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Julli [10]
4 years ago
14

What is the length of a rectangle whose perimeter is 140 ft if its width-to-length is 2 : 5

Mathematics
1 answer:
Sauron [17]4 years ago
6 0

Answer:

50 ft.

Step-by-step explanation:

It is given that the ratio of width-to-length is 2 : 5.

Let,

Width = 2x ft

Length = 5x ft

Perimeter of the rectangle is

Perimeter=2(lenght+width)

Perimeter=2(5x+2x)

Perimeter=2(7x)

Perimeter=14x

It is given that the perimeter of the rectangle is 140 ft.

14x=140

Divide both sides by 14.

x=10

The value of x is 10.

Length=5x=5\times 10=50

Therefore, the length of the rectangle is 50 ft.

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When an object is sold at a discount.
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Remember: The discount price can also be arrived at by multiplying the Original price with the difference of 100% and the discount rate, in its decimal form.

Original price * (100%-discount rate)/100 = Discount Price

Using the latest formula, we can compute for the original price by dividing the discount price by the difference of 100% and the discount rate, in its decimal form.

Original Price = Discount Price / (100%-discount rate)/100

Original Price = 24 / (100%-20%)/100
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(1 point) A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of
Paraphin [41]

1. dy/dt is the net rate of change of salt in the tank over time. As such, it's equal to the difference in the rates at which salt enters and leaves the tank.

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(0.4 kg/L) (6 L/min) = 2.4 kg/min

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(concentration of salt at time t) (4 L/min)

The concentration of salt is the amount of salt (in kg) per unit volume (in L). At any time t > 0, the volume of solution in the tank is

100 L + (6 L/min - 4 L/min) t = 100 L + (2 L/min) t

That is, the tank starts with 100 L of pure water, and every minute 6 L of solution flows in and 4 L is drained, so there's a net inflow of 2 L of solution per minute. The amount of salt at time t is simply y(t). So, the outflow rate is

(y(t)/(100 + 2t) kg/L) (4 L/min) = 2 y(t) / (50 + t) kg/min

and the differential equation for this situation is

\dfrac{dy}{dt} = 2.4 \dfrac{\rm kg}{\rm min} - \dfrac{2y}{50+t} \dfrac{\rm kg}{\rm min}

There's no salt in the tank at the start, so y(0) = 0.

2. Solve the ODE. It's linear, so you can use the integrating factor method.

\dfrac{dy}{dt} = 2.4 - \dfrac{2y}{50+t}

\dfrac{dy}{dt} + \dfrac{2}{50+t} y = 2.4

The integrating factor is

\mu = \displaystyle \exp\left(\int \frac{2}{50+t} \, dt\right) = \exp\left(2\ln|50+t|\right) = (50+t)^2

Multiply both sides of the ODE by µ :

(50+t)^2 \dfrac{dy}{dt} + 2(50+t) y = 2.4 (50+t)^2

The left side is the derivative of a product:

\dfrac{d}{dt}\left[(50+t)^2 y\right] = 2.4 (50+t)^2

Integrate both sides with respect to t :

\displaystyle \int \dfrac{d}{dt}\left[(50+t)^2 y\right] \, dt = \int 2.4 (50+t)^2 \, dt

\displaystyle (50+t)^2 y = \frac{2.4}3 (50+t)^3 + C

\displaystyle y = 0.8 (50+t) + \frac{C}{(50+t)^2}

Use the initial condition to solve for C :

y(0) = 0 \implies 0 = 0.8 (50+0) + \dfrac{C}{(50+0)^2} \implies C = -100,000

Then the amount of salt in the tank at time t is given by the function

y(t) = 0.8 (50+t) - \dfrac{10^5}{(50+t)^2}

so that after t = 50 min, the tank contains

y(50) = 0.8 (50+50) - \dfrac{10^5}{(50+50)^2} = \boxed{70}

kg of salt.

7 0
2 years ago
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