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katovenus [111]
3 years ago
6

Math question about percent error. Picture of the problem attached. PLease hurry.

Mathematics
2 answers:
kifflom [539]3 years ago
4 0
I think it is C. but Im not positive

KatRina [158]3 years ago
3 0
The answer is b and c your welcome for helping u don't need to thank me
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Terrence took a total of 16 pages of notes during 8 hours of class. In all, how many hours will
r-ruslan [8.4K]
16 pages/8 hrs

2 pages/ he

18/2=9

18 pages/9 hrs

9 hours is your answer.
3 0
3 years ago
Choose the correct simplification of the expression a2 ⋅ a3.
Andreyy89

Answer:

a^5

Step-by-step explanation:

( thats a to the 5th power. )

6 0
3 years ago
van someone please help me with this i dont need an explaination i just need a. b. c. or d. please and thank you
Talja [164]

Answer:

option C

f(x)=-\sqrt{x+3}+8

Step-by-step explanation:

we have

f(x)=-2\sqrt{x-3}+8

using a graphing tool

see the attached figure N1

The range is the interval--------> (-∞,8]

y\leq 8

case A) f(x)=\sqrt{x-3}-8

using a graphing tool

The range is the interval--------> [-8,∞)

y\geq -8

case B) f(x)=\sqrt{x-3}+8

using a graphing tool

The range is the interval--------> [8,∞)

y\geq 8

case C) f(x)=-\sqrt{x+3}+8

using a graphing tool

The range is the interval--------> (-∞,8]

y\leq 8

case D) f(x)=-\sqrt{x-3}-8

using a graphing tool

The range is the interval--------> (-∞,-8]

y\leq -8

5 0
3 years ago
What is the slope of the line that passes through the points (-5,9) and (5,11)?
Minchanka [31]
The slope is 1/5


I hope this helps!!!!
4 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
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