Jodi poured herself a cold soda that had an initial temperature 36 degrees f and immediately went outside to sunbathe where the temperature was a steady 99 degree f. after 5minutes the temperature of the soda was 46 degree f .jodi had to run back into the house to answer the phone . what is the expected temperature of the soda after an additional 13 minutes ?

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

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.