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pychu [463]
3 years ago
9

(-3; -3) and (8-3). What is the slope​

Mathematics
1 answer:
pentagon [3]3 years ago
6 0

Answer:

Step-by-step explanation:

(-3,-3) ; (8,-3)

Slope=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{-3-[-3]}{8-[-3]}\\\\=\frac{-3+3}{8+3}\\\\=0

Slope = 0. so, the line is parallel to x-axis

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A cube is shown.<br><br> What is the surface area of the cube?
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1350 cm^2.  First, we must find the area of one side.  The formula for the area of a square is S^2 or S*S, so plugging in the value of the side we get 15*15=225.  So each side of the cube is 225 cm^2, so since there are 6 sides, we just multiply by 6 (225*6=1350).

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If (-5x^5) (a) = 25x^7 + 100x^5, find a
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3 years ago
A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per lit
Veronika [31]

Answer:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

Step-by-step explanation:

1) Identify the problem

This is a differential equation problem

On this case the amount of liquid in the tank at time t is 1600−24t. (When the process begin, t=0 ) The reason of this is because the liquid is entering at 16 litres per second and leaving at 40 litres per second.

2) Define notation

y = amount of chlorine in the tank at time t,

Based on this definition, the concentration of chlorine at time t is y/(1600−24t) g/ L.

Since liquid is leaving the tank at 40L/s, the rate at which chlorine is leaving at time t is 40y/(1600−24t) (g/s).

For this we can find the differential equation

dy/dt = - (40 y)/ (1600 -24 t)

The equation above is a separable Differential equation. For this case the initial condition is y(0)=(1600L )(0.0125 gr/L) = 20 gr

3) Solve the differential equation

We can rewrite the differential equation like this:

dy/40y = -  (dt)/ (1600-24t)

And integrating on both sides we have:

(1/40) ln |y| = (1/24) ln (|1600-24t|) + C

Multiplying both sides by 40

ln |y| = (40/24) ln (|1600 -24t|) + C

And simplifying

ln |y| = (5/3) ln (|1600 -24t|) + C

Then exponentiating both sides:

e^ [ln |y|]= e^ [(5/3) ln (|1600-24t|) + C]

with e^c = C , we have this:

y(t) = C (1600-24t)^ (5/3)

4) Use the initial condition to find C

Since y(0) = 20 gr

20 = C (1600 -24x0)^ (5/3)

Solving for C we got

C = 20 / [1600^(5/3)] =  20 [1600^(-5/3)]

Finally the amount of chlorine in the tank as a function of time, would be given by this formula:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

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Step-by-step explanation:

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