Answer:
The initial mass of the sample was 16 mg.
The mass after 5 weeks will be about 0.0372 mg.
Step-by-step explanation:
We can write an exponential function to model the situation.
Let the initial amount be A. The standard exponential function is given by:

Where r is the rate of growth/decay.
Since the half-life of Palladium-100 is four days, r = 1/2. We will also substitute t/4 for t to to represent one cycle every four days. Therefore:

After 12 days, a sample of Palladium-100 has been reduced to a mass of two milligrams.
Therefore, when x = 12, P(x) = 2. By substitution:

Solve for A. Simplify:

Simplify:

Thus, the initial mass of the sample was:

5 weeks is equivalent to 35 days. Therefore, we can find P(35):

About 0.0372 mg will be left of the original 16 mg sample after 5 weeks.
You did not provide us with equations to select.
Find the slope m.
m = (1 - 2)/(3 - (-1))
m = -1/(3 + 1)
m = -1/4
Use the slope and one of the points and plug into the point-slope formula.
y - 1 = (-1/4)(x - 3)
Isolate y.
y - 1 = (-1/4)x + (3/4)
y = (-1/4)x + (3/4) + 1
y = (-1/4)x + (7/4)
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Answer:
variable, In algebra, a symbol (usually a letter) standing in for an unknown numerical value in an equation. Commonly used variables include x and y (real-number unknowns), z (complex-number unknowns), t (time), r (radius), and s (arc length).