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3 years ago

Find the exact (leave pi) and the approximate ( pi = 3.14 ) circumference for each circle

Mathematics

1 answer:

sineoko [7]3 years ago

6 0

Answer/Step-by-step explanation:

a. Circumference (C) = πd

d = 12 cm

π = 3.14

Exact value of C = π*12 = 12π cm

Approximate value of C = 12*3.14 = 37.68 cm

b. Circumference (C) = πd

d = 8 m

π = 3.14

Exact value of C = π*8 = 8π m

Approximate value of C = 8*3.14 = 25.12 m

c. Circumference (C) = πd

d = 6 km

π = 3.14

Exact value of C = π*6 = 6π km

Approximate value of C = 6*3.14 = 18.84 km

d. Circumference = πd

d = 9 ft

π = 22/7

Exact value of C = π*9 = 9π ft

Approximate value of C = 9*22/7 = 28.2857143 ≈ 28.29 ft (nearest hundredth)

e. Circumference = 2πr

r = 4 mm

π = 22/7

Exact value of C = 2*π*4 = 8π mm

Approximate value of C = 2*22/7*4 = 25.1428571 ≈ 25.14 mm (nearest hundredth)

f. Circumference = 2πr

r = 7 mi

π = 22/7

Exact value of C = 2*π*7 = 14π mi

Approximate value of C = 2*22/7*7 = 44 mi (nearest hundredth)

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3 years ago

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tino4ka555 [31]

Answer:

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Step-by-step explanation:

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2 years ago

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

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3 years ago

What is 5 to the power of 1 in expanded form

fomenos

Answer:

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hope it helps.

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