4. Draw Tre's position at the pitcher's mound as the point (42.78, 42.78) on your diagram above. (1 point)5. Using the distance formula, calculate how far Tre threw the ball. (4 points: 2 points for setup, 1 forcalculation, 1 for the answer). (0-42.78)2+(90-42.78)2, (-42.78)2+(47.22)2= 1,830.13+2,224.73= 63.72 ft 6. Calculate the coordinates for Hector's position. [Note: We can assume that 95 feet is an approximatelyhorizontal distance from the pitcher's mound to the grass line.] (2 points: 1 for x, 1 for y ) 7. Using the distance formula, calculate how far Hector threw the ball. (4 points: 2 points for setup, 1 forcalculation, 1 for the answer). (137.78-90)2+(45-90)2, (47.78)2+(-45)2, 2282.93+2,025= square root 8. Who threw the ball farther, and by how much? (1 point) the ball was thrown 1.9 ft further by hector 9. How far is this throw? (2 points) (322-0)2+(322-0)2, (322)2+(322)2, square root 207,368=455.37 ft