Question

The normd balling point of water is 100°C or 212°F and the freezing point of water is 0°C or 32°F.
1) Find the linear relationship between C and F.
ii) Find the value of C for 98.6°F and (iii) the value of F for 38°C.
step answer​

Answer 1

$\huge{\mathfrak{Answer:-}$

Let the x-axis represent the temperature in Celsius

and y-axis represent the temperature is Fahrenheit.

It is given that, the normal boiling point of water is (100, 212)

and the freezing point of water is (0,32)

(1) the equation of line between (100, 212) and (0, 32) is,

$y=mx+c...(1)\\\\where\ m = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} = \dfrac{32 -212}{0-100} = \dfrac{-180}{-100} = \dfrac{9}{5}\\\\Therefore\ equation\ (1)\ becomes\\\\y = \dfrac{9}{5}x+c_1\\\\which\ can\ also\ be\ written\ as,\\\\F = \dfrac{9}{5}C+C_{1}...(ii)\\\\Initially\ at\ C\ = 0, F = 32\\\\Substituting\ in\ (ii)\ we\ get,\\\\32= \frac{9}{5} (0)+c_1\\\\\Rightarrow c_1 = 32\\\\Hence, F = \dfrac{9}{5}C+32.\\\\or\ C = (F-32)\dfrac{5}{9}...(iii)$

$(ii) Given, F = 98.6. C = (9.8.6 - 32) \dfrac{5}{9} = 66.6 \times \dfrac{5}{9}\\\\Thus,\ C=37 (using\ equation\ (ii))\\\\(iii) Given,C = 38, F = \dfrac{9}{5}(38)+32 = 68.4+32 = 100.4\\\\Thus,\ F=100.4\ (using\ equation (iii))$

$\rule {130}{2}\ Be\ Brainly\ \star$

Answer 2

Answer:

\huge{\mathfrak{Answer:-}

Let the x-axis represent the temperature in Celsius

and y-axis represent the temperature is Fahrenheit.

It is given that, the normal boiling point of water is (100, 212)

and the freezing point of water is (0,32)

(1) the equation of line between (100, 212) and (0, 32) is,

\begin{lgathered}y=mx+c...(1)\\\\where\ m = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} = \dfrac{32 -212}{0-100} = \dfrac{-180}{-100} = \dfrac{9}{5}\\\\Therefore\ equation\ (1)\ becomes\\\\y = \dfrac{9}{5}x+c_1\\\\which\ can\ also\ be\ written\ as,\\\\F = \dfrac{9}{5}C+C_{1}...(ii)\\\\Initially\ at\ C\ = 0, F = 32\\\\Substituting\ in\ (ii)\ we\ get,\\\\32= \frac{9}{5} (0)+c_1\\\\\Rightarrow c_1 = 32\\\\Hence, F = \dfrac{9}{5}C+32.\\\\or\ C = (F-32)\dfrac{5}{9}...(iii)\\\\\end{lgathered}

y=mx+c...(1)

where m=

x

2

−x

1

y

2

−y

1

=

0−100

32−212

=

−100

−180

=

5

9

Therefore equation (1) becomes

y=

5

9

x+c

1

which can also be written as,

F=

5

9

C+C

1

...(ii)

Initially at C =0,F=32

Substituting in (ii) we get,

32=

5

9

(0)+c

1

⇒c

1

=32

Hence,F=

5

9

C+32.

or C=(F−32)

9

5

...(iii)

\begin{lgathered}(ii) Given, F = 98.6. C = (9.8.6 - 32) \dfrac{5}{9} = 66.6 \times \dfrac{5}{9}\\\\Thus,\ C=37 (using\ equation\ (ii))\\\\(iii) Given,C = 38, F = \dfrac{9}{5}(38)+32 = 68.4+32 = 100.4\\\\Thus,\ F=100.4\ (using\ equation (iii))\end{lgathered}

(ii)Given,F=98.6.C=(9.8.6−32)

9

5

=66.6×

9

5

Thus, C=37(using equation (ii))

(iii)Given,C=38,F=

5

9

(38)+32=68.4+32=100.4

Thus, F=100.4 (using equation(iii))