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2 years ago

HELP!!!! it’s due by 10 am and it’s 9:30am!!!!!!

Mathematics

2 answers:

Mashutka [201]2 years ago

6 0

Answer:

2010.62

Step-by-step explanation:

the formula to find a cylinders volume is r and h so you just divide 14 and get 8, then you get the 10 and you solve for how you would usually get your cylinder volume. which is V=πr2h=π·82·10≈2010.62

Yakvenalex [24]2 years ago

5 0

Answer:

V=1539.38

Step-by-step explanation:

V=πr2h=π·7^2·10≈1539.3804

or if you just want shortened version

V=π·7^2·10≈1539.3804

hope this helps!

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A highway engineer specifies that a certain section of roadway covering a horizontal distance of 2km should have a downgrade of

Zigmanuir [339]

Answer:

$0.16km$

Step-by-step explanation:

A highway engineer wants to compute the change in elevation of a section of road. The horizontal distance of this section of road is 2km and downgrade is 8%

The slope formula is given by

$m=\frac{rise}{run}=\frac{x}{y}$

m = 8% = 8/100

run = y = 2km

$m =\frac{x}{y}$

$\frac{8}{100} =\frac{x}{2}$

$\frac{2*8}{100} =x$

$x=\frac{16}{100} =0.16km$

Verification:

$m=\frac{0.16}{2}= 0.08=8%$ %

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3 years ago

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3 yards of cotton . 3/4 for each skirt. How many skirts

OLEGan [10]

Answer:

Step-by-step explanation:

3/3= 1 => 1 * 4 +4

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3 years ago

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Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is