<img src="https://tex.z-dn.net/?f=3%20%20%5Ctimes%20%20%7B9%7D%5Ex%20%3D%2027%20" id="TexFormula1" title="3 \times {9}^x = 27

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Verizon [17]

3 years ago

" alt="3 \times {9}^x = 27 " align="absmiddle" class="latex-formula">

Mathematics

2 answers:

Drupady [299]3 years ago

4 0

I think it’s 27x i don’t know for sure tho sorry if I’m wrong

LenKa [72]3 years ago

3 0

It’s 27 it’s in the photo

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Please answer

GuDViN [60]

The answer is going to be d

3 0

3 years ago

In a paint factory, an old conveyer line has filled 10 barrels of paint, and is filling more at a rate of 3 barrels per minute.

Andrews [41]

Answer:

40 barrels

Step-by-step explanation:

Given that:

Let x = number of minutes

Old conveyor line :

Filling rate = 3 barrels per minute

Already filled = 10 barrels

Equation :

10 + 3x - - - (1)

NEW conveyor line:

Filling rate = 4 barrels per minute

Equation :

4x - - - (2)

Equate equation (1) and (2)

10 + 3x = 4x

10 = 4x - 3x

10 = x

Hence, in 10 minutes, each line will have filmed same number of barrels which is

4x = 4*10 = 40 barrels

10 + 3x = 10 + 3(10) = 40 barrels

5 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

3 years ago

-6.6(3x+6)-1.7(2x-5)

melisa1 [442]

Answer:

Step-by-step explanation:

-6.6(3x+6)-1.7(2x-5)

open the bracket

-19.8x - 39.6 - 3.3x + 8.5

combine the like terms

-19.8x - 3.3x -39.8 + 8.5

-23.1x - 31.3

4 0

3 years ago

A RECTANGULAR FIELD IS 120 YARDS LONG AND 53 YARDS WIDE COVERS PART OF THE FIELD.HOW MANY SQUARE YARDS OF THE FIELD WILL NOT COV

lbvjy [14]

Answer:

6000 yard² of area is not covered by the tent.

Step-by-step explanation:

The question is "A rectangular field is 120 yards long and 53 yards wide. A tent that is 20 yards long and 18 yards wide covers part of the field. How many square yards of the field are not covered by the tent."

Given that,

Length of a rectangular field = 120 yards

Breadth of a rectangular field = 53 yards

Area of rectangular field = l×b

= 120 × 53

= 6360 yard²

Length of a tent = 20 yards

Breadth of the tent = 18 yards

Area of the tent = lb

= 20 × 18

= 360 yards²

Area not covered = 6360 yard² - 360 yards²

= 6000 yard²

So, 6000 yard² of area is not covered by the tent.

8 0

2 years ago