rocks) - 125.0 (rocks already placed) = 64 more rocks Mrs. Hilt has boxed{64} more rocks to complete her garden border.