Answer:
N = 74(1/2)^(t/2.8)
Step-by-step explanation:
The exponential function expressing a half-life relation can be written ...
amount = (initial amount) × (1/2)^(t/(half-life))
For the numbers given in this problem, this is ...
N = 74(1/2)^(t/2.8)
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Some folks like to express these relations in the form ...
N = 74e^(-kt)
In this form, the value of k is ...
k = ln(2)/(half-life) ≈ 0.693147/2.8 ≈ 0.24755
N = 74e^(-0.24755t)
Answer:
The correct answer would be the very first one: x > 9.
Step-by-step explanation:
Hope this helps!
Answer:
![f(d(x)) = \frac{9}{5}|x^{100}| + 32](https://tex.z-dn.net/?f=f%28d%28x%29%29%20%3D%20%5Cfrac%7B9%7D%7B5%7D%7Cx%5E%7B100%7D%7C%20%2B%2032)
Step-by-step explanation:
The function d(x) takes a value of x in degrees centigrade and provides the number of degrees that a container of water at that temperature x is far from the boiling point of water.
The function f(d) takes a value d in degrees centigrade and returns a value d(x) in degrees fahrenheit.
Therefore, by doing f(d(x)) we are introducing the function d(x) within the function f(d).
So the range of d(x) now is the domain of f(d(x))
This means that the function f(d(x)) shows the <em>number of degrees Fahrenheit</em> that a water container at a<em> temperature x in degrees Celsius</em> is far from the boiling point of water.
![f(d(x)) = \frac{9}{5}|x^{100}| + 32](https://tex.z-dn.net/?f=f%28d%28x%29%29%20%3D%20%5Cfrac%7B9%7D%7B5%7D%7Cx%5E%7B100%7D%7C%20%2B%2032)
Answer:
where is the table? needed to answer question
Step-by-step explanation: