Ask question

Ask question

All categories

Artyom0805 [142]

2 years ago

7

In circle O, central angle AOB measures StartFraction pi Over 3 EndFraction radians. Circle O is shown. Line segments A O and B

O are radii with length 18 centimeters. What is the length of arc AB? 6π cm 12π cm 18π cm 36π cm

2 answers:

Oksanka [162]2 years ago

7 0

Answer:

1) 6π cm

Step-by-step explanation:

on edg 2020

blondinia [14]2 years ago

6 0

Answer:

6(pie)

Step-by-step explanation:

trust me ; )

You might be interested in

The area of the figure

STALIN [3.7K]

Rectangle:

4*8=32

Semicircle:

3.14*4^2= 50.24
50.24/2=25.12

Add:

25.12+32= 57.12

Final answer: A

5 0

2 years ago

Read 2 more answers

Select the coordinates of two points on the line y= 8

notsponge [240]

The answer for this is (0,8) and (8,0)

7 0

2 years ago

Read 2 more answers

Agnes wants to buy a gallon of iced tea. If a 1-gallon jug costs $3.05 and a

AleksandrR [38]

Answer:

c

Step-by-step explanation:

8 pints in a gal

.45 x 8 = 3.60

3.60 - 3.05 = . 55

6 0

2 years ago

Landry earned $2,300 over the summer and decided to invest it in an account that earns 4% annual compound interest. If Landry do

solniwko [45]

Answer:

Landry will have $5039.58

Step-by-step explanation:

compound interest formula: amount = p(1 + \frac{r}{n})^{nt}

p= principal ($2,300)

r= interest rate as a decimal (4% = 0.04)

n= number of times the principal is compounded per year (annually = once per year so 1 time per year)

t= time in years (20 years)

new equation: amount = 2300(1+\frac{0.04}{1} )^{1*20}

That equation equals $2,739.58 which you add to the principal.

$2,739.58 + $2,300 = $5039.58

hope this helps :) the equations aren't showing up right :(

6 0

2 years ago

Read 2 more answers

Uranus has a mass of 8.68 1025 kg and a radius of 2.56 107 m. Assume it is a uniform solid sphere. The distance of Uranus from t

Lelechka [254]

Answer

Given,

Mass of the Uranus, M = 8.68 x 10²⁵ Kg

Radius of Uranus, R = 2.56 x 10⁷ m

Distance of Uranus, D = 2.87 x 10¹² days

a) Rotational Kinetic energy of the Uranus

moment of inertia of the Uranus

$I = \dfrac{2}{5}MR^2$

$I = \dfrac{2}{5}\times 8.68\times 10^{25}\times (2.56\times 10^7)^2$

I = 22.75 x 10³⁹ kg.m²

Angular speed

$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{17.3\times 3600}$ \

$\omega = 1 \times 10^{-4}$

Rotational Kinetic energy

$KE = \dfrac{1}{2}I\omega^2$

$KE = \dfrac{1}{2}\times 22.75\times 10^{39}\times (10^{-4})^2$

$KE = 11.38\times 10^{31}\ J$

b) Rotational Kinetic energy of Uranus in its orbit around sun

moment of inertia of the Uranus

$I = \dfrac{2}{5}MR^2+ Ma^2$

$I = 22.75\times 10^{39}+ 8.68\times 10^{25}\times (2.87\times 10^{12})^2$

I = 7.15 x 10⁵⁰ kg.m²

Angular speed

$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{3.08\times 10^4\times 3600\times 24}$ \

$\omega =2.36\times 10^{-9}$

Rotational Kinetic energy

$KE = \dfrac{1}{2}I\omega^2$

$KE = \dfrac{1}{2}\times 7.15\times 10^{50}\times (2.36\times 10^{-9})^2$

$KE = 1.99\times 10^{33}\ J$

6 0

2 years ago