Newton's cooling model is ΔT = ΔTo * e ^ (-k t)
ΔTo = 200°F - 70°F = 130°F
k = 0.6
t = 2 hours
=> ΔT = 130 * e ^ (-0.6 t) = 130 * e^ (-0.6 * 2) = 130 * e ^ (-1.2)
ΔT = 39.15°F
ΔT = T - Tenvironment => T = ΔT + Tenvironment = 39.15°F + 70°F = 109.15°F ≈ 109 °F.
Answer: T = 109 °F
To solve this problem we need to know the form of the general equation of the line. That is, y= mx+b where m is the slope and b is the intercept. We are already given the y-intercept (2) our next step is to fin the slope. Substitute the point give (1,1) in the equation y= mx+2 and then solve for m. The slope is then -6. The final equation is y = -x+2 or <span>x + y - 2 = 0</span>
Answer:
Option B
Step-by-step explanation:
