A relationship of conversion between Fahrenheit temperature (tf) and Celsius temperature (tc) is a straight line whose equation is

Question

A relationship of conversion between Fahrenheit temperature (tf) and Celsius temperature (tc) is a straight line whose equation is (A) (tc-32/100)=(tf/180) (B) (tf-32/100)=(tc-0/100-0) (C) (tf-32/180)=(tc/100) (D) (tc-32/180)=(tf/100)

Tardigrade Admin · Accepted Answer

The ice point in Fahrenheit scale is 32° F and the steam point in Fahrenheit scale is 212° F. Similarly, the ice point in Celsius scale is 0° C and the steam point in Celsius scale is 100° C. Thus, (tf-32/212-32)=(t c -0/100-0) ⇒ (tf-32/180)=(t c /100)

Q. A relationship of conversion between Fahrenheit temperature $(t_{f})$ and Celsius temperature $(t_{c})$ is a straight line whose equation is

$\frac{t _{c} - 32}{100} = \frac{t _{f}}{180}$

$\frac{t _{f} - 32}{100} = \frac{t _{c} - 0}{100 - 0}$

$\frac{t _{f} - 32}{180} = \frac{t _{c}}{100}$

$\frac{t _{c} - 32}{180} = \frac{t _{f}}{100}$

Q. A relationship of conversion between Fahrenheit temperature (tf​) and Celsius temperature (tc​) is a straight line whose equation is

100tc​−32​=180tf​​

100tf​−32​=100−0tc​−0​

180tf​−32​=100tc​​

180tc​−32​=100tf​​

The ice point in Fahrenheit scale is 32∘F and the steam point in Fahrenheit scale is 212∘F. Similarly, the ice point in Celsius scale is 0∘C and the steam point in Celsius scale is 100∘C. Thus, 212−32tf​−32​=100−0tc​−0​ ⇒180tf​−32​=100tc​​

Q. A relationship of conversion between Fahrenheit temperature $(t_{f})$ and Celsius temperature $(t_{c})$ is a straight line whose equation is

$\frac{t _{c} - 32}{100} = \frac{t _{f}}{180}$

$\frac{t _{f} - 32}{100} = \frac{t _{c} - 0}{100 - 0}$

$\frac{t _{f} - 32}{180} = \frac{t _{c}}{100}$

$\frac{t _{c} - 32}{180} = \frac{t _{f}}{100}$