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maxonik [38]
3 years ago
13

What is the value of 5x when x=3.2

Mathematics
2 answers:
Ivan3 years ago
8 0
The answer is 16! I just multiplied 3.2 by 5!
soldier1979 [14.2K]3 years ago
7 0

Answer: 16

Step-by-step explanation:

All you have to do is plug 3.2 into the expression. So 5 * 3.2  is 16.

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What is the best estimate for the answer to 321.786 + 240.273?
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4 years ago
Consider 7×10^3. Write a pattern to find the value of the expression
kkurt [141]

Answer:

7*10*10*10 = 7000

7*10 = 70

70*10 = 700

700*10 = 7000

Step-by-step explanation:

The given expression is:

7*10^3

Here 10^3 means that 10 will be multiplied 3 times:

7*10*10*10 = 7000

How did we get 7000?

7*10*10*10 = 7000

7*10 = 70

70*10 = 700

700*10 = 7000

We can also say that there are 7  1000s in 7000....

5 0
3 years ago
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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.

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3 years ago
Where is 2/4 on the number line with a number line of on 3 ticks?
KIM [24]
2 to the left is 3 your very welcome hope this helps
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