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REY [17]
3 years ago
9

Consider the expression 2 3 (14x − 3) − 3x( 4 5 + 5). Which statements are true about simplifying this expression? Check all tha

t apply. Using the distributive property, the Two-thirds would be multiplied to 14x and to –3. Combining Four-fifths + 5 can be done before multiplying it by 3x. Using the distributive property, 3x times Four-fifths is subtracted, but the 3x times 5 is added. Combining 14x – 3 can be done before multiplying by Two-thirds. Subtracting 3x is the same as adding –3x.
Mathematics
2 answers:
pav-90 [236]3 years ago
6 0

Answer:

A B E

Step-by-step explanation:

Using the distributive property, the Two-thirds would be multiplied to 14x and to –3.

Combining Four-fifths + 5 can be done before multiplying it by 3x.

Subtracting 3x is the same as adding –3x.

IgorLugansk [536]3 years ago
4 0

Answer:

1. using the distributive property, the two-thirds would be multiplied to 14x and to –3. 2. combining four-fifths + 5 can be done before multiplying it by 3x. 3. subtracting 3x is the same as adding –3x. (A, B, & E)

Step-by-step explanation:

just finished this lesson lol

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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
Maria scored 72, 97, and 82 on her first three math tests. She wants to have a mean score of 82 for the quarter. How many points
Rom4ik [11]

Answer:

<em>She needs </em><em>77 </em><em>on her last test to earn an 82 for the quarter.</em>

Step-by-step explanation:

Maria scored 72, 97, and 82 on her first three math tests.

She wants to have a mean score of 82 for the quarter.

Let us assume that she must score x on her last test to earn an 82 for the quarter.

So the average score will be,

=\dfrac{72+97+82+x}{4}

But the average score is given as 82, so

\Rightarrow \dfrac{72+97+82+x}{4}=82

\Rightarrow 72+97+82+x=82\times 4

\Rightarrow 72+97+82+x=328

\Rightarrow x=328-72-97-82

\Rightarrow x=77

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