Athermometer is taken from a room where the temperature is 24oc to the outdoors, where the temperature is −15oc. after one minute the thermometer reads 14oc. (a) what will the reading on the thermometer be after 3 more minutes?

Solution:

Use Newton's Law of Cooling.  

T = T_s + (T_0 - T_s)*e^(-kt)  

where  

T = temperature at any instant  

T_s = temperature of surroundings  

T_0 = original temperature  

t = elapsed time  

k = constant  

Now, we need to find this constant. We are given that after one hour, the temperature drops to 13° C in a 7°C Environment.  

T = 14, T_0 = 24, T_s = -15, t = 1, k = ?  

T = T_s + (T_0 - T_s)*e^(-kt)  

==> 14 = -15 + (24 - 7)*e^(-k)  

==> 14 = 7 + 17*e^(-k)  

==> 7 = 17*e^(-k)  

==> 7/13 = e^(-k)  

==> -k = ln(7/17)  

==> k = -ln(7/17) ≈ 0.774  

Now,

Let's calculate temperatures!  

T = ?, T_0 = 24, T_s = -15, k = 0.773, t = 3  

T = T_s + (T_0 - T_s)*e^(-kt)  

==> T = -15 + (24 –(-15))*e^[ -(0.774)(2) ]  

==> T = -15 + 39*e^(-1.548)  

==> T ≈ 15.72° C  

This the required answer.

the first question mark is 6 and the second question mark is 26 (the square root)

step-by-step explanation:

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.