What is the circumference of an circle of 3 cm

What would the answer to this question be? show work please

AURORKA [14]

Answer:

B

Step-by-step explanation:

The volume of a rectangular prism is the length times width times height.

V = LWH

V = (4√3)(3√6)W

V = 12√18 W

V = 36√2 W

If the volume is irrational, then W cannot have a radical that is half of a perfect square, because when multiplied by √2, that would yield a rational volume. For example, √18 × √2 = √36 = 6.

Therefore, the answer must be B, because 12 is not half of a perfect square.

V = 36√2 (4√12)

V = 144√24

V = 288√6

5 0

3 years ago

Plz help me plzzzzzzzzzzzzzzzzzz

12345 [234]

Answer:

3

Step-by-step explanation:

8 0

3 years ago

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What kind of property is this problem 3x(k+4)=3xk+12

ser-zykov [4K]

If i'm not wrong its distributing property of addition. <span />

4 0

3 years ago

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago

216q^3+15q^3 please help me solve it

katrin [286]

Answer:

231q^3

Step-by-step explanation:

Just add them together. They have the same expressions. Whenever you see a number with an ending similarly to another such as 3x and 2x, they can add together.

This is no different. If something is 6x^2 and another is 7x^2, you would add them to get 13x^2.

3 0

3 years ago

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