ASAPPP NEED IT NOWWW pleaseee

A linear equation is an equation whose graph is a straight line

DaniilM [7]

Answer:

True

Step-by-step explanation:

Plotting the point of any linear equation on a graph gives a straight line

7 0

2 years ago

Please help me with these questions

Sindrei [870]

1) 8^-24 = 1 / 8^24
2) 6^-1 = 1/6
3) 5^-8 = 1 / 5^8
4) -3^1 = -3
5) 4^-4 = 1 / 4^4
6) 5^9
7) 8^-3 = 1 / 8^3
8) 9^4
9) 7^1 = 7
10) 2^-4 = 1 / 2^4

6 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

. An amusement park bought some arcade games that cost $2,000 each and photo booths that cost $3,200 each. A total of 22 of both

geniusboy [140]

Answer:

14 arcade games and 8 photobooths were purchased

Step-by-step explanation:

x will represent the number of arcade games

y will represent the number of photobooths

We know that 22 total machines were bought which means that x+y=22

We also know that the total was 53600 which means that 2000x+3200y=53600

We can use substitution to solve this

x+y=22

x-22= -y

-x+22=y

Then plug this in for y in the next equation

2000x+3200(-x+22)=53600

Solve this and get x=14

Then plug in x=14 into any equation (I will choose the x+y one because that's easiest to solve)

14+y=22

this means y=8

This means that 14 arcade games and 8 photobooths were purchased

I double checked the math on desmos and it is correct

6 0

3 years ago

Please help I need it fast thanks!

madam [21]

Answer:

B

Step-by-step explanation:

8 0

3 years ago