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____ [38]
3 years ago
8

Point G is on line segment{FH}

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
8 0

Answer:

FG = 7

Step-by-step explanation:

5x+2+3x-1=9

8x+1=9

8x = 8

x = 1

FG = 5(1)+2 = 7

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solong [7]
6x - 3.5 = -15.5
Add 3.5 to both sides
6x = -12
Divide by 6
x = -2
Carly has the correct response
8 0
3 years ago
Help!<br><br><br> −160=22 + 8(3p − 4)<br><br> Whats the value of p
larisa86 [58]
-6.25 is the value of p. First distribute the 8 to the parentheses and then solve the equation
4 0
3 years ago
Read 2 more answers
A recipe for one full cake calls for one fourth cup of sugarIf three eighths of a cup of sugar is used, how many cakes were made
marin [14]

Answer:

1 1/2 cakes

Step-by-step explanation:

3/8 ÷ 1/4

3/8 * 4/1

12/8

3/2

1 1/2

5 0
3 years ago
Please give me the correct answer ​
Yuri [45]

Answer: 36/9

Step-by-step explanation:

8 0
2 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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