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3 years ago

What can be concluded about a cylinder and a cone with the same height and radius measures? Check all that apply.

2 answers:

5 0

The Formula's for a cylinder is V= $\pi$ r^2h and the volume of a Cone is V=1/3 $\pi$ r^2h. So you would know that the volume of a cone is 1/3 the volume of a cylinder.

zysi [14]3 years ago

3 0

Answer:

$\text{volume of cone=}\frac{1}{3}\text{volume of cylinder}$

Step-by-step explanation:

We have to choose the correct statement about the volume of cone and cylinder with same height and radius.

Let r and h be the radius and height of cylinder and cone

$\text{Volume of cone=}\frac{1}{3}\pi r^2h$ ....(1)

$\text{Volume of cylinder=}\pi r^2 h$ ...... (2)

$\frac{\text{Volume of cone}}{\text{Volume of cylinder}}=\frac{\frac{1}{3}\pi r^2h}{\pi r^2 h}$

$\frac{\text{Volume of cone}}{\text{Volume of cylinder}}=\frac{1}{3}$

$\text{volume of cone=}\frac{1}{3}\text{volume of cylinder}$

Option C is correct

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3 years ago

Read 2 more answers

A glass holds 189 milliliters of juice. How many liters will 36 glasses hold?

julsineya [31]

Answer:

A: 6,804

Step-by-step explanation:

189 x 36 = 6804

8 0

3 years ago

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

7 0

3 years ago

The distance around a rectangular cafe is 35m . The ratio of length of the cafe to the width is 3:2. Find the dimension of the

marusya05 [52]

Hi there! :)

Answer:

Length = 10.5 m, width = 7 m.

Step-by-step explanation:

Given:

Perimeter, or P = 35 m

Ratio of l to w = 3 : 2

Since the ratio is 3 : 2, let l = 3x, and w = 2x.

We know that the formula for the perimeter of a rectangle is P = 2l + 2w. Therefore:

35 = 2(3x) + 2(2x)

35 = 6x + 4x

35 = 10x

x = 3.5

Plug this value of "x" into each expression to solve for the dimensions:

2(3.5) = w

w = 7 m

3(3.5) = l

l = 10.5 m

Therefore, the dimensions are:

Length = 10.5 m, width = 7 m.

4 0

3 years ago

Consider that the length of rectangle A is 10 cm and its width is 6 cm. Which rectangle is similar to rectangle A? A) A rectangl

HACTEHA [7]

ANSWER

B) A rectangle with a length of 15 cm and a width of 9 cm.

EXPLANATION

The given rectangle, A, has length, 10cm and its width is 6cm.

The rectangle that is similar to rectangle A, has the corresponding sides in the same proportion.

For option A, the triangle has length 9cm and width 6cm.

The ratio of the corresponding sides are:

$\frac{10}{9} \ne \frac{6}{6}$

Since the ratios are not equal, the two triangles are not similar.

For the triangle in option B, the length is 15cm and the width is 9cm.

The ratio of the corresponding sides are:

$\frac{10}{15} = \frac{6}{9} = \frac{2}{3}$

Since the sides of the triangle are in the same proportion, the two triangles are similar.

For option C, the proportions are not the same.

$\frac{10}{14} \ne \frac{6}{7}$

The proportions are not the same for the triangle in option D.

$\frac{10}{12} \ne \frac{6}{8}$

8 0

3 years ago