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3 years ago

A circular picture is 8 inches in diameter.

Mathematics

1 answer:

tino4ka555 [31]3 years ago

4 0

For number one I would say c or b for part b I would say it would be b or d but I could be wrong I am sorry if I am

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Solve<br> -20y + 15 = 2 - 19y + 11

andrey2020 [161]

Answer:

the answer to this equation is 2

4 0

3 years ago

Read 2 more answers

For a class trip, the teachers would like to have one adult for every ten students. There are 190 students on the trip. How many

coldgirl [10]

You can solve this problem with the following

$\frac{1}{10} = \frac{a}{190}$
$10a = 190$
$a = 19$

There should be 19 adults on the trip.

6 0

3 years ago

Josiah bought 22 chicken wings for $26.40. How much would it cost for 9 wings?

cricket20 [7]

Answer:

it would cost $10.8 for 9 chicken wings.

Step-by-step explanation:

Cost of 22 chicken wings = $26.40

we have to find first, cost of 1 chicken wings

For that we divide both side by 22

cost of 22/22 chicken wings = $26.40/22 = 1.2

Thus, cost of 1 chicken wings = $1.2

to fins cost of 9 chicken wings

we multiply both sides by 9

cost of 1*9 chicken wings = $1.2*9

cost of 9 chicken wings = $10.8

Thus, it would cost $10.8 for 9 chicken wings.

7 0

3 years ago

Find the moment of inertia about the y-axis of the thin semicirular region of constant density

arlik [135]

The moment of inertia about the y-axis of the thin semicircular region of constant density is given below.

$\rm I_y = \dfrac{1}{8} \times \pi r^4$

<h3>What is rotational inertia?</h3>

Any item that can be turned has rotational inertia as a quality. It's a scalar value that indicates how complex it is to adjust an object's rotational velocity around a certain axis.

Then the moment of inertia about the y-axis of the thin semicircular region of constant density will be

$\rm I_x = \int y^2 dA\\\\I_y = \int x^2 dA$

x = r cos θ

y = r sin θ

dA = r dr dθ

Then the moment of inertia about the x-axis will be

$\rm I_x = \int _0^r \int _0^{\pi} (r\sin \theta )^2 \ r \ dr \ d\theta\\\\\rm I_x = \int _0^r \int _0^{\pi} r^3 \sin ^2\theta \ dr \ d\theta$

On integration, we have

$\rm I_x = \dfrac{1}{8} \times \pi r^4$

Then the moment of inertia about the y-axis will be

$\rm I_y = \int _0^r \int _0^{\pi} (r\cos\theta )^2 \ r \ dr \ d\theta\\\\\rm I_y = \int _0^r \int _0^{\pi} r^3 \cos ^2\theta \ dr \ d\theta$

On integration, we have

$\rm I_y = \dfrac{1}{8} \times \pi r^4$

Then the moment of inertia about O will be

$\rm I_o = I_x + I_y\\\\I_o = \dfrac{1}{8} \times \pi r^4 + \dfrac{1}{8} \times \pi r^4\\\\I_o = \dfrac{1}{4} \times \pi r^4$

More about the rotational inertia link is given below.

brainly.com/question/22513079

#SPJ4

3 0

2 years ago

At the beg ending of 2009 Marilyn invested $5000 in a savings account account pays 4% interest per year at the end of the year a

pychu [463]

Answer:Use the given functions to set up and simplify

2009

5000

0.04

Step-by-step explanation:

I Think

6 0

3 years ago