Please help me I want to pull up my grade.

Bonnie withdrew $1,200 from her savings account. She spent 1/5 of the money on travel arrangements. She spent 2/3 of the remaini

svetoff [14.1K]

Withdrew 1200
spent 1/5 on travel : 1/5(1200) = 1200/5 = 240
leaving her with : 1200 - 240 = 960
spent 2/3 on hardware : 2/3(960) = 1920/3 = 640
leaving her with : 960 - 640 = 320
320 - lunch = 300.50
320 - 300.50 = lunch
19.50 = lunch <==

5 0

3 years ago

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Which expression is equal to<br> to (-10 – 2i) + (3 + i)?<br> 0 -7 ti<br> o – 13 –<br> 07-<br> -7-

solong [7]

Answer:

- 7 - i

Step-by-step explanation:

Given

(- 10 - 2i) + (3 + i) ← remove parenthesis

= - 10 - 2i + 3 + i ← collect like terms

= - 7 - i

3 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

2 years ago

Math homework please help

Viktor [21]

the first on is -42

the second one is -80

the last on is -88

3 0

2 years ago

Alejandro solved the equation 5(7x^2-2)-8=3. He wrote 5(7x^2-2)=11 as his first step. Which property justifies his first step?

Scorpion4ik [409]

The addition property of equality. It says that if you add the same number to each side of the equation, the two sides of the equation will be equal. In this case, the number 8 was added to each side.

Hope this helps :)

3 0

3 years ago

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