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3 years ago

Solve the equation x^3-5x^2+2x+8=0 given that -1 is a zero of f(x)=x^3-5x^2+2x+8

Mathematics

1 answer:

liraira [26]3 years ago

3 0

-1^3-5(-1)^2+2(-1)+8
-1-5-2+8
-6-2+8
-8+8=0

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What is the range of possible sizes for side x? _ < x < _

allochka39001 [22]

Answer:

0.5 < x < 16.5

Step-by-step explanation:

The third side of the triangle must be longer than the difference of the other two sides:

x > (8.5 -8.0)

x > 0.5

And it must be shorter than their sum:

x < (8.5 +8.0)

x < 16.5

The third side must be in the range ...

0.5 < x < 16.5

_____

These limits are a direct consequence of the triangle inequality, which requires the sum of the two shortest sides exceed the length of the longest side.

3 0

3 years ago

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.