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Bingel [31]
3 years ago
8

Construct the matrix diagram for each of the following models. Drug diffusion:A drug taken orally eventually diffuses into the b

loodstream through the stomach.Let st be the amount of drug in the stomach at the times t and bt the amount that has been absorbed into the bloodsteam. each time step, 50% of the drug is absorbed from the stomach into the bloodstream and 80% of the drug in the blood stream is metabolized. we have st+1= 0.9st, bt+1=0.5st+0.2bt

Mathematics
1 answer:
yarga [219]3 years ago
8 0

Answer:

Please see attachment

Step-by-step explanation:

Please see attachment

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drag the fractions that is equivalent to each repeating decimal each fractions may be used only once​
gregori [183]

Answer:

0.23 = 7/30

0.4 = 4/9

1.6 = 5/3

3.5 = 32/9

Step-by-step explanation:

I'm 100% that this is correct

6 0
3 years ago
What is the purpose of using fraction strips and a 1srip
love history [14]
the one strip is a whole fraction strip and the others are just part of the whole
8 0
3 years ago
Ashley had a summer lemonade stand where she sold small cups of lemonade for $1.25 and large cups for $2.50. if ashley sold a to
Levart [38]
The answer is 98 small and 57 large cups.

s - the number of small cups
l - the number of large cups

<span>Ashley sold a total of 155 cups: s + l = 155
</span><span>Ashley earned</span><span>for $265: 1.25 * s + 2.50 * l = 265

</span>s + l = 155
1.25 * s + 2.50 * l = 265
________
s = 155 - l
1.25 * s + 2.50 * l = 265
________
1.25 * (155 - l) + 2.50 * l = 265
193.75 - 1.25 * l + 2.50 * l = 265
193.75 + 1.25 * l = 265
1.25 * l = 265 - 193.75
1.25 * l = 71.25
l = 71.25 / 1.25
l = 57

______
s = 155 - l
l = 57

s = 155 - 57
s = 98

5 0
3 years ago
54,000 families have incomes less than $20,000 per year. This number of families is 60% of the families that had this income lev
tangare [24]
54000/.6= 90000, so answer should be 90,000
4 0
3 years ago
(x^2y+e^x)dx-x^2dy=0
klio [65]

It looks like the differential equation is

\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0

Check for exactness:

\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0

*is* exact. If this modified DE is exact, then

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}

We have

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

5 0
3 years ago
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