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Pavel [41]
3 years ago
9

Neeeeeed helppp!!

rt{10^1^0}" align="absmiddle" class="latex-formula">
Thanks!
Mathematics
2 answers:
Orlov [11]3 years ago
7 0

Answer:

100000 is your answer

Step-by-step explanation:

Hope it helps...

Thanks for the points

galben [10]3 years ago
3 0

Answer:

100000

Step-by-step explanation:

\sqrt{10^{10} }

10^5

100000

---------------

Thanks, have a great day!!

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Combine the like terms to create am equivalent expression y-(-3y)
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Answer:

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Step-by-step explanation:

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3 years ago
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How is 36 classified?
Lostsunrise [7]

Answer: as a number?

Step-by-step explanation:

Be a little more specific

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3 years ago
Britany has a bag with 2 mint sticks, 4 jelly treats, and 14 fruit tart chews. If she eats one piece every 9 minutes, what is th
zheka24 [161]
That "9 minutes" doesn't affect the outcome!

How many pieces of candy are in the bag at the beginning?  How many of those are "fruit tart chews?"  Write a fraction involving these 2 counts.  Remember that Britany immediately eats what she draws from the bag, so the 2nd time around, there are only 19 pieces, not 20.  What is the prob. that she will pick a jelly treat on her second draw?
Because these experiments are independent, you can find the joint probability by multiplying the 2 probabilities together.  Please show your work.
3 0
3 years ago
Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen
Jobisdone [24]

Parameterize S by the vector function

\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k

so that the normal vector to S is given by

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k

with magnitude

\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}

In this case, the normal vector is

\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k

with magnitude \sqrt{1^2+2^2+1^2}=\sqrt6. The integral of f(x,y,z)=e^z over S is then

\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx

where T is the region in the x,y plane over which S is defined. In this case, it's the triangle in the plane z=0 which we can capture with 0\le x\le8 and 0\le y\le\frac{8-x}2, so that we have

\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}

5 0
3 years ago
Please HELP ME WITH THIS TY!
LenKa [72]
X=117 degrees
Y = 63 degrees
Z = 117 degrees
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3 years ago
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