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2 years ago

I need help please!!! thank you so much to the person who is going to help me.

Mathematics

2 answers:

garri49 [273]2 years ago

7 0

5x-3
When parentheses are open you get 2x+5+3x-8
Because both parts with x are positive we add them together and get
5x+5-8
+5-8
Is the same as -3 because when you add 5 to -8 you are making the number greater
Hence you get 5x-3

Verizon [17]2 years ago

5 0

Answer:

5x - 3

Step-by-step explanation:

Subtract the numbers

(2x+5) + (3x-8) ----> 2x-3 +3x

Combine like terms

2x-3 + 3x ----> 5x - 3

Answer

5x - 3

done! :)

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A helicopter leaves bristol and flies due east for 10 miles.Then the helicopter flies 8miles north before landing. What is the d

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The distance of the helicopter from the bristol is approximately 1<u>2.81 miles</u>

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Helicopter flies 10 miles east of bristol.

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In order to find the distance of the helicopter from the bristol before landing, we will trace the path of the helicopter

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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:

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

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

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$c$ - 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:

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

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This equation is solved as follows:

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

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

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Where t is measured in minutes.

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