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Anettt [7]
2 years ago
5

9. The half-life of aspirin in your bloodstream is 12 hours. Use exponential model to

Mathematics
1 answer:
monitta2 years ago
6 0

after 6.2 hours the dose will decay to 70% in your bloodstream.

<h3>How long you must wait for the dosage to decay to 70% in your bloodstream?</h3>

We know that the half-life is 12 hours. Then the exponential relation for an initial dosage of A is:

D(t) = A*e^{-t*ln(2)/12h}

If the dosage needs to decay to a 70% of the initial dosage, then we must have:

e^{-t*ln(2)/12h} = 0.7

Now we need to solve that for t:

If we apply the natural logarithm in both sides, we get:

ln(e^{-t*ln(2)/12h}) = ln(0.7)\\\\-t*ln(2)/12h = ln(0.7)\\\\t = -ln(0.7)*12h/ln(2) = 6.2h

So after 6.2 hours the dose will decay to 70% in your bloodstream.

If you want to learn more about half-life:

brainly.com/question/11152793

#SPJ1

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rjkz [21]
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\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle

where 0\le u\le2\pi and 0\le v\le\dfrac\pi/2. We then have

x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v
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Then the surface integral is equivalent to

\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du

We have

\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle
\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle
\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle
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So the surface integral is equivalent to

\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du
=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv
=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw

where w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv.

=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}
=\dfrac{243}2\pi
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3 years ago
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