Suppose than an object that is originally at room temperature of 30 Celsius is placed in a freezer. The temperature T (x) of the object can be approximated by the model T (x) = 360 / xˆ2 + 4x + 12, where x is the time in hours after the object is placed in the freezer.a) What is the horizontal asymptote of the graph of this function and what does it represent in the context of this problem?b) A chemist needs a compound cooled to less than 5 Celsius. Determine the amount of time required for the compound to cool so that its temperature is less than 5 Celsius.

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Asked Nov 2, 2019
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Suppose than an object that is originally at room temperature of 30 Celsius is placed in a freezer. The temperature T (x) of the object can be approximated by the model T (x) = 360 / xˆ2 + 4x + 12, where x is the time in hours after the object is placed in the freezer.

a) What is the horizontal asymptote of the graph of this function and what does it represent in the context of this problem?

b) A chemist needs a compound cooled to less than 5 Celsius. Determine the amount of time required for the compound to cool so that its temperature is less than 5 Celsius.

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Expert Answer

Step 1

Given:

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360 where x is the time in hours The temperature function is T(x)= 2 x4x12

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Step 2

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360 lim T (x)lm x4x12 = 0 Thus, the horizontal asymptote is T 0, that is the temperature becomes zero as the time value gets bigger

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Step 3

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360 <5 x4x12 x4x1 72 x+4x-60>0 Solve for x х? +10х — 6х - 60 > 0 (x+10) (x-6)0 x> -10 or x > 6 Time cannot be negative and for x less than or equal to 6, the inequality (x+10) (x-6)> 0 does not hold. Hence, the value ofx must be greater than 6

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Algebra