answer:1. Let's draw the diameter ( AE ) of the circumscribed circle around the cyclic quadrilateral ( ABCD ). 2. Because the diagonals of ( ABCD ) are perpendicular, we have: [ angle AEB = angle BCP = 90^circ. ] 3. Also, since ( O ) is the circumcenter, ( angle ABE = angle BPC = 90^circ ) because ( AE ) is a diameter. 4. Therefore, we have: [ angle EAB = angle CBP. ] 5. Since the angles subtended by the chords ( EB ) and ( CD ) are equal, it implies that the lengths of these chords are equal: [ EB = CD. ] 6. Considering ( angle EBA = 90^circ ), this means the line ( EB ) acts as a perpendicular dropped from point ( E ) to the side ( AB ). 7. Now observe that in the right triangle formed, ( E ) is the midpoint of ( EB ) because it subtends the diameter: [ text{The distance from } O text{ to } AB text{ is half of } EB. ] 8. Finally, because ( EB = CD ), [ text{The distance from } O text{ to } AB = frac{EB}{2} = frac{CD}{2}. ] 9. Thus, we have demonstrated that the distance from point ( O ) to the side ( AB ) is half the length of the side ( CD ). Conclusion: [ boxed{frac{CD}{2}} ]

answer:1. **Understand the problem and define variables:** - Let ( x ) be the length of the pipe in meters. - Let ( a = 0.8 ) meters be the length of one step of Gavrila. - Let ( y ) be the distance the pipe moves with each step. - Given that Gavrila takes ( m = 210 ) steps when walking in the same direction as the tractor and ( n = 100 ) steps when walking in the opposite direction. 2. **Formulate equations from given conditions:** - When Gavrila walks in the same direction as the tractor: [ x = m(a - y) ] - When Gavrila walks in the opposite direction as the tractor: [ x = n(a + y) ] 3. **Solve the system of equations:** - From the first equation: [ x = 210 (0.8 - y) ] - From the second equation: [ x = 100 (0.8 + y) ] 4. **Express both equations in terms of ( x ):** - From equation 1: [ x = 210 (0.8 - y) implies x = 168 - 210y ] - From equation 2: [ x = 100 (0.8 + y) implies x = 80 + 100y ] 5. **Set equations equal to each other:** [ 168 - 210y = 80 + 100y ] - Combine like terms: [ 168 - 80 = 100y + 210y implies 88 = 310y implies y = frac{88}{310} implies y = frac{44}{155} ] 6. **Substitute ( y ) back into the equation to find ( x ):** - Using ( x = 210(0.8 - y) ): [ x = 210 left(0.8 - frac{44}{155}right) = 210 left(frac{124 - 44}{155}right) = 210 left(frac{80}{155}right) = frac{16800}{155} approx 108.3871 text{ meters} ] 7. **Round to the nearest whole number:** [ x approx 108 text{ meters} ] # Conclusion: [ boxed{108 text{ meters}} ]

answer:To find out the height of the mosaic, we need to determine the total area of the mosaic in square inches and then convert the height into feet. First, we know that each tile is 1 inch square, so the area of each tile is 1 square inch. Mary needs 21600 tiles for the mosaic, so the total area of the mosaic in square inches is: 21600 tiles * 1 square inch/tile = 21600 square inches The mosaic is 15 feet long. To convert the length into inches, we multiply by 12 (since there are 12 inches in a foot): 15 feet * 12 inches/foot = 180 inches Now we have the length of the mosaic in inches and the total area in square inches. To find the height in inches, we divide the total area by the length: 21600 square inches / 180 inches = 120 inches Finally, to convert the height from inches to feet, we divide by 12: 120 inches / 12 inches/foot = 10 feet So, the mosaic will be boxed{10} feet tall.

answer:We start by finding the largest power of 2 less than 312. We see that 2^8 = 256 is the largest power of 2 less than 312, leaving us with 312 - 256 = 56. The next largest power of 2 less than 56 is 2^5 = 32, leaving a remainder of 56 - 32 = 24. Proceeding further, 2^4 = 16 is less than 24, and subtracting gives 24 - 16 = 8. Finally, 2^3 = 8 exactly equals the remainder, so we use this and get a remainder of zero. Thus, 312_{10} = 1cdot2^8 + 0cdot2^7 + 0cdot2^6 + 1cdot2^5 + 1cdot2^4 + 0cdot2^3 + 0cdot2^2 + 0cdot2^1 + 0cdot2^0 = 101110000_2. The sum of the digits in this binary number is 1 + 0 + 0 + 1 + 1 + 0 + 0 + 0 + 0 = boxed{3}.