I hope you can see it just try to zoom in
Mc012-2.jpg ........................ 10/90%= 11.1111111111111
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
</span>150<span>°FxN=10x260
</span>150N=2600
N=2600/150
N=17.33
N<span>≈20 minutes
Thus, it takes approximately 20 minutes </span><span>for the cake to cool to 90°F</span>
The answer is 0o45.......
