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Vlad [161]
3 years ago
11

Robert and Katie Masterson bought 99 shares of oil stock at a price

Mathematics
1 answer:
anastassius [24]3 years ago
8 0
Cost of oil stock
99×55=5,445+(25) commission=5470
Cost of steel stock
150×24=3600+150×0.20=3630

total combined amount
3,630+5,470=9,100
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Are these ratios equivalent?
masya89 [10]

Answer:

A, yes.

Step-by-step explanation:

63:60, 21:20:

The difference between 60 and 20 is *3. If you do that to 21, you get 63. So a , yes.

3 0
3 years ago
I need help please help. 2 Questions.
Colt1911 [192]

Answer:

Step-by-step explanation:

8 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
Find the slope of the line passing through the points (-6,2), (0,-6).
Serhud [2]

C. -4/3

slope =  \frac{(y _{2} - y _{1}) }{(x _{2} - x _{1} )}  \\  =  \frac{( - 6 - 2)}{(0 -( -  6)}  \\  =  \frac{ - 8}{ 6}  \\  =   - \frac{4}{3}

3 0
3 years ago
In circle K shown below, points B, C, D, and E lie on the circle with secants HBD and HCE drawn. Prove:
Alexxx [7]

Answer:

By exterior angle theorem, we have;

∠DBE = ∠H + ∠HEB = ∠ECD = ∠H + ∠HDC

∴ ∠H + ∠HEB = ∠H + ∠HDC

By addition property of equality, we have

∠HEB = ∠HDC

∠H = ∠H by reflexive property

∴ ΔHCD ~ ΔHEB by Angle Angle AA similarity postulate

∴ HE/HD = EB/DC, by the definition of similarity

Therefore, by cross multiplication, we have;

HE × DC = EB × HD

Therefore, by commutative property of multiplication, we have;

HE × DC = HD × EB

Step-by-step explanation:

3 0
3 years ago
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