MATH SOLVE

2 months ago

Q:
# WILL GIVE BRAINLIEST If a hospital patient is given 100 milligrams of medicine which leaves the bloodstream at 14% per hour, how many milligrams of medicine will remain in the system after 10 hours? Use the function A(t) = Iert. 24.66 mg 86.94 mg 90.48 mg 405.52 mg

Accepted Solution

A:

All you have to do is plug in the given values into the given equation and evaluate.

The expression is,

[tex]A(t) = l{e}^{rt} [/tex]

But we have to analyze the problem carefully. This is a natural phenomenon that can be modelled by a decay function. The reason is that, after every hour we expect the medicine in the blood to keep reducing.

Therefore we use the decay function rather. This is given by,

[tex]A(t) = l{e}^{-rt}[/tex]

where,

[tex]l = 100 \: milligrams[/tex]

[tex]r = \frac{14}{100} = 0.14[/tex]

and

[tex]t = 10 \: hours[/tex]

On substitution, we obtain;

[tex]A(10) = 100 \times {e}^{ - 0.14 \times 10} [/tex]

[tex]A(10) = 100 \times {e}^{ - 1.4} [/tex]

Now, we take our calculators and look for the constant

[tex]e[/tex]

,then type e raised to exponent of -1.4. If you are using a scientific or programmable calculator you will find this constant as a secondary function. Remember it is the base of the Natural logarithm.

If everything goes well, you should obtain;

[tex]A(10) = 100 \times 0.24659639[/tex]

This implies that,

[tex]A(10) = 24.66[/tex]

Therefore after 10 hours 24.66 mg of the medicine will still remain in the system.

The expression is,

[tex]A(t) = l{e}^{rt} [/tex]

But we have to analyze the problem carefully. This is a natural phenomenon that can be modelled by a decay function. The reason is that, after every hour we expect the medicine in the blood to keep reducing.

Therefore we use the decay function rather. This is given by,

[tex]A(t) = l{e}^{-rt}[/tex]

where,

[tex]l = 100 \: milligrams[/tex]

[tex]r = \frac{14}{100} = 0.14[/tex]

and

[tex]t = 10 \: hours[/tex]

On substitution, we obtain;

[tex]A(10) = 100 \times {e}^{ - 0.14 \times 10} [/tex]

[tex]A(10) = 100 \times {e}^{ - 1.4} [/tex]

Now, we take our calculators and look for the constant

[tex]e[/tex]

,then type e raised to exponent of -1.4. If you are using a scientific or programmable calculator you will find this constant as a secondary function. Remember it is the base of the Natural logarithm.

If everything goes well, you should obtain;

[tex]A(10) = 100 \times 0.24659639[/tex]

This implies that,

[tex]A(10) = 24.66[/tex]

Therefore after 10 hours 24.66 mg of the medicine will still remain in the system.