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nikitadnepr [17]

2 years ago

3

The Lengths of the shorter altitude and the shorter side of parallelogram are 9cm and root 82 cm, respectively. The Length of a

longer diagonal is 15 cm. What is the area of this parallelogram

1 answer:

levacccp [35]2 years ago

7 0

Consider parallelogram ABCD. The shorter sides are AB and CD, so $AB=CD=\sqrt{82}$ cm.

The shorter altitudes are BH=CF=9 cm.

1. Consider right triangle FCD:

$CD^2=DF^2+CF^2,$

$\sqrt{82}^2=9^2+FD^2,\\ \\FD^2=82-81=1,\\ \\FD=1 cm.$

2. Consider right triangle FCA:

$CA^2=DF^2+FA^2,$

$15^2=9^2+FA^2,\\ \\FA^2=225-81=144,\\ \\FA=12 cm.$

3.

$FA=AD+FD,\\ \\AD=12-1=11 cm.$

4. Therefore, the area of parallelogram ABCD is

$A_{ABCD}=AD\cdot BH=11\cdot 9=99$ sq. cm.

Answer: 99 sq. cm.

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Solve for q. 22q+73 > 52q + 63

sammy [17]

Hey!

Let's write the problem.
$22q+73\ \textgreater \ 52q+63$
Subtract $73$ from both sides.
$22q+73-73\ \textgreater \ 52q+63-73$
$22q\ \textgreater \ 52q-10$
Subtract $52q$ from both sides.
$22q-52q\ \textgreater \ 52q-10-52q$
$-30q\ \textgreater \ -10$
Since we don't want negative, we will multiply both sides by $-1$ .
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$30q\ \textless \ 10$
Divide both sides by $30$ .
$\frac{30q}{30}\ \textless \ \frac{10}{30}$

Our final answer would be,
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3 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

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