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PilotLPTM [1.2K]
2 years ago
8

Can u answer these for me thank u. first 2 only

Mathematics
2 answers:
8_murik_8 [283]2 years ago
3 0
I think it’s r or q I’m not sure I tried pls brainlest
g100num [7]2 years ago
3 0
It’s R (happy b-day)
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Which of these expressions is equal to 6+(2+3)x5
Bumek [7]
<span>6+(2+3)x5 is equal to 31.
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3 0
3 years ago
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Lina's green pepper plant was 32.65 cm tall in June. During the months of July, August, and September the plant grew 2.6 cm ever
butalik [34]

Answer:

40.45 cm

Step-by-step explanation:

Length of Lina's green pepper plant = 32.65 cm

It grows 2.6 cm during the months of July, August, and September.

Length at the end of July = 32.65 cm + 2.6 cm = 35.25  cm

Length at the end of August = 35.25  cm + 2.6 cm = 37.85  cm

Length at the end of September = 37.85 cm + 2.6 cm = 40.45 cm

So, its length will become 40.45 cm at the end of September.

8 0
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
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
2 years ago
Find the slope: (-2,5) and (2,1
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-1,4 I think so, good luck
7 0
3 years ago
multiply (x-4) (2x+3) using the distributive property. select the answer choice showing the correct distribution
lawyer [7]

Answer:

2x'2-3x-12

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(x-4) (2x+3)

2x'2 + 3x-6x-12

2x'2-3x-12

* this " '2 is squared

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3 years ago
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