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Karolina [17]
2 years ago
9

Write the quadratic equation whose roots are six and -2 and who’s leading coefficient is 3 use the letter X to represent the var

iable
Mathematics
1 answer:
Artist 52 [7]2 years ago
6 0

Answer:

3X^{2} - 12X - 36 = 0

Step-by-step explanation:

6 and - 2 are the only two solutions to the required quadratic equation.

So, if the variable is represented by X then (X - 6) and (X + 2) will be the only two factors of the polynomial function.

Therefore, the equation is  

(X - 6)(X + 2) = 0

⇒ X^{2} - 4X - 12 = 0

If the leading coefficient of the equation is 3 then we can write the equation as 3X^{2} - 12X - 36 = 0 (Answer)

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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
1 year ago
David made a class banner out of a large rectangular piece of paper. He cut a
Irina18 [472]

Answer:

100 in²

Step-by-step explanation:

The area of the banner is equal to the area of the initial rectangle minus the area of the cutout triangle.

The rectangle has a height of 8 inches and width of 14 inches, so its area is:

A = (8 in) (14 in) = 112 in²

The triangle has a base of 8 inches and a height of 3 inches, so its area is:

A = ½ (8 in) (3 in) = 12 in²

So the area of the banner is 112 in² − 12 in² = 100 in².

6 0
2 years ago
For a quadrilateral to be a parallelogram, its opposite sides and _____ must be congruent.
Natalija [7]
<span>opposite sides and _____ </span>and angles 
8 0
2 years ago
Read 2 more answers
A hot air balloon is flying at a constant speed of 20 mi/h at a bearing of N 36° E. There is a 10mi/h crosswind blowing due east
crimeas [40]

Answer:

Step-by-step explanation:

v = {[(20sin36°)i + (20cos36°)j] + 10i} mi/h

 

vE = 20sin36º + 10 = 21.76 mi/h

 

vN = 20cos36° = 16.18 mi/h

 

v = √(vE2 + vN2) = √(21.762 + 16.182) mi/h = 27.12 mi/h

 

θ = tan-1(vN/vE) = tan-1(16.18/21.76) = 36.6º north of east

3 0
1 year ago
Repeated student samples. Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class p
Dima020 [189]

Answer:

a) p-hat (sampling distribution of sample proportions)

b) Symmetric

c) σ=0.058

d) Standard error

e) If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).

Step-by-step explanation:

a) This distribution is called the <em>sampling distribution of sample proportions</em> <em>(p-hat)</em>.

b) The shape of this distribution is expected to somewhat normal, symmetrical and centered around 16%.

This happens because the expected sample proportion is 0.16. Some samples will have a proportion over 0.16 and others below, but the most of them will be around the population mean. In other words, the sample proportions is a non-biased estimator of the population proportion.

c) The variability of this distribution, represented by the standard error, is:

\sigma=\sqrt{p(1-p)/n}=\sqrt{0.16*0.84/40}=0.058

d) The formal name is Standard error.

e) If we divided the variability of the distribution with sample size n=90 to the variability of the distribution with sample size n=40, we have:

\frac{\sigma_{90}}{\sigma_{40}}=\frac{\sqrt{p(1-p)/n_{90}} }{\sqrt{p(1-p)/n_{40}}}}= \sqrt{\frac{1/n_{90}}{1/n_{40}}}=\sqrt{\frac{1/90}{1/40}}=\sqrt{0.444}= 0.667

If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).

0 0
2 years ago
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