We know that the Fahrenheit scale can be converted to the centigrade scale as
t(∘C)=(t(∘F)−32)⋅5/9.t(^\circ\mathrm{C}) = (t(^\circ\mathrm{F}) - 32)\cdot 5/9.t(∘C)=(t(∘F)−32)⋅5/9.
t(∘C)t(^\circ\mathrm{C})t(∘C) should be equal to 3t(∘F)3t(^\circ\mathrm{F})3t(∘F) , therefore we obtain an equation
3t(∘F)=(t(∘F)−32)⋅5/9.3t(^\circ\mathrm{F}) = (t(^\circ\mathrm{F}) - 32)\cdot 5/9.3t(∘F)=(t(∘F)−32)⋅5/9.
27t(∘F)=5t(∘F)−160, t(∘F)=−7311∘F≈−7.3∘F.27t(^\circ\mathrm{F}) = 5t(^\circ\mathrm{F})-160, \;\; t(^\circ\mathrm{F}) = -7\dfrac{3}{11}^\circ F \approx-7.3^\circ F.27t(∘F)=5t(∘F)−160,t(∘F)=−7113∘F≈−7.3∘F.
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