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andreyandreev [35.5K]

2 years ago

5

Edna says that when (x - 2)2 = 9, that x - 2 = 3. Use complete sentences to explain whether Edna is correct. Use specific detail

s in your explanation.

2 answers:

Maslowich2 years ago

7 0

Answer:x=5,-1

Step-by-step explanation:

Edna says that when \left ( x-2\right )^2=9[/tex]

then x-2=3

such that x=5

but this is not true as when we put the value of x in equation then it will not satisfy the equation.

It can be solved by taking the terms either on LHS or RHS

$\left ( x-2\right )^2-9=0$

$x-2=\pm 3$

x=5

x=-1

nataly862011 [7]2 years ago

4 0

Assuming the given equation is (x-2)^2 = 9, Edna is partially correct. Saying x-2 = 3 is halfway there in terms of getting the ful solution. She forgot about the minus form of the plus minus. If (x-2)^2 = 9, then applying the square root to both sides leads to these two equations: x-2 = 3 or x-2 = -3. So we have a plus 3 and then a minus 3.

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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

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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 :

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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

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1 year ago

How many times is 0.02 contained in 80

Brrunno [24]

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2 years ago

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attashe74 [19]

Answer:

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photoshop1234 [79]

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One of the properties of an isosceles trapezoid is that the angles on either side of the two bases are equal. Since line AD is equal to line BC, then angle D is equal to angle C. It also implies that angle A is equal to angle B.

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