write a 5 paragraph essay on how the character sqeaky changed over the course of raymonds run. if your not going to answer the question then don't bother I would wrather nobody answer then someone just take the points ​

Squeaky like other people throughout life are changing so much

because they have new experiences that make the way they see things

different. Taking care of her brother Raymond changes Squeaky

because she dont want anyone to hurt his feelings. Squeaky likes to

be a winner and who can blame her. She wants to beat everyone in

the big race because she works hard to be the most fastest runner

around. She wants to be the best and gets a little mad at a girl in

the story. She’s good runner too so maybe Squeaky is jealous or

worried that she could win the race instead. Squeaky likes this girl

better at the end of story because Squeaky is in first place and the

other one just comes in second

Raymond changes in the story and Squeaky changes right along

with him. At first you think Squeaky is the only runner in the family

but later on you find out that Raymond could be a pretty good runner

hisself with Squeaky acting as his coach. So a brother and sister

change the way they see other people and themselves.

because she don’t want anyone to hurt his feelings. Squeaky likes to

be a winner and who can blame her. She wants to beat everyone in

the big race because she works hard to be the most fastest runner

around. She wants to be the best and gets a little mad at a girl in

the story. She’s good runner too so maybe Squeaky is jealous or

worried that she could win the race instead. Squeaky likes this girl

better at the end of story because Squeaky is in first place and the

other one just comes in second. This is a change Squeaky has that

makes her a better person because now she might even be freinds

with the girl.

I really like stories that tell about characters who learn new

things about themselves and change for the better. Raymond’s Run was

one of those stories. I wish more people I know would change like

Squeaky did, becoming more honest and more understanding of others

like people should be.

It is the fifty-yard dash that changes Squeaky. During the race

she becomes aware of Raymond’s own race. She sees him running

outside the fence. Winning the race makes Squeaky a different person.

Her views of others has changed. She smiles a smile of respect to her

opponent Gretchen, and she even considers the idea of Gretchen as an

assistant coach for Raymond. During the race, watching Raymond run

along the side, Squeaky decided to coach him to run. No longer does

Raymond have to be simply in her care because now she is able to

see him as having a fuller life.

In conclusion, the recognition and pride of winning the race makes

Squeaky become more grown up. She now knows herself better. She can

be do well at other things besides running and discovers new ways to

improve herself and make things better for her brother too.

Squeaky is a girl living in Harlem. She doesn’t have to do chores

or work for extra spending money. She just cares for Raymond, which

is her main responsability. Raymond is her bigger and older brother.

She thinks of him as her little brother though because Raymond

relies on her. Squeaky is sensitive about Raymond’s differences. He

has a large head and is mentally handicapped, so that is why he

seems like a younger brother.

queaky, the young narrator of “Raymond’s Run” by Tony Cade

Bambara, changes over the course of the story. In her Harlem

neighborhood, Squeaky has earned a reputation as a champion runner

who takes great pride in her speed. Competing in the big May Day

race has a major impact on her. This event alters how Squeaky views

herself, her mentally handicapped brother, Raymond, and her main

rival, Gretchen.

Competing in the May Day race changes Squeaky’s social relationships as well as her family relationships. After the race, she seems

more likely to make friends with other girls, especially Gretchen,

Squeaky’s most challenging competitor. Squeaky no longer treats

Gretchen rudely and criticizes her. Instead, Squeaky recognizes

Gretchen’s athletic skills, gains respect for her, and gives her a warm,

sincere smile. Squeaky wonders if Gretchen would like to help coach

Raymond too, which is Squeaky’s way of paying Gretchen a very high

compliment.

The May Day race is a turning point in Squeaky’s life. She

discovers new sides of herself, her brother, and Gretchen. These

changes transform Squeaky into a more mature young girl with a

greater awareness of her own and other people’s talents

Early in the story, Squeaky announces, “I run. That is what I am

all about.” However, after the May Day race is over, Squeaky’s selfimage broadens. She sees herself as more than just a budding track

star and begins to consider other talents she might develop. For

example, she thinks about taking piano lessons, studying more, and

winning a spelling bee. Because Squeaky is a champion at heart, she

feels confident that she can succeed in other areas of her life and

welcomes new challenges.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Form- 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalmente factorizada: f (x) = (x - 3) (x + 3) [x - (2 - i)] [x - (2 + i)] La función tiene raíces reales y raíces imaginarias.

Forma totalment