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just olya [345]

3 years ago

6

Write the radical expression in exponential form.

x%7D%20%7D%20" id="TexFormula1" title=" \frac{1}{ \sqrt[2]{x} } " alt=" \frac{1}{ \sqrt[2]{x} } " align="absmiddle" class="latex-formula">

1 answer:

Len [333]3 years ago

5 0

1/x = x^-1
sqrt(x) = x^(1\2)

Therefore, answer is:

x^(-1/2)

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A square room has a tiled floor with 81 square tiles. How many tiles are along an edge of the room?

Strike441 [17]

As there are 81 tiles total, that makes your area 81. Being a square, that makes your area equation A=s^2. As you need a side, this makes your equation s=√A. Plugging in 81 as A, we get s=√81=9, so an edge has 9 tiles.

4 0

3 years ago

adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

3 years ago

If the ratio of beakers to test tubes in a lab is 3:8, and there are 64 test tubes, how many beakers are there?

OLga [1]

3:8
Beakers: test tubes

3+8=15

64/15= 4

3:4. X4
12:16

12 beakers

Please mark brainiest :)

3 0

3 years ago

Read 2 more answers

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago

Find the equation of the line through point (-5,5) and perpendicular to y = 5/9x - 4

tankabanditka [31]

Answer:

$9x+5y=-20$

Step-by-step explanation:

$y = \frac{5}{9} x - 4$

compare with that y=mx+c

where m=slope .

So after comparing we get m= $\frac{5}{9}$

As the lines are perpendicular

$m1*m2=\ -1$

$m2 \ =\ \frac{-1}{m1}$

therefore m2= $\frac{-9}{5}$

Now from the equation

$y - b = m(x - a )$ this equation passing to the point (-5,5) with m= $\frac{-9}{5}$

$y-5=\frac{-9}{5} (x+5)\\5y-25=-9x\ -45\\9x+5y=-20$

Therefore equation is

$9x+5y=-20$

7 0

3 years ago