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3 years ago

I really need help with this geometry ASAP

Mathematics

1 answer:

SIZIF [17.4K]3 years ago

4 0

I believe A, C, D, E, and G. I don't know if I am right, I am just making an educated guess. Please tell me if I am right!

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Mitchell solves the following system of equations using the elimination method.

antiseptic1488 [7]

Multiply the first equation by 2 or divide the second one my 2, to get the #X opposites, to cancel each other out

-4x+6y=12
4x-6y=-6

Or

-2x+3y=6
2x-3y=-3

there’s no solution either way

7 0

2 years ago

10 rubber stamps cost

polet [3.4K]

Answer:

Step-by-step explanation:

each one is $1.03

so if you want 2 stamps they would be $2.06

7 0

3 years ago

How do you simplify this

Gala2k [10]

15 + 1 because you are using apsolute value so that negative 1 becomes a positive

8 0

3 years ago

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,

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

...(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

Urgent please hell..

vovikov84 [41]

8 x (3^2/4^2)
=4.5
........................

6 0

2 years ago