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

total amount to be paid = $19.5

step-by-step explanation:

given ,

original price = $25.99

discount = 25%

discount amount = 25% of original price

discount amount = 25/100   * 25.99

discount amount = $6.49

hence , the total amount to be paid = original price - discount amount

                                                            = $25.99 - $6.49

                                                            =$19.5

total amount to be paid = $19.5

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!