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denis-greek [22]
3 years ago
12

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~
Mathematics
1 answer:
BARSIC [14]3 years ago
8 0
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>
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\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}

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y(x)=\cos(\ln x)+8\sin(\ln x)
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