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3 years ago

I need help with this problem does anyone know how to solve it

Mathematics

1 answer:

kirill115 [55]3 years ago

6 0

Answer:

i cant see it

Step-by-step explanation:

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Given the slope of 3 and the y-intercept 6, write the equation in slope-intercept form.

Maslowich

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Answer:

y = 3x +6

Step-by-step explanation:

Put the numbers in the formula in their corresponding places.

y = mx + b . . . . slope-intercept form with slope m, intercept b

y = 3x +6 . . . . . slope-intercept form with slope 3, intercept 6

8 0

3 years ago

If f(x)=3[x-2],what is f(5.9)? 9 10 11 12

inna [77]

f(x)=3[x-2]
So, f(5.9) = ?
f(5.9) = 3(5.9 - 2)
=3(3.9)
=11.7 = 12
Thus, the answer is 12.

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3 years ago

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–4(5x) simplify .......

Monica [59]

Answer:

-20x.....

Step-by-step explanation:

–4(5x)

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3 years ago

A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like

mixas84 [53]

Answer:

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

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

c) the steady state mass of the drug is 2000 mg

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

The inflow rate is 0.06 L/min.

Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant.

The objective of the question is to calculate the following :

a) Write an initial value problem that models the mass of the drug in the blood for t ≥ 0.

From above information given :

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

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

Therefore;

$\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}$

with respect to x(0) = 0

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

b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.

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

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

By Using Integration Method:

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

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

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

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

= 1800

Hence;

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

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

$ln(0.1) = -0.015t$

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

t = 153.5056729

t ≅ 153.51 minutes

4 0

3 years ago

On Tuesday the dollar worth of merchandise was sold on Wednesday

____ [38]

Answer:

and?

Step-by-step explanation:

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3 years ago