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kifflom [539]
4 years ago
12

Graph y= - x - 6 on a graph

Mathematics
2 answers:
iogann1982 [59]4 years ago
5 0

Answer:

(Picture below)

Step-by-step explanation:

To solve this you will first need to make a coordinate plane that is 10 by 10.

After you have done that you will need to know what slope intercept form is. Slope intercept form is y=mx+b and the equation that was given was y=-x-6. In the formula y=mx+b, m is the slope or rise over run and b is where the line crosses the y axis.

Now that you know the basics of slope intercept form we can start graphing. so b in this equation is -6, so we will put the first point on -6. after we have done that we can start putting the points of the line. So since the slope is just -1x  we will have to go down one and go to the right one about 3 times and just to finish it up go up and left on 3 times to. And there you have it.

Ede4ka [16]4 years ago
4 0
Yeah yeah singing challenge if I win I get to shave your head
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8.58 divided by 6 estimate the quotient
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3 years ago
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Given the function, f(x) = {x-1-6. choose the correct transformation.
Artemon [7]

Answer:

16

Step-by-step explanation:

4 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
What is the domain of g?
katrin2010 [14]
What graph are you exactly referring to??
There are no pictures….. so I’m not sure
7 0
3 years ago
Ashely found 2 boxes of sugar in the kitchen. The green boxe is 1.26 kg and the red box is 1.026 kg. Which box contains more sug
USPshnik [31]

Answer:

The green box by 0.234kg of sugar

Step-by-step explanation:

Given data

The green box weight= 1.26 kg

The red box weight=  1.026 kg

comparing the two values

the green box contains 0.234kg more sugar

=1.26-1.026

=0.234kg

6 0
3 years ago
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