Anybody know the answer ?

A rope of length 18 feet is arranged in the shape of a sector of a circle with central angle O radians, as shown in the

creativ13 [48]

Answer:

$A(\theta)=\frac{162 \theta}{(\theta+2)^2}$

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Let

r ---> the radius of the sector

s ---> the arc length of sector

Find the radius r

we know that

$2r+s=18$

$s=r \theta$

$2r+r \theta=18$

solve for r

$r=\frac{18}{2+\theta}$

step 2

Find the value of s

$s=r \theta$

substitute the value of r

$s=\frac{18}{2+\theta}\theta$

step 3

we know that

The area of complete circle is equal to

$A=\pi r^{2}$

The complete circle subtends a central angle of 2π radians

so

using proportion find the area of the sector by a central angle of angle theta

Let

A ---> the area of sector with central angle theta

$\frac{\pi r^{2} }{2\pi}=\frac{A}{\theta} \\\\A=\frac{r^2\theta}{2}$

substitute the value of r

$A=\frac{(\frac{18}{2+\theta})^2\theta}{2}$

$A=\frac{162 \theta}{(\theta+2)^2}$

Convert to function notation

$A(\theta)=\frac{162 \theta}{(\theta+2)^2}$

6 0

3 years ago

The value of a gold coin picturing the head of the Roman Emperor Vespasian is increasing at the rate of 5% per year. If the coin

bazaltina [42]

Answer:

255.75

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

During an Apollo moon landing, reflecting panels were placed on the moon. This allowed Earth-based astronomers to shoot laser be

rusak2 [61]

Answer:

The distance between the astronomers and the moon was $7.56*10^{8]$ meters.

Step-by-step explanation:

We have that the speed is the distance divided by the time, so:

$v = \frac{d}{t}$

In this problem, we have that:

The reflected laser beam was observed by the astronomers 2.52 s after the laser pulse was sent. This means that $t = 2.52$ .

If the speed of light is 3.00 times 10^8 m/s, what was the distance between the astronomers and the moon?

We have that $v = 3*10^{8}$ m/s.

We have to find d. So:

$v = \frac{d}{t}$

$d = vt$

$d = 3*10^{8}*2.52$

$7.56*10^{8]$

The distance between the astronomers and the moon was $7.56*10^{8]$ meters.

7 0

3 years ago

At noon, ship A is 170 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 20 km/h. How fast is

garik1379 [7]

Check the picture below.

now, keep in mind that ship B is going at 20kph, thus from noon to 4pm, is 4 hours, so it has travelled by then 20 * 4 or 80 kilometers, thus b = 80.

whilst the ship B is moving north, the distance "a" is not really changing, and thus is a constant, that matters because the derivative of a constant is 0.

$\bf c^2=a^2+b^2\implies \stackrel{chain~rule}{2c\cfrac{dc}{dt}}=0+2b\cfrac{db}{dt}\implies \cfrac{dc}{dt}=\cfrac{b\frac{db}{dt}}{c}
\\\\\\
\begin{cases}
\frac{db}{dt}=20\\
c=10\sqrt{353}\\
b=80
\end{cases}\implies \cfrac{dc}{dt}=\cfrac{80\cdot 20}{10\sqrt{353}}\implies \cfrac{dc}{dt}=\cfrac{160}{\sqrt{353}}
\\\\\\
\textit{and rationalizing the denominator}\implies \cfrac{160\sqrt{353}}{353}$

8 0

2 years ago

Hello please help ty

Amiraneli [1.4K]

Answer:

i think a but not sure

Step-by-step explanation:

i always second guess myself but a was first answer for me

7 0

1 year ago

Read 2 more answers