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3 years ago

The perimeter of a rectangular pool is 312 m.

Mathematics

1 answer:

expeople1 [14]3 years ago

3 0

The perimeter of a rectangle is calculated by:

$\boxed{P=width +width+length+ length}$

Remember: In a rectangle, opposite sides are equals.

If we know how much is the width, then it's just apply it to the expression:

$P=width+width+length+length\\312 =68 + 68 + length + length\\312 = 136+(2\cdot length)\\312 - 136 = 2\cdot length\\2\cdot length = 176\\length = \frac{176}{2} \\length = 88$

Therefore, the length of the pool is 88m.

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What is the value of 7

maw [93]

7 can be written as :-

7 + 0 = 7

7 - 0 = 7

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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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3 years ago

Last month, Lana sold 35 necklaces a week at $15 each. However, throughout the month, 6 customers returned their necklaces for a

Lubov Fominskaja [6]

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3 years ago

Read 2 more answers

How many cubic centimeters of water can this paper cone cup hold?

RUDIKE [14]

By definition, the volume of a cone is given by:
V = (1/3) * (pi) * (r ^ 2) * (h)
Where,
r: cone radius
h: height of the cone
Substituting values:
V = (1/3) * (3.14) * ((8/2) ^ 2) * (9)
V = 150.72 cm ^ 3

Answer:
this paper cone cup can hold:
V = 150.72 cm ^ 3

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4 years ago

ONLY! people that can answer this please do so!

trapecia [35]

Answer:

2.59 feet

Step-by-step explanation:

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3 years ago