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3 years ago

(3x3 + 2x2 – 2x – 1) = (3x – x-1)

2 answers:

6 0

(3x3 + 2x2 - 2x - 1) = (3x - x - 1)

Add 1 to both sides

(3x3 + 2x2 - 2x) = (3x - x)

Divide both sides by 3

(x3 + 2x2 - 2x) = (x - x)

alisha [4.7K]3 years ago

3 0

Answer:

x=0 or x=0.8685170918213297 or x=−1.5351837584879966

Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Find the length of the curve x=e^t e^{-t},\;\;y=5-2t,\;\;0 \le t \le 3.

Oksana_A [137]

$\begin{cases}x(t)=e^t+e^{-t}\\y(t)=5-2t\end{cases}\implies\begin{cases}x'(t)=e^t-e^{-t}\\y'(t)=-2\end{cases}$

The length of the curve is given by the integral

$\displaystyle\int_0^3\sqrt{(e^t-e^{-t})^2+(-2)^2}\,\mathrm dt$

Expand and rewrite the integrand:

$(e^t-e^{-t})^2+(-2)^2=e^{2t}+2+e^{-2t}$
$=e^{-2t}(e^{4t}+2e^{2t}+1)$
$=e^{-2t}(e^{2t}+1)^2$
$\implies\sqrt{(e^t-e^{-t})^2+(-2)^2}=\dfrac{e^{2t}+1}{e^t}$

Now the integral is

$\displaystyle\int_0^3\dfrac{e^{2t}+1}{e^t}\,\mathrm dt=\int_0^3(e^t+e^{-t})\,\mathrm dt$
$=2\displaystyle\int_0^3\cosh t\,\mathrm dt$
$=2\sinh t\bigg|_{t=0}^{t=3}$
$=2\sinh 3$

7 0

3 years ago