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kolbaska11 [484]
3 years ago
14

What is 20/3 equal to?

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0
20/3 is equal to ------?


First we will divide numerator by the denominator.

20 ÷ 3 = <span>6.66666666667
</span>
Ok, Now we will see how many times 3 goes in to 20. 

3 × 6 = 18.
 
There will be 2 left. SO we will set our fraction like a mixed number:-



6 2/3


Hope I helped:P



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12,00 equals how many hundreds?
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Factor 3x^4+5x^3-11x^2+3x
Andreas93 [3]

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Step-by-step explanation:

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3 years ago
One leg of a right triangle is 12 inches, and the measure of the angle opposite that leg is $30^\circ$. what is the number of in
Alik [6]
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in a 30-60-90 right triangle, if the leg oposite the 30 degree angle is x then the hypontuse is 2x

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4 0
4 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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