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3 years ago

A storage container at the arena is 3 metres wide, 6 metres long and 4 metres high.

Mathematics

2 answers:

OverLord2011 [107]3 years ago

7 0

17.36 boxes hope im right

saveliy_v [14]3 years ago

7 0

Answer:

576 boxes

Step-by-step explanation:

Given that a storage container at the arena is 3 metres wide, 6 metres long and 4 metres high.

Hence volume of the container $= 3x6x4 = 72m cube.$

Volume of boxes = $50x50x50 = 125000 cm cube= 0.125 m^3$

No of boxes that fit into storage container

= $\frac{72}{0.125} \\=576$

576 boxes can be fit inside the container.

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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,

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

...(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

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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

2 years ago

Calculate the mean deviation by median (by using the mid-point(x) of leaf weights. b) A fair die is rolled. Find the probability

SVETLANKA909090 [29]

Answer:

hello attached below is the required table that is missing and the completed table as well

A) mean deviation = 0.1645

B) 2/3

Step-by-step explanation:

From the table we calculated the midpoint value (x) and c.f

N = 27

median class = 2.35 to 2.45

median = l + [ (N/2) - cf / F ] * H

= 2.35 + [ (13.5 - 13) / 6 ] *0.1 = 2.3583

hence mean deviation by median

= summation of fi |xi -M| / summation of fi

= 4.4417 / 27 = 0.1645

B ) probability of getting an odd number or prime number or both

the probability of an odd number = 1/2

also the probability of an even number = 1/2

while the probability of getting neither of them = 1/3

hence the probability of getting odd or prime or both

= 1/2 + 1/2 - 1/3 = 2/3

6 0

3 years ago

How do I solve this?

blagie [28]

Answer:

We'll solve that in parts.

The upper part is a right triangle that is 4 by (2 + 2 +6)

Area = (4 * 10) /2 = 20

The bottom of that figure is made of 3 parts.

A) a 2 by 2 right triangle. area equals (2*2)/2 = 2

B) a 2 by 2 square whose area equals 2*2 = 4

C) a 6 by 2 right triangle whose area equals (6*2) / 2 = 6

Total Area = 20 + 2 + 4 + 6 = 32

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

How many questions do you have to answer before you can message people on brainly?

Kaylis [27]

Answer:

1000

Step-by-step explanation:

4 0

3 years ago

When reading a book, Charlie made a list by writing down the page number of the last page he finished reading at the end of each

Assoli18 [71]

Answer:

No. of pages he read by the end of the 8th day = 425

Step-by-step explanation:

Charlie started from page 1 and read the same number of pages each day. Suppose x is the number of pages he read each day, the last page he read on the first day would be:

This would be the first number he noted on the list.

The second page number can be determined by adding the number of pages he read to the last page for day 1.

Last page for day 2 = 1 + x + x

= 1 + 2x

Similarly, the last page he read on the third day is:

Last page for day 3 = 1 + 3x. Similarly,

Last page for day 4 = 1 + 4x

Last page for day 5 = 1 + 5x

Last page for day 6 = 1 + 6x

Last page for day 7 = 1 + 7x

Last page for day 8 = 1 + 8x

His Mom added all of the page numbers and got a total of 432. So,

432 = 1+x + 1+2x + 1+3x + 1+4x + 1+5x + 1+6x + 1+7x + 1+8x

432 = 8 + 8x

8x = 432-8

8x = 424

x = 424/8

x = 53

By the end of the 8th day he actually read 1 + 8x pages. Substituting the value of x in this expression, we get:

No. of pages he read by the end of the 8th day = 1 + 8(53)

= 1 + 424

No. of pages he read by the end of the 8th day = 425

4 0

3 years ago