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evablogger [386]
2 years ago
8

Lana is the oldest of four sisters. Her youngest sister is half her age. The other two sisters are twins $2$ years younger than

Lana. When the four sisters' ages are added together, the result is $45$. If the youngest sister is $n$ years old, what is $n$?Lana is the oldest of four sisters. Her youngest sister is half her age. The other two sisters are twins $2$ years younger than Lana. When the four sisters' ages are added together, the result is $45$. If the youngest sister is $n$ years old, what is $n$?
Mathematics
1 answer:
vagabundo [1.1K]2 years ago
8 0
Let the youngest sister be n years
Lana will be 2n years, as her youngest sister is half her age
The age of the twins separately will be 2n-2 as they are 2 years younger than Lana
So, the equation looks like this:
n + 2n + 2(2n-2) = 45
Now solve for n:
3n + 4n - 4 = 45
7n - 4 = 45
7n = 49
n = 7
So the youngest sister is 7, Lana is 14 and the twins are 12. 14 + 12 + 12 + 7 =45
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Answer:

285  boxes are in the display

Step-by-step explanation:

Given data

top layer box = 1

last row box = 81

to find out

how many box

solution

we know that every row is a square so that if the bottom layer has 81 squares it mean this is 9² and every row has one lesser box

so that next row will have 8^2 and than 7² and so on till 1²

so we can say that cubes in the rows as that

Sum of all Squares = 9² + 8² +..........+ 1²

Sum of Squares positive Consecutive Integers formula are

Sum of Squares of Consecutive Integers = (1/6)(n)(n+1)(2n+1)  

here n = 9 so equation will be

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2 years ago
The admission fee at a movie theater is $5 for children and $9 for adults. If 3200 people go to the movies and $24000 is collect
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For this problem, let x be the number of children and y for adults. Formulate the equations: 1st equation, x + y = 3,200 and 2nd equation 5x + 9y = 24,000. Re-arrange 1st equation into x = 3200 - y. Then, substitute into 2nd equation, 5(3,200-y) + 9y = 24,000. Then, solve for y. The 16,000 - 5y + 9y = 24000. Final answer is, y = 2000 adults went to watch the movie.
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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)}

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\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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