What number is one hundred thousand more than 75,000?

kap26 [50]

2. y=2x+3

When x =0 and y= something, the 'something' is your y-intercept which is b in the y=mx+b formula. Then we look at what is changed in the x and y value to determine the m part. the x increase by 1 and y increase by 2 so the m part is 2.

3. y=-7x
Same as the last one, there is no y-inter this time because the y=0 when x=0/ So the y deceased by 7 each time and x incease by 1 so it is -7 for the m part.

6 0

2 years ago

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Cos2a= 2cos^2a-1 for all values of a<br><br> true or false?

Law Incorporation [45]

True
<span>Cos(A+B)=CosACosB-SinASinB
therefore Cos(A+A)= CosACosA - SinASinA
= Cos^2A - Sin^2A</span>

5 0

3 years ago

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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

TRUE OR FALSE:

enot [183]

<span>It is possible to have an obtuse triangle that also contains a 35° angle.
Yes, this is true.
That's because an obtuse angle just means it's more than 90 but less than 180 degrees.

</span><span>It is possible to have a right triangle that also contains a 110° angle.
</span>No, this is false.
That's because a right triangle already has 90. That means that the two remaining angles must add up to 90 and not more or less.

<span>It is possible to have an acute triangle that also contains a 70° angle.
Yes, this is true.
An acute triangle means that all of the angles are more than 0 but less than 90.

Hope this helps :)</span>

7 0

3 years ago

1. Solve the equation.<br>4(x + 5) = 24<br><br>

bixtya [17]

Answer:

x=1

Step-by-step explanation:

4(x+5)= 24

4x+20=24

-20 -20

____________

4x = 4

___ ____

4 4

x= 1

7 0

3 years ago

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