this can be solve using newtons heating of cooling
(Ts – T) =(Ts – To)*e^(-kt)
Where Ts is the ambient temperature
To is initial temperature
T is the temperature at time t
t is the time
k is constant
fisrt solve the constant k for the given first scenario
(99 – 36) = (99 – 46)*e(-5k)
K = -0.0346
Using k, solve T at t = 13 min
(99 – 46) = (99 – T)*e(-13*(-0.0346)
T = 58.82 degree F
slope = (y2-y1)/(x2-x1)
= (70-52)/(8-1)
= 18/7
=2.6
point slope form
y-y1 = m(x-x1)
y-52 = 2.6(x-1)
distribute
y-52 = 2.6x-2.6
add 52 to each side
y = 2.6x+49.4
y = 2.6x +49
the initial temperature is 49.4 or about 49
and it increases about 2.6 degrees each week
Answer:
t = 6/5
Step-by-step explanation:
Step 1: Define
k(t) = 10t - 19
k(t) = -7
Step 2: Substitute and Evaluate
-7 = 10t - 19
12 = 10t
t = 6/5
2 and 4 are the answers that apply to this statement
