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3 years ago

What is the volume of the water contained in the cylinder?

Mathematics

1 answer:

Margarita [4]3 years ago

3 0

Answer:

Vf = 118.365 m3

Step-by-step explanation:

Vcil = Pi*[(20)^2]*(100=

Vcil = 125.600 m3

Vesf = (4/3)*Pi*(12)^3

Vesf = 7235 m3

Then

Vf = Vci - Vesf

Vf = 125.600 - 7235

Vf = 118365 m3

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Match each expression on the left with an equivalent expression on the right

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Answer: See explanation
Step-by-step explanation:
You didn't give the expressions but here are some expressions
a. ✓4x²y^4. 1. 2x✓y
b. ✓8x²y. 2. 2y✓2x
c. ✓4x²y. 3. 2xy²
d. ✓16xy². 4. 2x✓2y
e. ✓8xy². 5. 4y✓x
a. ✓4x²y^4 = ✓4 × ✓x² × ✓y^4
= 2 × x × y²
= 2xy²
Therefore, ✓4x²y^4 = 2xy²
b. ✓8x²y = ✓8 × ✓x² × ✓y
= ✓4 × ✓2 × ✓x² × ✓y
= 2 × ✓2 × x × ✓y
= 2x✓2y
Therefore, ✓8x²y = 2x✓2y
c. ✓4x²y = ✓4 × ✓x² × ✓y
= 2 × x × ✓y
= 2x✓y
Therefore, ✓4x²y = 2x✓y
d. ✓16xy² = ✓16 × ✓x × ✓y²
= 4 × ✓x × y
= 4y✓x
Therefore, ✓16xy² = 4y✓x
e. ✓8xy² = ✓8 × ✓x × ✓y²
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3 years ago

Read 2 more answers

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

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