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3 years ago

(-11x^2 +6) - (14x^2 +2)

Mathematics

1 answer:

Nataly [62]3 years ago

8 0

Answer:

−1(5−2)(5+2)

Step-by-step explanation:

Eliminate redundant parentheses

(−11²+6)−1(14² +2)

Distribute

−11² + 6 -- 1(14² +2)

−11² + 6 - 14² − 2

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Bishwant and sunayana cimplete a oiece of a work in 6 days working together.if bishwant alime can complete it n 10 days.in how m

alina1380 [7]

Step-by-step explanation:

sunaina's alona work=

1/6-1/10

10-6/60

4/60

1/15

therefore sunaina can alone do the work in 15 days

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3 years ago

Find the inequality represented by the graph.

Mkey [24]

Answer:

y > 4x - 2

Step-by-step explanation:

y-intercept is at -2

slope is rise by 4, run by 1

if you don't know where to shade, select a point and plug it in to see if T or F

example (0, 0)

is 0 > 4(0) - 2

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3 years ago

Read 2 more answers

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.

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3 years ago

V (t)=32t <br> If a rock dropped off a bridge, how fast will it be falling after 3 seconds

kondor19780726 [428]

V(t) = 32t

when t = 3 seconds,

V = 32*3

V = 96

So velocity will be 96 units/s after 3 seconds.

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3 years ago

I didn’t learn to do this please help

Naddika [18.5K]

Find the area of the blue square including the white square then find the area of the white square and subtract that from the area of the blue square.

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3 years ago