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3 years ago

Center at the origin

Mathematics

1 answer:

Lunna [17]3 years ago

3 0

The answer is:
-4x+6y=c

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A recipe for orange water says, “Mix 3 teaspoons yellow water with 1 teaspoon red water.” For this recipe, we might say: “The ra

Ivahew [28]

Answer:6:2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Bc-ad = 3c+r solve for c

Annette [7]

Answer:

-ad-r=C

-------- Sorry thats the best I could do for a fraction symbol

3-b

Step-by-step explanation:

In order to solve for C we need to get it C on one side of the equation all by itself. To start we can subtract R from both sides and subtract bc from both sides to get both C's on one side. That gets us to

-ad-r=3c-bc

Since the side with C has a common factor, C between 3c and bc we can pull that out, and we get

-ad-r=c(3-b)

and then we just divide 3-b on both sides to get C completely by itself

-ad-r=C

--------

3-b

I hope this helps and please don't hesitate to ask if there is anything still unclear!

4 0

3 years ago

Find the laplace transform by intergration f(t)=tcosh(3t)

Shkiper50 [21]

$\mathcal L_s\{t\cosh3t\}=\displaystyle\int_0^\infty t\cosh3t e^{-st}\,\mathrm dt$

Integrate by parts, setting

$u_1=t\implies\mathrm du_1=\mathrm dt$
$\mathrm dv_1=\cosh3t e^{-st}\,\mathrm dt\implies v_1=\displaystyle\int\cosh3t e^{-st}\,\mathrm dt$

To evaluate $v_1$ , integrate by parts again, this time setting

$u_2=\cosh3t\implies\mathrm du_2=3\sinh3t\,\mathrm dt$
$\mathrm dv_2=\displaystyle\int e^{-st}\,\mathrm dt\implies v_2=-\frac1se^{-st}$

$\implies\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}+\frac3s\int \sinh3te^{-st}$

Integrate by parts yet again, with

$u_3=\sinh3t\implies\mathrm du_3=3\cosh3t\,\mathrm dt$
$\mathrm dv_3=e^{-st}\,\mathrm dt\implies v_3=-\dfrac1se^{-st}$

$\implies\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}+\frac3s\left(-\frac1s\sinh3te^{-st}+\frac3s\int\cosh3te^{-st}\,\mathrm dt\right)$
$\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}-\frac3{s^2}\sinh3te^{-st}+\frac9{s^2}\int\cosh3te^{-st}\,\mathrm dt$
$\displaystyle\frac{s^2-9}{s^2}\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}-\frac3{s^2}\sinh3te^{-st}$
$\implies\displaystyle\underbrace{\int\cosh3te^{-st}\,\mathrm dt}_{v_1}=-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}$

So we have

$\displaystyle\int_0^\infty t\cosh3t e^{-st}\,\mathrm dt=u_1v_1\big|_{t=0}^{t\to\infty}-\int_0^\infty v_1\,\mathrm du_1$
$=\displaystyle-\frac{t(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\bigg|_{t=0}^{t\to\infty}-\int_0^\infty \left(-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\right)\,\mathrm dt$
$=\displaystyle\frac1{s^2-9}\int_0^\infty(s\cosh3t+3\sinh3t)e^{-st}\,\mathrm dt$

We already have the antiderivative for the first term:

$\displaystyle\frac s{s^2-9}\int_0^\infty \cosh3te^{-st}\,\mathrm dt=\frac s{s^2-9}\left(-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\right)\bigg|_{t=0}^{t\to\infty}$
$=\dfrac{s^2}{(s^2-9)^2}$

And we can easily find the remaining term's antiderivative by integrating by parts (for the last time!), or by simply exchanging $\cosh$ with $\sinh$ in the derivation of $v_1$ , so that we have

$\displaystyle\frac3{s^2-9}\int_0^\infty\sinh3te^{-st}\,\mathrm dt=\frac3{s^2-9}\left(-\frac{(s\sinh3t+3\cosh3t)e^{-st}}{s^2-9}\right)\bigg|_{t=0}^{t\to\infty}$
$=\dfrac9{(s^2-9)^2}$

(The exchanging is permissible because $(\sinh x)'=\cosh x$ and $(\cosh x)'=\sinh x$ ; there are no alternating signs to account for.)

And so we conclude that

$\mathcal L_s\{t\cosh3t\}=\dfrac{s^2+9}{(s^2-9)^2}$