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Effectus [21]
3 years ago
6

WILL MARK BRAINLIEST!!! The men's costumes for an upcoming play cost $25.67 each, while the women's costumes cost $41.39 each. I

f there were four men and six women who needed costumes and the overall costume budget was $400.00, was there any money left over? If so, how much was left?
Mathematics
1 answer:
NeX [460]3 years ago
4 0

Answer:

There is money left over. They have 48.98 dollars left.

Step-by-step explanation:

Men costumes in total cost $102.68 and women costumes in total cost $248.34. There budget it $400 so we add both the men and women costumes together which is $351.02 which means that they do have left over money. To find the left over money, we have to subtract $400 from $351.02 which is equal to $48.98.

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Can someome please help me solve this my teacher hasn't done a good with teaching us this.
scoundrel [369]

Answer:

  30500 = 3.05·10^4

Step-by-step explanation:

Your calculator can do this for you. You may need to set the display to scientific notation, if that's the form of the answer you want.

__

This can be computed by converting both numbers to standard form:

  (5·10^2) +(3·10^4)

  = 500 +30000 = 30500 = 3.05·10^4

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Addition of numbers in scientific notation in general requires that they have the same power of 10. It may be convenient to convert both numbers to the highest power of 10.

  5·10^2 + 3·10^4

  = 0.05·10^4 +3·10^4 . . . . now both have multipliers of 10^4

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2 years ago
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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.

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