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Kipish [7]
3 years ago
9

Which graph best represents the equation -5x+2y=2

Mathematics
1 answer:
masha68 [24]3 years ago
8 0
This is what the graph would look like. Hope this helps:)

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About how tall is a two-story building? 6 inches 20 feet 25 inches 6 feet
lara [203]

The most likely measurement for a two-story building would be 20 feet. 6 inches, 25 inches, and 6 feet all are way too short for a two-story building.

8 0
3 years ago
Which of the following ratios is not equivalent to the following statement: The ratio of right-handed students to left-handed st
sukhopar [10]

In (A), we have 9 : 2.

Multiply by 2, we get 18 : 4

In (B), we have 36 : 8.

Divide by 2, we get, 18 : 4.

In (D), we have 54 : 12.

Divide by 3, we get, 18 : 4.

Hence, (C) is not equivalent to the given statement.

6 0
3 years ago
A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and
natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: E=k\frac{q}{r^{2} }; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: p=\frac{q}{A}; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: S=4\pi r^{2}; which gives us the surface of a sphere.

The inner surface: Si=4\pi (0.06m)^{2} = 0.04 m^{2}

The outer surface: S=4\pi (0.08m)^{2}=0.08m^{2}

Now we can calculate the charges,

Inner charge: Qi=pA=(-100\frac{nC}{m^{2} } )(0.04m^{2} )=-4nC

Outer charge: Qo=pA=100\frac{nC}{m^{2} } )(0.08m^{2} )=8nC

Then, we are able to calculate both fields:

Inner field: Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}

Outer field:  Eo=k\frac{Qo}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{8x10^{-9} }{0.08m^{2} }=11250\frac{N}{C}

The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

3 0
3 years ago
Explain how you would find the height of a rectangular prism if you know that the volume is 60 cubic cm and that the area of the
Oksi-84 [34.3K]
Base = length * width.

We know that the base has an area of 10
And 60 is the volume. Then 60 / 10 = 6 Which is the height.
7 0
3 years ago
11.
yKpoI14uk [10]

\\ \tt\leadsto tan\theta=\dfrac{Perpendicular}{Base}

\\ \tt\leadsto tan29=\dfrac{h}{400}

\\ \tt\leadsto h=400tan29

\\ \tt\leadsto h=221.7ft

6 0
1 year ago
Read 2 more answers
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