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2 years ago

Express 3.72727272727... as a rational number, in the form pq

1 answer:

6 0

Hi there!

Let x = 3. 72727.... ---(i)

Multiply 100 in eqn. (i)

100x = 372. 72727... ---(ii)

Subtract Eqn. (i) from (ii)

100x - x = 372. 7272... - 3. 7272...

99x = 369

x = $\dfrac {369}{99}$

x = $\dfrac {41}{11}$

Hence, $\dfrac {41}{11}$ is th' required p-q form.

~ Hope it helps!

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Conor has a roll of fence that is 25 feet long. He needs 8 yards of fence to go around a flower bed. Does he have enough fence t

vlabodo [156]

Answer:

Yes he has enough fence to go round the flower bed.

Step-by-step explanation:

1 yard = 3 feet

25 feet = 25/3 = 8.33 yards

Yes he has enough fence to go round the flower bed.

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2 years ago

$290, 12.5%, 6 months

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2 years ago

If the egress rate for a stadium is 5,500 fans per hour and it takes fans about one-third less time for fans to exit a game as i

Marina CMI [18]

Answer:

1833 fans per hour

Step-by-step explanation:

Ingress is entering and Egress is exiting.

Since exiting time is 1/3rd less than entering time, we have 5500 people exiting, so ingessing would be 1/3rd that.

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4 0

2 years ago

Read 2 more answers

Given: ΔABC, AC = BC, AB = 3 CD⊥ AB, CD = 3 Find: AC

viva [34]

Answer:

3 $\sqrt{5}$ /2

Step-by-step explanation:

As given, AC = BC, its an isosceles triangle.

Now, CD is perpendicular to AB, therefore AD = DB = 0.5AB = 3/2

Now take triangle ACD,

using Pythagoras theorem,

AC = $\sqrt{AD^{2} + CD^{2} }$

AC = 3 $\sqrt{5}$ /2

:-)

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2 years ago

Read 2 more answers

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.

6 0

2 years ago