Mine

What is temperature in the mine at a depth of 1160 m, where at depth 9 m is 11°C and every 100 m, the temperature increases by 0.7°C?

Result

t =  19.06 °C

Solution:

t=11+0.7100(11609)=19.06Ct = 11 + \dfrac{ 0.7}{ 100} \cdot (1160-9) = 19.06 ^\circ \text{C}



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