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xxMikexx [17]
3 years ago
10

5a - 24b =-15 10a + 45c =21 48b + 35c = 49

Mathematics
1 answer:
andrew-mc [135]3 years ago
6 0

5a + 15 = 24b so,

10a + 30 = 48b,

sub to third equation,

10a + 30 + 35c = 49

10a + 35c = 19

perform elimination with second equation

10a + 45c = 21

10a + 35c = 19

________________-

10c = 2

c = 1/5

sub to second equation,

10a + (45)(1/5) = 21

10a + 9 = 21

10a = 12

a = 6/5

sub to first equation

(5)(6/5) - 24b = -15

6 + 15 = 24b

21 = 24b

b = 7/8

redo the calculations, correct me if I'm wrong

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P(0) =\frac{1200}{1+11*e^-{0.2*0}}

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t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

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P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

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The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

5 0
2 years ago
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