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
A=(-36). When finding variables, try multiplying the quotient and divisor to get your dividend. Also, when multiplying/dividing negative integers, if there is one negative integer, the answer will be negative. the only exception is when multiplying/dividing two negative integers, then it would be positive. -36/9=(-4)
|2x-3|=2x-3 ;x≥ 3/2
|2x-3|=-2x+3;x≤3/2
