Find the circumference and the area of a circle with diameter 2ft

2 answers:

8 0

C = Pi x diameter.
C = 3.14 x 2
C = 6. 28

A = Pi x radius^2
Radius = 1/2diameter
Radius = 1
Area = (3.14)(1^2)
Area = 3.14

Finding the area and circumference of a circle is pretty easy, actually. You just need to be comfortable with the formulas. Hope this helped!

rosijanka [135]3 years ago

3 0

Circumference: 6.28 feet ~rounded to the nearest hundredth

To find circumference, multiply the diameter by pi.

Area: 3.14 feet squared ~rounded to the nearest hundredth

To find area, divide the diameter by 2 and multiply by pi.

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Need help with this first one to answer gets branliest

mel-nik [20]

Answer:

-27

Step-by-step explanation:

Given: f(x)= -x^2+10x-16, g(x)= |11-2x^2|

f(9)= -(9)^2+10(9)-16

= -7

g(-5)= |11-2(-5)^2|

=39

Solve: 2[f(9)]-1/3[g(-5)]

= 2[(-7)] - 1/3[39]

= -27

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3 years ago

WILL GIVE BRAINLIEST!!

kvv77 [185]

Answer:

Step-by-step explanation:

i think so,it is 19

have a nice day

8 0

2 years ago

Read 2 more answers

Surface integrals using a parametric description. evaluate the surface integral \int \int_{s} f(x,y,z)dS using a parametric desc

DiKsa [7]

You can parameterize $S$ using spherical coordinates by

$\vec s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial\vec v}\times\dfrac{\partial\vec s}{\partial\vec u}=36\langle\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v\rangle$

(I use $\vec s_v\times\vec s_u$ to avoid negative signs. The orientation of the normal vector doesn't matter for a scalar surface integral; you could just as easily use $\vec s_u\times\vec s_v=-(\vec s_v\times\vec s_u)$ .)

Then

$f(x,y,z)=f(6\cos u\sin v,6\sin u\sin v,6\cos v)=36\sin^2v$

and the integral of $f$ over $S$ is

$\displaystyle\iint_Sf(x,y,z)\,\mathrm dS=\int_0^{\pi/2}\int_0^{2\pi}36\sin^2v\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$