To show that the perimeter of AABL is constant, we need to prove that AB + BA + AL + BL is constant.

First, let us label the angles in the figure as follows:

∠NLM = ∠LNM = α (since LM and LN are tangents from an external point, they are equal in measure)
∠MAB = β (angle between the tangent AB and the chord MC)
∠NBA = β (angle between the tangent AB and the chord NC)
∠ALN = γ (angle between the tangent LN and the chord MN)
∠BLM = γ (angle between the tangent LM and the chord MN)

Using the fact that the sum of angles in a triangle is 180 degrees, we can write:

∠LAM = 180 - α - β
∠LBN = 180 - α - β

Now, let us consider the perimeter of AABL:

AB + BA + AL + BL

Using the fact that angles in the same segment of a circle are equal, we can write:

∠MCN = ∠MAB + ∠NBA = 2β

Using the fact that the opposite angles in a cyclic quadrilateral add up to 180 degrees, we can write:

∠ALB = 180 - ∠MCN = 180 - 2β

Now, let us consider the triangles LAM and LBN. Using the fact that angles in a triangle add up to 180 degrees, we can write:

∠LAM + ∠ALN + ∠ALM = 180
∠LBN + ∠BLM + ∠BLN = 180

Substituting the values we have calculated for ∠LAM and ∠LBN, and using the fact that ∠ALN = ∠BLM = γ, we get:

(180 - α - β) + γ + ∠ALM = 180
(180 - α - β) + γ + ∠BLN = 180

Simplifying, we get:

∠ALM = α + β - γ
∠BLN = α + β - γ

Finally, let us consider the perimeter of AABL again:

AB + BA + AL + BL = AB + BA + 2AL sin γ + 2BL sin γ

Using the fact that AL = AM and BL = BN, and using the sine rule in triangles LAM and LBN, we get:

AB + BA + 2AM sin γ + 2BN sin γ = AB + BA + 2LM sin (α + β - γ) + 2LN sin (α + β - γ)

Using the fact that LM = LN (since they are tangents from an external point), we get:

AB + BA + 2LM sin (α + β - γ) + 2LM sin (α + β - γ)

Simplifying, we get:

AB + BA + 4LM sin (α + β - γ)

Using the fact that sin (α + β - γ) = sin (180 - γ) = sin γ, we get:

AB + BA + 4LM sin γ

Substituting the value of LM (which is equal to LN), we get:

AB + BA + 4LN sin γ

But we know that sin γ is a constant, since it is determined by the position of point L and the circle MN. Therefore, the perimeter of AABL is constant, since it does not depend on the position of point C on the arc MN.

Hence, we have proved that the perimeter of AABL is constant.

Read-Only Memory (ROM) is a storage area that is used to store instructions. The contents of read-only memory are permanent. That means, switching off the power supply will not erase the data stored in ROM. The data stored in ROM is either unchangeable or requires special operations to change the contents. A normal user cannot write anything in read-only memory.
Random Access Memory (RAM) also called the main memory of the computer, is a storage area that acts as a rough working area for the programs It is a read-write memory. That means a user can read the contents stored in it as well as he writes the contents on to the memory. RAM temporary storage area that means, the con cannot be permanently stored in random a memory. The contents of random access to me are lost when the computer is switched off. That is why it is also called "volatile memory".

Trivial things can have great impact on our lives

It is quite true, after the Wavewalker was irreparably damaged by the colossal wave, everyone on-board contributed to save it.  

Not only the elders on the boat worked hard, kept nerves, and cooperated with one another, the six year old son, and the seven year old daughter behaved courageously and bravely also in the face of the storm. The six year old son Jon reassured his dad that he was not afraid to die if he could be together with his family. So these fearless words motivated the author to do his best to save the damaged boat. The author was really a great sailor as he repaired the damaged boat, steered it and made it steady in the raging storm.  

Larry Vigil and Herb Seigler, the professional sailors hired in S. Africa helped the narrator in tackling the Southern Indian Ocean. They proved to be real saviours when the ship had been flooded with the sea water. They pumped like madmen to keep the water levels reasonably low. Each and every person on Wavewalker did their best to save the boat. Their courageous efforts were nothing but sheer hard work, keeping nerves, and cooperation. So, they survived the disaster.

Solution is given below

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The future is bleak because the reserves of oil would last only 50 years more.

Trade is essentially buying and selling of items produced elsewhere. All the services in retail and wholesale trading or commerce are specifically intended for profit. All this work takes place in towns and cities also known as trading centres.
Characteristics of rural marketing centre:

1. Rural marketing centres cater to nearby settlements. These are quasi-urban centres.
2. They serve as trading centres of the most rudimentary type. Here personal and professional services are not well-developed. These form local collecting and distributing centres. Most of these have mandis (wholesale markets) and also retailing areas. They are not urban centres per se but are significant centres for making available goods and services which are most frequently demanded by rural folk.
3. Periodic markets in rural areas are found where there are no regular markets and local periodic markets are organised at different temporal intervals. These may be weekly, bi-weekly markets from where people from the surrounding areas meet their temporally accumulated demand. These markets are held on specified dates and move from one place to another. The shopkeepers thus remain busy on all the days while a large area is served by them.

Who hasn't heard of the great American inventor Thomas Edison!

His inventions include the gramophone and the electric lamp.  

He invented the gramophone at the age of thirty-one only.  

Later, he invented the electric bell.  

One of his jobs was to improve the telephone first invented by Graham Ball.  

Do you know how much he earned by improving it?  

A company called Western Union paid 1,00,000 for him.  

He had to carry out six thousand tests to make a good electric bulb.  

How much patience it must have needed!  

Try to get inspiration from the life of Edison.

Santhali

Tourism is a highly labour intensive activity of a unique kind in the world. Many local persons are employed to provide services like accommodation, meals, transport, entertainment and special shops serving the tourists.

When tectonics plates move past each other, they cause large earthquakes, which tilt, offset, or displace large areas of the ocean floor from a few kilometers to as much as a 1,000 km or more.  The sudden vertical displacements over such large areas, disturb the ocean's surface, displace water, and generate destructive tsunami waves.  The waves can travel great distances from the source region, spreading destruction along their path.  For example, the Great 1960 Chilean tsunami was generated by a magnitude 9.5 earthquake that had a rupture zone of over 1,000 km.   Its waves were destructive not only in Chile, but also as far away as Hawaii, Japan and elsewhere in the Pacific. 

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AB = 3 cm, AC = 4 cm
In {tex}\triangle{/tex}BAC, by pythagoras theorem
BC² = AB² + AC² 
{tex}\Rightarrow{/tex}BC² = 3² + 4² 
{tex}\Rightarrow{/tex}BC² = 25 
{tex}\Rightarrow{/tex}BC = {tex}\sqrt {25} {/tex} = 5 cm
In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}CAB
{tex}\angle{/tex}ABO = {tex}\angle{/tex}ABC [common]
{tex}\angle{/tex}AOB = {tex}\angle{/tex}BAC [each 90^o] 
Then, {tex}\triangle{/tex}AOB - {tex}\triangle{/tex}CAB [by AA similarity]
{tex}\therefore{/tex} {tex}\frac { A O } { C A } = \frac { O B } { A B } = \frac { A B } { C B }{/tex} [c.p.s.t]
{tex}\Rightarrow{/tex} {tex}\frac { A O } { 4 } = \frac { O B } { 3 } = \frac { 3 } { 5 }{/tex}
Then, AO = {tex}\frac{{4 \times 3}}{5}{/tex} and OB = {tex}\frac{{3 \times 3}}{5}{/tex}
{tex}\Rightarrow{/tex} AO = {tex}\frac{12}{5}{/tex} cm and OB = {tex}\frac{9}{5}{/tex} cm
{tex}\therefore{/tex}OC = 5 - {tex}\frac{9}{5}{/tex} = {tex}\frac{16}{5}{/tex}cm
{tex}\therefore{/tex} Volume of double cone thus generated = volume of first cone + volume of second cone
{tex}= \frac { 1 } { 3 } \pi ( A O ) ^ { 2 } \times B O + \frac { 1 } { 3 } \pi ( A O ) ^ { 2 } \times O C{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 12 } { 5 } \right) ^ { 2 } \times \frac { 9 } { 5 } + \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \left( \frac { 12 } { 5 } \right) ^ { 2 } \times \frac { 16 } { 5 }{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \frac { 12 } { 5 } \times \frac { 12 } { 5 } \left[ \frac { 9 } { 5 } + \frac { 16 } { 5 } \right]{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times \frac { 12 } { 5 } \times \frac { 12 } { 5 } \times 5{/tex}
={tex}\frac{1056}{35}{/tex} = {tex}30 \frac { 6 } { 35 } \mathrm { cm } ^ { 3 }{/tex}.

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