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3 years ago

Divide x^4 + 2x^2 + 1 by x+1 .

Mathematics

1 answer:

Ivahew [28]3 years ago

6 0

Answer:

$=x^3-x^2+3x-3+\frac{4}{x+1}$

Step-by-step explanation:

$\frac{x^4+2x^2+1}{x+1}\\\mathrm{Divide}\:\frac{x^4+2x^2+1}{x+1}:\quad \frac{x^4+2x^2+1}{x+1}=x^3+\frac{-x^3+2x^2+1}{x+1}\\=x^3+\frac{-x^3+2x^2+1}{x+1}\\\mathrm{Divide}\:\frac{-x^3+2x^2+1}{x+1}:\quad \frac{-x^3+2x^2+1}{x+1}=-x^2+\frac{3x^2+1}{x+1}\\=x^3-x^2+\frac{3x^2+1}{x+1}\\\mathrm{Divide}\:\frac{3x^2+1}{x+1}:\quad \frac{3x^2+1}{x+1}=3x+\frac{-3x+1}{x+1}\\=x^3-x^2+3x+\frac{-3x+1}{x+1}\\\mathrm{Divide}\:\frac{-3x+1}{x+1}:\quad \frac{-3x+1}{x+1}=-3+\frac{4}{x+1}\\=x^3-x^2+3x-3+\frac{4}{x+1}$

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2 years ago

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Tju [1.3M]

Answer:-

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Step-by-step explanation:

It's quadrilateral so it's total sum degree is 360

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3 years ago

Which expressions are equivalent to the one below? check all that apply. log(10^3)

Mademuasel [1]

Answer:

They are a and b.

Step-by-step explanation:

We have got logarithms to the base 10 here.

We know that log 10^3 = 3 log 10 ( by the law of logarithms).

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4 years ago

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Solve the given DE dx/dt=3x+2y+1 dy/dt=-2x-y+1

kirill115 [55]

We can solve for $x(t)$ first by rewriting the system of first-order ODEs as a single second-order ODE in $x(t)$ :

Taking the derivative of the first ODE gives

$\dfrac{\mathrm dx}{\mathrm dt}=3x+2y+1\implies\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}+2\dfrac{\mathrm dy}{\mathrm dt}$

while solving for $2y$ gives

$\dfrac{\mathrm dx}{\mathrm dt}=3x+2y+1\implies2y=\dfrac{\mathrm dx}{\mathrm dt}-3x-1$

Then

$\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}+2(-2x-y+1)$

$\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}-4x-2y+2$

$\dfrac{\mathrm d^2x}{\mathrm dt^2}=3\dfrac{\mathrm dx}{\mathrm dt}-4x-\left(\dfrac{\mathrm dx}{\mathrm dt}-3x-1\right)+2$

$\implies\dfrac{\mathrm d^2x}{\mathrm dt^2}-2\dfrac{\mathrm dx}{\mathrm dt}+x=3$

which is linear with constant coefficients, so it's trivial to solve; the corresponding homogeneous ODE

$x''-2x'+x=0$

has characteristic equation

$r^2-2r+1=(r-1)^2=0$

with root $r=1$ (multiplicity 2), so the characteristic solution is

$x_c=C_1e^t+C_2te^t$

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$x_p=a\implies{x_p}'={x_p}''=0$

Substituting these into the ODE gives

$0-2\cdot0+a=3\implies a=3$

Then the general solution for $x(t)$ is

$\boxed{x(t)=C_1e^t+C_2te^t+3}$

From here, we find

$\dfrac{\mathrm dx}{\mathrm dt}=C_1e^t+C_2(t+1)e^t$

so that

$2y=(C_1e^t+C_2(t+1)e^t)-3(C_1e^t+C_2te^t+3)-1$

$\implies\boxed{y(t)=\left(\dfrac{C_2}2-C_1\right)e^t-C_2te^t-5}$

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4 years ago

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Sladkaya [172]

Answer:

Option c. $2,250

Step-by-step explanation:

we know that

The minimum household income needed for the given budget, must be equal at least to the sum of all categories values of the monthly budget

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therefore

The minimum household income needed for this budget is $2,250

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3 years ago

Divide x^4 + 2x^2 + 1 by x+1 .​

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