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Please help!!!!!!!!!!!!!!!!!!!! Due tomorrow

Please help!!!!!!!!!!!!!!!!!!!! Due tomorrow-example-1

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9a. From 6 yards deep to 15 yards deep, the temperature is constant. So, the temperature at 10 yards deep is the same as it is 6 yards below the surface. The temperature has dropped 2° from the surface temperature for each of those 6 yards, so is (2°/yd)×(6 yd) = 12° less than the surface temperature.

At 10 yards deep, the temperature is 70° -12° = 58 °F

9b. From part A, we know that the temperature at 15 yards (45 feet) below the surface is 58 °F. For each of the 5 feet from 45 feet deep to 50 feet deep, the temperature is increasing 3°. At 50 ft deep, it will be (3°/ft)×(5 ft) = 15° higher than at 45 ft deep.

At 50 feet deep, the temperature is 58° +15° = 73 °F

9c. Based on the result from part B, we can write the expression for the temperature 20 yards (60 ft) deep as

... 73 °F + (3 °F/ft)×(10 ft) . . . . temperature 60 ft below the surface

_____

Evaluating this expression, we get the source temperature to be ...

... = 103 °F

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