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Mnenie [13.5K]
2 years ago
5

ANSWER AND STEPS PLEASE. HELP

Mathematics
1 answer:
Ilia_Sergeevich [38]2 years ago
8 0

Answer:

A. 75.36 in^3

Step-by-step explanation:

r =3 in

h=8 in

v=1/3 π r^2h

= 1/3×3.14×3×8

=75.36 in^2

hope it helps...

have a great day!!

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Find the total area of the entire figure. *
bonufazy [111]

Answer:

392.5 in²

Step-by-step explanation:

This figure is consists of a trapezium and a rectangle.

So, to find the total area of the figure you have to find the area of the trapezium and the rectangle separately  and add them together.

Let us find now.

<u>Area of the Trapezium</u>

Area = \frac{1}{2} × ( sum of the parallel sides ) × height

Area = \frac{1}{2} × (  9 + 20 ) × 5

Area = \frac{1}{2} × 29 × 5

Area = \frac{1}{2} × 145

Area = 72.5 in²

<u>Area of the rectangle</u>

Area = Length × Width

Area = 20 × 16

Area = 320 in²

<u>Total area of the figure</u>

Total area = Area of the trapezium + Area of the rectangle

Total area = 72.5 in² + 320 in²

Total area = 392.5 in²

Hope this helps you :-)

Let me know if you have any other questions :-)

6 0
2 years ago
Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g
AysviL [449]

Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

The initial concentration in the tank is:

c_{o} = \frac{10\,pd}{60\,gal}

c_{o} = 0.167\,\frac{pd}{gal}

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

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Answer:

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Step-by-step explanation:

absolute value is always positive

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