At the instant that a cake is removed from the oven, the temperature of the cake is 350°F. After 10 minutes, the cake's temperat
ure is 200°F. If the temperature of the room is 70°F, how much longer will it take for the cake to cool to 90°F? (Round your answer to the nearest 5 minutes.) a) 15 min
b) 20 min
c) 25 min
d) 30 min
e) 35 min
Please show your work. Thanks~
Once the cake is removed from the oven, it has been exposed to the room temperature of 70°F. After 10 minutes, the cake's temperature decreased to 200°F, which is 150°F cooler than it initially was (350°F). As for the what is asked, 90°F is 260°F cooler than its original temperature (350°F).
This problem can be expressed in a ratio: 10mins:150<span>°F = N:260</span><span>°F where N is how long it will take for</span><span> cake to cool to 90°F
I cant really show you how to graph this but i can try to explain it. you would make a 4 quadrant graph and do rise over run. 1)start by making the graph 2) on the line of the graph marked Y is where you would put -4 3) From -4, you would go up (rise) by 2 and go over to the right 1 where you end is the slop
