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2 years ago

Find the length of side X

Mathematics

1 answer:

mel-nik [20]2 years ago

5 0

Solve using the Pythagorean theorem.

The hypotenuse is c.

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What is 4+5????????????????

eimsori [14]

Answer:

Step-by-step explanation:

4+5=9

6 0

2 years ago

Read 2 more answers

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

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

8 0

2 years ago

Mr.Afzal bought an article for Rs4400 for 3years at 8 percent what was the interest charge?

zimovet [89]

$\large\underline{\underline{\red{\rm\blue{\longmapsto} Step-by-step\: Explanation:-}}}$

Here , the given data is :

Principal = Rs 4400 . ( P )
Time = 3 years . ( T )
Rate of Internet = 8% . ( R )

We can calculate Simple Interest using ,

$\boxed{\purple{\bf Simple\: Interest=\dfrac{P\times\:R\:\times\:T}{100}}}$

On substituting the respective values ,

⇒ SI = P × R × T / 100.

⇒ SI = Rs 4400 × 8 × 3 / 100 .

⇒ SI = Rs 44 × 8 × 3 .

⇒ SI = Rs 1,056 .

Hence the required Interest is Rs 1,056.

5 0

2 years ago

PLEASE HELP

zzz [600]

$(x,\ y)\\\text{Put the values of x and y to the inequality}\ 3y-4x\geq-6\\\\A.\ (-3,\ 8)\\3(8)-4(-3)\geq-6\\24+12\geq-6\\36\geq-6\qquad TRUE\\\\B.\ (5,\ 10)\\3(10)-4(5)\geq-6\\30-20\geq-6\\10\geq-6\qquad TRUE\\\\C.\ (1,\ 2)\\(3)(2)-(4)(1)\geq-6\\6-4\geq-6\\2\geq-6\qquad\ TRUE\\\\D.\ (-5,\ 2)\\(3)(2)-(4)(-5)\geq-6\\6+20\geq-6\\26\geq-6\qquad TRUE\\\\Answer:\ a,\ b,\ c\ and\ d\ are\ solutions\ of\ this\ inequality.$

5 0

2 years ago

2 2/3 as a improper fraction

wel

Answer:

8/3

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers