Joanna Mueller

Answered

2022-07-18

Water boils as 100C(derees Celsius) which is 212F(degrees Fahrenheit) and freezes at 0C which is 32 F.Let x denote the temperature of an object in: F and y the temperature of the same object in: C

(a) Express x in terms of y (F)

(b) Express y in terms of y (C)

(c) Use (b) to convert 70 F into Celsius

(d) Use (a) to convert your answer to (c) back into Fahrenheit.

(e) What temperature is the same in both Celsius and Fahrenheit?

(a) Express x in terms of y (F)

(b) Express y in terms of y (C)

(c) Use (b) to convert 70 F into Celsius

(d) Use (a) to convert your answer to (c) back into Fahrenheit.

(e) What temperature is the same in both Celsius and Fahrenheit?

Answer & Explanation

sweetwisdomgw

Expert

2022-07-19Added 20 answers

Let $x=ay+b$

When $x={212}^{\circ}F,y=100\phantom{\rule{0ex}{0ex}}\therefore 212=100a+b\text{}\text{}\text{}\text{}(1)$

When $x={212}^{\circ}F,y=100\phantom{\rule{0ex}{0ex}}\therefore 212=100a+b\text{}\text{}\text{}\text{}(1)$

(1) $\Rightarrow 212=100a+32\Rightarrow a=\frac{212-32}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{180}{100}=\frac{9}{5}\phantom{\rule{0ex}{0ex}}\therefore x(y)=\frac{9}{5}y+32$

a) $x(y)=\frac{9}{5}y+32F$

b) $\therefore \frac{9}{5}y=x-32\Rightarrow y=\frac{5}{9}(x-32)\phantom{\rule{0ex}{0ex}}\Rightarrow y(x)=\frac{5x}{9}-\frac{160}{9}C$

c) When $x=70F,y=\frac{5(70)}{9}-\frac{160}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{350-160}{9}=\frac{190}{9}C$

d) $x=\frac{9}{5}(\frac{190}{9})+32\phantom{\rule{0ex}{0ex}}=\frac{190}{5}+32=\frac{190+160}{5}=\frac{350}{5}=70F$

e) When x=y, from (a), $x=\frac{9}{5}$

$\Rightarrow x-\frac{9}{5}x=32\Rightarrow \frac{-4}{5}x=32\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{32\times -5}{4}=-40C=-40F$

When $x={212}^{\circ}F,y=100\phantom{\rule{0ex}{0ex}}\therefore 212=100a+b\text{}\text{}\text{}\text{}(1)$

When $x={212}^{\circ}F,y=100\phantom{\rule{0ex}{0ex}}\therefore 212=100a+b\text{}\text{}\text{}\text{}(1)$

(1) $\Rightarrow 212=100a+32\Rightarrow a=\frac{212-32}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{180}{100}=\frac{9}{5}\phantom{\rule{0ex}{0ex}}\therefore x(y)=\frac{9}{5}y+32$

a) $x(y)=\frac{9}{5}y+32F$

b) $\therefore \frac{9}{5}y=x-32\Rightarrow y=\frac{5}{9}(x-32)\phantom{\rule{0ex}{0ex}}\Rightarrow y(x)=\frac{5x}{9}-\frac{160}{9}C$

c) When $x=70F,y=\frac{5(70)}{9}-\frac{160}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{350-160}{9}=\frac{190}{9}C$

d) $x=\frac{9}{5}(\frac{190}{9})+32\phantom{\rule{0ex}{0ex}}=\frac{190}{5}+32=\frac{190+160}{5}=\frac{350}{5}=70F$

e) When x=y, from (a), $x=\frac{9}{5}$

$\Rightarrow x-\frac{9}{5}x=32\Rightarrow \frac{-4}{5}x=32\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{32\times -5}{4}=-40C=-40F$

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