What do the following two equations represent?

Find the equation of a line which is perpendicular to y = 1/2 x - 5 and goes throught the point (0, 0). Write your answer in the

Strike441 [17]

Answer:

Y8jdrupkgyipojhfdddfiii

4 0

4 years ago

Must the difference between two rational numbers be a rational number? Explain

Alona [7]

Claim: The difference between two rational numbers always is a rational number

Proof: You have a/b - c/d with a,b,c,d being integers and b,d not equal to 0.

Then:

a/b - c/d ----> ad/bd - bc/bd -----> (ad - bc)/bd

Since ad, bc, and bd are integers since integers are closed under the operation of multiplication and ad-bc is an integer since integers are closed under the operation of subtraction, then (ad-bc)/bd is a rational number since it is in the form of 1 integer divided by another and the denominator is not eqaul to 0 since b and d were not equal to 0. Thus a/b - c/d is a rational number.

8 0

4 years ago

Read 2 more answers

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

Tell me Dark (appropriate) facts, best dark fact gets brainliest (repost since Echo1255 got it taken down)

podryga [215]

Answer:

A human head remains conscious for around 20 seconds after being decapitated.

5 0

3 years ago

Factorise this expression as fully as possible<br> 2xy2 + 8y

zysi [14]

Is the second two an exponent?

5 0

3 years ago