Bookmark On the Isle of Man off the coast of Ireland, a curious cat .pdf
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Bookmark On the Isle of Man off the coast of Ireland, a curious cat referred to as a manx cat was described in 1845. It had long hind legs and a short tail. Further work showed that the manx characteristics were the result of a dominant M allele at a single locus. An MM zygote would not survive, but Mm genotypes produced the manx characteristics. Imagine a scenario where the M allele arises by mutation on the Isle of Man. Explain how genetic drift could cause the M allele to reach a relatively high frequency (freq(M) > 0.25), even though it has harmful effects on fitness. Solution When the M allele is dominant and also deleterious when it is in the homozygous state. In the heterozygote stage Mm, the deleterious M allele is hidden due to which the selection will not eliminate it. The Mm genotypes are successful, so the frequency of M allele rises in the population. So the genetic drift selection cannot eliminate the M allele completely..
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Discuss the seven factors that analysts consider where selecting a f.pdf
Discuss the seven factors that analysts consider where selecting a file organization. Evaluate
each file organization with respect to each of the seven factors. Review Table 9-3 to review how
each of these factors can easily be summarized.
Solution
1. Frequency of update: A file that needs to be updated every now and then needs an organisation
method that will allow easy retrieval of information and ease of updating, example of such a file
is the transaction file.
2. File activity: Different files have different activities, example a sort file is used to sort data in
sequential order and therefore sequential method would be appropriate for such a file.
3. File access method: Definately different files have different methods of being accessed,
example a reference file is accessed using random method for easy retrieval of data.
4. Nature of the system: Files that are used in a particular system will depend on the nature of
the system i.e the suitable organisation method for that particular system.
5. Masterfile medium: The masterfile is the main file for keeping permanent updates of records
from transaction files and other sources, the medium by which it is updated will determine the
organisation method to be used..
Consider a 4-Link robot manipulator shown below. Use the forward kine.pdf
Consider a 4-Link robot manipulator shown below. Use the forward kinematic D-H table and
write an m file that plots the manipulator. The instructions are given in the module 6. Submit
your solutions by the due date, in a single MATLAB m file.
Solution
Please give the kinetic D-H table else it would be difficult to code as we need to know the
rotation spin axis and other momentum of manipulator
Stating a general example code for manipulator with data
function X = fwd_kin(q,x)
% given a position in the configuration space, calculate the position of
% the end effector in the workspace for a two-link manipulator.
% q: vector of joint positions
% x: design vector (link lengths)
% X: end effector position in cartesian coordinates
% configuration space coordinates:
q1 = q(1); % theta 1
q2 = q(2); % theta 2
% manipulator parameters:
l1 = x(1); % link 1 length
l2 = x(2); % link 2 length
% calculate end effector position:
X = [l1*cos(q1) + l2*cos(q1+q2)
l1*sin(q1) + l2*sin(q1+q2)];
% SimulateTwolink.m uses inverse dynamics to simulate the torque
% trajectories required for a two-link planar robotic manipulator to follow
% a prescribed trajectory. It also computes total energy consumption. This
% code is provided as supplementary material for the paper:
%
% \'Engineering System Co-Design with Limited Plant Redesign\'
% Presented at the 8th AIAA Multidisciplinary Design Optimization
% Specialist Conference, April 2012.
%
% The paper is available from:
%
% http://systemdesign.illinois.edu/publications/All12a.pdf
%
% Here both the physical system design and control system design are
% considered simultaneously. Manipulator link length and trajectory
% specification can be specified, and torque trajectory and energy
% consumption are computed based on this input. It was found that maximum
% torque and total energy consumption calculated using inverse dynamics
% agreed closely with results calculated using feedback linearization, so
% to simplify optimization problem solution an inverse dynamics approach
% was used, which reduces the control design vector to just the trajectory
% design.
%
% In the conference paper several cases are considered, each with its own
% manipulator task, manipulator design, and trajectory design. The
% specifications for each of these five cases are provided here, and can be
% explored by changing the case number variable (cn).
%
% This code was incorporated into a larger optimization project. The code
% presented here includes only the analysis portion of the code, no
% optimization.
%
% A video illustrating the motion of each of these five cases is available
% on YouTube:
%
% http://www.youtube.com/watch?v=OR7Y9-n5SjA
%
% Author: James T. Allison, Assistant Professor, University of Illinois at
% Urbana-Champaign
% Date: 4/10/12
clear;clc
% simulation parameters:
p.dt = 0.0005; % simulation step size
tf = 2; p.tf = tf; % final time
p.ploton = 0; % turn off additional plotting capabilities
p.ploton2 = 0;
p.Tallow = 210; % maximum .
Conduct some basic research on the recessive disorder Albinism.Des.pdf
Conduct some basic research on the recessive disorder Albinism.
Describe the disease and its genetic basis.
Explain how this disease is inherited and why it is not always inherited.
Why is genetic testing controversial?
Provide references for your research
Solution
The term recessive disorder Albinism isdefined a is a congenital disorder which usually happens
due to complete or partial absence of the pigment which is present in the skin, hair and eyes.
Moreover it isr elated to a number of vision defects.These include: photophobia, nystagmus, and
amblyopia. Another important aspect is the Lack of skin pigmentation can usually lead to
sunburn .
This disorder is usually passes from parents to their children through the concept of genes. The
most importaant aspect for inhertiance is that both the parents should carry an albinism gene in
order to have a child who suffers from albinism disorder.Moreover, the parents can also have
normal pigmentation but they still are able to carry the gene. Finally as both parents are able to
carry the gene, and none of the parent has albinism.This means that there is a one-in-four chance
which can happen at pregnancy that the baby will be born with albinism disorder.
Genetic testing is controversial because after determining the sex of the child ,some parents
decide to abort it.So,in order to prevent this it has become controversial in some countries like
india where it is illegal.
References:
1.https://www.aapos.org/terms/conditions/12
2..
calculate the next annual dividend payment for stock currently selli.pdf
calculate the next annual dividend payment for stock currently selling for $50 per share. with a
16% expected total return and a 10% expected constant growth rate?
Solution
Expected return=(D1/Current price)+Growth rate
0.16=(D1/50)+0.1
Hence D1=(0.16-0.1)*50
which is equal to
=$3=next annual dividend payment..
Answer the questions below based on the database model found on the n.pdf
Answer the questions below based on the database model found on the next page: 2. What titles
are available in Digital Video format? 3. What titles and years are available for the Pop-
Alternative genre? What is the result of the SQL statement below? SELECT from USIC where 4.
MEDIA FOorrat \"Analog\" s. Which is the correct SQL statement to obtain a list of all music
(titles and artists) in CD format? (select one) a. SELECT Artist, Title from MEDIA where
Mediaype CD b. SELECT Artist, Title from MUSIC where Meaanp9,“ \"CD\" c. SELECT
MUSIC where xediatse *CD Write valid SQL statements for the following actions: 6. Retrieve
all the free Titles from the MUSIC relation: 7. Remove the Cassette Tape row from the MEDIA
table: 8. Change the Year on \"Somebody 1 Used to Know to be 2012: 9. Add a new Artist row
to the ARTIST Table (add any Artist and Genre you like):
Solution
2. From media relation:
MP4 corresponds to Digital Video Format
Hence, Titles which are MP4:
High School
Eyes Wide Open
Payphone
3. Artists that are associated with Pop-alternative genre are Gotye, Coldplay, Fiest
Hence, the titles and years that correspond to these artists:
The Reminder, 2007
Open Season, 2006
Mylo Xyloto, 2011
Somebody I used to know, 2011
Eyes wide open, 2012
4.
We need to select music whose media format is analog. Hence, it will return music records with
media type vinyl and cassette tape
Hence, it will return the following records:
Born to run, Bruce Springstein, 1981, Vinyl, 15.95
Greatest Hits, Ella Flitzgerald, 1972, Vinyl, 15.95
Quadrophenia, The Who, 1973, Vinyl, 25.95
Rolling Stones Box Set, Rolling Stones, 2013, Vinyl, 50.00
5. b. SELECT Artist, Title from MUSIC where MediaType = \"CD\"
6. SELECT Title from MUSIC;
7. DELETE FROM MEDIA where MediaType=\"Cassette Tape\";
8. UPDATE MUSIC
SET Year=2012
WHERE Title=\"Somebody I used to know\";
9. INSERT INTO ARTIST
Values (\"Adam Young\", \"Pop\");.
You are reviewing basic network protocols with a new tech in your co.pdf
You are reviewing basic network protocols with a new tech in your company. Of the following,
what do you say are true about network protocols? (Choose two.)
A. NetBIOS is used primarily to connect Windows computers to the Internet.
B. TCP/IP is used exclusively to allow computers to connect to each other on the local area
network.
C. TCP/IP is the de facto communications protocol for the Internet.
D. NetBIOS is used in Ethernet networks.
Solution
C and D. TCP/IP, or Transmission Control Protocol/Internet Protocol, is the basic
communication protocol for the Internet and is also used on private networks such as intranets
(web environments within a company) or extranets (web environments between two or more
business partners or between a vendor and customers). NetBIOS, or Network Basic Input/Output
System, is a program that allows computers to communicate on local area networks, or LANs,
but does not support transport into wide area networks, or WANs, such as the Internet..
Why might a gene have more than 2 different allelesa. DNA seq.pdf
Why might a gene have more than 2 different alleles?
a. DNA sequence can deviate in many places (base-pairs)
b. DNA sequence can only deviate in one place (base-pair)
c. DNA sequence cannot deviate
d. Genes can only have 2 possible alleles
Solution
Answer:
a. DNA sequence can deviate in many places (base-pairs)
Explanation:
Multiple alleles means that there are three or more forms of a gene for a trait. That doesn\'t mean
that a person can have all three of them..
Which of the following are true statementsI. sin2+cos2=1II. sec.pdf
Which of the following are true statements?
I. sin2+cos2=1
II. sec2+1=tan2
III. 1+csc2=cot2
IV. 1csc2+1sec2=1
I am aware I have to re-learn the identies.
Solution
Given the statements:
I. True for Sin2x + Cos2x =1.
II. False, because Sec2x = 1 + tan2x
III. False, because csc2x = 1 + cot2x
iV. False, because is is different to 1.
Let me know if you have any question..
Recommended
Discuss the seven factors that analysts consider where selecting a f.pdf
Discuss the seven factors that analysts consider where selecting a file organization. Evaluate
each file organization with respect to each of the seven factors. Review Table 9-3 to review how
each of these factors can easily be summarized.
Solution
1. Frequency of update: A file that needs to be updated every now and then needs an organisation
method that will allow easy retrieval of information and ease of updating, example of such a file
is the transaction file.
2. File activity: Different files have different activities, example a sort file is used to sort data in
sequential order and therefore sequential method would be appropriate for such a file.
3. File access method: Definately different files have different methods of being accessed,
example a reference file is accessed using random method for easy retrieval of data.
4. Nature of the system: Files that are used in a particular system will depend on the nature of
the system i.e the suitable organisation method for that particular system.
5. Masterfile medium: The masterfile is the main file for keeping permanent updates of records
from transaction files and other sources, the medium by which it is updated will determine the
organisation method to be used..
Consider a 4-Link robot manipulator shown below. Use the forward kine.pdf
Consider a 4-Link robot manipulator shown below. Use the forward kinematic D-H table and
write an m file that plots the manipulator. The instructions are given in the module 6. Submit
your solutions by the due date, in a single MATLAB m file.
Solution
Please give the kinetic D-H table else it would be difficult to code as we need to know the
rotation spin axis and other momentum of manipulator
Stating a general example code for manipulator with data
function X = fwd_kin(q,x)
% given a position in the configuration space, calculate the position of
% the end effector in the workspace for a two-link manipulator.
% q: vector of joint positions
% x: design vector (link lengths)
% X: end effector position in cartesian coordinates
% configuration space coordinates:
q1 = q(1); % theta 1
q2 = q(2); % theta 2
% manipulator parameters:
l1 = x(1); % link 1 length
l2 = x(2); % link 2 length
% calculate end effector position:
X = [l1*cos(q1) + l2*cos(q1+q2)
l1*sin(q1) + l2*sin(q1+q2)];
% SimulateTwolink.m uses inverse dynamics to simulate the torque
% trajectories required for a two-link planar robotic manipulator to follow
% a prescribed trajectory. It also computes total energy consumption. This
% code is provided as supplementary material for the paper:
%
% \'Engineering System Co-Design with Limited Plant Redesign\'
% Presented at the 8th AIAA Multidisciplinary Design Optimization
% Specialist Conference, April 2012.
%
% The paper is available from:
%
% http://systemdesign.illinois.edu/publications/All12a.pdf
%
% Here both the physical system design and control system design are
% considered simultaneously. Manipulator link length and trajectory
% specification can be specified, and torque trajectory and energy
% consumption are computed based on this input. It was found that maximum
% torque and total energy consumption calculated using inverse dynamics
% agreed closely with results calculated using feedback linearization, so
% to simplify optimization problem solution an inverse dynamics approach
% was used, which reduces the control design vector to just the trajectory
% design.
%
% In the conference paper several cases are considered, each with its own
% manipulator task, manipulator design, and trajectory design. The
% specifications for each of these five cases are provided here, and can be
% explored by changing the case number variable (cn).
%
% This code was incorporated into a larger optimization project. The code
% presented here includes only the analysis portion of the code, no
% optimization.
%
% A video illustrating the motion of each of these five cases is available
% on YouTube:
%
% http://www.youtube.com/watch?v=OR7Y9-n5SjA
%
% Author: James T. Allison, Assistant Professor, University of Illinois at
% Urbana-Champaign
% Date: 4/10/12
clear;clc
% simulation parameters:
p.dt = 0.0005; % simulation step size
tf = 2; p.tf = tf; % final time
p.ploton = 0; % turn off additional plotting capabilities
p.ploton2 = 0;
p.Tallow = 210; % maximum .
Conduct some basic research on the recessive disorder Albinism.Des.pdf
Conduct some basic research on the recessive disorder Albinism.
Describe the disease and its genetic basis.
Explain how this disease is inherited and why it is not always inherited.
Why is genetic testing controversial?
Provide references for your research
Solution
The term recessive disorder Albinism isdefined a is a congenital disorder which usually happens
due to complete or partial absence of the pigment which is present in the skin, hair and eyes.
Moreover it isr elated to a number of vision defects.These include: photophobia, nystagmus, and
amblyopia. Another important aspect is the Lack of skin pigmentation can usually lead to
sunburn .
This disorder is usually passes from parents to their children through the concept of genes. The
most importaant aspect for inhertiance is that both the parents should carry an albinism gene in
order to have a child who suffers from albinism disorder.Moreover, the parents can also have
normal pigmentation but they still are able to carry the gene. Finally as both parents are able to
carry the gene, and none of the parent has albinism.This means that there is a one-in-four chance
which can happen at pregnancy that the baby will be born with albinism disorder.
Genetic testing is controversial because after determining the sex of the child ,some parents
decide to abort it.So,in order to prevent this it has become controversial in some countries like
india where it is illegal.
References:
1.https://www.aapos.org/terms/conditions/12
2..
calculate the next annual dividend payment for stock currently selli.pdf
calculate the next annual dividend payment for stock currently selling for $50 per share. with a
16% expected total return and a 10% expected constant growth rate?
Solution
Expected return=(D1/Current price)+Growth rate
0.16=(D1/50)+0.1
Hence D1=(0.16-0.1)*50
which is equal to
=$3=next annual dividend payment..
Answer the questions below based on the database model found on the n.pdf
Answer the questions below based on the database model found on the next page: 2. What titles
are available in Digital Video format? 3. What titles and years are available for the Pop-
Alternative genre? What is the result of the SQL statement below? SELECT from USIC where 4.
MEDIA FOorrat \"Analog\" s. Which is the correct SQL statement to obtain a list of all music
(titles and artists) in CD format? (select one) a. SELECT Artist, Title from MEDIA where
Mediaype CD b. SELECT Artist, Title from MUSIC where Meaanp9,“ \"CD\" c. SELECT
MUSIC where xediatse *CD Write valid SQL statements for the following actions: 6. Retrieve
all the free Titles from the MUSIC relation: 7. Remove the Cassette Tape row from the MEDIA
table: 8. Change the Year on \"Somebody 1 Used to Know to be 2012: 9. Add a new Artist row
to the ARTIST Table (add any Artist and Genre you like):
Solution
2. From media relation:
MP4 corresponds to Digital Video Format
Hence, Titles which are MP4:
High School
Eyes Wide Open
Payphone
3. Artists that are associated with Pop-alternative genre are Gotye, Coldplay, Fiest
Hence, the titles and years that correspond to these artists:
The Reminder, 2007
Open Season, 2006
Mylo Xyloto, 2011
Somebody I used to know, 2011
Eyes wide open, 2012
4.
We need to select music whose media format is analog. Hence, it will return music records with
media type vinyl and cassette tape
Hence, it will return the following records:
Born to run, Bruce Springstein, 1981, Vinyl, 15.95
Greatest Hits, Ella Flitzgerald, 1972, Vinyl, 15.95
Quadrophenia, The Who, 1973, Vinyl, 25.95
Rolling Stones Box Set, Rolling Stones, 2013, Vinyl, 50.00
5. b. SELECT Artist, Title from MUSIC where MediaType = \"CD\"
6. SELECT Title from MUSIC;
7. DELETE FROM MEDIA where MediaType=\"Cassette Tape\";
8. UPDATE MUSIC
SET Year=2012
WHERE Title=\"Somebody I used to know\";
9. INSERT INTO ARTIST
Values (\"Adam Young\", \"Pop\");.
You are reviewing basic network protocols with a new tech in your co.pdf
You are reviewing basic network protocols with a new tech in your company. Of the following,
what do you say are true about network protocols? (Choose two.)
A. NetBIOS is used primarily to connect Windows computers to the Internet.
B. TCP/IP is used exclusively to allow computers to connect to each other on the local area
network.
C. TCP/IP is the de facto communications protocol for the Internet.
D. NetBIOS is used in Ethernet networks.
Solution
C and D. TCP/IP, or Transmission Control Protocol/Internet Protocol, is the basic
communication protocol for the Internet and is also used on private networks such as intranets
(web environments within a company) or extranets (web environments between two or more
business partners or between a vendor and customers). NetBIOS, or Network Basic Input/Output
System, is a program that allows computers to communicate on local area networks, or LANs,
but does not support transport into wide area networks, or WANs, such as the Internet..
Why might a gene have more than 2 different allelesa. DNA seq.pdf
Why might a gene have more than 2 different alleles?
a. DNA sequence can deviate in many places (base-pairs)
b. DNA sequence can only deviate in one place (base-pair)
c. DNA sequence cannot deviate
d. Genes can only have 2 possible alleles
Solution
Answer:
a. DNA sequence can deviate in many places (base-pairs)
Explanation:
Multiple alleles means that there are three or more forms of a gene for a trait. That doesn\'t mean
that a person can have all three of them..
Which of the following are true statementsI. sin2+cos2=1II. sec.pdf
Which of the following are true statements?
I. sin2+cos2=1
II. sec2+1=tan2
III. 1+csc2=cot2
IV. 1csc2+1sec2=1
I am aware I have to re-learn the identies.
Solution
Given the statements:
I. True for Sin2x + Cos2x =1.
II. False, because Sec2x = 1 + tan2x
III. False, because csc2x = 1 + cot2x
iV. False, because is is different to 1.
Let me know if you have any question..
What scientific mechanism for evolution did Charles Darwin and Alfred.pdf
What scientific mechanism for evolution did Charles Darwin and Alfred Wallace propose in
1858? Describe how this concept works. What is the difference between microevolution and
macroevolution?
Solution
1. Charles Darwin during 1858 had proposed the Theory of Evolution by Natural Selection,
where he has said that organisms change over time as a result of changes in heritable physical or
behavioral traits. Changes that allow an organism to better adapt to its environment will help it
survive and have more offspring.
Alfred Wallace who is also called the Father of Biogeography, studied the geographical
distribution of animal species and said the evolution is caused due to environmental pressures on
varieties of species it forces different species to adapt themselves to the local conditions, leading
to different populations in different locations.
Both groups of scientists have said that natural selection is the single most and important factor
in evolutionary changes seen in species.
Scientifically, This concept works due to different phenomenon such as:
a)Mutation- It is the Sudden change in the genetic makeup of an organism.And it is the driving
force of evolution, which influences the population’s gene pool.The change in the genetic
makeup of organism which favors the life of species makes it the fittest to survive!
b)Genetic Drift-It can occur when a small group of individuals leaves a population and
establishes a new one in a geographically isolated region.It occurs in a large population and helps
the population to survive.
c)Natural selection-This happens when populations of organisms are subjected to the
environment. The fittest creatures are more likely to survive and pass their genes to their
offspring, producing a population that is better adapted to the environment.
2.MicroEvolution-Microevolution is defined as changes in gene frequency in a population from
one generation to the next.These small changes in species occur by recombining existing genetic
material within the group of same species.
Micro Evolution is caused due to Mutation, Migration or Gene Flow, Genetic Drift and Selection
(natural and artificial).
Macro Evolution: It refers to major evolutionary changes over time, the origin of new types of
organisms from previously existing, but different, ancestral types.It is an evolution on a larger
scale where we find the descent of many species from one common ancestor over billions of
years.It is very difficult to see the Macro Evolution happening because it takes a longer time to
happen. But Instead, we can reconstruct the history of life using all available evidence like
geology, fossils, and living organisms..
What is the main difference between doing business process analysis .pdf
What is the main difference between doing business process analysis for ERP and other types of
systems
Solution
The key differences in between the erp and other type of analysis can be explained by taking
three basic concepts and bt these we can understand that and it is
system implementation: In ERP systems it requires data conversations to estimate the analysis
which contains seperate shells so thet it will be very helpful whereas in other systems there is no
special software which is used it is designed manually and gone through the source code and not
so conversions are required.
Technological scope: In ERP systems it is homogenous environments and eith common data
structures and tranfer overflows and it is data centric mainly whereas in other systems it is
process centric and supports IT application and supports hetrogenous environments.
Domain specific: In ERP domain is specific that is static which contains international sttings
whereas in other systems it is quite opposite it does not contain domains and not having any
international settings..
What are the genetic differences between a processed pseudogene and .pdf
What are the genetic differences between a processed pseudogene and a mutated pseudogene and
how do their typical locations relative to their functional paralogous genes differ?
Solution
Pseudogenes are the genes that have lost some part or entire of its original ability, of their real
gene relative and thus they have a altered or no gene expression activity. Most commonly,
pseudogenes are formed as a result of accumulation of multiple mutations in a single gene,
whose gene product is not essential for survival. Pseudogenes may not be fully functional and
may perform regulatory functions. Thus a pseudogene has its association with a functional gene,
and both of them share a common ancestor.
Processed pseudogenes are the result of retrotransposition, i.e, a part of the mRNA or hnRNA
transcript from a gene spontaneously reverses back (transcribed again) into DNA and gets
inserted into chromosomal DNA.Thus they are processed once and then reverses back to result in
a pseudogene. As a result, random copies of genes are produced in vitro. When such
pseudogenes are inserted into the genome, they usually contain a poly-A tail,they are without
introns (introns are spliced out) and also lack the upstream promoters. Thus after transposition
event, they become non-functional pseudogenes. Such pseudogenes are randomly inserted into
the genome.
On the other hand, mutated pseudogenes arise after multiple mutation events which prevent a
gene from normal transcription and make them less functional or non-functional; like gene
duplication event caused by homologous recombination (repititive sequences on misaligned
chromosomes become non-functional) or one allele of gene having deleterious mutation resulting
in loss of function, non-sense mutations, unequal cross overs and sister chromatid exchanges etc.
In such pseudogenes, the location of genes remain same as their functional paralogous genes..
True or False_____ Cells Composed of chemicals and structures, ma.pdf
True or False?
_____ Cells Composed of chemicals and structures, make up all living things, and vary in size
and shape.
_____ Cells Work together to allow for proper functioning of processes necessary for life.
_____ Every cell in the human body contain Almost all human cells contain a nucleus,
organelles, cytoplasm and a cell membrane.
_____ The plasma membrane acts as a protective covering, maintains the cell’s integrity and
allows it to survive.
_____ Osmosis, filtration, facilitated diffusion, endocytosis and exocytosis are examples of
passive transport mechanisms.
_____ Diffusion is when a substance moves from high concentration to low concentration (down
the concentration gradient).
_____ Refering to osmotic pressure: water tends to travel across the plasma membrane from
areas that have a high concentration of a solute to a lower concentration until the concentration is
equal on both sides.
_____Cystic Fibrosis and Diabetes Mellitus are examples of passive transport disorders.
_____Ribosomes are the site for production of enzymes and other proteins needed for cell repair.
_____Mitochondria is referred to as the “power house of the cell” and provides up to 95% of the
body’s energy needs for cellular repair, movement and reproduction.
_____Lysosomes have a pH of 7 and contain powerful enzymes that clean up waste from the
cytoplasm.
_____ Mitosis is the process of cellular reproduction and the process of sorting chromosomes so
the new cell gets all the genetic material.
_____Organs are groups of specialized cells that are similar in structure and that perform
common functions.
_____Skin, lining of the mouth, bladder, lungs, and blood vessels are examples of epithelial
tissue.
_____Simple epithelium tissues are a single layer of cells.
_____A basement membrane is located directly beneath the cells of connective tissue and is a
supporting non-cellular layer.
_____Connective tissue Support the softer organs of the body against the forces of gravity.
_____Muscle tissue consists of cells that are specialized to shorten and contract resulting in
movement of some kind.
_____Skeletal tissue and cardiac tissue are examples of connective tissue.
_____Nervous tissue consists primarily of cells that are specialized for generating and
transmitting electrical impulses.
_____Serous, mucous, synovial and cutaneous are examples of tissue membranes.
_____The Integumentary system consists of skin and all its accessory components (hair, nails,
sweat and sebaceous glands).
_____Intact skin is the best protection from most infections.
_____Skin is the largest organ of the body.
_____The epidermis layer of the skin is vascular and contains nerve cells.
_____Cells of the epidermis level are constantly shedding replaced with new cells from the
stratum basale.
_____Melanocytes are specialized cells located deep in the epidermis and are responsible for
skin color.
_____Albinism is a condition in which individual has too much pigment in their skin, hair and
eyes.
___.
The picture above shows the life cycle of a moss. Mark all true stat.pdf
The picture above shows the life cycle of a moss. Mark all true statements about the structures
indicated on the picture above.
a. “B” is haploid
b. “B” is part of a sporophyte
c. “A” is haploid
d. Cells that make up “C” and “B” are genetically identical
e. “A” is part of a gametophyte
f. “A” is part of a sporophyte
g. “C” is the reproductive structure of the offspring of plant “B” (the next generation)
h. “C” is part of a sporophyte
i. “B” is part of a gametophyte
j. “C” is the reproductive structure of plant “B”
k. “C” is part of a gametophyte
l. “A” is diploid
m. “B” is diploid
n. Cells that make up “A” and “B” are genetically identical
a. “B” is haploid
b. “B” is part of a sporophyte
c. “A” is haploid
d. Cells that make up “C” and “B” are genetically identical
e. “A” is part of a gametophyte
f. “A” is part of a sporophyte
g. “C” is the reproductive structure of the offspring of plant “B” (the next generation)
h. “C” is part of a sporophyte
i. “B” is part of a gametophyte
j. “C” is the reproductive structure of plant “B”
k. “C” is part of a gametophyte
l. “A” is diploid
m. “B” is diploid
n. Cells that make up “A” and “B” are genetically identical U
Solution
Some moss plants have both male and female plants as different , but others have both male and
female parts in the same plant. At matured stage during wet conditions male plant releases sperm
cells from antheridium, which fertilises the egg cell enclosed in female organs ( archegonia ) , a
spore producing plant develops with capsule and lid. When spre capsule matured , it releases
spores . Spores will germinate during wet conditions developed into moss plant . Then again the
cycle repeats.
Diagram explanation
In the diagram on the left side labelled plant is spore producing plant with capsule .
Capsule releasing spores
Spores developed. Into separate female and male plants.( upper plant male plant , lower plant
female plant)
Sperm cells entry into the female organs ( archegonia)
Fertilisation of egg
Development of spore producing plant..
The conversion of pyruvate to acetyl-CoA. A. requires the addition o.pdf
The conversion of pyruvate to acetyl-CoA. A. requires the addition of CO2 B. occurs in the
mitochondria of the cell and is irreversible. C. occurs in he nucleus of the cell. D. is needed for
red blood cell production.
Solution
The conversion of pyruate to acetyl COA (B) occurs in the mitochondria of the cell and is
irriversible. The process is carriedout with the help of an enzyme pyruate dehydrogenase, which
is located in mitochondrial matrix. During the conversion process each pyruate molecule loses
one carbon atom with the release of CO2..
Suppose S is a set of n + 1 integers. Prove that there exist distinct.pdf
Suppose S is a set of n + 1 integers. Prove that there exist distinct a, b S such that a - b is a
multiple of n.
Solution
There n distinct remainders modulo n ie 0,1,2,...,n-1
So let there be n boxes each corresponding to one of these remainders
We need to put these n+1 integers into these n boxes
So two of them must be in the same box ie two of them give the same remainder modulo n
Let those two integers be a,b
SO,a=b modulo n
a-b=0 modulo n
ie a-b is a multiple of n
HEnce proved.
Structural features of fungi Complete the following paragraph to desc.pdf
Structural features of fungi Complete the following paragraph to describe the structural features
of fungi.
Solution
The thallus of most fungi is known as a(n) mycelium. which is divided into a network of
filaments called hyphae
These hyphae increase the surface area, maximizing the absorption of nutrients.
Some species of fungi have hyphae that are partitioned by walls of tissue called septae
Fungal cells have cell walls comprised of chitin
Similar to animals they store energy in the form of glycogen
Fungi tend to be Nonmotile, although some do contain flagella at some point in their life cycle.
Fungi reproduce sexually and asexually during their life cycles.
Cells produced during the Asexual phase are called spores. These are haploid as they contain
only one set of chromosomes.
State some successful predictions from Einsteins theory of special.pdf
State some successful predictions from Einstein\'s theory of special relativity
Solution
Special relativity implies a wide range of consequences, which have been experimentally
verified, including length contraction, time dilation, relativistic mass, mass–energy equivalence,
a universal speed limit and relativity of simultaneity.The theory is \"special\" in that it only
applies in the special case where the curvature of spacetime due to gravity is negligible. In order
to include gravity, Einstein formulated general relativity in 1915
Several experiments predating Einstein\'s 1905 paper are now interpreted as evidence for
relativity.Particle accelerators routinely accelerate and measure the properties of particles
moving at near the speed of light, where their behavior is completely consistent with relativity
theory and inconsistent with the earlier Newtonian mechanics. These machines would simply not
work if they were not engineered according to relativistic principles. In addition, a considerable
number of modern experiments have been conducted to test special relativity. Some examples
1.) According to special relativity, the properties of particles moving approximately at thespeed
of light significantly deviate from the predictions of Newtonian mechanics. For instance, the
speed of lightcannot be reached by massive particles.
Today, those relativistic expressions for particles close to the speed of light are routinely
confirmed inundergraduate laboratories, and necessary in the design and theoretical evaluation of
collision experiments inparticle accelerators.
2.) The Ives–Stilwell experiment tested the contribution of relativistic time dilation to the
Doppler shift of light. The result was in agreement with the formula for the transverse Doppler
effect, and was the first direct, quantitative confirmation of the time dilation factor.
3.) Time dilation of moving particles as predicted by special relativity can be measured in
particle lifetime experiments. According to special relativity, the rate of clock C traveling
between two synchronized laboratory clocks A and B is slowed with respect to the laboratory
clock rates.
4.) The Kennedy–Thorndike experiment, first conducted in 1932, is a modified form of the
Michelson–Morley experimental procedure, testing special relativity.The modification is to make
one arm of the classical Michelson–Morley (MM) apparatus shorter than the other one. While
the Michelson–Morley experiment showed that the speed of light is independent of the
orientation of the apparatus, the Kennedy–Thorndike experiment showed that it is also
independent of the velocity of the apparatus in different inertial frames
5.) Hughes–Drever experiments (also clock comparison-, clock anisotropy-, mass isotropy-, or
energy isotropy experiments) are spectroscopic tests of the isotropy of mass and space.Unlike
Michelson–Morley type experiments, Hughes–Drever experiments test the isotropy of the
interactions of matter itself, that is, of .
QUESTION 7 Forward and futures markets provide insurance or hedging a.pdf
QUESTION 7 Forward and futures markets provide insurance or hedging against different types
of risk. True O False
Solution
False, As Forward and future market both gives hedgeing mainly against price fluctuation risk or
interest rate risk but still are risk associated to them, Like in Forward contract we have default
risk.
Problem 14-1A On January 1, 2017, Geffrey Corporation had the followi.pdf
Problem 14-1A On January 1, 2017, Geffrey Corporation had the following stockholders\'
equity accounts. Common Stock ($20 par value, 50,000 shares issued and outstanding) Paid-in
Capital in Excess of Par-Common Stock Retained Earnings $1,000,000 190,000 550,000 During
the year, the following transactions occurred. Feb. 1 Declared a $1 cash dividend per share to
stockholders of record on February 15, payable March 1. Mar. 1 Paid the dividend declared in
February. Apr. 1 Announced a 2-for-1 stock split. Prior to the split, the market price per share
was $35 July 1 Declared a 10% stock dividend to stockholders of record on July 15 distributable
uly 31 On uly 1, the market price of the stock was SA er share. 31 Dec. 1 31 Issued the shares for
the stock dividend. Declared a $0.30 per share dividend to stockholders of record on December
15, payable January 5, 2018. Determined that net income for the year was $300,000
Solution
a) Journal Entries Date Accounts and Titles Debit Credit Feb.1 Retained Earnings (50,000
X $1) $50,000.00 Dividends Payable $50,000.00 Mar.1 Dividends Payable $50,000.00
Cash $50,000.00 Apr.1 No Entry Two-for-one stock split increases number of shares to
100,000 = (50,000 X 2) and reduces par value to $10 per share Jul.1 Retained Earnings
(100,000 x 10% x $14) $140,000.00 Common Stock Dividends Distributable (10,000 X $10)
$100,000.00 Paid-in Capital in Excess of Par Value (10,000 X $4) $40,000.00 Jul.31
Common Stock Dividends Distributable $100,000.00 Common Stock $100,000.00 Dec.
1 Retained Earnings (110,000 X $.30) $33,000.00 Dividends Payable $33,000.00 Dec.
31 Income Summary $300,000.00 Retained Earnings $300,000.00 b) Geffrey
Corporation Balance Sheet (Partial) December 31, 2017 Stockholders’ equity Paid-
in capital Capital stock Common stock, $10 par value, 110,000 shares issued and
outstanding $1,100,000.00 Additional paid-in capital In excess of par value ($190,000 +
$40,000) $230,000.00 Total paid-in capital $1,330,000.00 Retained earnings ($300,000 +
33000 + 140000 + 50000) $523,000.00 Total stockholders’ equity $1,853,000.00.
on the way to the cerebrum 15) Almost all somatic sensory signals pas.pdf
on the way to the cerebrum 15) Almost all somatic sensory signals pass through the A)
cerebellum B) thalamus C) hypothalamus D) corpus callosum E) corticospinal tracts
Solution
A large number of pathways travel through the thalamus before reaching the cortex. All sensory
signals except for olfaction also pass through the thalamus first. Why so?
Because the thalamus is considered as the relay station. The thalamus is made up of a number of
nuclei that possess functional specialization for dealing with particular type of information. This
means that all the sensory information is first travelled to the thalamus which is then routed to a
nucleus tailored to dealing with that type of sensory data. The information is then sent from the
nucleus to the appropriate area in the cortex where it is further processed. In this way, the nuceli
of the thalamus make sure to transfer the signal to the area which is specialized to handle that
information for further processing..
Please recreate a simple game of flappy birds using two classes and .pdf
Please recreate a simple game of flappy birds using two classes and incorporate images and ez
java
The project must use EZ and incorporate at least 2 of the following concepts:
Array Lists
Finite State Machines
File reading / parsing
File writing
2D arrays
Inheritance
Private, public member variables and member functions
Solution
Java code for flappy bird are below:
/* * Added frameRate(30) in because the program apparently started going 60fps when it was
configured to run at 30
* */
frameRate(30);
// variables for the width and space in the middle of pipes
var pipeW = 40;
var space = 100;
var endW = pipeW + 10;
var endH = 20;
// space between pipes
var pipeSpace = width / (4 - 1);
// max and min height of pipes
var pipeMaxY = 250;
var pipeMinY = 100;
// speed the pipe\'s move left
var pipeSpeed = 4;
// variables for bird\'s location and movement
var birdX = 100; // x is constant
var birdY = 200;
var birdR = 10;
var birdDY = 0;
// var birdDX = 0;
// PHYSICS
var grav = 1.5;
var jump = 0.15;
var groundY = 350;
var pipeStartLocation = 500;
var pipes = [[pipeStartLocation, random(pipeMinY, pipeMaxY)],
[pipeStartLocation + pipeSpace, random(pipeMinY, pipeMaxY)],
[pipeStartLocation + pipeSpace*2, random(pipeMinY, pipeMaxY)],
[pipeStartLocation + pipeSpace*3, random(pipeMinY, pipeMaxY)]];
noStroke();
var inGame = false;
var inStart = true;
var inGameOver = false;
var score = 0;
var gotFirstPoint = false;
var drawBird = function() {
pushMatrix();
translate(birdX, birdY);
// angle the bird
rotate(-40);
rotate(max(-0, min(120, 20*birdDY)));
// body
stroke(0, 0, 0);
strokeWeight(1);
fill(255, 221, 0);
ellipse(0, 0, birdR*2 - 2, birdR*1.6 - 2);
// belly
fill(255, 170, 0);
arc(0, 0, birdR*2 - 2, birdR*1.6 - 2, 30, 150);
// wing
fill(254, 255, 178);
pushMatrix();
rotate(sin(frameCount * 80) * 15); // flap rotation
ellipse(-birdR*4/5, 0, birdR * 0.9, birdR * 0.6);
popMatrix();
// eye
fill(255, 255, 255);
pushMatrix();
rotate(22);
ellipse(birdR/2, -birdR/3, birdR, birdR * 0.8);
popMatrix();
fill(0, 0, 0);
ellipse(birdR*0.9, -birdR/20, birdR/5, birdR / 3);
// lips
fill(255, 0, 72);
ellipse(birdR/2, birdR*3/5, birdR*0.8, birdR / 4);
ellipse(birdR*0.6, birdR*2/5, birdR, birdR / 4);
noStroke();
popMatrix();
};
var drawGround = function() {
// dirt
fill(214, 189, 145);
rect(0, groundY, 400, 400 - groundY);
// dark green grass
fill(68, 143, 62);
rect(0, groundY, 400, 5);
stroke(82, 179, 85);
// diagonal grass
strokeWeight(3.0);
for(var x = 0; x < 440; x += 10) {
pushMatrix();
translate(x, groundY + 5);
if(!inGameOver) {
translate(-(pipeSpeed*frameCount % 10), 0);
}
rotate(225);
line(0, 2, 0, 5);
popMatrix();
}
noStroke();
};
var drawPipe = function(x, y) {
strokeWeight(4.0);
// pipe body
fill(65, 186, 79);
stroke(0, 0, 0, 180);
rect(x - pipeW / 2, 0, pipeW, y - space / 2); // top
rect(x - pipeW / 2, y + space / 2, pipeW, 400 - y - space / 2 + 10); // bottom
noStroke();
// pipe stripes
colorMode(HSB);
var bodyColW = 5;
for(var c = x - pipeW / 2; c < x + pipeW / 2; .
My C proggram is having trouble in the switch in main. Also the a co.pdf
My C proggram is having trouble in the switch in main. Also the a couple of funtion like delete
and display tree are failing
#include
#include
#include
#include
#include
#include
struct tree_node {
int data;
struct tree_node *left, *right;
};
typedef struct nodeT {
int key;
struct nodeT *left, *right;
int data;
} nodeT, *treeT;
//struct nodeT *t1, *t2;
void DisplayTree(nodeT*t)
{
if (t != NULL) {
DisplayTree(t->left);
printf(\"%c \", t->key);
DisplayTree(t->right);
}
}
nodeT *FindNode(nodeT *t, int key){
if(t==NULL) return 0;
if (key == t->key)
return t;
if(key < t->key){
return FindNode(t->left, key);;
} else {
return FindNode(t->right, key);;
}
}
int max(nodeT *p){
nodeT *t, *tmp;
while(tmp->right != NULL)
tmp=tmp->right;
return tmp->data;
}
int min(nodeT *t){
nodeT *tmp;
while(tmp->left != NULL)
tmp = tmp->left;
return tmp->data;
}
void ModifyPreOrderWalk(nodeT *t, int h)
{
int i;
if (t == NULL) return;
for(i=0; i<-1;i++){
printf(\"| \");
}
if(h>1)printf(\"+---\");
printf(\"%d\ “\", t->key);
ModifyPreOrderWalk(t->left, h+1);
ModifyPreOrderWalk(t->right, h+1);
}
void PrintTree(nodeT *t)
{
ModifyPreOrderWalk(t, 1);
}
void delete(nodeT **p){
nodeT *target, *imd_r, *plmd_r;
target=*p;
if (target->left==NULL && target->right==NULL) {
*p=NULL;
} else if (target->left == NULL) {
*p=target->right;
} else
if (target->right == NULL) {
*p=target->left;
} else {
plmd_r = target;
imd_r = target->right;
while( imd_r->left != NULL){
plmd_r = imd_r;
imd_r = imd_r->left;
}
if(plmd_r == target)
plmd_r->right = imd_r->right;
else
plmd_r->left = imd_r->right;
imd_r->left = target->left;
imd_r->right = target->right;
*p = imd_r;
}
free(target);
}
void listInorder(nodeT *t){
if (t != NULL) {
DisplayTree(t->left);
printf(\"%d \", t->key);
DisplayTree(t->right);
}
}
void listPreorder(nodeT *t) {
if (t != NULL) {
printf(\"%d \", t->key);
DisplayTree(t->left);
DisplayTree(t->right);
}
}
void listPostOrder(nodeT *t){
if (t != NULL) {
DisplayTree(t->left);
DisplayTree(t->right);
printf(\"%d\", t->key);
}
}
void InsertNode(nodeT **tptr, int key){
nodeT *t;
nodeT *tmp;
t=*tptr;
if (t == NULL) {
tmp= malloc(sizeof(nodeT));
tmp->key = key;
tmp->left=tmp->right=NULL;
*tptr=tmp;
return;
}
if (key < t->key) {
InsertNode
(&t->left, key);
} else {
InsertNode(&t->right, key);
}
}
int height(nodeT *t){
int lDepth = height(t->left);
int rDepth = height(t->right);
if(lDepth > rDepth)
return(lDepth+1);
else
return(rDepth+1);
}
int sum(nodeT *p){
if (p == NULL)
return 0;
else
return (p->data + sum(p->left) +sum(p->right) );
}
int count(nodeT *t){
if(t == NULL)
return 0;
int i = 1+count(t->left)+count(t->right);
return i;
}
/*nodeT *New(){
nodeT *tmp;
tmp = (nodeT *)malloc(sizeof(nodeT));
if (tmp==NULL) return NULL;
return tmp;
}*/
int main(){
nodeT *node;
nodeT *t=NULL, *findA;
int f;
int d, a, b, c, j, k, e, s, y;
while (1) {
printf(\"\ I - insert\ F - find node\ D - delete\ L - list in order\ R - list preorder\ P - list
PorstOrder\ M - max\ N - min\ H - height\ C - count\ S - sum.
Need help on creating code using cart. The output has to show multip.pdf
Need help on creating code using cart. The output has to show multiple lines of the users
information.
So far this is what I have:
index.html:
chapter8.jsp
<%@page contentType=\"text/html\" pageEncoding=\"UTF-8\"%>
JSP Page
<%@ page import=\"java.io.*, business.*, data.*, list.*\" %>
<%
// get parameters from the request
String firstName = request.getParameter(\"firstName\");
String lastName = request.getParameter(\"lastName\");
String email = request.getParameter(\"email\");
String phoneNumber = request.getParameter(\"phoneNumber\");
// get a relative file name
// ServletContext sc = this.getServletContext();
// String path = sc.getRealPath(\"/WEB-INF/EmailList.txt\");
File tempdir = (File) this.getServletContext().getAttribute(\"javax.servlet.context.tempdir\");
String temppath = tempdir.toString();
String path = temppath + \"/EmailList.txt\";
// use regular Java objects
User1 user = new User1(firstName, lastName, email, phoneNumber);
UserIO.add(user, path);
%>
Thanks for joining our email list
Here is the information that you entered:First NameLast NameEmailPhone Number<%=
user.getFirstName() %><%= user.getLastName() %><%= user.getEmailAddress()
%><%=user.getPhoneNumber() %>
User1.java
package business;
public class User1
{
private String firstName;
private String lastName;
private String email;
private String phoneNumber;
public User1()
{
firstName = \"\";
lastName = \"\";
email = \"\";
phoneNumber = \"\";
}
public User1(String firstName, String lastName, String email, String phoneNumber)
{
this.firstName = firstName;
this.lastName = lastName;
this.email = email;
this.phoneNumber = phoneNumber;
}
public void setFirstName(String firstName)
{
this.firstName = firstName;
}
public String getFirstName()
{
return firstName;
}
public void setLastName(String lastName)
{
this.lastName = lastName;
}
public String getLastName()
{
return lastName;
}
public void setEmailAddress(String email)
{
this.email = email;
}
public String getEmailAddress()
{
return email;
}
public void setPhoneNumber(String phoneNumber)
{
this.phoneNumber = phoneNumber;
}
public String getPhoneNumber()
{
return phoneNumber;
}
}
UserIO.java
package data;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.*;
import business.User1;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class UserIO extends HttpServlet {
public static void add(User1 user, String filepath) throws IOException
{
File file = new File(filepath);
PrintWriter out = new PrintWriter(
new FileWriter(file, true));
out.println(user.getEmailAddress()+ \"|\"
+ user.getFirstName() + \"|\"
+ user.getLastName() + \"|\"
+ user.getPhoneNumber());
out.close();
}
}
*NEED HELP!*
cart.java
package list;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
.
Name key elements of Object-Oriented systems. Note .pdf
Name key elements of Object-Oriented systems.
Note : no plagiarism please
Name key elements of Object-Oriented systems.
Name key elements of Object-Oriented systems.
Name key elements of Object-Oriented systems.
Note : no plagiarism please
Solution
there are 3 basic elements of object oriented system.they are:-
1.inheritance 2.polymorphism 3.encapsulation
Inheritance:-
inheritance is a concept where a class or an object can get the properties of its parent or sibling
class,for example let us consider A is a parent class and it has a method called read_me().and B
is another class inherited from class A.so using inheritance B class also has access rigths or can
explicitly use the read_me() method from A.
polymorphism:- (poly-many) (morph-forms)
polymorphism is an ability of an object to take on many forms.in general sense polymorphism
means a parent class refering the child class object
encapsulation:-
encapsulation in oop is nothing but binding or grouping of methods and member function in a
class.
Match the item on the left with the meaning on the right. Just show .pdf
Match the item on the left with the meaning on the right. Just show
Solution
17) A (makes cout print floats with a fixed number of decimals)
18) Flag to left justify output, for right justify ios::right
19) C
20) F Used to format floating point numbers
21) B Sets the width in order to insert next value.
22) E.
Let f be holomorphic on and inside a Jordan curve C and let f have m.pdf
Let f be holomorphic on and inside a Jordan curve C and let f have maximum value M on C
(such a number exists by the Extreme Value Theorem). Suppose that |f(z0)| = M for some z0
ins(C). Show that f is constant over ins(C).
Solution
By maximum modulus principle, if |f| attains the maximum value at the interior , it should be the
constant function. Hence the result.
Linear algebra. Do both part i and part ii. I have found part i and .pdf
Linear algebra. Do both part i and part ii. I have found part i and wanted to check my work. Part
ii im severly confused about. 3. Let a= | 3 (i) Compute the perpendicular projection of the vector
b1on the line in the direction 2 of a. (ii) Write a formula for the distance of the point Xa to the
point b. Show that this distance attains the minimum value when a_ p is the perpendicular
projection found in (i).
Solution
Projection of b in the direction of a=(a dot b)
=(1 3 -2) dot (0 1 2)
=(1*0)+(3*1)+(-2*2)
=0+3-4
=-1
a dot a=(1^2)+(3^2)+(-2^2)
=1+9+4
=14
now [(a dot b)/a dot a]=-1/14
now {[(a dot b)/a dot a]}b=(-1/14)(0, 1, 2)
=(0,-1/14,-2/14).
Let A = {Leah, Dora, Mitch, Andy} which of the following are corr.pdf
Let A = {Leah, Dora, Mitch, Andy} which of the following are correct statements? D A Leah
A A Mitch D A Mitch A
Solution
A={Leah, Dora, Mitch, Andy}
Leah is belongs to A
So correct statements is second statement ( Leah belongs to A ) and fourth statement ( D not
belongs to A )
Remaining are in correct because D not in A ( first is wrong ), Mitch in A ( fifth is wrong) and A
is a super set, so third is wrong..
1.4 modern child centered education - mahatma gandhi-2.pptx
Child centred education is an educational approach that priorities the interest, needs and abilities of the child in the learning process.
Home assignment II on Spectroscopy 2024 Answers.pdf
Answers to Home assignment on UV-Visible spectroscopy: Calculation of wavelength of UV-Visible absorption
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What scientific mechanism for evolution did Charles Darwin and Alfred.pdf
What scientific mechanism for evolution did Charles Darwin and Alfred Wallace propose in
1858? Describe how this concept works. What is the difference between microevolution and
macroevolution?
Solution
1. Charles Darwin during 1858 had proposed the Theory of Evolution by Natural Selection,
where he has said that organisms change over time as a result of changes in heritable physical or
behavioral traits. Changes that allow an organism to better adapt to its environment will help it
survive and have more offspring.
Alfred Wallace who is also called the Father of Biogeography, studied the geographical
distribution of animal species and said the evolution is caused due to environmental pressures on
varieties of species it forces different species to adapt themselves to the local conditions, leading
to different populations in different locations.
Both groups of scientists have said that natural selection is the single most and important factor
in evolutionary changes seen in species.
Scientifically, This concept works due to different phenomenon such as:
a)Mutation- It is the Sudden change in the genetic makeup of an organism.And it is the driving
force of evolution, which influences the population’s gene pool.The change in the genetic
makeup of organism which favors the life of species makes it the fittest to survive!
b)Genetic Drift-It can occur when a small group of individuals leaves a population and
establishes a new one in a geographically isolated region.It occurs in a large population and helps
the population to survive.
c)Natural selection-This happens when populations of organisms are subjected to the
environment. The fittest creatures are more likely to survive and pass their genes to their
offspring, producing a population that is better adapted to the environment.
2.MicroEvolution-Microevolution is defined as changes in gene frequency in a population from
one generation to the next.These small changes in species occur by recombining existing genetic
material within the group of same species.
Micro Evolution is caused due to Mutation, Migration or Gene Flow, Genetic Drift and Selection
(natural and artificial).
Macro Evolution: It refers to major evolutionary changes over time, the origin of new types of
organisms from previously existing, but different, ancestral types.It is an evolution on a larger
scale where we find the descent of many species from one common ancestor over billions of
years.It is very difficult to see the Macro Evolution happening because it takes a longer time to
happen. But Instead, we can reconstruct the history of life using all available evidence like
geology, fossils, and living organisms..
What is the main difference between doing business process analysis .pdf
What is the main difference between doing business process analysis for ERP and other types of
systems
Solution
The key differences in between the erp and other type of analysis can be explained by taking
three basic concepts and bt these we can understand that and it is
system implementation: In ERP systems it requires data conversations to estimate the analysis
which contains seperate shells so thet it will be very helpful whereas in other systems there is no
special software which is used it is designed manually and gone through the source code and not
so conversions are required.
Technological scope: In ERP systems it is homogenous environments and eith common data
structures and tranfer overflows and it is data centric mainly whereas in other systems it is
process centric and supports IT application and supports hetrogenous environments.
Domain specific: In ERP domain is specific that is static which contains international sttings
whereas in other systems it is quite opposite it does not contain domains and not having any
international settings..
What are the genetic differences between a processed pseudogene and .pdf
What are the genetic differences between a processed pseudogene and a mutated pseudogene and
how do their typical locations relative to their functional paralogous genes differ?
Solution
Pseudogenes are the genes that have lost some part or entire of its original ability, of their real
gene relative and thus they have a altered or no gene expression activity. Most commonly,
pseudogenes are formed as a result of accumulation of multiple mutations in a single gene,
whose gene product is not essential for survival. Pseudogenes may not be fully functional and
may perform regulatory functions. Thus a pseudogene has its association with a functional gene,
and both of them share a common ancestor.
Processed pseudogenes are the result of retrotransposition, i.e, a part of the mRNA or hnRNA
transcript from a gene spontaneously reverses back (transcribed again) into DNA and gets
inserted into chromosomal DNA.Thus they are processed once and then reverses back to result in
a pseudogene. As a result, random copies of genes are produced in vitro. When such
pseudogenes are inserted into the genome, they usually contain a poly-A tail,they are without
introns (introns are spliced out) and also lack the upstream promoters. Thus after transposition
event, they become non-functional pseudogenes. Such pseudogenes are randomly inserted into
the genome.
On the other hand, mutated pseudogenes arise after multiple mutation events which prevent a
gene from normal transcription and make them less functional or non-functional; like gene
duplication event caused by homologous recombination (repititive sequences on misaligned
chromosomes become non-functional) or one allele of gene having deleterious mutation resulting
in loss of function, non-sense mutations, unequal cross overs and sister chromatid exchanges etc.
In such pseudogenes, the location of genes remain same as their functional paralogous genes..
True or False_____ Cells Composed of chemicals and structures, ma.pdf
True or False?
_____ Cells Composed of chemicals and structures, make up all living things, and vary in size
and shape.
_____ Cells Work together to allow for proper functioning of processes necessary for life.
_____ Every cell in the human body contain Almost all human cells contain a nucleus,
organelles, cytoplasm and a cell membrane.
_____ The plasma membrane acts as a protective covering, maintains the cell’s integrity and
allows it to survive.
_____ Osmosis, filtration, facilitated diffusion, endocytosis and exocytosis are examples of
passive transport mechanisms.
_____ Diffusion is when a substance moves from high concentration to low concentration (down
the concentration gradient).
_____ Refering to osmotic pressure: water tends to travel across the plasma membrane from
areas that have a high concentration of a solute to a lower concentration until the concentration is
equal on both sides.
_____Cystic Fibrosis and Diabetes Mellitus are examples of passive transport disorders.
_____Ribosomes are the site for production of enzymes and other proteins needed for cell repair.
_____Mitochondria is referred to as the “power house of the cell” and provides up to 95% of the
body’s energy needs for cellular repair, movement and reproduction.
_____Lysosomes have a pH of 7 and contain powerful enzymes that clean up waste from the
cytoplasm.
_____ Mitosis is the process of cellular reproduction and the process of sorting chromosomes so
the new cell gets all the genetic material.
_____Organs are groups of specialized cells that are similar in structure and that perform
common functions.
_____Skin, lining of the mouth, bladder, lungs, and blood vessels are examples of epithelial
tissue.
_____Simple epithelium tissues are a single layer of cells.
_____A basement membrane is located directly beneath the cells of connective tissue and is a
supporting non-cellular layer.
_____Connective tissue Support the softer organs of the body against the forces of gravity.
_____Muscle tissue consists of cells that are specialized to shorten and contract resulting in
movement of some kind.
_____Skeletal tissue and cardiac tissue are examples of connective tissue.
_____Nervous tissue consists primarily of cells that are specialized for generating and
transmitting electrical impulses.
_____Serous, mucous, synovial and cutaneous are examples of tissue membranes.
_____The Integumentary system consists of skin and all its accessory components (hair, nails,
sweat and sebaceous glands).
_____Intact skin is the best protection from most infections.
_____Skin is the largest organ of the body.
_____The epidermis layer of the skin is vascular and contains nerve cells.
_____Cells of the epidermis level are constantly shedding replaced with new cells from the
stratum basale.
_____Melanocytes are specialized cells located deep in the epidermis and are responsible for
skin color.
_____Albinism is a condition in which individual has too much pigment in their skin, hair and
eyes.
___.
The picture above shows the life cycle of a moss. Mark all true stat.pdf
The picture above shows the life cycle of a moss. Mark all true statements about the structures
indicated on the picture above.
a. “B” is haploid
b. “B” is part of a sporophyte
c. “A” is haploid
d. Cells that make up “C” and “B” are genetically identical
e. “A” is part of a gametophyte
f. “A” is part of a sporophyte
g. “C” is the reproductive structure of the offspring of plant “B” (the next generation)
h. “C” is part of a sporophyte
i. “B” is part of a gametophyte
j. “C” is the reproductive structure of plant “B”
k. “C” is part of a gametophyte
l. “A” is diploid
m. “B” is diploid
n. Cells that make up “A” and “B” are genetically identical
a. “B” is haploid
b. “B” is part of a sporophyte
c. “A” is haploid
d. Cells that make up “C” and “B” are genetically identical
e. “A” is part of a gametophyte
f. “A” is part of a sporophyte
g. “C” is the reproductive structure of the offspring of plant “B” (the next generation)
h. “C” is part of a sporophyte
i. “B” is part of a gametophyte
j. “C” is the reproductive structure of plant “B”
k. “C” is part of a gametophyte
l. “A” is diploid
m. “B” is diploid
n. Cells that make up “A” and “B” are genetically identical U
Solution
Some moss plants have both male and female plants as different , but others have both male and
female parts in the same plant. At matured stage during wet conditions male plant releases sperm
cells from antheridium, which fertilises the egg cell enclosed in female organs ( archegonia ) , a
spore producing plant develops with capsule and lid. When spre capsule matured , it releases
spores . Spores will germinate during wet conditions developed into moss plant . Then again the
cycle repeats.
Diagram explanation
In the diagram on the left side labelled plant is spore producing plant with capsule .
Capsule releasing spores
Spores developed. Into separate female and male plants.( upper plant male plant , lower plant
female plant)
Sperm cells entry into the female organs ( archegonia)
Fertilisation of egg
Development of spore producing plant..
The conversion of pyruvate to acetyl-CoA. A. requires the addition o.pdf
The conversion of pyruvate to acetyl-CoA. A. requires the addition of CO2 B. occurs in the
mitochondria of the cell and is irreversible. C. occurs in he nucleus of the cell. D. is needed for
red blood cell production.
Solution
The conversion of pyruate to acetyl COA (B) occurs in the mitochondria of the cell and is
irriversible. The process is carriedout with the help of an enzyme pyruate dehydrogenase, which
is located in mitochondrial matrix. During the conversion process each pyruate molecule loses
one carbon atom with the release of CO2..
Suppose S is a set of n + 1 integers. Prove that there exist distinct.pdf
Suppose S is a set of n + 1 integers. Prove that there exist distinct a, b S such that a - b is a
multiple of n.
Solution
There n distinct remainders modulo n ie 0,1,2,...,n-1
So let there be n boxes each corresponding to one of these remainders
We need to put these n+1 integers into these n boxes
So two of them must be in the same box ie two of them give the same remainder modulo n
Let those two integers be a,b
SO,a=b modulo n
a-b=0 modulo n
ie a-b is a multiple of n
HEnce proved.
Structural features of fungi Complete the following paragraph to desc.pdf
Structural features of fungi Complete the following paragraph to describe the structural features
of fungi.
Solution
The thallus of most fungi is known as a(n) mycelium. which is divided into a network of
filaments called hyphae
These hyphae increase the surface area, maximizing the absorption of nutrients.
Some species of fungi have hyphae that are partitioned by walls of tissue called septae
Fungal cells have cell walls comprised of chitin
Similar to animals they store energy in the form of glycogen
Fungi tend to be Nonmotile, although some do contain flagella at some point in their life cycle.
Fungi reproduce sexually and asexually during their life cycles.
Cells produced during the Asexual phase are called spores. These are haploid as they contain
only one set of chromosomes.
State some successful predictions from Einsteins theory of special.pdf
State some successful predictions from Einstein\'s theory of special relativity
Solution
Special relativity implies a wide range of consequences, which have been experimentally
verified, including length contraction, time dilation, relativistic mass, mass–energy equivalence,
a universal speed limit and relativity of simultaneity.The theory is \"special\" in that it only
applies in the special case where the curvature of spacetime due to gravity is negligible. In order
to include gravity, Einstein formulated general relativity in 1915
Several experiments predating Einstein\'s 1905 paper are now interpreted as evidence for
relativity.Particle accelerators routinely accelerate and measure the properties of particles
moving at near the speed of light, where their behavior is completely consistent with relativity
theory and inconsistent with the earlier Newtonian mechanics. These machines would simply not
work if they were not engineered according to relativistic principles. In addition, a considerable
number of modern experiments have been conducted to test special relativity. Some examples
1.) According to special relativity, the properties of particles moving approximately at thespeed
of light significantly deviate from the predictions of Newtonian mechanics. For instance, the
speed of lightcannot be reached by massive particles.
Today, those relativistic expressions for particles close to the speed of light are routinely
confirmed inundergraduate laboratories, and necessary in the design and theoretical evaluation of
collision experiments inparticle accelerators.
2.) The Ives–Stilwell experiment tested the contribution of relativistic time dilation to the
Doppler shift of light. The result was in agreement with the formula for the transverse Doppler
effect, and was the first direct, quantitative confirmation of the time dilation factor.
3.) Time dilation of moving particles as predicted by special relativity can be measured in
particle lifetime experiments. According to special relativity, the rate of clock C traveling
between two synchronized laboratory clocks A and B is slowed with respect to the laboratory
clock rates.
4.) The Kennedy–Thorndike experiment, first conducted in 1932, is a modified form of the
Michelson–Morley experimental procedure, testing special relativity.The modification is to make
one arm of the classical Michelson–Morley (MM) apparatus shorter than the other one. While
the Michelson–Morley experiment showed that the speed of light is independent of the
orientation of the apparatus, the Kennedy–Thorndike experiment showed that it is also
independent of the velocity of the apparatus in different inertial frames
5.) Hughes–Drever experiments (also clock comparison-, clock anisotropy-, mass isotropy-, or
energy isotropy experiments) are spectroscopic tests of the isotropy of mass and space.Unlike
Michelson–Morley type experiments, Hughes–Drever experiments test the isotropy of the
interactions of matter itself, that is, of .
QUESTION 7 Forward and futures markets provide insurance or hedging a.pdf
QUESTION 7 Forward and futures markets provide insurance or hedging against different types
of risk. True O False
Solution
False, As Forward and future market both gives hedgeing mainly against price fluctuation risk or
interest rate risk but still are risk associated to them, Like in Forward contract we have default
risk.
Problem 14-1A On January 1, 2017, Geffrey Corporation had the followi.pdf
Problem 14-1A On January 1, 2017, Geffrey Corporation had the following stockholders\'
equity accounts. Common Stock ($20 par value, 50,000 shares issued and outstanding) Paid-in
Capital in Excess of Par-Common Stock Retained Earnings $1,000,000 190,000 550,000 During
the year, the following transactions occurred. Feb. 1 Declared a $1 cash dividend per share to
stockholders of record on February 15, payable March 1. Mar. 1 Paid the dividend declared in
February. Apr. 1 Announced a 2-for-1 stock split. Prior to the split, the market price per share
was $35 July 1 Declared a 10% stock dividend to stockholders of record on July 15 distributable
uly 31 On uly 1, the market price of the stock was SA er share. 31 Dec. 1 31 Issued the shares for
the stock dividend. Declared a $0.30 per share dividend to stockholders of record on December
15, payable January 5, 2018. Determined that net income for the year was $300,000
Solution
a) Journal Entries Date Accounts and Titles Debit Credit Feb.1 Retained Earnings (50,000
X $1) $50,000.00 Dividends Payable $50,000.00 Mar.1 Dividends Payable $50,000.00
Cash $50,000.00 Apr.1 No Entry Two-for-one stock split increases number of shares to
100,000 = (50,000 X 2) and reduces par value to $10 per share Jul.1 Retained Earnings
(100,000 x 10% x $14) $140,000.00 Common Stock Dividends Distributable (10,000 X $10)
$100,000.00 Paid-in Capital in Excess of Par Value (10,000 X $4) $40,000.00 Jul.31
Common Stock Dividends Distributable $100,000.00 Common Stock $100,000.00 Dec.
1 Retained Earnings (110,000 X $.30) $33,000.00 Dividends Payable $33,000.00 Dec.
31 Income Summary $300,000.00 Retained Earnings $300,000.00 b) Geffrey
Corporation Balance Sheet (Partial) December 31, 2017 Stockholders’ equity Paid-
in capital Capital stock Common stock, $10 par value, 110,000 shares issued and
outstanding $1,100,000.00 Additional paid-in capital In excess of par value ($190,000 +
$40,000) $230,000.00 Total paid-in capital $1,330,000.00 Retained earnings ($300,000 +
33000 + 140000 + 50000) $523,000.00 Total stockholders’ equity $1,853,000.00.
on the way to the cerebrum 15) Almost all somatic sensory signals pas.pdf
on the way to the cerebrum 15) Almost all somatic sensory signals pass through the A)
cerebellum B) thalamus C) hypothalamus D) corpus callosum E) corticospinal tracts
Solution
A large number of pathways travel through the thalamus before reaching the cortex. All sensory
signals except for olfaction also pass through the thalamus first. Why so?
Because the thalamus is considered as the relay station. The thalamus is made up of a number of
nuclei that possess functional specialization for dealing with particular type of information. This
means that all the sensory information is first travelled to the thalamus which is then routed to a
nucleus tailored to dealing with that type of sensory data. The information is then sent from the
nucleus to the appropriate area in the cortex where it is further processed. In this way, the nuceli
of the thalamus make sure to transfer the signal to the area which is specialized to handle that
information for further processing..
Please recreate a simple game of flappy birds using two classes and .pdf
Please recreate a simple game of flappy birds using two classes and incorporate images and ez
java
The project must use EZ and incorporate at least 2 of the following concepts:
Array Lists
Finite State Machines
File reading / parsing
File writing
2D arrays
Inheritance
Private, public member variables and member functions
Solution
Java code for flappy bird are below:
/* * Added frameRate(30) in because the program apparently started going 60fps when it was
configured to run at 30
* */
frameRate(30);
// variables for the width and space in the middle of pipes
var pipeW = 40;
var space = 100;
var endW = pipeW + 10;
var endH = 20;
// space between pipes
var pipeSpace = width / (4 - 1);
// max and min height of pipes
var pipeMaxY = 250;
var pipeMinY = 100;
// speed the pipe\'s move left
var pipeSpeed = 4;
// variables for bird\'s location and movement
var birdX = 100; // x is constant
var birdY = 200;
var birdR = 10;
var birdDY = 0;
// var birdDX = 0;
// PHYSICS
var grav = 1.5;
var jump = 0.15;
var groundY = 350;
var pipeStartLocation = 500;
var pipes = [[pipeStartLocation, random(pipeMinY, pipeMaxY)],
[pipeStartLocation + pipeSpace, random(pipeMinY, pipeMaxY)],
[pipeStartLocation + pipeSpace*2, random(pipeMinY, pipeMaxY)],
[pipeStartLocation + pipeSpace*3, random(pipeMinY, pipeMaxY)]];
noStroke();
var inGame = false;
var inStart = true;
var inGameOver = false;
var score = 0;
var gotFirstPoint = false;
var drawBird = function() {
pushMatrix();
translate(birdX, birdY);
// angle the bird
rotate(-40);
rotate(max(-0, min(120, 20*birdDY)));
// body
stroke(0, 0, 0);
strokeWeight(1);
fill(255, 221, 0);
ellipse(0, 0, birdR*2 - 2, birdR*1.6 - 2);
// belly
fill(255, 170, 0);
arc(0, 0, birdR*2 - 2, birdR*1.6 - 2, 30, 150);
// wing
fill(254, 255, 178);
pushMatrix();
rotate(sin(frameCount * 80) * 15); // flap rotation
ellipse(-birdR*4/5, 0, birdR * 0.9, birdR * 0.6);
popMatrix();
// eye
fill(255, 255, 255);
pushMatrix();
rotate(22);
ellipse(birdR/2, -birdR/3, birdR, birdR * 0.8);
popMatrix();
fill(0, 0, 0);
ellipse(birdR*0.9, -birdR/20, birdR/5, birdR / 3);
// lips
fill(255, 0, 72);
ellipse(birdR/2, birdR*3/5, birdR*0.8, birdR / 4);
ellipse(birdR*0.6, birdR*2/5, birdR, birdR / 4);
noStroke();
popMatrix();
};
var drawGround = function() {
// dirt
fill(214, 189, 145);
rect(0, groundY, 400, 400 - groundY);
// dark green grass
fill(68, 143, 62);
rect(0, groundY, 400, 5);
stroke(82, 179, 85);
// diagonal grass
strokeWeight(3.0);
for(var x = 0; x < 440; x += 10) {
pushMatrix();
translate(x, groundY + 5);
if(!inGameOver) {
translate(-(pipeSpeed*frameCount % 10), 0);
}
rotate(225);
line(0, 2, 0, 5);
popMatrix();
}
noStroke();
};
var drawPipe = function(x, y) {
strokeWeight(4.0);
// pipe body
fill(65, 186, 79);
stroke(0, 0, 0, 180);
rect(x - pipeW / 2, 0, pipeW, y - space / 2); // top
rect(x - pipeW / 2, y + space / 2, pipeW, 400 - y - space / 2 + 10); // bottom
noStroke();
// pipe stripes
colorMode(HSB);
var bodyColW = 5;
for(var c = x - pipeW / 2; c < x + pipeW / 2; .
My C proggram is having trouble in the switch in main. Also the a co.pdf
My C proggram is having trouble in the switch in main. Also the a couple of funtion like delete
and display tree are failing
#include
#include
#include
#include
#include
#include
struct tree_node {
int data;
struct tree_node *left, *right;
};
typedef struct nodeT {
int key;
struct nodeT *left, *right;
int data;
} nodeT, *treeT;
//struct nodeT *t1, *t2;
void DisplayTree(nodeT*t)
{
if (t != NULL) {
DisplayTree(t->left);
printf(\"%c \", t->key);
DisplayTree(t->right);
}
}
nodeT *FindNode(nodeT *t, int key){
if(t==NULL) return 0;
if (key == t->key)
return t;
if(key < t->key){
return FindNode(t->left, key);;
} else {
return FindNode(t->right, key);;
}
}
int max(nodeT *p){
nodeT *t, *tmp;
while(tmp->right != NULL)
tmp=tmp->right;
return tmp->data;
}
int min(nodeT *t){
nodeT *tmp;
while(tmp->left != NULL)
tmp = tmp->left;
return tmp->data;
}
void ModifyPreOrderWalk(nodeT *t, int h)
{
int i;
if (t == NULL) return;
for(i=0; i<-1;i++){
printf(\"| \");
}
if(h>1)printf(\"+---\");
printf(\"%d\ “\", t->key);
ModifyPreOrderWalk(t->left, h+1);
ModifyPreOrderWalk(t->right, h+1);
}
void PrintTree(nodeT *t)
{
ModifyPreOrderWalk(t, 1);
}
void delete(nodeT **p){
nodeT *target, *imd_r, *plmd_r;
target=*p;
if (target->left==NULL && target->right==NULL) {
*p=NULL;
} else if (target->left == NULL) {
*p=target->right;
} else
if (target->right == NULL) {
*p=target->left;
} else {
plmd_r = target;
imd_r = target->right;
while( imd_r->left != NULL){
plmd_r = imd_r;
imd_r = imd_r->left;
}
if(plmd_r == target)
plmd_r->right = imd_r->right;
else
plmd_r->left = imd_r->right;
imd_r->left = target->left;
imd_r->right = target->right;
*p = imd_r;
}
free(target);
}
void listInorder(nodeT *t){
if (t != NULL) {
DisplayTree(t->left);
printf(\"%d \", t->key);
DisplayTree(t->right);
}
}
void listPreorder(nodeT *t) {
if (t != NULL) {
printf(\"%d \", t->key);
DisplayTree(t->left);
DisplayTree(t->right);
}
}
void listPostOrder(nodeT *t){
if (t != NULL) {
DisplayTree(t->left);
DisplayTree(t->right);
printf(\"%d\", t->key);
}
}
void InsertNode(nodeT **tptr, int key){
nodeT *t;
nodeT *tmp;
t=*tptr;
if (t == NULL) {
tmp= malloc(sizeof(nodeT));
tmp->key = key;
tmp->left=tmp->right=NULL;
*tptr=tmp;
return;
}
if (key < t->key) {
InsertNode
(&t->left, key);
} else {
InsertNode(&t->right, key);
}
}
int height(nodeT *t){
int lDepth = height(t->left);
int rDepth = height(t->right);
if(lDepth > rDepth)
return(lDepth+1);
else
return(rDepth+1);
}
int sum(nodeT *p){
if (p == NULL)
return 0;
else
return (p->data + sum(p->left) +sum(p->right) );
}
int count(nodeT *t){
if(t == NULL)
return 0;
int i = 1+count(t->left)+count(t->right);
return i;
}
/*nodeT *New(){
nodeT *tmp;
tmp = (nodeT *)malloc(sizeof(nodeT));
if (tmp==NULL) return NULL;
return tmp;
}*/
int main(){
nodeT *node;
nodeT *t=NULL, *findA;
int f;
int d, a, b, c, j, k, e, s, y;
while (1) {
printf(\"\ I - insert\ F - find node\ D - delete\ L - list in order\ R - list preorder\ P - list
PorstOrder\ M - max\ N - min\ H - height\ C - count\ S - sum.
Need help on creating code using cart. The output has to show multip.pdf
Need help on creating code using cart. The output has to show multiple lines of the users
information.
So far this is what I have:
index.html:
chapter8.jsp
<%@page contentType=\"text/html\" pageEncoding=\"UTF-8\"%>
JSP Page
<%@ page import=\"java.io.*, business.*, data.*, list.*\" %>
<%
// get parameters from the request
String firstName = request.getParameter(\"firstName\");
String lastName = request.getParameter(\"lastName\");
String email = request.getParameter(\"email\");
String phoneNumber = request.getParameter(\"phoneNumber\");
// get a relative file name
// ServletContext sc = this.getServletContext();
// String path = sc.getRealPath(\"/WEB-INF/EmailList.txt\");
File tempdir = (File) this.getServletContext().getAttribute(\"javax.servlet.context.tempdir\");
String temppath = tempdir.toString();
String path = temppath + \"/EmailList.txt\";
// use regular Java objects
User1 user = new User1(firstName, lastName, email, phoneNumber);
UserIO.add(user, path);
%>
Thanks for joining our email list
Here is the information that you entered:First NameLast NameEmailPhone Number<%=
user.getFirstName() %><%= user.getLastName() %><%= user.getEmailAddress()
%><%=user.getPhoneNumber() %>
User1.java
package business;
public class User1
{
private String firstName;
private String lastName;
private String email;
private String phoneNumber;
public User1()
{
firstName = \"\";
lastName = \"\";
email = \"\";
phoneNumber = \"\";
}
public User1(String firstName, String lastName, String email, String phoneNumber)
{
this.firstName = firstName;
this.lastName = lastName;
this.email = email;
this.phoneNumber = phoneNumber;
}
public void setFirstName(String firstName)
{
this.firstName = firstName;
}
public String getFirstName()
{
return firstName;
}
public void setLastName(String lastName)
{
this.lastName = lastName;
}
public String getLastName()
{
return lastName;
}
public void setEmailAddress(String email)
{
this.email = email;
}
public String getEmailAddress()
{
return email;
}
public void setPhoneNumber(String phoneNumber)
{
this.phoneNumber = phoneNumber;
}
public String getPhoneNumber()
{
return phoneNumber;
}
}
UserIO.java
package data;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.*;
import business.User1;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class UserIO extends HttpServlet {
public static void add(User1 user, String filepath) throws IOException
{
File file = new File(filepath);
PrintWriter out = new PrintWriter(
new FileWriter(file, true));
out.println(user.getEmailAddress()+ \"|\"
+ user.getFirstName() + \"|\"
+ user.getLastName() + \"|\"
+ user.getPhoneNumber());
out.close();
}
}
*NEED HELP!*
cart.java
package list;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
.
Name key elements of Object-Oriented systems. Note .pdf
Name key elements of Object-Oriented systems.
Note : no plagiarism please
Name key elements of Object-Oriented systems.
Name key elements of Object-Oriented systems.
Name key elements of Object-Oriented systems.
Note : no plagiarism please
Solution
there are 3 basic elements of object oriented system.they are:-
1.inheritance 2.polymorphism 3.encapsulation
Inheritance:-
inheritance is a concept where a class or an object can get the properties of its parent or sibling
class,for example let us consider A is a parent class and it has a method called read_me().and B
is another class inherited from class A.so using inheritance B class also has access rigths or can
explicitly use the read_me() method from A.
polymorphism:- (poly-many) (morph-forms)
polymorphism is an ability of an object to take on many forms.in general sense polymorphism
means a parent class refering the child class object
encapsulation:-
encapsulation in oop is nothing but binding or grouping of methods and member function in a
class.
Match the item on the left with the meaning on the right. Just show .pdf
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Solution
17) A (makes cout print floats with a fixed number of decimals)
18) Flag to left justify output, for right justify ios::right
19) C
20) F Used to format floating point numbers
21) B Sets the width in order to insert next value.
22) E.
Let f be holomorphic on and inside a Jordan curve C and let f have m.pdf
Let f be holomorphic on and inside a Jordan curve C and let f have maximum value M on C
(such a number exists by the Extreme Value Theorem). Suppose that |f(z0)| = M for some z0
ins(C). Show that f is constant over ins(C).
Solution
By maximum modulus principle, if |f| attains the maximum value at the interior , it should be the
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ii im severly confused about. 3. Let a= | 3 (i) Compute the perpendicular projection of the vector
b1on the line in the direction 2 of a. (ii) Write a formula for the distance of the point Xa to the
point b. Show that this distance attains the minimum value when a_ p is the perpendicular
projection found in (i).
Solution
Projection of b in the direction of a=(a dot b)
=(1 3 -2) dot (0 1 2)
=(1*0)+(3*1)+(-2*2)
=0+3-4
=-1
a dot a=(1^2)+(3^2)+(-2^2)
=1+9+4
=14
now [(a dot b)/a dot a]=-1/14
now {[(a dot b)/a dot a]}b=(-1/14)(0, 1, 2)
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Let A = {Leah, Dora, Mitch, Andy} which of the following are correct statements? D A Leah
A A Mitch D A Mitch A
Solution
A={Leah, Dora, Mitch, Andy}
Leah is belongs to A
So correct statements is second statement ( Leah belongs to A ) and fourth statement ( D not
belongs to A )
Remaining are in correct because D not in A ( first is wrong ), Mitch in A ( fifth is wrong) and A
is a super set, so third is wrong..
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