Teach Yourself - Algebra A complete introduction E6.2 Q13

Linear, quadratic, module, parametric equations

Teach Yourself - Algebra A complete introduction E6.2 Q13

Postby Guest » Sat Aug 11, 2018 1:02 pm

Can someone explain how to get to the answer?

Thanks!
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Re: Teach Yourself - Algebra A complete introduction E6.2 Q1

Postby phw » Sat Aug 11, 2018 10:39 pm

You're given the conversion formula

[tex]x ^\circ F = \left( \frac{9x}{5}+32\right)^\circ C[/tex]

If you substitute [tex]x=86[/tex] into that formula you get

[tex]86 ^\circ F = \left( \frac{9 \cdot 86}{5}+32\right)^\circ C = 186.8^\circ C[/tex]

This where you go, "Bullshit, 86F is a pleasant sunny day. 186C is the temperature for baking a cake." That's because your book made a mistake and got the conversion reversed. They should have written

[tex]x ^\circ C = \left( \frac{9x}{5}+32\right)^\circ F[/tex]

In this case, [tex]x[/tex] represents degrees Celsius which is what you want to solve for, and you are given that [tex]\frac{9x}{5}+32 = 86[/tex]. Subtract 32 for [tex]\frac{9x}{5} = 54[/tex] and multiply by 5/9 for [tex]x=54*5/9 = 30[/tex]. 86F = 30C is much more plausible and is in fact the correct conversion.

phw
 
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