Question

A pizza pan is removed at 9:00 PM from an oven whose temperature is fixed at 425°F in a room that is constant 74°F. After 5 minutes, the pizza pan is at 300 °F

A) at what time is the temperature of the pan 130°F ?


B) determine the time that needs to elapse before the pan is 180°

Answer 1

9:11:48 p.m

9.8 minutes

Explanation:

We Obtian the slope of the relation :

Slope = Rise / Run

(0, 425) ; (5, 300)

x2 = 0 ; y2 = 425 ; x1 = 5 ; y1 = 300

(425 - 300) / (0 - 5)

125 / - 5

= - 25

The intercept :

y = mx + c

425 = - 25(0) + c

425 = 0 + c

c = 425

The equation becomes :

y = - 25x + 425

Temperature at 130

130 = - 25x + 425

130 - 425 = - 25x

-295 = - 25x

x = 11.8

9:00 + 11.8 = 9:11:48 p.m

B.)

180 = - 25x + 425

180 - 425 = - 25x

-245 = - 25x

x = 9.8 minutes