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.
Answer:
6.5 cm
Step-by-step explanation:
The computation of the length of the apothem is shown below:
Given that
The side length is 9.4 centimeters
And, the radius is 8 centimeters
Now based on the above information
As per the attached figure
AB = 8 cm
BC = 9.4 ÷ 2 = 4.7 cm
Now apply the pythagoras theorem
AB^2 = BC^2 + AC^2
8^2 = 4.7^2 + AC^2
AC^2 = 41.91
AC = 6.47 cm
= 6.5 cm
Answer: y= 3/2x+8
Step-by-step explanation:
H = 3b+2
A = (h*b)/2 28 = (3b+2)b/2 56 = 3b²+2b 0 = 3b² + 2b - 56
⊕
![\left \{ {{y=2} \atop {x=2}} \right. \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta \\ \\ \\ x^{2} \sqrt{x} \sqrt[n]{x} \frac{x}{y} x_{123} x^{123} \leq \geq \pi \alpha \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] x_{123} \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}}](https://tex.z-dn.net/?f=%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%20%5Cbeta%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20x%5E%7B2%7D%20%20%5Csqrt%7Bx%7D%20%20%5Csqrt%5Bn%5D%7Bx%7D%20%20%5Cfrac%7Bx%7D%7By%7D%20%20x_%7B123%7D%20%20x%5E%7B123%7D%20%20%5Cleq%20%20%5Cgeq%20%20%5Cpi%20%20%5Calpha%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%20x_%7B123%7D%20%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D)
ω
l
∩
C= 2000 + 3(2500)
C= 2000 + 7,500
C= $9,500