C = 5/9(F - 32)
C = -30
-30 = 5/9(F - 32)
-30 = 5/9F - 160/9
-30 + 160/9 = 5/9F
-270/9 + 160/9 = 5/9F
-110/9 = 5/9F
-110/9 * 9/5 = F
- 990/45 = F
- 22 = F
C = 5/9(F - 32)
C = 130
130 = 5/9F - 160/9
130 + 160/9 = 5/9F
1170/9 + 160/9 = 5/9F
1330/9 = 5/9F
1330/9 * 9/5 = F
11970/45 = F
266 = F
So in Fahrenheit temp, the the car is protected between -22 F and 266 F
I have no clue man maybe ask the teacher
<u><em>Answer:</em></u>
The slope of the line is 77
<u><em>Explanation:</em></u>
We know the speed is constant and that the mile-marker location over time can be represented by a line. This means that the equation is linear.
<u>The slope can, therefore, be calculated as follows:</u>

where (x₁ , y₁) and (x₂ , y₂) are two points on the line
We are given that the two points (1, 123) and (3, 277) are two points in the line
<u>Substitute with them in the above equation to get the slope as follows:</u>

Hope this helps :)
-4.8 here’s in decimal form