Worth lots of extra points + a friend if you answer
two stoves were turned off at the same time, and they both started to cool down.
the following equation gives the temperature of the first stove (in degrees celsius) as a function of time (in minutes)
t= 180 - 1.55m
the graph of the temperature of the second stove (in degrees celsius) as a function of time (in minutes) is shown below.
which stove started at a higher temperature?
a. first stove
b. second stove
c. the stove started at the same temperature
which stove cooled down at a greater rate?
a. first stove
b. second stove
c. the stoves cooled done at same rate

Which stove started at a higher temperature?

Second stove

Which stove cooled down at a greater rate?

First stove

Step-by-step explanation:

Which stove started at a higher temperature?

"Started" ---> x = 0

T= 180 - 1.55(0)

T = 180 °C

T = 200 °C

180 °C < 200 °C     ----->     32 °F < 392 °F

Which stove cooled down at a greater rate?

x = 40

slope (rate) = -1.55 °C/min

P1(0,200)     and   P2 (40,130)

-1.55 > -1.75

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.