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Zorluk: OrtaLinear Equations and Graphing

A scientist is monitoring the temperature of a chemical solution. At time t=13t = \frac{1}{3} hours, the temperature is 2∘C2^\circ\text{C}. At time t=12t = \frac{1}{2} hours, the temperature is 5∘C5^\circ\text{C}. Assuming the temperature increases at a constant rate, what was the initial temperature of the solution, in degrees Celsius?

  1. -4Cevap
  2. B
    29\frac{2}{9}
  3. C
    3
  4. D
    8
  5. E
    30

Cevap

-4
The linear relationship is modeled by T(t)=mt+bT(t) = mt + b. Calculating the slope mm gives 5−21/2−1/3=31/6=18\frac{5 - 2}{1/2 - 1/3} = \frac{3}{1/6} = 18. Substituting the point (1/3,2)(1/3, 2) into the equation yields 2=18(1/3)+b2 = 18(1/3) + b, which simplifies to 2=6+b2 = 6 + b, giving b=−4b = -4. Thus, the initial temperature at t=0t = 0 is −4∘C-4^\circ\text{C}.

Adım Adım Çözüm

1
Find the constant rate of change (slope, mm) using the points (13,2)(\frac{1}{3}, 2) and (12,5)(\frac{1}{2}, 5).
m=5−212−13=316=18m = \frac{5 - 2}{\frac{1}{2} - \frac{1}{3}} = \frac{3}{\frac{1}{6}} = 18
The rate of change represents the slope of the linear relationship between time and temperature.
2
Use the slope-intercept form T(t)=mt+bT(t) = mt + b and the point (13,2)(\frac{1}{3}, 2) to solve for the vertical intercept (bb), which represents the initial temperature.
2=18(13)+b⇒2=6+b⇒b=−42 = 18(\frac{1}{3}) + b \Rightarrow 2 = 6 + b \Rightarrow b = -4
The initial temperature occurs at time t=0t = 0, which corresponds to the yy-intercept (bb) of the linear equation.

Anahtar Kavram

Determining a linear equation from two points to find the initial value (vertical intercept)
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Konular

Linear Equations and Graphing

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