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3 years ago

1. Solve proportion 12/15 = 18/b

1 answer:

3 0

We solve the as follows:

1. Solve proportion 12/15 = 18/b
( 12/15 = 18/b ) b
( 12b/15 = 18 ) 15/12
b = 18(15/12)
b = 45/2

2. Solve equations

A. -x + 4 = x +6
-x + 4 - x = x +6 - x
-2x + 4 - 4 = 6 - 4
-2x = 2
x = -1

B. 5n + 7 =7(n+1) -2n
5n + 7 = 7n + 7 - 2n
5n -7n + 2n = 7 - 7
0 = 0

C. -4(p+2) + 8 = 2(p-1) - 7p + 15
-4p - 8 + 8 = 2p - 2 - 7p + 15
-4p + 7p - 2p = -2 + 15 + 8 - 8
p =13

3. Solve a/b x - c = 0 for x
a/b x - c = 0
( a/b x = c ) b/a
x = bc / a

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Answer:

a) $\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) $\mathbf{x = 2000 - 2000e^{-0.015t}}$

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From the given information;

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$Rate _{(out)}=\dfrac{x}{4} \ mg/L \times 0.06 \ L/min = 0.015x \ mg/min$

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$\dfrac{dx}{dt} = -0.015(x - 2000)$

$\dfrac{dx}{(x - 2000)} = -0.015 \times dt$

By Using Integration Method:

$ln(x - 2000) = -0.015t + C$

$x -2000 = Ce^{(-0.015t)$

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However; if x(0) = 0 ;

Then

C = -2000

Therefore

$\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) What is the steady-state mass of the drug in the blood?

the steady-state mass of the drug in the blood when t = infinity

$\mathbf{x = 2000 - 2000e^{-0.015 \times \infty }}$

x = 2000 - 0

x = 2000

Thus; the steady state mass of the drug is 2000 mg

d) After how many minutes does the drug mass reach 90% of its stead-state level?

After 90% of its steady state level; the mas of the drug is 90% × 2000

= 0.9 × 2000

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Hence;

$\mathbf{1800 = 2000 - 2000e^{(-0.015t)}}$

$0.1 = e^{(-0.015t)$

$ln(0.1) = -0.015t$

$t = -\dfrac{In(0.1)}{0.015}$

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