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KATRIN_1 [288]
3 years ago
15

Please answer ASAP!!!.........................................................

Mathematics
2 answers:
Olin [163]3 years ago
6 0
L=11.5m
I really hope this will help u
Mnenie [13.5K]3 years ago
3 0
11.5 is your answer


hope this helps 
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H ( x ) { − 2 + 3 , i f x < 1 1 /2 x + 1 if x ≥ 1 }
hoa [83]

Answer:????

Step-by-step explanation:

What

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3 years ago
Anne owns a used car lot and in order to get cars to sell, she buys used cars at an
azamat

Answer:

It would be 2,000x+200

Step-by-step explanation:

okay it costs $200 to attend so that's a cost you are going to have to pay regardless. The $2,000 depends on how many cars you buy. So it would be 2,000 (times how ever many cars you buy) plus the $200.

3 0
3 years ago
Please help I think it's A but im unsure
Likurg_2 [28]

Answer:

b is the correct answer I am also not sure

5 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
7kg :420g simplest form​
DENIUS [597]

Answer:

50:3

Step-by-step explanation:

7000g:420  both are divisible by 70

1000:60 both are divisible by 10

100:6  both are divisible by 2

50:3

please mark me as brainliest!!<3

8 0
2 years ago
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