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satela [25.4K]
4 years ago
8

Parabola passes through the points (0,5), (1,4), and (2,5). What function does the graph represent?

Mathematics
1 answer:
AfilCa [17]4 years ago
7 0

Answer:

x^2 - 2x + 5 = 0.

Step-by-step explanation:

The (0, 5) is the point where the parabola passes through the y axis (where x = 0), so we can write the equation as

y = ax^2 + bx + 5   where a and b are constants to be found.

Also, since (1, 4) and (2, 5) are points on the curve, substituting, we have the system:

a(1)^2 + 1b + 5 = 4

a(2)^2 + 2b + 5 = 5

Simplify these 2 equations:

a  + b  =  -1  .................(1)

4a + 2b = 0..................(2)

Multiply the first equation by -2:

-2a - 2b = 2 .................(3)

Add (2) + (3):

2a = 2

a = 1.

Substitute a = 1 into (2):-

4*1 + 2b = 0

2b = -4

b = -2.

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Nezavi [6.7K]

Answer:

d

Step-by-step explanation:

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7 0
3 years ago
(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.

The inflow rate is

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

and the outflow rate is

(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 :

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so that after t = 50 min, the tank contains

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marishachu [46]

Answer:

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Step-by-step explanation:

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Oduvanchick [21]

Answer:

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Step-by-step explanation:

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