A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F. Af
1 answer:
Answer:
53.3324
Step-by-step explanation:
given that a thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F.
By Newton law of cooling we have
T(t) =
where T (t) is temperature at time t,T =surrounding temperature = 40, T0 =70 = initial temperature
After half minute thermometer reads 60° F. Using this we can find k
So equation is
When t=1,
we get
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Answer:
a) 1/27
b) 16
c) 1/8
Step-by-step explanation:
a)
One of the properties of the exponents tells us that when we have a negative exponent we can express it in terms of its positive exponent by turning it into the denominator (and changing its sign), so we would have:
And now, solving for x = 9 we have:
b)
This is already a positive rational exponent so we are just going to substitute the value of y = 8 into the expression
c)
Using the same property we used in a) we have:
And now, solving for z = 16 we have:
Answer:
The shadow is decreasing at the rate of 3.55 inch/min
Step-by-step explanation:
The height of the building = 60ft
The shadow of the building on the level ground is 25ft long
Ѳ is increasing at the rate of 0.24°/min
Using SOHCAHTOA,
Tan Ѳ = opposite/ adjacent
= height of the building / length of the shadow
Tan Ѳ = h/x
X= h/tan Ѳ
Recall that tan Ѳ = sin Ѳ/cos Ѳ
X= h/x (sin Ѳ/cos Ѳ)
Differentiate with respect to t
dx/dt = (-h/sin²Ѳ)dѲ/dt
When x= 25ft
tanѲ = h/x
= 60/25
Ѳ= tan^-1(60/25)
= 67.38°
dѲ/dt= 0.24°/min
Convert the height in ft to inches
1 ft = 12 inches
Therefore, 60ft = 60*12
= 720 inches
Convert degree/min to radian/min
1°= 0.0175radian
Therefore, 0.24° = 0.24 * 0.0175
= 0.0042 radian/min
Recall that
dx/dt = (-h/sin²Ѳ)dѲ/dt
= (-720/sin²(67.38))*0.0042
= (-720/0.8521)*0.0042
-3.55 inch/min
Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min
