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Serjik [45]
2 years ago
14

Which expression is equivalent to StartFraction (3 m Superscript negative 2 Baseline n) Superscript negative 3 Baseline Over 6 m

n Superscript negative 2 Baseline EndFraction? Assume m not-equals 0, n not-equals 0.
StartFraction m Superscript 5 Baseline Over 162 n EndFraction
StartFraction 1 Over 2 m cubed n EndFraction
StartFraction 8 m Superscript 9 Baseline Over n Superscript 9 Baseline EndFraction
StartFraction 4 m Superscript 8 Baseline Over 3 n cubed EndFraction
Mathematics
2 answers:
Dovator [93]2 years ago
8 0

Option a: \frac{m^{5} }{162n} is the equivalent expression.

Explanation:

The expression is \frac{(3m^{-2} n)^{-3}}{6mn^{-2} } where m\neq 0, n\neq 0

Let us simplify the expression, to determine which expression is equivalent from the four options.

Multiplying the powers, we get,

\frac{3^{-3}m^{6} n^{-3}}{6mn^{-2} }

Cancelling the like terms, we have,

\frac{3^{-3}m^{5} n^{-1}}{6 }

This equation can also be written as,

\frac{m^{5}}{3^{3}6 n^{1} }

Multiplying the terms in denominator, we have,

\frac{m^{5} }{162n}

Thus, the expression which is equivalent to \frac{(3m^{-2} n)^{-3}}{6mn^{-2} } is \frac{m^{5} }{162n}

Hence, Option a is the correct answer.

egoroff_w [7]2 years ago
6 0

Answer:

D

Step-by-step explanation:

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A=16500(1-0.0575)^5
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3 0
1 year ago
If pq=9 and qr=28, find PR
Ksenya-84 [330]
PQ = 9
QR = 28

Note, there is a Q in each line

Combine the two lines: PQ + QR = PR

plug in the numbers to corresponding variables

9 + 28 = PR

PR = 37

hope this helps
8 0
2 years ago
What is the solution to the system of equations graphed below?
murzikaleks [220]

Answer: C. (1, 4)

Step-by-step explanation:

The point where the two lines meet or intersect is the solution to the system of equations graphed. And in this case, the lines intersect at (1, 4).

7 0
1 year 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
1 year ago
The library is 1.75 miles directly north from the school. The park is 0.6 miles directly south of the school. How far away is th
Kryger [21]

Answer:

2.35 miles

Step-by-step explanation:

Please find the attachment for visual understanding.

We are told that the library is 1.75 miles directly north from the school and the park is 0.6 miles directly south of the school.

To find distance between library and park we will add distance of library from school to the distance of park from school.

(1.75+0.6)\text{ miles}=2.35\text{ miles}  

Therefore, library is 2.35 miles away from park.  

7 0
2 years ago
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