# 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?

Correct result:

t =  19.06 °C

#### Solution:

$t=11+\frac{0.7}{100}\cdot \left(1160-9\right)=19.0{6}^{\circ }\text{C}$

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