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vampirchik [111]

3 years ago

8

Suppose a camping tent has a center pole that is 6 feet high. If the sides of the tent make 40° angles with the ground, how wide

, in feet, is the tent?

1 answer:

Montano1993 [528]3 years ago

8 0

Answer:

14.3 feet

Step-by-step explanation:

See attached photo for work...

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In 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B. If 70 houses wer

Stella [2.4K]

Answer:

The number of houses built in Town B is 56.

Step-by-step explanation:

We are given that in 2010, the number of houses built in Town A was 25 percent greater than the number of houses built in Town B.

Also, 70 houses were built in Town A during 210.

Let the number of houses built in Town B be 'x'.

So, according to the question;

Number of houses built in Town A = Number of houses built in Town B + 25% of the houses built in Town B

$70 = x + (25\% \times x)$

$70 = x(1+0.25)$

$x=\frac{70}{1.25}$

x = 56

Hence, the number of houses built in Town B is 56.

8 0

2 years ago

Factor mn - 4m - 5n + 20

anastassius [24]

Mn - 4m - 5n + 20

In this, mn and -4m have a common factor; m. So we can take that out to make it:

m(n-4) -5n + 20

Now, -5n + 20 also has a common factor which is 5, because 20/5 gives you 4, and 5/5 equals 1. So we can talk out that -5 (it doesn't need to be negative, but negatives on variable, no gusta). So you'd get

m(n-4)-5(n-4) OR m(n-4)+5(-n+4)

4 0

3 years ago

Read 2 more answers

I will Give brainliest to who ever can show me how to solve this killer!!!!!!! Using Descartes Rule and the rational zeros of po

Angelina_Jolie [31]

see if other people has already answered this question

7 0

3 years ago

Read 2 more answers

Find the laplace transform by intergration<br> 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

Pepper Jamie filled up her 10 Gallon gas tank for 54.50 how much did the gas cost per gallon

dangina [55]

Answer:

$5.45

Step-by-step explanation:

54.50÷10

4 0

3 years ago