A submarine is exploring the Pacific Ocean. At 502.5 feet below sea level, the water temperature is 63 1/4°F. The submarine dives down 115 feet deeper, and the water temperature drops by 2 1/5 °F. What were the elevation and the water temperature after the submarine dove? Assume that sea level is at 0 feet of elevation. Write an expression to represent the water temperature after the submarine’s final dive. Then simplify the expression to find the temperature.

At the elevation 617.5 feet below the sea level temperature of the water was 61\frac{1}{20}°F

Step-by-step explanation:

At 502.5 feet below the sea level temperature of the water is 63\frac{1}{4}°F

When submarine dives down 115 feet deeper, water temperature drops by 2\frac{1}{5}°F

Elevation after the submarine dove = -502.5 - 115 = -617.5 feet

This negative sign represents the elevation below the sea level.

And the temperature will be = 63\frac{1}{4}-2\frac{1}{5}

                                              = \frac{253}{4}-\frac{11}{5}

                                              = \frac{1265-44}{20}

                                              = \frac{1221}{20}

                                              = 61\frac{1}{20}°F

Therefore, at the elevation 617.5 feet below the sea level temperature of the water was 61\frac{1}{20}°F

at the ellevation5.17

Step-by-step explanation:

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

answer: 8d

step-by-step explanation:

45+5= 50+5=55+5=60