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Zorluk: KolayLinear Functions and Graphs

A scientist is measuring the temperature of a sample that is being heated at a constant rate. At a starting time of 00 minutes, the temperature of the sample is 12∘C12^\circ\text{C}. After 88 minutes of heating, the temperature of the sample is 44∘C44^\circ\text{C}. If the temperature of the sample increases linearly with time, what is the temperature of the sample, in degrees Celsius, after 1515 minutes of heating?

Cevap: 72 degrees Celsius

Cevap

72
The rate of temperature change is 44−128−0=4∘C\frac{44 - 12}{8 - 0} = 4^\circ\text{C} per minute. Using the slope-intercept form, the temperature TT after tt minutes can be represented by the function T(t)=4t+12T(t) = 4t + 12. Substituting 1515 for tt yields T(15)=4(15)+12=60+12=72T(15) = 4(15) + 12 = 60 + 12 = 72.

Adım Adım Çözüm

1
Calculate the constant rate of temperature change (slope).
4 degrees Celsius per minute
To find how much the temperature increases each minute, divide the total change in temperature by the total change in time: 44−128−0=328=4\frac{44 - 12}{8 - 0} = \frac{32}{8} = 4.
2
Set up the linear function for temperature T(t)T(t) over time tt.
T(t)=4t+12T(t) = 4t + 12
The initial temperature at t=0t = 0 is 12∘C12^\circ\text{C}, which represents the vertical intercept. The constant rate of change is 4∘C4^\circ\text{C} per minute.
3
Find the temperature at t=15t = 15 minutes.
72
Substitute 1515 for tt in the equation: T(15)=4(15)+12=60+12=72T(15) = 4(15) + 12 = 60 + 12 = 72.

Anahtar Kavram

Linear Functions and Rates of Change
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Linear Functions and Graphs

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