Think of all the things Gulliver learns about the Lilliputians. What is your opinion of these little people?

Jonathan Swift satirizes English society—particularly the political and theological disputes of his own period—through his description of the Lilliputians.

Gulliver learns that despite their miniature stature, Lilliputians are extremely arrogant, pretentious, greedy, aggressive individuals, which ironically makes them completely human. The Lilliputian politicians are split into two factions, the Low-heelers and High-heelers, which satirizes the Whig and Tory political parties in England. Politicians are forced to perform dangerous stunts, in the form of rope-dancing, to...

Jonathan Swift satirizes English society—particularly the political and theological disputes of his own period—through his description of the Lilliputians.

Gulliver learns that despite their miniature stature, Lilliputians are extremely arrogant, pretentious, greedy, aggressive individuals, which ironically makes them completely human. The Lilliputian politicians are split into two factions, the Low-heelers and High-heelers, which satirizes the Whig and Tory political parties in England. Politicians are forced to perform dangerous stunts, in the form of rope-dancing, to retain their political office.

The Lilliputians are also engaged in a continual war with the Blefuscudians over a theological difference regarding what end of an egg one should break first. The debate between the Big-Endians and Small-Endians satirizes the religious schism between the Protestants and Catholics.

Gulliver also learns that the Lilliputians are treacherous, brutal individuals who are unabashed hypocrites. When the Lilliputian Emperor decides to blind and starve Gulliver, he is praised as being merciful and benevolent. Overall, the Lilliputians are portrayed as corrupt, competitive individuals who are callous and treacherous at their core.

How do George and Hazel react to the televised murder of their son Harrison Bergeron?

George and Hazel show an unnatural lack of concern regarding the televised murder of their son.

Because of his handicap, George is prevented from experiencing any emotions that are considered subversive in the dystopian world he lives in. In this world, all those who are considered of above average intelligence and beauty are required to wear "handicaps" to prevent them from taking undue advantage of these attributes. George is required to wear a mental handicap...

George and Hazel show an unnatural lack of concern regarding the televised murder of their son.

Because of his handicap, George is prevented from experiencing any emotions that are considered subversive in the dystopian world he lives in. In this world, all those who are considered of above average intelligence and beauty are required to wear "handicaps" to prevent them from taking undue advantage of these attributes. George is required to wear a mental handicap radio that is tuned to a government transmitter. The handicap ensures that any unacceptably intelligent or cogent thoughts are dispelled by the transmitter sending sharp jolts of noise to his brain.

Hazel is only of average intelligence and thus considered no threat to the existing order. She is not required to wear a mental handicap of any sort.

When their son is shot by Diana Moon Glampers, the Handicapper General, Hazel doesn't know what to think about it. She experiences confusion and sadness all at the same time, but she can't explain her feelings to George. Meanwhile, George is unable to feel pain, anger, or sadness after witnessing his son's televised execution; this is because the mental handicap he wears prevents him from doing so. Due to the government's machinations, neither Hazel nor George can muster up any normal feelings regarding their son's execution.

At the beginning of Section 2, what does the narrator say Dexter wants?

At the beginning of Section 2, the narrator says that Dexter wants to possess the material accoutrements (the trappings or outward signs of success) that come with being affluent.

He wanted not association with glittering things and glittering people--he wanted the glittering things themselves.

In other words, Dexter longs for luxury, and he works to get it. At twenty-three years old, he borrows a thousand dollars to purchase "a partnership in a laundry." His laundry...

At the beginning of Section 2, the narrator says that Dexter wants to possess the material accoutrements (the trappings or outward signs of success) that come with being affluent.

He wanted not association with glittering things and glittering people--he wanted the glittering things themselves.

In other words, Dexter longs for luxury, and he works to get it. At twenty-three years old, he borrows a thousand dollars to purchase "a partnership in a laundry." His laundry business eventually becomes well-known for its expertise in laundering English woolen golf stockings, Shetland hosiery, delicate sweaters, and women's lingerie. Before the age of twenty-seven, Dexter is the proud owner of five laundry branches, making him the owner of "the largest string of laundries in his section of the country." After making his money, Dexter sells his business and heads to New York.

As can be seen from the beginning of Section 2, Dexter is a man who prizes wealth and its material trappings above all else in life.

Two astronauts find themselves floating beside International Space Station after their spacesuit thrusters are damaged by a solar flare and...

The overall theme for this question is that in the absence of any external influences, total momentum (as well as total internal energy) is always conserved.  The basic principle behind getting both astronauts back to the ISS is for the first astronaut (5 yards away from the ISS) to rescue the second astronaut (10 yards from the ISS) before they both can return to the ISS together.   

A) If we consider the system of reference to be the ISS, Astronaut 1, and Astronaut 2, then both astronauts posses 0 momentum as they are both floating alongside the ISS, neither moving towards or away from it in any direction.  Astronaut 1 should take one of the tools from their belt and throw it in the opposite direction as Astronaut 2.  This will impart momentum (inertia in motion) to the tool as it will begin moving towards the ISS.  However, according to Newton's 3rd Law, Astronaut 1 will now possess that same amount of momentum, but in the opposite direction (towards Astronaut 2).  In short, this is like an elastic collision, though the total momentum before was 0.  According to the momentum equation, the larger mass of the astronaut will result in a much smaller velocity than the tool of choice would have received.  Astronaut 1, depending on the availability of tools, can continue this process until they reach Astronaut 2.  Once they collide, Astronaut 2 will need to catch Astronaut 1 and hold on tight, resulting in an inelastic collision.  The result will be Astronaut 1 slowing down as Astronaut 2 will have, in essence, increased his/her mass.  The original process of discarding tools should be repeated, though this time opposite the ISS.  As each tool is thrown away from the station, the astronaut pair will slow down, stop, and eventually achieve a velocity in the direction of the ISS.

B)  The conservation of energy in this action is larger than simply throwing the hammer.  The Law of Conservation of Energy states that the total internal energy of a system remains constant, but it can be transferred between objects and transformed to different types of energy.  The astronaut has an amount of potential energy stored inside of their bodies.  The action of throwing the hammer will utilize that chemical potential energy and convert it into kinetic energy, enabling the astronaut to move.  That motion will then cause the hammer to move, imparting kinetic energy to the hammer.  Overall, the total energy of the astronaut and the hammer remains constant, but the total energy of the astronaut decreases while the energy of the hammer increases.  Potential energy of the astronaut is converted to kinetic energy and then transferred as kinetic energy to the hammer.

C) When bodies rotate, inertia in motion is referred to as angular momentum.  Just as in a linear system, angular momentum is also conserved during encounters.  When the astronaut throws the hammer with a clockwise rotation, he imparts angular momentum on the hammer.  Before he threw it, the total rotational momentum of the hammer and he combined was 0.  After he throws it, the total rotational momentum will still be 0.  As a result, he will be rotating in the opposite direction as the hammer, but with a much slower rotation due to a larger mass.

a) showing details of the work, write an expansion in Fourier series of the signal f(x) which is assumed to have the period 2`pi`f(x) = "Please...

This function is odd, thus its Fourier expansions contains only `sin(nx)` terms, i.e. `a_n=0, ngt=0.` This expansion has the form  `sum_(n=1)^(oo) b_n sin(nx)`  where  `b_n = 1/pi int_(-pi)^(pi) f(x) sin(nx) dx.`

Find these coefficients:

`b_n = 1/pi int_(-pi)^(pi) f(x) sin(nx) dx = 2/pi int_0^(pi) f(x) sin(nx) dx =`

`= 2/pi int_0^(pi/2) f(x) sin(nx) dx + 2/piint_(pi/2)^(pi) f(x) sin(nx) dx) =`  

`= 2/pi int_0^(pi/2) x sin(nx) dx + 2/piint_(pi/2)^(pi) pi/2 sin(nx) dx.`

The second integral...

This function is odd, thus its Fourier expansions contains only `sin(nx)` terms, i.e. `a_n=0, ngt=0.` This expansion has the form  `sum_(n=1)^(oo) b_n sin(nx)`  where  `b_n = 1/pi int_(-pi)^(pi) f(x) sin(nx) dx.`

Find these coefficients:

`b_n = 1/pi int_(-pi)^(pi) f(x) sin(nx) dx = 2/pi int_0^(pi) f(x) sin(nx) dx =`

`= 2/pi int_0^(pi/2) f(x) sin(nx) dx + 2/piint_(pi/2)^(pi) f(x) sin(nx) dx) =`  

`= 2/pi int_0^(pi/2) x sin(nx) dx + 2/piint_(pi/2)^(pi) pi/2 sin(nx) dx.`

The second integral is obviously

`-(cos(nx))|_(x=pi/2)^pi = -1/n (cos(n pi)-cos((n pi)/2)).`

To find the second, use integration by parts:

`2/pi int_0^(pi/2) x sin(nx) dx =`

`= |u=x, du=dx, dv=sin(nx)dx, v=-1/n cos(nx)| =`  

`= -2/(n pi) (x cos(nx))|_(x=0)^(pi/2) + 2/(n pi) int_0^(pi/2) cos(nx) dx =`

`= -1/n cos((n pi)/2) + 2/(n^2 pi) (sin(nx))|_(x=0)^(pi/2) =`

`= 2/(n^2 pi)sin((n pi)/2)-1/n cos((n pi)/2).`

This way  `b_n =2/(n^2 pi)sin((n pi)/2)-1/n cos(n pi).`   Therefore  `b_1=1+2/pi, b_3=1/3 -2/(9 pi), b_5=1/5+2/(25 pi).`

The graphs are attached (the function is in blue, the approximation is in green). They are not so close but are somewhat similar. Note that on `(-4pi,4pi)` the function is `2pi` -periodic.

Why is he traveling light without many supplies?

To some extent, this is an opinion-based question because the story does not expressly say why the man was travels light. It is up to readers to determine why he is traveling that way based on the details given. I have some possible reasons that I think are all responsible for why the man was traveling light.  

1. He is ignorant of what he might need in the frigid Yukon. Readers are told that he...

To some extent, this is an opinion-based question because the story does not expressly say why the man was travels light. It is up to readers to determine why he is traveling that way based on the details given. I have some possible reasons that I think are all responsible for why the man was traveling light.  

1. He is ignorant of what he might need in the frigid Yukon. Readers are told that he is new to the area, and this is his first winter. The man simply doesn't know that he should be better prepared.  

But all this—the distant trail, no sun in the sky, the great cold, and the strangeness of it all—had no effect on the man. It was not because he was long familiar with it. He was a newcomer in the land, and this was his first winter.

2.  He only plans for what is right now. He does not know how to plan for what might happen. The story tells readers that he doesn't have the ability to imagine. He knows it's cold, but he can't imagine exactly how the cold might cause him problems.

The trouble with him was that he was not able to imagine. He was quick and ready in the things of life, but only in the things, and not in their meanings.

3. His companions already made the trip, and the man is counting on an equally uneventful trip. He is counting on having no problems, and he is expecting to arrive at a camp that already has supplies and food. He sees no reason to bring extra.

He would be in camp by six o’clock that evening. It would be a little after dark, but the boys would be there, a fire would be burning, and a hot supper would be ready.

How can I find the probability (in fractions)? S={3,5,6,8,9,12,13,14,15,16} A={3,12,5,13} B={14,15,6,3} C={12,5,8,9,16} 1. P(B|A)= ? and are...

I am assuming that we are selecting one item from a sample space of

s={3,5,6,8,9,12,13,14,15,16}.

We have event spaces:

A={3,5,12,13}

B={3,6,14,15}

C={5,8,9,12,16}

We can compute the probabilities for each event: P(A)=P(B)=2/5 and P(C)=1/2. (We find these by taking the size of the event space divided by the size of the sample space.)

(1) P(B|A) is the probability that the item selected is in B given that it is in A. If the item selected is...

I am assuming that we are selecting one item from a sample space of

s={3,5,6,8,9,12,13,14,15,16}.

We have event spaces:

A={3,5,12,13}

B={3,6,14,15}

C={5,8,9,12,16}

We can compute the probabilities for each event: P(A)=P(B)=2/5 and P(C)=1/2. (We find these by taking the size of the event space divided by the size of the sample space.)

(1) P(B|A) is the probability that the item selected is in B given that it is in A. If the item selected is in A then it is one of 3,5,12,13. Only 3 is in B, so the probability that an item is in B given that it is in A is 1/4.

P(B|A)=1/4. Using the formula `P(B|A)=(P(A and B))/(P(A)) ` , and noting that the probability that an item is in A and in B is 1/10 (only 3 is in both A and B, so there is 1 item out of the 10 total items.)

So P(B|A)=(1/10)/(2/5)=1/4 as above.

The events are not independent since `P(A and B) != P(A)*P(B) `

(2) P(A or B) These events are not mutually exclusive (it is possible for an item to be in A and B) so P(A or B)=P(A)+P(B)-P(A and B) or

P(A or B)=2/5+2/5-1/10=7/10

(3) P(A|C)

`P(A|C)=(P(A and C))/(P(C))=(1/5)/(1/2)=2/5 `

(If the item is known to be in C then it is one of 5,8,9,12,16; two of these numbers are in A so the probability is 2/5 as above.)