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3 years ago

8

A circle with radius 9 has a sector with a central angle of 120°. What is the area of the sector. Provide step by step explanati

on please

1 answer:

inn [45]3 years ago

6 0

Okay, so you find the area of the full circle, 81pi and then 120 is 1/3 of the circle so you divide 81pi by 3 and get 27pi.

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While placing a compact disc into a CD player, you notice a small chip on it\'s edge. But you attempt to play the CD anyway by p

Vera_Pavlovna [14]

Answer:

(x, y) ≈ (-5.89, 1.16)

Step-by-step explanation:

We assume the CD is still accelerating at the same rate at the end of the given time period. The angular position will be found from ...

θ(t) = θ₀ + (1/2)a·t²

Filling in the numbers given, we get

θ(3.51) = (10π/180) + (1/2)(2.49)(3.51²) = 15.51306 . . . radians

The x-y coordinates will be found from ...

(x, y) = radius·(cos(θ), sin(θ)) ≈ 6·(-0.981066, 0.193674)

(x, y) ≈ (-5.89, 1.16)

_____

For the trig functions, the calculator must be in radians mode, unless you have converted the angles to degrees.

8 0

3 years ago

Write the statement for him a f(n).

Effectus [21]

Answer:

180f + 23(n)

Step-by-step explanation:

just make an equation

7 0

3 years ago

enot [183]

The triangles are shown to be similar, therefore

$\displaystyle \frac{54}{12} = \frac{x}{2} \\ \\
x = 2 \cdot \frac{54}{12} \\
x = 9$

The answer is 9 units

8 0

3 years ago

Read 2 more answers

find the sum of a 9-term geometric sequence when the first term is 4 and the last term is 1,024 and select the correct answer be

olasank [31]

Hello,

$u_{1} =4\\
 u_{2} =4*q\\
 u_{3} =4*q^2\\
...\\

 u_{9} =4*q^8\\\\
==\textgreater\ 4*q^8=1024\\
==\textgreater\ q^8=256\\
==\textgreater\ q=2\mbox{( and there is a problem in the question) } \ or \ q=-2\\
if\ q=2 \ then\ \sum_{i=0}^{8}\ 4*(2)^i= 4*\dfrac{1-2^9}{1-2} =2044\\

if \ q=-2 \ then\ \sum_{i=0}^{8}\ 4*(-2)^i= 4*\dfrac{1-(-2)^9}{1+2} =684\\



$

Answer B (but see the problem)

7 0

3 years ago

Read 2 more answers

D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx

sineoko [7]

Answer:

$\frac{1}{2\sqrt{5} }$

Step-by-step explanation:

Let, $\text{sin}^{-1}(\frac{x}{6})$ = y

sin(y) = $\frac{x}{6}$

$\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})$

$\frac{d}{dx}\text{sin(y)}=\frac{1}{6}$

$\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx}$ ---------(1)

$\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}$

cos(y) = $\sqrt{1-\text{sin}^{2}(y) }$

= $\sqrt{1-(\frac{x}{6})^2}$

= $\sqrt{1-(\frac{x^2}{36})}$

Therefore, from equation (1),

$\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

Or $\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

At x = 4,

$\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}$

$\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}$

$=\frac{1}{6\sqrt{\frac{36-16}{36}}}$

$=\frac{1}{6\sqrt{\frac{20}{36} }}$

$=\frac{1}{\sqrt{20}}$

$=\frac{1}{2\sqrt{5}}$

4 0

3 years ago