Answer:
This is a linear function.
Answer:
B. 31.62
Step-by-step explanation:
First, find the distance of the line using the distance formula,
. Plug the two points (-1,0) and (5,2) into the formula and solve. This gives you 6.32, then this is only one line of a pentagon, which has 5 sides, so you need to multiply. Since you are finding the perimeter, which is all of the sides added together, multiply the one side by 5. 6.32*5=31.62.
Answer:
Step-by-step explanation:
Part 1:
Let
Q₁ = Amount of the drug in the body after the first dose.
Q₂ = 250 mg
As we know that after 12 hours about 4% of the drug is still present in the body.
For Q₂,
we get:
Q₂ = 4% of Q₁ + 250
= (0.04 × 250) + 250
= 10 + 250
= 260 mg
Therefore, after the second dose, 260 mg of the drug is present in the body.
Now, for Q₃ :
We get;
Q₃ = 4% of Q2 + 250
= 0.04 × 260 + 250
= 10.4 + 250
= 260.4
For Q₄,
We get;
Q₄ = 4% of Q₃ + 250
= 0.04 × 260.4 + 250
= 10.416 + 250
= 260.416
Part 2:
To find out how large that amount is, we have to find Q₄₀.
Using the similar pattern
for Q₄₀,
We get;
Q₄₀ = 250 + 250 × (0.04)¹ + 250 × (0.04)² + 250 × (0.04)³⁹
Taking 250 as common;
Q₄₀ = 250 (1 + 0.04 + 0.042 + ⋯ + 0.0439)
= 2501 − 0.04401 − 0.04
Q₄₀ = 260.4167
Hence, The greatest amount of antibiotics in Susan’s body is 260.4167 mg.
Part 3:
From the previous 2 components of the matter, we all know that the best quantity of the antibiotic in Susan's body is regarding 260.4167 mg and it'll occur right once she has taken the last dose. However, we have a tendency to see that already once the fourth dose she had 260.416 mg of the drug in her system, that is simply insignificantly smaller. thus we will say that beginning on the second day of treatment, double every day there'll be regarding 260.416 mg of the antibiotic in her body. Over the course of the subsequent twelve hours {the quantity|the quantity|the number} of the drug can decrease to 4% of the most amount, that is 10.4166 mg. Then the cycle can repeat.
last one. because we don't have equality sign and it can't be the third one
I FOUND YOUR COMPLETE QUESTION IN OTHER SOURCES.
SEE ATTACHED IMAGE.
Part A:
The differential equation is:
y '= 0.08y
The initial condition is:
P (0) = 500 Part B:
The formula for this case is:
P (t) = 500exp (0.08 * t) Part C:
After five days we have to evaluate t = 5 in the equation.
We have then:
P (5) = 500 * exp (0.08 * 5)
P (5) = 745.9123488
Rounding:
P (5) = 746