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ella [17]
4 years ago
9

A bedroom wall is to be painted around a window as shown below. A rectangle with length 11 feet and width 10 feet. A smaller rec

tangle with length 3 feet and width 2 feet is cut out of the larger rectangle. What is the area of the wall that will be painted? 6 feet squared 104 feet squared 110 feet squared 116 feet squared
Mathematics
2 answers:
GREYUIT [131]4 years ago
5 0

Answer:

104 ft Squared

Step-by-step explanation:

You are taking away the room area. So 11x10=110 (Area of the whole wall) and you are cutting off 3x2=6 (Area of the wall you are cutting off). So you subtract 110-6= 104 ft2.

goldenfox [79]4 years ago
5 0

Answer:

104

Step-by-step explanation:

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Maria buys 15 apples at the store and places them into bags she puts 5 apples into each bag how many bags does Maria use for all
VMariaS [17]

it's 3 bags 5 times 3 is 15

6 0
3 years ago
Read 2 more answers
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
Compute: 5 ∙ (2 + 3i)
dimulka [17.4K]

Answer:

It will be equal to 40

Step-by-step explanation:

We have to compute 5\times (2+3!)

Let first find 3 !

For this we have to use factorial concept

So 3! will be equal to 3! = 3×2×1 = 6

Now According to 6+2 = 8

And now after solving bracket we have to multiply with 5

So 5×8 = 40

So after computation 5\times (2+3!)=40

So the final answer will be 40

4 0
4 years ago
Question down below. Thank you
Ann [662]

Answer:

N/A.

Step-by-step explanation:

I will not answer your question and I will be flagging it due to you adding a test/exam question.

8 0
3 years ago
The volume of Cylinder A is 189 ft and the
andrezito [222]

Answer:

27:8

Step-by-step explanation:

189/56 = 27/8

6 0
3 years ago
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