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VARVARA [1.3K]
3 years ago
5

I need help. What is -8+y=-3y-2

Mathematics
2 answers:
Bond [772]3 years ago
8 0
-8+y=-3y-2
-8+2=3y-y
-6=2y
y=-3
PtichkaEL [24]3 years ago
4 0
Your answer is either 1.5 or 1/2
-8+y=-3y-2        y=-3y+6    4y=6            y=1.5  or 1/2
+8          +8   +3y    +3y    ---- ----
                                           4    4


hope this helps
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Let and be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.
elena-14-01-66 [18.8K]

The given identities are verified by using operations of the del operator such as divergence and curl of the given vectors.

<h3>What are the divergence and curl of a vector field?</h3>

The del operator is used for finding the divergence and the curl of a vector field.

The del operator is given by

\nabla=\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}

Consider a vector field F=x\^i+y\^j+z\^k

Then the divergence of the vector F is,

div F = \nabla.F = (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(x\^i+y\^j+z\^k)

and the curl of the vector F is,

curl F = \nabla\times F = \^i(\frac{\partial Fz}{\partial y}- \frac{\partial Fy}{\partial z})+\^j(\frac{\partial Fx}{\partial z}-\frac{\partial Fz}{\partial x})+\^k(\frac{\partial Fy}{\partial x}-\frac{\partial Fx}{\partial y})

<h3>Calculation:</h3>

The given vector fields are:

F1 = M\^i + N\^j + P\^k and F2 = Q\^i + R\^j + S\^k

1) Verifying the identity: \nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2

Consider L.H.S

⇒ \nabla.(aF1+bF2)

⇒ \nabla.(a(M\^i + N\^j + P\^k) + b(Q\^i + R\^j + S\^k))

⇒ \nabla.((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

Applying the dot product between these two vectors,

⇒ \frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z} ...(1)

Consider R.H.S

⇒ a\nabla.F1+b\nabla.F2

So,

\nabla.F1=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(M\^i + N\^j + P\^k)

⇒ \nabla.F1=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}

\nabla.F2=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(Q\^i + R\^j + S\^k)

⇒ \nabla.F1=\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z}

Then,

a\nabla.F1+b\nabla.F2=a(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z})+b(\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z})

⇒ \frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z} ...(2)

From (1) and (2),

\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2

2) Verifying the identity: \nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

Consider L.H.S

⇒ \nabla\times(aF1+bF2)

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times(a(M\^i+N\^j+P\^k)+b(Q\^i+R\^j+S\^k))

⇒ (\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times ((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)

Applying the cross product,

\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y}) ...(3)

Consider R.H.S,

⇒ a\nabla\times F1+b\nabla\times F2

So,

a\nabla\times F1=a(\nabla\times (M\^i+N\^j+P\^k))

⇒ \^i(\frac{\partial aP\^k}{\partial y}- \frac{\partial aN\^j}{\partial z})+\^j(\frac{\partial aM\^i}{\partial z}-\frac{\partial aP\^k}{\partial x})+\^k(\frac{\partial aN\^j}{\partial x}-\frac{\partial aM\^i}{\partial y})

a\nabla\times F2=b(\nabla\times (Q\^i+R\^j+S\^k))

⇒ \^i(\frac{\partial bS\^k}{\partial y}- \frac{\partial bR\^j}{\partial z})+\^j(\frac{\partial bQ\^i}{\partial z}-\frac{\partial bS\^k}{\partial x})+\^k(\frac{\partial bR\^j}{\partial x}-\frac{\partial bQ\^i}{\partial y})

Then,

a\nabla\times F1+b\nabla\times F2 =

\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y})

...(4)

Thus, from (3) and (4),

\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

Learn more about divergence and curl of a vector field here:

brainly.com/question/4608972

#SPJ4

Disclaimer: The given question on the portal is incomplete.

Question: Let F1 = M\^i + N\^j + P\^k and F2 = Q\^i + R\^j + S\^k be differential vector fields and let a and b arbitrary real constants. Verify the following identities.

1)\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2\\2)\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2

8 0
1 year ago
Round 60488 to the nearest ten thousand
ANTONII [103]
60,000 would be the answer because it stays the same if under 5 it stays if above 5 rounds up once
4 0
3 years ago
Read 2 more answers
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