See image help me as fast as possible, please

Mathematics

2 answers:

Ira Lisetskai [31]3 years ago

5 0

Answer:

11.4

Step-by-step explanation:

Area=area of rectangle-area of circle

=6*4 -pi(2^2)

=11.4

area of circle =pi (r^2)

r-radius of circle

Arada [10]3 years ago

3 0

Answer: 11.4

Step-by-step explanation:

area of rect. = 24

area of circle (half circle+half circle = full circle) = pi*r^2 = pi*(2*2) = pi*4 = about 12.57

2 = radius since 4 = diameter

24-12.57 = 11.43

round = 11.4

You might be interested in

The lens equation is 1/f=1/p+1/q , where f is the focal length of the lens, p is the distance of the object from the lens, and q

vichka [17]

1/f=1/p + 1/q
least comom multiple=pfq
(pq)/(pfq)=(fq)/(pfq)+(pf)/(pfq)
Because all denominators are the same, we can eliminate the denominators.
pq=fq+pf
pq-fq=pf
q(p-f)=pf
q=pf / (p-f)

Answer: a. q=pf / (p-f)

5 0

3 years ago

Read 2 more answers

2) Use spherical coordinates to find the integral of f(x,y,z) = z over x^2 + y^2 + z^2 < 4 and x,y,z < 0

777dan777 [17]

1. Denote by $\mathcal D$ the region bounded by the cylinder $x^2+y^2=9$ and the planes $z=0$ and $z=6$ . Converting to cylindrical coordinates involves setting

$\begin{cases}x=r\cos u\\y=r\sin u\\z=v\end{cases}$

and $\mathcal D$ is obtained by varying the parameters over $0\le r\le3$ , $0\le u\le2\pi$ , and $0\le v\le6$ . The volume element is

$\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm du\,\mathrm dv$

so the integral becomes

$\displaystyle\iiint_{\mathcal D}xz\,\mathrm dV=\int_{v=0}^{v=6}\int_{u=0}^{u=2\pi}\int_{r=0}^{r=3}r^2v\cos u\,\mathrm dr\,\mathrm du\,\mathrm dv=0$

2. Now let $\mathcal D$ denote the region bounded by the sphere $x^2+y^2+z^2=4$ and the coordinate planes in the octant in which each coordinate of the point $(x,y,z)$ is negative.

Converting to spherical coordinates, we set

$\begin{cases}x=\rho\cos u\sin v\\y=\rho\sin u\sin v\\z=\rho\cos v\end{cases}$

where we obtain $\mathcal D$ by varying over $0\le\rho\le2$ , $\pi\le u\le\dfrac{3\pi}2$ , and $\dfrac\pi2\le v\le\pi$ . The volume element is now

$\mathrm dV=\rho^2\sin v\,\mathrm d\rho\,\mathrm du\,\mathrm dv$

so the integral becomes

$\displaystyle\iiint_{\mathcal D}z\,\mathrm dV=\int_{v=\pi/2}^{v=\pi}\int_{u=\pi}^{u=3\pi/2}\int_{\rho=0}^{\rho=2}\rho^3\cos v\sin v\,\mathrm d\rho\,\mathrm du\,\mathrm dv=-\pi$