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allochka39001 [22]
3 years ago
6

Consider the inequality -5(x+7) <-10. write an inequalityrepresenting the soulution for x.

Mathematics
1 answer:
Gwar [14]3 years ago
6 0

Answer:

x > -5

Step-by-step explanation:

You multiply both sides by -1

Simplify

Divide both sides by 5

Simplify

Subtract 7 from both sides

Simplify

Please give me BRAINLIEST and THANKS lol <( ̄︶ ̄)>

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zalisa [80]

Answer:

A will be the answer hope this will help u

3 0
3 years ago
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What is an angle that measures 1/2 turn
Ad libitum [116K]
An angle that measures 1/2 turn is 180 degree angle 
4 0
3 years ago
What is the surface area of a sphere with radius 2
Gelneren [198K]

Answer:

A

Step-by-step explanation:

The surface area (A) of a sphere is calculated as

A = 4πr² ← r is the radius

here r = 2, hence

A = 4π × 2² = 4π × 4 = 16π units² → A

3 0
3 years ago
Having serious block because of the units The density of mercury is 13.6 g/cm3. What approximate mass of mercury is required to
mart [117]
I would approach the problem like this:

1 oz bottle.............................30cm^3
33 oz bottle.............................xcm^3

in a 33 oz bottle will be x=33*30=990cm^3
density=mm/v
13.6g/cm^3= m/ 990cm^3
m= 13.6 * 990=13464 g
or 13.464 kg in a 33 oz bottle
3 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
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