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
D=√(1.5)(1.4)
d= √6
d= 2.44948.....
Rounded to the nearest tenth would be c. 2.4 mi
Parallel:
y
=
2
3
x
−
11
3
y=
3
2
x−
3
11
perpendicular:
y
=
−
3
2
x
+
5
y=−
2
3
x+5
Answer:
The answer to the first question is B.) The student should have subtracted the 4 first, instead of adding, because it is positive originally. And the answer to the second one is D.) Distribute the -2.
Step-by-step explanation:
Hope this helps! :)
Answer:
A. moving forward for 10m at a steady pace of 2m/s
B. moving backwards 8m at a steady pace of 2.67m/s
C. staying in the same spot for 4 seconds
D. moving 9m at a steady pace of 4.5m/s
E. moving 4.5m with an <u><em>average</em></u><em> </em>pace of 1.5m/s
