What is the length and width of a room that is 60 square meters and has a perimeter of 32 meters

1 answer:

7 0

L = lenght
W = width

$L * W = 60 \\
2L + 2W=32\\\\
L = \frac{32-2W}{2}\\
\frac{32-2W}{2}*W = 60\\
(16-W)*W = 60\\
-W^{2} + 16W -60 = 0\\\\

d = 16^{2} - 4*1*60 = 256 - 240 = 16 \\\\
W1 = \frac{-16+\sqrt{16}}{-2} = \frac{-12}{-2} = 6\\
W2 = \frac{-16-\sqrt{16}}{-2} = \frac{-20}{-2} = 10\\

L1 = 10\\
L2 = 6\\
$

so

{L,W} = {6,10} or {10,6}

but L > W so

L = 10
W = 6

You might be interested in

Doblando un alambre de 40 cm formamos un rectángulo. Hallar la expresión algebraica que define el área del rectángulo y calcula

Lady bird [3.3K]

Answer:

$A = 128\,cm^{2}$

Step-by-step explanation:

El área del rectángulo se halla a través de la siguiente fórmula: (The area of the rectangle is obtained by the following formula:)

$A = (40\,cm - 2\cdot x)\cdot x$

Donde el área y $x$ se miden en centímetros cuadrados y centímetros, respectivamente. (Where area and $x$ are measured in square centimeters and centimeters, respectively)

Ahora, $x$ se reemplaza y se encuentra el valor numérico del área: (Now, $x$ is substituted and numerical value of area is finally determined:)

$A = [40\,cm - 2\cdot (4\,cm)]\cdot (4\,cm)$

$A = 128\,cm^{2}$

5 0

2 years ago

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

Novosadov [1.4K]

Answer:

$A(t)=240-220e^{-\frac{t}{40}}$

Step-by-step explanation:

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved.

Volume of the tank = 240 liters
Initial Amount of Salt in the tank, A(0)=20 grams

Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min

$R_{in}=$ (concentration of salt in inflow)(input rate of fluid)

$R_{in}=(1\frac{gram}{liter})( 6\frac{Liter}{min})=6\frac{gram}{min}$

$R_{out}=$ (concentration of salt in outflow)(output rate of fluid)

$R_{out}=(\frac{A(t)}{240})( 6\frac{Liter}{min})\\R_{out}=\frac{A}{40}$

Rate of change of the amount of salt in the tank:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{40}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{40}=6\\$The integrating factor: e^{\int \frac{1}{40}dt} =e^{\frac{t}{40}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{40}}+\dfrac{A}{40}e^{\frac{t}{40}}=6e^{\frac{t}{40}}\\(Ae^{\frac{t}{40}})'=6e^{\frac{t}{40}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{40}})'=\int 6e^{\frac{t}{40}} dt\\Ae^{\frac{t}{40}}=6*40e^{\frac{t}{40}}+C, $(C a constant of integration)\\Ae^{\frac{t}{40}}=240e^{\frac{t}{40}}+C\\$Divide all through by e^{\frac{t}{40}}\\A(t)=240+Ce^{-\frac{t}{40}}$