Preston is attending a school orchestra concert. He sees his math teacher seated 6.8 feet ahead of him and his science teacher seated 7.3 feet to his right. How far apart are the two
teachers? If necessary, round to the nearest tenth.
Preston is attending a school orchestra concert. He sees his math teacher seated 6.8 feet
ahead of him and his science teacher seated 7.3 feet to his right. How far apart are the two
teachers? If necessary, round to the nearest tenth.
2 answers
We can use the Pythagorean Theorem to solve this problem. Let's label the distance between Preston and his math teacher as "a" and the distance between Preston and his science teacher as "b". The distance between the math and science teachers can then be labeled as "c", which is what we are trying to find:
a = 6.8 feet
b = 7.3 feet
c^2 = a^2 + b^2
c^2 = (6.8 feet)^2 + (7.3 feet)^2
c^2 = 46.24 feet^2 + 53.29 feet^2
c^2 = 99.53 feet^2
c ≈ sqrt(99.53) feet ≈ 9.98 feet
Therefore, the distance between the math and science teachers is approximately 9.98 feet. Rounded to the nearest tenth, the answer is 10.0 feet.
a = 6.8 feet
b = 7.3 feet
c^2 = a^2 + b^2
c^2 = (6.8 feet)^2 + (7.3 feet)^2
c^2 = 46.24 feet^2 + 53.29 feet^2
c^2 = 99.53 feet^2
c ≈ sqrt(99.53) feet ≈ 9.98 feet
Therefore, the distance between the math and science teachers is approximately 9.98 feet. Rounded to the nearest tenth, the answer is 10.0 feet.