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3 years ago

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Mathematics

1 answer:

Ulleksa [173]3 years ago

6 0

The answer : x= one half

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Find the volume of these rctangular prisms length=6.4mm width=2.5mm hight=7mm<br> step by step pls

Soloha48 [4]

Answer:

The volume of this rectangular prism is 112mm.

Step-by-step explanation:

$volume \: = \: length \times width \times height \\ v = 6.4mm \times 2.5mm \times 7mm \\ v = 112mm$

4 0

2 years ago

Can u answer number 10?

KonstantinChe [14]

5.7(0.59)= B
2.8(1.99)= A

8 0

3 years ago

(5 – 3x)4 – 2<br><br> Please I need help. Show steps im so confused

RSB [31]

Answer:

Step-by-step explanation:

P (5-3) =2

E X

M 2 x 4 =8

D X

A X

S 8 - 2 =6

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3 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.

7 0

3 years ago

Find the midpoint of X (26,56) and Y (26,1)

Svetradugi [14.3K]

Answer:

(26, $\frac{57}{2}$ )

Step-by-step explanation:

Hi there!

We are given the coordinates (26, 56) and (26, 1)

We want to find the midpoint of these two points

The midpoint can be found using the formula $(\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2})$ , where $(x_1, y_1)$ & $(x_2, y_2)$ are points

We have 2 points, which is what we need to find the midpoint, but let's label the values of the points to avoid confusion and mistakes when actually calculating

$x_1=26\\y_1=56\\x_2=26\\y_2=1$