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leva [86]
3 years ago
13

you buy seven bags of gummy bears and 3 bag of chocolate and your total is $22 your friend buys a your friend buys three bags of

gummy bears in it and one bag of the chocolate and her total is $8 what is the price of a bag of Gummy Bears and a bag of chocolate
Mathematics
1 answer:
tino4ka555 [31]3 years ago
8 0
Let the cost of 3 bags of chocolate be x and that of gummy be y. Thus to solve the question we form systems of equation as:
3x+7y=22...............i
x+3y=8.....ii
from ii
x=8-3y........iii
substituting iii in i we get:
3(8-3y)+7y=22
24-9y+7y=22
-2y=-2
y=1
thus the value of x=8-3*1=5
thus the price of each bag of chocolate is $5 and that of gummy is $1
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2 years ago
A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like
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Answer:

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

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d) t ≅ 153.51  minutes

Step-by-step explanation:

From the given information;

At time t= 0

an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500

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Rate _{(in)}= 500 \ mg/L  \times 0.06 \  L/min = 30 mg/min

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

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

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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)

x = 2000 + Ce^{(-0.015t)}

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C = -2000

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

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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

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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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Step-by-step explanation:

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