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Dmitrij [34]
3 years ago
11

I dont understand how to do this please help me i want to get a good grade​

Mathematics
2 answers:
horsena [70]3 years ago
7 0

Answer:

y =20

Step-by-step explanation:

you have to get the y by itself so what you do on one side you do to the other. subtract 12 on both sides and you get y=20

dedylja [7]3 years ago
3 0
You subtract 12 from 32 and you get the answer from y. 20=y
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Division is basically
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\frac{18}{ \frac{2}{3} } times  \frac{ \frac{3}{2} }{ \frac{3}{2} } = \frac{ \frac{54}{2} }{ \frac{6}{6} }= \frac{27}{1}=27

the answe ris 27
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In a certain region of space, the potential is given by v=2x−5x2y 3yz2. How much is the magnitude of the electric field at point
Masja [62]

The magnitude of the electric field at the point (-2, 2, 0) is 46.52 V/m

<h3>How to calculate the electric field?</h3>

Since in a certain region of space, the potential is given by v = 2x − 5x²y 3yz²,

The electric field, E = -grad(V) where grad(V) = (dV/dx)i + (dV/dy)j + (dV/dz)k

Since V = 2x − 5x²y + 3yz²

dV/dx = d(2x − 5x²y + 3yz²)/dx

= d2x/dx - d5x²y/dx + d3yz²/dx

= 2 - 10xy + 0

= 2 - 10xy

dV/dy = d(2x − 5x²y + 3yz²)/dy

= d2x/dy - d5x²y/dy + d3yz²/dy

= 0 - 5x² + 3z²

=  - 5x² + 3z²

dV/dz = d(2x − 5x²y + 3yz²)/dz

= d2x/dz - d5x²y/dz + d3yz²/dz

= 0 - 0 + 6z

=  6z

<h3>The electric field</h3>

So, E = -grad(V)

= -[(dV/dx)i + (dV/dy)j + (dV/dz)k]

= -[(2 - 10xy)i + (- 5x² + 3z²)j + (6z)k]

So, the electric field at (-2, 2, 0) is

E = -[(2 - 10xy)i + (- 5x² + 3z²)j + (6z)k]

E = -[(2 - 10(-2)(2))i + (- 5(2)² + 3(0)²)j + (6(0))k]

E = -[(2 + 40)i + (- 5(4) + 0)j + (0)k]

E = -[42i + (-20 + 0)j + (0)k]

E = -[42i - 20j + 0k]

E = -42i + 20j - 0k

<h3>The magnitude of the electric field</h3>

So, the magnitude of E at (-2, 2, 0) is

|E| = √[(-42)² + 20² + 0²]

=  √[1764 + 400 + 0]

= √2164

= 46.52 V/m

So, the magnitude of the electric field at the point (-2, 2, 0) is 46.52 V/m

Learn more about electric field here:

brainly.com/question/25751825

#SPJ12

3 0
1 year ago
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