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3 years ago

Can someone help me

Mathematics

1 answer:

Vikki [24]3 years ago

7 0

Answer:

5x + (x+12) = 90

x = 13

m∡1 = 5(13) = 65°

m∡2 = 13 + 12 = 25°

Step-by-step explanation:

6x + 12 = 90

6x = 78

x = 13

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-2.3z = -13.8 I need to show work lol

dsp73

Here are your answers.

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3 years ago

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Which statement about y = x2 + 3x - 18 is true?

Anestetic [448]

Answer:

Step-by-step explanation:

first of all, the correct factorization is

y = (x - 3)(x + 6)

because when we do the multiplication

y = x² + 6x - 3x - 18 = x² + 3x - 18

and then, the factorization tends us also directly the zero points, because y is only 0, when one of the factors is 0.

so, either x-3 or x+6 has to be 0 to make the whole expression 0.

and that is only for x=3 or x=-6

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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

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

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3 years ago

A square field has an area of 479ft^2. What is the approximate length of a side of the field? Give your answer to the nearest fo

ikadub [295]

Area of a square= length^2
479=length^2
√479= length
so length= 21.89ft rounded to 22ft

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3 years ago

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GREYUIT [131]

0.31 = 31%
% is out of 100 so 31% is 31 out of 100 which is 31/100 :)

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3 years ago

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