If one-third of a number, x, is greater than five less than twice the number, which of the following is true?
1 answer:
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Let:
Vj = Speed of jane
Vm = Speed of mike
d = distance
t = time
Jane travels 7 mph less than 2 times as fast as Mike, therefore:
Vj = 2Vm - 7
Remeber:
distance = speed*time
Distance traveled by mike:
d=Vm*t = Vm*6
Distance traveled by jane:
d + 198 = Vj*6
where:
Vj = 2Vm - 7
d + 198 = (2Vm-7)*6
Now, let:
d=Vm*6 (1)
d + 198 = (2Vm-7)*6 (2)
Replace (1) into (2)
6Vm + 198 = 12Vm - 42
Subtract 6Vm from both sides:
6Vm + 198 - 6Vm = 12Vm - 42 - 6Vm
198 = 6Vm - 42
Add 42 to both sides:
198 + 42 = 6Vm - 42 + 42
240 = 6Vm
Divide both sides by 6:
240/6 = 6Vm/6
40 = Vm
Vm = 40mph
Replace Vm into this equation Vj = 2Vm - 7 :
Vj = 2(40) - 7 = 80 - 7 = 73mph
Answer:
Newton's law of cooling says that:
T(t) = Tₐ + (T₀ - Tₐ)*e^(k*t)
or:
in the differential form.
where:
T is the temperature as a function of time
Tₐ is the ambient temperature, in this case, 70F
T₀ is the initial temperature of the object, in this case, 150F
k is a constant, and we want to find the value of k.
Then our equation is:
T = 70F + (150F - 70F)*e^(k*t)
Now we also know that after a minute, or 60 seconds, the temperature was 135F
then:
135F = 70F + (150F - 70F)*e^(k*60s)
We can solve this for k:
135F = 70F + 80F*e^(k*60s)
135F - 70F = 80F*e^(k*60s)
65F = 80F*e^(k*60s)
(65/80) = e^(k*60s)
Now we can apply the Ln(x) function to both sides to get:
Ln(65/80) = Ln(e^(k*60s))
Ln(65/80) = k*60s
Ln(65/80)/60s = k = -0.0035 s^-1
Then the differential equation is: