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2 years ago

Will mark brainliest! Please help!

Mathematics

1 answer:

ValentinkaMS [17]2 years ago

4 0

Answer:

The dimensions are:

$3(2x+1)$ by $(7x+5)$

Step-by-step explanation:

The area of the rectangle is given as

$A=42x^2+51x+15$

The factored form of this quadratic trinomial gives the dimensions of the rectangle.

We factor 3 first to obtain;

$A=3(14x^2+17x+5)$

We split the middle term to get;

$A=3(14x^2+10x+7x+5)$

We factor within the parenthesis to get;

$A=3(2x(7x+5)+1(7x+5))$

We factor further to get;

$A=3(2x+1)(7x+5)$

The dimensions are:

$3(2x+1)$ by $(7x+5)$

Then the perimeter will be

$2(6x+3+7x+5)=26x+16\:\:\:\boxed{\sqrt{} }$

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3 years ago

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3 years ago

4x^2-7x=15<br> Solve by factoring and please please please explain or show the work

vodomira [7]

Answer:

x=3 and x=-1.25

Step-by-step explanation:

4x^2-7x=15

subtract 15 from each side:

4x^2-7x-15=0

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(2x-6 )(2x+2.5 )=0

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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.

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3 years ago

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klemol [59]

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3 years ago