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gtnhenbr [62]
3 years ago
7

Plz help me :,( also show the work too /:

Mathematics
1 answer:
ZanzabumX [31]3 years ago
6 0

Answer:

<em>Surface Area for cube: 263 inches^3</em>

Step-by-step explanation:

<em>Formula: 2lw + 2lh + 2wh</em>

<em>2(6 1/2)(3) + 2(6 1/2)(7) + 2(3)(7) =</em>

<em>39 + 182 + 42 =</em>

<em>263 inches</em>

<em>Hope This Helps! :-)</em>

You might be interested in
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
{1, 2, 6, 24, 120, ...}
enyata [817]
This is a factorial sequence that can be modeled by An = n!. As you may see, the increasing numbers are factorials of 1,2,3,4, and 5. Factorial means multiplying backwards and is represented by !. For example, 1! is 1*1 =1. 2! is 2*1, 3! is 3*2*1, 4! is 4*3*2*1 etc.
8 0
3 years ago
How do I find the variable?<br><br> N x 1/5 = 2/15<br><br> N / 1/5 = 1/3
algol [13]
First you need to get all variables on one side. You do that by multiplying the reciprocal of the fraction to everything.
N(5/1) x 1/5(5/1)=2/15(5/1)
5n=2/15 x 5/1
Then solve the multiplication problem
5n= 10/15
Then you should reduce the fraction
5n=2/3
Then divide both sides by 5
5n/5=N
2/3 divided by 5 =2/15

N=2/15




4 0
3 years ago
Will give brainliest-H(t)=-16t^2+28t
____ [38]
For this case we have the following equation:
 H (t) = - 16t ^ 2 + 28t
 When the dolphin finishes a jump, then the height is equal to zero.
 We have then:
 -16t ^ 2 + 28t = 0
 We look for the roots of the polynomial:
 t1 = 0
 t2 = 1.75 s
 Therefore, the dolphin takes 1.75 seconds to make a jump.
 Answer:
 
it will take for the dolphin to make one jump about:
 
t = 1.75 s
4 0
3 years ago
Find the missing side. round to the nearest tenth.
horrorfan [7]

Answer:

24) x = 9.2

25) x = 30.8

Step-by-step explanation:

Given

See attachment for triangles

Solving (24)

To solve for x, we make use of cosine formula

i.e.

cos(40) = adjacent ÷ hypotenuse

So, we have:

cos(40) = x ÷ 12

Multiply both sides by 12

12 cos(40) = x

12 * 0.7660 = x

x = 9.2

Solving (25)

To solve for x, we make use of sine formula

i.e.

sin(25) = opposite ÷ hypotenuse

So, we have:

sin(25) = 13 ÷ x

Multiply both sides by

x sin(25) = 13

Divide by sin(25)

x = 13 ÷ sin(25)

Using a calculator

x = 30.8

5 0
2 years ago
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