1. using the equation convert the following temperatures (2 decimal place accuracy unless exact answer):

$c = \frac { 5 } { 9 } \cdot f - \frac { 160 } { 9 }$

a. $40 ^ { \circ } f$ to celsius
b. $80 ^ { \circ } f$ to celsius

Question

1. using the equation convert the following temperatures (2 decimal place accuracy unless exact answer):

$c = \frac { 5 } { 9 } \cdot f - \frac { 160 } { 9 }$

a. $40 ^ { \circ } f$ to celsius
b. $80 ^ { \circ } f$ to celsius

Accepted Answer

# Explanation:
## Step1: Substitute $ F = 40 $ into the formula $ C=\frac{5}{9}\cdot F-\frac{160}{9} $
$$
\begin{align*}
C&=\frac{5\times40}{9}-\frac{160}{9}\
&=\frac{200}{9}-\frac{160}{9}\
&=\frac{200 - 160}{9}\
&=\frac{40}{9}\
&\approx4.44
\end{align*}
$$
## Step2: Substitute $ F = 80 $ into the formula $ C=\frac{5}{9}\cdot F-\frac{160}{9} $
$$
\begin{align*}
C&=\frac{5\times80}{9}-\frac{160}{9}\
&=\frac{400}{9}-\frac{160}{9}\
&=\frac{400-160}{9}\
&=\frac{240}{9}\
&=\frac{80}{3}\
&\approx26.67
\end{align*}
$$

# Answer:
a. $ 4.44^{\circ}C $
b. $ 26.67^{\circ}C $

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QUESTION IMAGE

Question

Explanation:

Step1: Substitute \( F = 40 \) into the formula \( C=\frac{5}{9}\cdot F-\frac{160}{9} \)

Step2: Substitute \( F = 80 \) into the formula \( C=\frac{5}{9}\cdot F-\frac{160}{9} \)

Answer: