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(7x14x)^12=(277x^2)^12=7x(2)^12Solution(7x14x)^12=(2.pdf
The document shows the step-by-step solution to simplify the expression (7x*14x)^1/2. It first writes the expression as (7x*14x)^1/2, then simplifies the terms inside the radical to get (2*7*7*x^2)^1/2, and finally further simplifies the terms to get the final expression of 7x(2)^1/2.
You can think of H as being heat energy added ( + ), or heat removed.pdf
You can think of H as being heat energy added ( + ), or heat removed ( - ) .
Entropy refers to the degree of disorder or randomness. Gases are more disordered (more
entropy) than liquids, and liquids are more disordered (more entropy) than solids. Solids have the
least entropy.
1. Sublimation is a change from solid directly to a gas. Heat energy has to be added for this
process to occur, so H is +. Gases are more disordered, so S is + also.
2. Freezing is a change from liquid to a solid. Heat energy has to be removed, so H is - . Solids
have less entropy so S is - .
3. Condensation is a change from a gas to a liquid. Heat energy has to be removed, so H is - .
Liquid entropy is lower, so S is - .
4. Deposition is a change from a gas directly to a solid. Heat energy must be removed, so H is -
.
S is - also.
5. Boiling is a change from liquid to a gas. Heat energy must be added, so H is +. S is + since
gases have higher entropy.
6. Melting is a change from a solid to a liquid. Heat energy must be added , so H is + . S is +
since liquids have more disorder than solids.
Solution
You can think of H as being heat energy added ( + ), or heat removed ( - ) .
Entropy refers to the degree of disorder or randomness. Gases are more disordered (more
entropy) than liquids, and liquids are more disordered (more entropy) than solids. Solids have the
least entropy.
1. Sublimation is a change from solid directly to a gas. Heat energy has to be added for this
process to occur, so H is +. Gases are more disordered, so S is + also.
2. Freezing is a change from liquid to a solid. Heat energy has to be removed, so H is - . Solids
have less entropy so S is - .
3. Condensation is a change from a gas to a liquid. Heat energy has to be removed, so H is - .
Liquid entropy is lower, so S is - .
4. Deposition is a change from a gas directly to a solid. Heat energy must be removed, so H is -
.
S is - also.
5. Boiling is a change from liquid to a gas. Heat energy must be added, so H is +. S is + since
gases have higher entropy.
6. Melting is a change from a solid to a liquid. Heat energy must be added , so H is + . S is +
since liquids have more disorder than solids..
wait, i am posting the solution.Solutionwait, i am posting the.pdf
This very short document does not provide any substantive information to summarize in 3 sentences or less. It simply states that the author is waiting to post a solution, but does not provide any details on the problem or nature of the solution.
There are many other contributing factors to the norms of culture th.pdf
There are many other contributing factors to the norms of culture that assert the statement
Solution
There are many other contributing factors to the norms of culture that assert the statement.
Thermal dimorphism is the capacity to form different structures at d.pdf
Thermal dimorphism is the capacity to form different structures at different temperatures. For
example, some fungi exhibit thermal dimorphism. They exist as molds when outside the human
body and as yeasts when they are in the human body. For example Penicillium marneffei grows
at room temperature as a mold and, it grows as yeast at body temperature.
The morphologic shift between hyphae or molds and yeast is the significant step for the
virulence, pathogenesis, and lifecycle of the dimorphic fungi. In the soil, the thermally dimorphic
fungi grow at 22–25°C as mycelia that produce infectious spores. Following soil disturbance,
aerosolized spores inhaled into the lungs of a mammalian host at 37°C convert to yeast to cause
pneumonia and various disorders.
Solution
Thermal dimorphism is the capacity to form different structures at different temperatures. For
example, some fungi exhibit thermal dimorphism. They exist as molds when outside the human
body and as yeasts when they are in the human body. For example Penicillium marneffei grows
at room temperature as a mold and, it grows as yeast at body temperature.
The morphologic shift between hyphae or molds and yeast is the significant step for the
virulence, pathogenesis, and lifecycle of the dimorphic fungi. In the soil, the thermally dimorphic
fungi grow at 22–25°C as mycelia that produce infectious spores. Following soil disturbance,
aerosolized spores inhaled into the lungs of a mammalian host at 37°C convert to yeast to cause
pneumonia and various disorders..
The prime factor of 210=2357 Anyways, numbers can only be divid.pdf
The prime factor of 210=2*3*5*7
Anyways, numbers can only be divided by their prime factors or combinations of their prime
factors and it is divisible by all of them
Now let n = n_1 * n_2 * ... * n_k
Then n * n=n^2 =( n_1 * n_2 * ... * n_k) *( n_1 * n_2 * ... * n_k) are all the prime numbers that
divide n^2 and no other prime number can divide them (or it would be in that multiplication).
Because 210 is a multiplication of prime factors (2*3*5*7) that divides n^2, then 210 must be in
that multiplication. Hence 210 divides n.
Solution
The prime factor of 210=2*3*5*7
Anyways, numbers can only be divided by their prime factors or combinations of their prime
factors and it is divisible by all of them
Now let n = n_1 * n_2 * ... * n_k
Then n * n=n^2 =( n_1 * n_2 * ... * n_k) *( n_1 * n_2 * ... * n_k) are all the prime numbers that
divide n^2 and no other prime number can divide them (or it would be in that multiplication).
Because 210 is a multiplication of prime factors (2*3*5*7) that divides n^2, then 210 must be in
that multiplication. Hence 210 divides n..
The OSI Reference Model layers, in order from top to bottomD. Appl.pdf
The OSI Reference Model layers, in order from top to bottom
D. Application, Presentation, Session, Transport, Network, Data Link, Physical
1. Application Layer
topmost layer of the OSI reference model, is the application layer. This layer directly support
user applications, like software for file transfers, database access, and e-mail. It serves as a
window through which application processes can access network services.
2.Presentation Layer
It defines the format used to exchange data among networked computers. It is a network\'s
translator. When dissimilar systems want to communicate, certain translation and byte reordering
is done. It translates data from the format from the application layer into a recognized format. At
the receiving computer, this layer translates intermediary format into format that can be useful to
that computer\'s application layer.
3. Session Layer
It allows two applications on different computers to open, use, and close a connection called a
session. It performs name-recognition and other functions, such as security, that are needed to
allow two applications to communicate over the network.
4. Transport Layer
It provides an additional connection level beneath the session layer. It ensures that packets are
delivered error free, in sequence, and without losses or duplications. At the sending computer,
this layer repackages messages, dividing long messages into several packets and collecting small
packets together in one package. T
5. Network Layer
It is responsible for addressing messages and translating logical addresses and names into
physical addresses. This layer also determines the route from the source to the destination
computer. It determines which path the data should take based on network conditions
6. Data-Link Layer
It sends data frames from the network layer to the physical layer. It controls the electrical
impulses that enter and leave the network cable. On the receiving end, the data-link layer
packages raw bits from the physical layer into data frames.
7. Physical Layer
This layer transmits the unstructured, raw bit stream over a physical medium (such as the
network cable). The physical layer is totally hardware-oriented and deals with all aspects of
establishing and maintaining a physical link between communicating computers. The physical
layer also carries the signals that transmit data generated by each of the higher layers.
Solution
The OSI Reference Model layers, in order from top to bottom
D. Application, Presentation, Session, Transport, Network, Data Link, Physical
1. Application Layer
topmost layer of the OSI reference model, is the application layer. This layer directly support
user applications, like software for file transfers, database access, and e-mail. It serves as a
window through which application processes can access network services.
2.Presentation Layer
It defines the format used to exchange data among networked computers. It is a network\'s
translator. When dissimilar systems want .
Synchronous IO CPU waits till the IO proceeds.Asynchronous IO.pdf
Synchronous I/O : CPU waits till the I/O proceeds.
Asynchronous I/O : It executes cuncurrently with CPU execution.
Special Instruction I/O : They uses the CPU instructions for controlling the I/O devices.
Memory-mapped I/O : They share the same address space with the memory. Devices are
connected directly to the main memory so that they can transfer block of data
Direct memory access (DMA) : It ensures that CPU grants I/O module authority to read from or
write to memory without involvement. This module is responsible for
exchanging data between I/O device and main memory. It uses the hardware DMAC for data
transfers.
Polling I/O : It periodically checks the status of the device to check if the device performs any
further I/O operation.
Interrupt I/O : It works on interrupt signal where a device controller puts an interrupt signal on
the bus when it requires CPU’s attention.
Programmed I/O :
* It is used only in low end microcomputers.
* It has only single output and single input instruction.
* In this procees each instructions selects one I/O device and transfers a single character (byte).
Working:
I/O operation is requested by CPU.
Operation is performed by I/O module.
Status bits is being set by I/O module
Status bits are checked periodically by CPU.
CPU is not directly informed by I/O module.
CPU is not interrupted by I/O module.
CPU may wait or come back after some time.
Advantages :
It is easy to program and understand.
Disadvantages :
It has single input and single output instruction.
CPU spends most of its time in a tight loop waiting for the device to become ready.(Busy
Waiting)
Interrupt-driven I/O
Advantages :
Fast and efficient
Disadvantages :
* Interrupting a running process is an expensive business (requires saving context).
* Requires extra hardware (DMA controller chip).
Working :
CPU is interupted bye the I/O module.
Data is requested by CPU.
The data is trasferres by I/O module.
DMA ensures that CPU grants I/O module authority to read from or write to memory without
involvement. This module is responsible for
exchanging data between I/O device and main memory. It uses the hardware DMAC for data
transfers.
Solution
Synchronous I/O : CPU waits till the I/O proceeds.
Asynchronous I/O : It executes cuncurrently with CPU execution.
Special Instruction I/O : They uses the CPU instructions for controlling the I/O devices.
Memory-mapped I/O : They share the same address space with the memory. Devices are
connected directly to the main memory so that they can transfer block of data
Direct memory access (DMA) : It ensures that CPU grants I/O module authority to read from or
write to memory without involvement. This module is responsible for
exchanging data between I/O device and main memory. It uses the hardware DMAC for data
transfers.
Polling I/O : It periodically checks the status of the device to check if the device performs any
further I/O operation.
Interrupt I/O : It works on interrupt signal where a device controller puts an in.
Recommended
(7x14x)^12=(277x^2)^12=7x(2)^12Solution(7x14x)^12=(2.pdf
The document shows the step-by-step solution to simplify the expression (7x*14x)^1/2. It first writes the expression as (7x*14x)^1/2, then simplifies the terms inside the radical to get (2*7*7*x^2)^1/2, and finally further simplifies the terms to get the final expression of 7x(2)^1/2.
You can think of H as being heat energy added ( + ), or heat removed.pdf
You can think of H as being heat energy added ( + ), or heat removed ( - ) .
Entropy refers to the degree of disorder or randomness. Gases are more disordered (more
entropy) than liquids, and liquids are more disordered (more entropy) than solids. Solids have the
least entropy.
1. Sublimation is a change from solid directly to a gas. Heat energy has to be added for this
process to occur, so H is +. Gases are more disordered, so S is + also.
2. Freezing is a change from liquid to a solid. Heat energy has to be removed, so H is - . Solids
have less entropy so S is - .
3. Condensation is a change from a gas to a liquid. Heat energy has to be removed, so H is - .
Liquid entropy is lower, so S is - .
4. Deposition is a change from a gas directly to a solid. Heat energy must be removed, so H is -
.
S is - also.
5. Boiling is a change from liquid to a gas. Heat energy must be added, so H is +. S is + since
gases have higher entropy.
6. Melting is a change from a solid to a liquid. Heat energy must be added , so H is + . S is +
since liquids have more disorder than solids.
Solution
You can think of H as being heat energy added ( + ), or heat removed ( - ) .
Entropy refers to the degree of disorder or randomness. Gases are more disordered (more
entropy) than liquids, and liquids are more disordered (more entropy) than solids. Solids have the
least entropy.
1. Sublimation is a change from solid directly to a gas. Heat energy has to be added for this
process to occur, so H is +. Gases are more disordered, so S is + also.
2. Freezing is a change from liquid to a solid. Heat energy has to be removed, so H is - . Solids
have less entropy so S is - .
3. Condensation is a change from a gas to a liquid. Heat energy has to be removed, so H is - .
Liquid entropy is lower, so S is - .
4. Deposition is a change from a gas directly to a solid. Heat energy must be removed, so H is -
.
S is - also.
5. Boiling is a change from liquid to a gas. Heat energy must be added, so H is +. S is + since
gases have higher entropy.
6. Melting is a change from a solid to a liquid. Heat energy must be added , so H is + . S is +
since liquids have more disorder than solids..
wait, i am posting the solution.Solutionwait, i am posting the.pdf
This very short document does not provide any substantive information to summarize in 3 sentences or less. It simply states that the author is waiting to post a solution, but does not provide any details on the problem or nature of the solution.
There are many other contributing factors to the norms of culture th.pdf
There are many other contributing factors to the norms of culture that assert the statement
Solution
There are many other contributing factors to the norms of culture that assert the statement.
Thermal dimorphism is the capacity to form different structures at d.pdf
Thermal dimorphism is the capacity to form different structures at different temperatures. For
example, some fungi exhibit thermal dimorphism. They exist as molds when outside the human
body and as yeasts when they are in the human body. For example Penicillium marneffei grows
at room temperature as a mold and, it grows as yeast at body temperature.
The morphologic shift between hyphae or molds and yeast is the significant step for the
virulence, pathogenesis, and lifecycle of the dimorphic fungi. In the soil, the thermally dimorphic
fungi grow at 22–25°C as mycelia that produce infectious spores. Following soil disturbance,
aerosolized spores inhaled into the lungs of a mammalian host at 37°C convert to yeast to cause
pneumonia and various disorders.
Solution
Thermal dimorphism is the capacity to form different structures at different temperatures. For
example, some fungi exhibit thermal dimorphism. They exist as molds when outside the human
body and as yeasts when they are in the human body. For example Penicillium marneffei grows
at room temperature as a mold and, it grows as yeast at body temperature.
The morphologic shift between hyphae or molds and yeast is the significant step for the
virulence, pathogenesis, and lifecycle of the dimorphic fungi. In the soil, the thermally dimorphic
fungi grow at 22–25°C as mycelia that produce infectious spores. Following soil disturbance,
aerosolized spores inhaled into the lungs of a mammalian host at 37°C convert to yeast to cause
pneumonia and various disorders..
The prime factor of 210=2357 Anyways, numbers can only be divid.pdf
The prime factor of 210=2*3*5*7
Anyways, numbers can only be divided by their prime factors or combinations of their prime
factors and it is divisible by all of them
Now let n = n_1 * n_2 * ... * n_k
Then n * n=n^2 =( n_1 * n_2 * ... * n_k) *( n_1 * n_2 * ... * n_k) are all the prime numbers that
divide n^2 and no other prime number can divide them (or it would be in that multiplication).
Because 210 is a multiplication of prime factors (2*3*5*7) that divides n^2, then 210 must be in
that multiplication. Hence 210 divides n.
Solution
The prime factor of 210=2*3*5*7
Anyways, numbers can only be divided by their prime factors or combinations of their prime
factors and it is divisible by all of them
Now let n = n_1 * n_2 * ... * n_k
Then n * n=n^2 =( n_1 * n_2 * ... * n_k) *( n_1 * n_2 * ... * n_k) are all the prime numbers that
divide n^2 and no other prime number can divide them (or it would be in that multiplication).
Because 210 is a multiplication of prime factors (2*3*5*7) that divides n^2, then 210 must be in
that multiplication. Hence 210 divides n..
The OSI Reference Model layers, in order from top to bottomD. Appl.pdf
The OSI Reference Model layers, in order from top to bottom
D. Application, Presentation, Session, Transport, Network, Data Link, Physical
1. Application Layer
topmost layer of the OSI reference model, is the application layer. This layer directly support
user applications, like software for file transfers, database access, and e-mail. It serves as a
window through which application processes can access network services.
2.Presentation Layer
It defines the format used to exchange data among networked computers. It is a network\'s
translator. When dissimilar systems want to communicate, certain translation and byte reordering
is done. It translates data from the format from the application layer into a recognized format. At
the receiving computer, this layer translates intermediary format into format that can be useful to
that computer\'s application layer.
3. Session Layer
It allows two applications on different computers to open, use, and close a connection called a
session. It performs name-recognition and other functions, such as security, that are needed to
allow two applications to communicate over the network.
4. Transport Layer
It provides an additional connection level beneath the session layer. It ensures that packets are
delivered error free, in sequence, and without losses or duplications. At the sending computer,
this layer repackages messages, dividing long messages into several packets and collecting small
packets together in one package. T
5. Network Layer
It is responsible for addressing messages and translating logical addresses and names into
physical addresses. This layer also determines the route from the source to the destination
computer. It determines which path the data should take based on network conditions
6. Data-Link Layer
It sends data frames from the network layer to the physical layer. It controls the electrical
impulses that enter and leave the network cable. On the receiving end, the data-link layer
packages raw bits from the physical layer into data frames.
7. Physical Layer
This layer transmits the unstructured, raw bit stream over a physical medium (such as the
network cable). The physical layer is totally hardware-oriented and deals with all aspects of
establishing and maintaining a physical link between communicating computers. The physical
layer also carries the signals that transmit data generated by each of the higher layers.
Solution
The OSI Reference Model layers, in order from top to bottom
D. Application, Presentation, Session, Transport, Network, Data Link, Physical
1. Application Layer
topmost layer of the OSI reference model, is the application layer. This layer directly support
user applications, like software for file transfers, database access, and e-mail. It serves as a
window through which application processes can access network services.
2.Presentation Layer
It defines the format used to exchange data among networked computers. It is a network\'s
translator. When dissimilar systems want .
Synchronous IO CPU waits till the IO proceeds.Asynchronous IO.pdf
Synchronous I/O : CPU waits till the I/O proceeds.
Asynchronous I/O : It executes cuncurrently with CPU execution.
Special Instruction I/O : They uses the CPU instructions for controlling the I/O devices.
Memory-mapped I/O : They share the same address space with the memory. Devices are
connected directly to the main memory so that they can transfer block of data
Direct memory access (DMA) : It ensures that CPU grants I/O module authority to read from or
write to memory without involvement. This module is responsible for
exchanging data between I/O device and main memory. It uses the hardware DMAC for data
transfers.
Polling I/O : It periodically checks the status of the device to check if the device performs any
further I/O operation.
Interrupt I/O : It works on interrupt signal where a device controller puts an interrupt signal on
the bus when it requires CPU’s attention.
Programmed I/O :
* It is used only in low end microcomputers.
* It has only single output and single input instruction.
* In this procees each instructions selects one I/O device and transfers a single character (byte).
Working:
I/O operation is requested by CPU.
Operation is performed by I/O module.
Status bits is being set by I/O module
Status bits are checked periodically by CPU.
CPU is not directly informed by I/O module.
CPU is not interrupted by I/O module.
CPU may wait or come back after some time.
Advantages :
It is easy to program and understand.
Disadvantages :
It has single input and single output instruction.
CPU spends most of its time in a tight loop waiting for the device to become ready.(Busy
Waiting)
Interrupt-driven I/O
Advantages :
Fast and efficient
Disadvantages :
* Interrupting a running process is an expensive business (requires saving context).
* Requires extra hardware (DMA controller chip).
Working :
CPU is interupted bye the I/O module.
Data is requested by CPU.
The data is trasferres by I/O module.
DMA ensures that CPU grants I/O module authority to read from or write to memory without
involvement. This module is responsible for
exchanging data between I/O device and main memory. It uses the hardware DMAC for data
transfers.
Solution
Synchronous I/O : CPU waits till the I/O proceeds.
Asynchronous I/O : It executes cuncurrently with CPU execution.
Special Instruction I/O : They uses the CPU instructions for controlling the I/O devices.
Memory-mapped I/O : They share the same address space with the memory. Devices are
connected directly to the main memory so that they can transfer block of data
Direct memory access (DMA) : It ensures that CPU grants I/O module authority to read from or
write to memory without involvement. This module is responsible for
exchanging data between I/O device and main memory. It uses the hardware DMAC for data
transfers.
Polling I/O : It periodically checks the status of the device to check if the device performs any
further I/O operation.
Interrupt I/O : It works on interrupt signal where a device controller puts an in.
SolutionSolutionSolution.pdf
The document discusses a solution to an unspecified problem. It states that the solution involves taking certain steps, but provides no details on what the actual problem or steps are. The summary is unable to extract any meaningful information due to the lack of context and specifics provided in the document.
slope = f(x) x= 12012= 10Solutionslope = f(x) x= 120.pdf
The document shows how to calculate the slope of a linear function by taking the change in y-values (f(x)) over the change in x-values. It gives an example where f(x) is 120 and x is 12, so the slope is 120/12 = 10. The slope is calculated to be 10.
Please find the required program along with the comments against eac.pdf
Please find the required program along with the comments against each line, also find the output
at last :
#include
#include
#include
#include
using namespace std;
void listCodonPositions(string dna, string codon){
vector positions; // holds all the positions that codon occurs within dna
size_t pos = dna.find(codon, 0); // finds the first occurance of codon
if(pos != string::npos) //if no codon is found
cout << pos << \" \" << flush;
else
cout << \"No occurance\" << flush;
while(pos != string::npos) //iterate till no occurance is found in a dna
{
pos = dna.find(codon,pos+1); //find next occurance
if(pos != string::npos) //if no codon is found
cout << pos << \" \" << flush;
}
cout << endl;
}
int main()
{
string humanDNA =
\"CGCAAATTTGCCGGATTTCCTTTGCTGTTCCTGCATGTAGTTTAAACGAGATTGCCA
GCACCGGGTATCATTCACCATTTTTCTTTTCGTTAACTTGCCGTCAGCCTTTTCTTTGA
CCTCTTCTTTCTGTTCATGTGTATTTGCTGTCTCTTAGCCCAGACTTCCCGTGTCCTTT
CCACCGGGCCTTTGAGAGGTCACAGGGTCTTGATGCTGTGGTCTTCATCTGCAGGTG
TCTGACTTCCAGCAACTGCTGGCCTGTGCCAGGGTGCAGCTGAGCACTGGAGTGGAG
TTTTCCTGTGGAGAGGAGCCATGCCTAGAGTGGGATGGGCCATTGTTCATG\";
string mouseDNA =
\"CGCAATTTTTACTTAATTCTTTTTCTTTTAATTCATATATTTTTAATATGTTTACTAT
TAATGGTTATCATTCACCATTTAACTATTTGTTATTTTGACGTCATTTTTTTCTATTTC
CTCTTTTTTCAATTCATGTTTATTTTCTGTATTTTTGTTAAGTTTTCACAAGTCTAATA
TAATTGTCCTTTGAGAGGTTATTTGGTCTATATTTTTTTTTCTTCATCTGTATTTTTAT
GATTTCATTTAATTGATTTTCATTGACAGGGTTCTGCTGTGTTCTGGATTGTATTTTTC
TTGTGGAGAGGAACTATTTCTTGAGTGGGATGTACCTTTGTTCTTG\";
string unknownDNA =
\"CGCATTTTTGCCGGTTTTCCTTTGCTGTTTATTCATTTATTTTAAACGATATTTATAT
CATCGGGTTTCATTCACTATTTTTCTTTTCGATAAATTTTTGTCAGCATTTTCTTTTAC
CTCTTCTTTCTGTTTATGTTAATTTTCTGTTTCTTAACCCAGTCTTCTCGATTCTTATCT
ACCGGACCTATTATAGGTCACAGGGTCTTGATGCTTTGGTTTTCATCTGCAAGAGTCT
GACTTCCTGCTAATGCTGTTCTGTGTCAGGGTGCATCTGAGCACTGATGTGGAGTTTT
CTTGTGGATATGAGCCATTCATAGTGTGGGATGTGCCATAGTTCATG\";
string codon;
cout << \"Enter codon : \" << endl;
cin >> codon;
char exit[] = \"*\";
while(strcmp(codon.c_str(),exit)!=0){
cout << \"Occurances in humanDNA : \"<< endl;
listCodonPositions(humanDNA,codon);
cout << \"Occurances in mouseDNA : \"<< endl;
listCodonPositions(mouseDNA,codon);
cout << \"Occurances in unknownDNA : \"<< endl;
listCodonPositions(unknownDNA,codon);
cout << \"Enter codon : \" << endl;
cin >> codon;
}
}
--------------------------------------------------------------------------------
OUTPUT :
Enter codon :
AAA
Occurances in humanDNA :
3 43
Occurances in mouseDNA :
No occurance
Occurances in unknownDNA :
43 91
Enter codon :
TGC
Occurances in humanDNA :
8 22 31 52 95 141 208 224 248 258 266 310
Occurances in mouseDNA :
269
Occurances in unknownDNA :
8 22 208 224 242 248 266 326
Enter codon :
*
Solution
Please find the required program along with the comments against each line, also find the output
at last :
#include
#include
#include
#include
using namespace std;
void listCodonPositions(string dna, string codon){
vector positions; // holds all the positions that codon occurs within dna
size_t pos = dna.find(codon, 0); // finds the first oc.
Order the steps of the viral life cycle.1. viral entry into host.pdf
Order the steps of the viral life cycle.
1. viral entry into host
2. disassembly of virus
3. replication of virus
4. transcription of virus
5. assembly of new viral units
6. release of new virus
Order these human vaccines by year of introduction from earliest to most recent.
1. Smallpox - 1796
2. plague - 1890
3. tetanus - 1930
4. polio - 1955
5. rubella - 1963
Question 10.
Gamma rays and X rays are ionizing radiations
Question 13.
The disadvantages of intravenous injection for drug administration are:
Risk of overdose
Risk of infection
Solution
Order the steps of the viral life cycle.
1. viral entry into host
2. disassembly of virus
3. replication of virus
4. transcription of virus
5. assembly of new viral units
6. release of new virus
Order these human vaccines by year of introduction from earliest to most recent.
1. Smallpox - 1796
2. plague - 1890
3. tetanus - 1930
4. polio - 1955
5. rubella - 1963
Question 10.
Gamma rays and X rays are ionizing radiations
Question 13.
The disadvantages of intravenous injection for drug administration are:
Risk of overdose
Risk of infection.
Listing of the type of alleles present in an organisms cellGenotyp.pdf
Listing of the type of alleles present in an organisms cell
Genotype
The allele for polydactylism (extra finger or toe) is expressed to different digrees in different
individuals that have it
Variable expressivity
Height in humans is determined by the interaction of genes on different loci
Polygenic inheritance
Individuals that are heterozygous for alleles for A antigen and B antigen have blood type AB
Codominance
All the genes in a cell of organism
Genome
Solution
Listing of the type of alleles present in an organisms cell
Genotype
The allele for polydactylism (extra finger or toe) is expressed to different digrees in different
individuals that have it
Variable expressivity
Height in humans is determined by the interaction of genes on different loci
Polygenic inheritance
Individuals that are heterozygous for alleles for A antigen and B antigen have blood type AB
Codominance
All the genes in a cell of organism
Genome.
An Operating System (OS) is an interface between a computer user and.pdf
An Operating System (OS) is an interface between a computer user and computer hardware. An
operating system is a software which performs all the basic tasks like file management, memory
management, process management, handling input and output.
The kernel is a used for operating system need to be protected from user programs to maintain
the proper operation of the computer system .The kernel is a computer program that constitutes
the central core of a computer\'s operating system. It has complete control over everything that
occurs in the system.As such, it is the first program loaded on startup, and then manages the
remainder of the startup, as well as input/output requests from software, translating them into
data processing instructions for the central processing unit.
kernel is usually loaded into a protected area of memory, which prevents it from being
overwritten by other, less frequently used parts of the operating system or by applications. The
kernel performs its tasks, such as executing processes and handling interrupts, in kernel space,
whereas everything a user normally does, such as writing text in a text editor or running
programs in a GUI, is done in user space. This separation prevents user data and kernel data from
interfering with each other and thereby diminishing performance or causing the system to
become unstable.
Solution
An Operating System (OS) is an interface between a computer user and computer hardware. An
operating system is a software which performs all the basic tasks like file management, memory
management, process management, handling input and output.
The kernel is a used for operating system need to be protected from user programs to maintain
the proper operation of the computer system .The kernel is a computer program that constitutes
the central core of a computer\'s operating system. It has complete control over everything that
occurs in the system.As such, it is the first program loaded on startup, and then manages the
remainder of the startup, as well as input/output requests from software, translating them into
data processing instructions for the central processing unit.
kernel is usually loaded into a protected area of memory, which prevents it from being
overwritten by other, less frequently used parts of the operating system or by applications. The
kernel performs its tasks, such as executing processes and handling interrupts, in kernel space,
whereas everything a user normally does, such as writing text in a text editor or running
programs in a GUI, is done in user space. This separation prevents user data and kernel data from
interfering with each other and thereby diminishing performance or causing the system to
become unstable..
Internal Evidence General.pdf
Internal Evidence General Ledger Files canceled Payrolled
Checks Payroll time records Receiving Reports Minutes of the Board of
Directors Cancelled Notes Payable
External Evidence Vendors Invoices Bank Statements
Purchase Requisitions Remittance Advices Signed Leased Agreements
Duplicate Copies of bills of Lading Notes Receivable
External Evidence are more reliable than the internal evidences because External evidences
are the evidences which we gets from the third party there is no involvement of the
management or employee of the organisations. Internal Evidences are prepared internally
within the organisation and the chances of manipulation in internal evidences are more than
the external evidences.
Solution
Internal Evidence General Ledger Files canceled Payrolled
Checks Payroll time records Receiving Reports Minutes of the Board of
Directors Cancelled Notes Payable
External Evidence Vendors Invoices Bank Statements
Purchase Requisitions Remittance Advices Signed Leased Agreements
Duplicate Copies of bills of Lading Notes Receivable
External Evidence are more reliable than the internal evidences because External evidences
are the evidences which we gets from the third party there is no involvement of the
management or employee of the organisations. Internal Evidences are prepared internally
within the organisation and the chances of manipulation in internal evidences are more than
the external evidences..
In reaction # 3 either the cis or trans diol may .pdf
In reaction # 3 either the cis or trans diol may be used as a reactant. These isomers
are rapidly interconverted under the rearrangement conditions, indicating that the initial water
loss is reversible; a result confirmed by isotopic oxygen exchange. The clear preference for a
methylene group shift versus a methyl group shift may reflect inherent migratory aptitudes, or
possibly group configurations in the 3º-carbocation intermediate. In the conformation shown
here both methyl and methylene groups may shift, or an epoxide ring may be formed reversibly.
An alternative chair-like conformation having an equatorial methyl group should be more stable,
but would not be suitable for a methyl shift. The predominant ring contraction is therefore
understandable. Reaction # 4 is an unusual case in which a strained ring contracts to an even
smaller ring. Phenyl groups generally have a high migratory aptitude, so the failure to obtain 2,2-
diphenylcyclobutanone as a product might seem surprising. However, the carbocation resulting
from a phenyl shift would be just as strained as its precursor; whereas the shift of a ring
methylene group generates an unstrained cation stabilized by phenyl and oxygen substituents.
Conjugative stabilization of the phenyl ketone and absence of sp2 hybridized carbon atoms in the
small ring may also contribute to the stability of the observed product. Finally, reaction # 5
clearly shows the influence of reaction conditions on product composition, but explaining the
manner in which different conditions perturb the outcome is challenging. Treatment with cold
sulfuric acid should produce the more stable diphenyl 3º-carbocation, and a methyl group shift
would then lead to the observed product. The action of a Lewis acid in acetic anhydride, on the
other hand, may selectively acetylate the less hindered dimethyl carbinol. In this event the
acetate becomes the favored leaving group (presumably coordinated with acid), followed by a
1,2-phenyl shift. It is reported that the symmetrically substituted isomeric diol (drawn in the
shaded box) rearranges exclusively by a methyl shift, but the configuration of the starting
material was not stated (two diastereomers are possible). Because of the influence of other
factors (above), it has not been possible to determine an unambiguous migratory order for
substituents in the pinacol rearrangement. However, some general trends are discernible.
Benzopinacol, (C6H5)2C(OH)C(OH)(C6H5)2, undergoes rapid rearrangement to
(C6H5)3CCOC6H5 under much milder conditions than required for pinacol. Indeed, it is often
the case that phenyl or other aromatic substituents adjacent to a forming carbocation will
facilitate that ionization in the course of their migration to the cationic site. In non-aromatic
compounds of the type (CH3)2C(OH)C(OH)RCH3, migration of R increases in the manner R =
CH3 < R = C2H5 << R = (CH3)3C. Since the shifting alkyl group must carry part of the overall
positive charge, alkyl substit.
Intermolecular forces exist between molecules and.pdf
Intermolecular forces exist between molecules and influence the physical
properties. We can think of H2O in its three forms, ice, water and steam. ... However, the
physical properties of H2O are very different in the three states.
Solution
Intermolecular forces exist between molecules and influence the physical
properties. We can think of H2O in its three forms, ice, water and steam. ... However, the
physical properties of H2O are very different in the three states..
H2 concentration will decrease and H2O concentrat.pdf
H2 concentration will decrease and H2O concentration will increase
Solution
H2 concentration will decrease and H2O concentration will increase.
elemental form of sodium and hydrogen is very rea.pdf
elemental form of sodium and hydrogen is very reactive ...that\'s why it is not
advisable to use them in their elemental form
Solution
elemental form of sodium and hydrogen is very reactive ...that\'s why it is not
advisable to use them in their elemental form.
C. Both A and B are correct .pdf
Both statements A and B provided in the previous question are deemed to be correct according to the solution. The solution simply states that both of the options given for the question are right.
C gained 2 hydrogen. It was reduced. It is theref.pdf
C gained 2 hydrogen. It was reduced. It is therefore an oxidizing agent.
Solution
C gained 2 hydrogen. It was reduced. It is therefore an oxidizing agent..
A is stronger acid because it give more H+ to sol.pdf
A is stronger acid because it give more H+ to solution
Solution
A is stronger acid because it give more H+ to solution.
Error message number = int(input(Enter a number )) Name.pdf
Error message
number = int(input(\"Enter a number: \")) NameError: name \'C\' is not defined
Solution
Error message
number = int(input(\"Enter a number: \")) NameError: name \'C\' is not defined.
Dear,The answer is .7.SolutionDear,The answer is .7..pdf
The document is a short note stating that the answer to an unknown problem or question is 0.7. It provides no context or explanation for how the answer was determined. In just two sentences, it directly states that the answer is 0.7 without any additional supporting details or calculations.
Crayfish contain an exoskeleton which means it needs to get rid of i.pdf
Crayfish contain an exoskeleton which means it needs to get rid of it in order to grow in size.
This is called molting. If the food is abundant they will molt several times quite often early in
life. The crayfish hides and crawl out of its exoskeleton through a slit along its dorsal surface. It
will and wait for a while until the new skeleton hardens.
In preparation for molting the crayfish withdraws most of the calcium from its shell, and stores
it in two white \"tablets\"in the sides of its head.
An incomplete cycle of molting can be caused due to mutation in a certain transcription factor
such that it could not bind to the DNA and transcribed the m RNA that encodes ecdysone. So,
the answer of the above qujestion wil be Ans. C.
Solution
Crayfish contain an exoskeleton which means it needs to get rid of it in order to grow in size.
This is called molting. If the food is abundant they will molt several times quite often early in
life. The crayfish hides and crawl out of its exoskeleton through a slit along its dorsal surface. It
will and wait for a while until the new skeleton hardens.
In preparation for molting the crayfish withdraws most of the calcium from its shell, and stores
it in two white \"tablets\"in the sides of its head.
An incomplete cycle of molting can be caused due to mutation in a certain transcription factor
such that it could not bind to the DNA and transcribed the m RNA that encodes ecdysone. So,
the answer of the above qujestion wil be Ans. C..
BasicPizza.javapublic class BasicPizza { String type; String c.pdf
BasicPizza.java
public class BasicPizza {
String type;
String crust;
String ingredients;
double cost;
public BasicPizza(String type) {
super();
this.type = type;
}
public String getType() {
return type;
}
public void setType(String type) {
this.type = type;
}
public String getCrust() {
return crust;
}
public void setCrust(String crust) {
this.crust = crust;
}
public BasicPizza() {
super();
}
public String getIngredients() {
return ingredients;
}
public void setIngredients(String ingredients) {
this.ingredients = ingredients;
}
public double getCost() {
return cost;
}
public void setCost() {
this.cost = 5;
}
@Override
public String toString() {
return \"BasicPizza [type=\" + type + \", crust=\" + crust + \", ingredients=\"
+ ingredients + \", cost=\" + cost + \"]\";
}
}
________________________________________________________________________
LiFiCheese.java
public class LiFiCheese extends BasicPizza{
private String crust;
private double cost;
private String ingredients;
public LiFiCheese() {
super(\"Cheese\");
this.cost=5;
}
public void setCrust()
{
this.crust=\"thin\";
}
public String getCrust()
{
return crust;
}
public double getCost() {
return cost;
}
public void setCost(double cost) {
}
public String getIngredients() {
return ingredients;
}
public void setIngredients(String ingredients) {
this.ingredients = ingredients;
}
@Override
public String toString() {
System.out.println(\"You Ordered :\");
System.out.println(getType());
setCrust();
System.out.println(getCrust());
System.out.println(\"Total Cost of :\"+getCost());
return \"\";
}
}
_______________________________________________________________________
LiFiPizza.java
import java.util.Scanner;
public class LiFiPizza extends BasicPizza {
private String type;
private double cost;
private String crust;
private String ingredients;
public LiFiPizza()
{
super();
this.type = \"Meat\";
this.cost=5;
}
public String getType() {
return type;
}
public void setType(String type) {
this.type = type;
}
public double getCost() {
return cost;
}
public void setCost() {
this.cost = this.cost+2;
}
public String getCrust() {
return crust;
}
public void setCrust(String crust) {
if(crust.equals(\"Thin\"))
{
this.crust=\"Thin\";
}
else if(crust.equals(\"Thick\"))
{
this.crust=\"Thick\";
}
}
public String getIngredients() {
return ingredients;
}
public void setIngredients(String ingredients) {
this.ingredients = ingredients;
}
@Override
public String toString() {
System.out.println(\"You Ordered :\");
System.out.println(getType());
System.out.println(getIngredients()+\"<+$2.00>\");
System.out.println(getCrust());
System.out.println(\"Total Cost of :\"+getCost());
return \"\";
}
}
_____________________________________________________________________
LiFiUnit5Ch14.java
import java.util.Scanner;
public class LiFiUnit5Ch14 {
public static void main(String[] args) {
BasicPizza bp=null;
String typeOfPizza;
Scanner sc=new Scanner(System.in);
System.out.print(\"What type of pizza would you like :\");
typeOfPizza=sc.nex.
BIOLOGY NATIONAL EXAMINATION COUNCIL (NECO) 2024 PRACTICAL MANUAL.pptx
Practical manual for National Examination Council, Nigeria.
Contains guides on answering questions on the specimens provided
Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering
Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
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SolutionSolutionSolution.pdf
The document discusses a solution to an unspecified problem. It states that the solution involves taking certain steps, but provides no details on what the actual problem or steps are. The summary is unable to extract any meaningful information due to the lack of context and specifics provided in the document.
slope = f(x) x= 12012= 10Solutionslope = f(x) x= 120.pdf
The document shows how to calculate the slope of a linear function by taking the change in y-values (f(x)) over the change in x-values. It gives an example where f(x) is 120 and x is 12, so the slope is 120/12 = 10. The slope is calculated to be 10.
Please find the required program along with the comments against eac.pdf
Please find the required program along with the comments against each line, also find the output
at last :
#include
#include
#include
#include
using namespace std;
void listCodonPositions(string dna, string codon){
vector positions; // holds all the positions that codon occurs within dna
size_t pos = dna.find(codon, 0); // finds the first occurance of codon
if(pos != string::npos) //if no codon is found
cout << pos << \" \" << flush;
else
cout << \"No occurance\" << flush;
while(pos != string::npos) //iterate till no occurance is found in a dna
{
pos = dna.find(codon,pos+1); //find next occurance
if(pos != string::npos) //if no codon is found
cout << pos << \" \" << flush;
}
cout << endl;
}
int main()
{
string humanDNA =
\"CGCAAATTTGCCGGATTTCCTTTGCTGTTCCTGCATGTAGTTTAAACGAGATTGCCA
GCACCGGGTATCATTCACCATTTTTCTTTTCGTTAACTTGCCGTCAGCCTTTTCTTTGA
CCTCTTCTTTCTGTTCATGTGTATTTGCTGTCTCTTAGCCCAGACTTCCCGTGTCCTTT
CCACCGGGCCTTTGAGAGGTCACAGGGTCTTGATGCTGTGGTCTTCATCTGCAGGTG
TCTGACTTCCAGCAACTGCTGGCCTGTGCCAGGGTGCAGCTGAGCACTGGAGTGGAG
TTTTCCTGTGGAGAGGAGCCATGCCTAGAGTGGGATGGGCCATTGTTCATG\";
string mouseDNA =
\"CGCAATTTTTACTTAATTCTTTTTCTTTTAATTCATATATTTTTAATATGTTTACTAT
TAATGGTTATCATTCACCATTTAACTATTTGTTATTTTGACGTCATTTTTTTCTATTTC
CTCTTTTTTCAATTCATGTTTATTTTCTGTATTTTTGTTAAGTTTTCACAAGTCTAATA
TAATTGTCCTTTGAGAGGTTATTTGGTCTATATTTTTTTTTCTTCATCTGTATTTTTAT
GATTTCATTTAATTGATTTTCATTGACAGGGTTCTGCTGTGTTCTGGATTGTATTTTTC
TTGTGGAGAGGAACTATTTCTTGAGTGGGATGTACCTTTGTTCTTG\";
string unknownDNA =
\"CGCATTTTTGCCGGTTTTCCTTTGCTGTTTATTCATTTATTTTAAACGATATTTATAT
CATCGGGTTTCATTCACTATTTTTCTTTTCGATAAATTTTTGTCAGCATTTTCTTTTAC
CTCTTCTTTCTGTTTATGTTAATTTTCTGTTTCTTAACCCAGTCTTCTCGATTCTTATCT
ACCGGACCTATTATAGGTCACAGGGTCTTGATGCTTTGGTTTTCATCTGCAAGAGTCT
GACTTCCTGCTAATGCTGTTCTGTGTCAGGGTGCATCTGAGCACTGATGTGGAGTTTT
CTTGTGGATATGAGCCATTCATAGTGTGGGATGTGCCATAGTTCATG\";
string codon;
cout << \"Enter codon : \" << endl;
cin >> codon;
char exit[] = \"*\";
while(strcmp(codon.c_str(),exit)!=0){
cout << \"Occurances in humanDNA : \"<< endl;
listCodonPositions(humanDNA,codon);
cout << \"Occurances in mouseDNA : \"<< endl;
listCodonPositions(mouseDNA,codon);
cout << \"Occurances in unknownDNA : \"<< endl;
listCodonPositions(unknownDNA,codon);
cout << \"Enter codon : \" << endl;
cin >> codon;
}
}
--------------------------------------------------------------------------------
OUTPUT :
Enter codon :
AAA
Occurances in humanDNA :
3 43
Occurances in mouseDNA :
No occurance
Occurances in unknownDNA :
43 91
Enter codon :
TGC
Occurances in humanDNA :
8 22 31 52 95 141 208 224 248 258 266 310
Occurances in mouseDNA :
269
Occurances in unknownDNA :
8 22 208 224 242 248 266 326
Enter codon :
*
Solution
Please find the required program along with the comments against each line, also find the output
at last :
#include
#include
#include
#include
using namespace std;
void listCodonPositions(string dna, string codon){
vector positions; // holds all the positions that codon occurs within dna
size_t pos = dna.find(codon, 0); // finds the first oc.
Order the steps of the viral life cycle.1. viral entry into host.pdf
Order the steps of the viral life cycle.
1. viral entry into host
2. disassembly of virus
3. replication of virus
4. transcription of virus
5. assembly of new viral units
6. release of new virus
Order these human vaccines by year of introduction from earliest to most recent.
1. Smallpox - 1796
2. plague - 1890
3. tetanus - 1930
4. polio - 1955
5. rubella - 1963
Question 10.
Gamma rays and X rays are ionizing radiations
Question 13.
The disadvantages of intravenous injection for drug administration are:
Risk of overdose
Risk of infection
Solution
Order the steps of the viral life cycle.
1. viral entry into host
2. disassembly of virus
3. replication of virus
4. transcription of virus
5. assembly of new viral units
6. release of new virus
Order these human vaccines by year of introduction from earliest to most recent.
1. Smallpox - 1796
2. plague - 1890
3. tetanus - 1930
4. polio - 1955
5. rubella - 1963
Question 10.
Gamma rays and X rays are ionizing radiations
Question 13.
The disadvantages of intravenous injection for drug administration are:
Risk of overdose
Risk of infection.
Listing of the type of alleles present in an organisms cellGenotyp.pdf
Listing of the type of alleles present in an organisms cell
Genotype
The allele for polydactylism (extra finger or toe) is expressed to different digrees in different
individuals that have it
Variable expressivity
Height in humans is determined by the interaction of genes on different loci
Polygenic inheritance
Individuals that are heterozygous for alleles for A antigen and B antigen have blood type AB
Codominance
All the genes in a cell of organism
Genome
Solution
Listing of the type of alleles present in an organisms cell
Genotype
The allele for polydactylism (extra finger or toe) is expressed to different digrees in different
individuals that have it
Variable expressivity
Height in humans is determined by the interaction of genes on different loci
Polygenic inheritance
Individuals that are heterozygous for alleles for A antigen and B antigen have blood type AB
Codominance
All the genes in a cell of organism
Genome.
An Operating System (OS) is an interface between a computer user and.pdf
An Operating System (OS) is an interface between a computer user and computer hardware. An
operating system is a software which performs all the basic tasks like file management, memory
management, process management, handling input and output.
The kernel is a used for operating system need to be protected from user programs to maintain
the proper operation of the computer system .The kernel is a computer program that constitutes
the central core of a computer\'s operating system. It has complete control over everything that
occurs in the system.As such, it is the first program loaded on startup, and then manages the
remainder of the startup, as well as input/output requests from software, translating them into
data processing instructions for the central processing unit.
kernel is usually loaded into a protected area of memory, which prevents it from being
overwritten by other, less frequently used parts of the operating system or by applications. The
kernel performs its tasks, such as executing processes and handling interrupts, in kernel space,
whereas everything a user normally does, such as writing text in a text editor or running
programs in a GUI, is done in user space. This separation prevents user data and kernel data from
interfering with each other and thereby diminishing performance or causing the system to
become unstable.
Solution
An Operating System (OS) is an interface between a computer user and computer hardware. An
operating system is a software which performs all the basic tasks like file management, memory
management, process management, handling input and output.
The kernel is a used for operating system need to be protected from user programs to maintain
the proper operation of the computer system .The kernel is a computer program that constitutes
the central core of a computer\'s operating system. It has complete control over everything that
occurs in the system.As such, it is the first program loaded on startup, and then manages the
remainder of the startup, as well as input/output requests from software, translating them into
data processing instructions for the central processing unit.
kernel is usually loaded into a protected area of memory, which prevents it from being
overwritten by other, less frequently used parts of the operating system or by applications. The
kernel performs its tasks, such as executing processes and handling interrupts, in kernel space,
whereas everything a user normally does, such as writing text in a text editor or running
programs in a GUI, is done in user space. This separation prevents user data and kernel data from
interfering with each other and thereby diminishing performance or causing the system to
become unstable..
Internal Evidence General.pdf
Internal Evidence General Ledger Files canceled Payrolled
Checks Payroll time records Receiving Reports Minutes of the Board of
Directors Cancelled Notes Payable
External Evidence Vendors Invoices Bank Statements
Purchase Requisitions Remittance Advices Signed Leased Agreements
Duplicate Copies of bills of Lading Notes Receivable
External Evidence are more reliable than the internal evidences because External evidences
are the evidences which we gets from the third party there is no involvement of the
management or employee of the organisations. Internal Evidences are prepared internally
within the organisation and the chances of manipulation in internal evidences are more than
the external evidences.
Solution
Internal Evidence General Ledger Files canceled Payrolled
Checks Payroll time records Receiving Reports Minutes of the Board of
Directors Cancelled Notes Payable
External Evidence Vendors Invoices Bank Statements
Purchase Requisitions Remittance Advices Signed Leased Agreements
Duplicate Copies of bills of Lading Notes Receivable
External Evidence are more reliable than the internal evidences because External evidences
are the evidences which we gets from the third party there is no involvement of the
management or employee of the organisations. Internal Evidences are prepared internally
within the organisation and the chances of manipulation in internal evidences are more than
the external evidences..
In reaction # 3 either the cis or trans diol may .pdf
In reaction # 3 either the cis or trans diol may be used as a reactant. These isomers
are rapidly interconverted under the rearrangement conditions, indicating that the initial water
loss is reversible; a result confirmed by isotopic oxygen exchange. The clear preference for a
methylene group shift versus a methyl group shift may reflect inherent migratory aptitudes, or
possibly group configurations in the 3º-carbocation intermediate. In the conformation shown
here both methyl and methylene groups may shift, or an epoxide ring may be formed reversibly.
An alternative chair-like conformation having an equatorial methyl group should be more stable,
but would not be suitable for a methyl shift. The predominant ring contraction is therefore
understandable. Reaction # 4 is an unusual case in which a strained ring contracts to an even
smaller ring. Phenyl groups generally have a high migratory aptitude, so the failure to obtain 2,2-
diphenylcyclobutanone as a product might seem surprising. However, the carbocation resulting
from a phenyl shift would be just as strained as its precursor; whereas the shift of a ring
methylene group generates an unstrained cation stabilized by phenyl and oxygen substituents.
Conjugative stabilization of the phenyl ketone and absence of sp2 hybridized carbon atoms in the
small ring may also contribute to the stability of the observed product. Finally, reaction # 5
clearly shows the influence of reaction conditions on product composition, but explaining the
manner in which different conditions perturb the outcome is challenging. Treatment with cold
sulfuric acid should produce the more stable diphenyl 3º-carbocation, and a methyl group shift
would then lead to the observed product. The action of a Lewis acid in acetic anhydride, on the
other hand, may selectively acetylate the less hindered dimethyl carbinol. In this event the
acetate becomes the favored leaving group (presumably coordinated with acid), followed by a
1,2-phenyl shift. It is reported that the symmetrically substituted isomeric diol (drawn in the
shaded box) rearranges exclusively by a methyl shift, but the configuration of the starting
material was not stated (two diastereomers are possible). Because of the influence of other
factors (above), it has not been possible to determine an unambiguous migratory order for
substituents in the pinacol rearrangement. However, some general trends are discernible.
Benzopinacol, (C6H5)2C(OH)C(OH)(C6H5)2, undergoes rapid rearrangement to
(C6H5)3CCOC6H5 under much milder conditions than required for pinacol. Indeed, it is often
the case that phenyl or other aromatic substituents adjacent to a forming carbocation will
facilitate that ionization in the course of their migration to the cationic site. In non-aromatic
compounds of the type (CH3)2C(OH)C(OH)RCH3, migration of R increases in the manner R =
CH3 < R = C2H5 << R = (CH3)3C. Since the shifting alkyl group must carry part of the overall
positive charge, alkyl substit.
Intermolecular forces exist between molecules and.pdf
Intermolecular forces exist between molecules and influence the physical
properties. We can think of H2O in its three forms, ice, water and steam. ... However, the
physical properties of H2O are very different in the three states.
Solution
Intermolecular forces exist between molecules and influence the physical
properties. We can think of H2O in its three forms, ice, water and steam. ... However, the
physical properties of H2O are very different in the three states..
H2 concentration will decrease and H2O concentrat.pdf
H2 concentration will decrease and H2O concentration will increase
Solution
H2 concentration will decrease and H2O concentration will increase.
elemental form of sodium and hydrogen is very rea.pdf
elemental form of sodium and hydrogen is very reactive ...that\'s why it is not
advisable to use them in their elemental form
Solution
elemental form of sodium and hydrogen is very reactive ...that\'s why it is not
advisable to use them in their elemental form.
C. Both A and B are correct .pdf
Both statements A and B provided in the previous question are deemed to be correct according to the solution. The solution simply states that both of the options given for the question are right.
C gained 2 hydrogen. It was reduced. It is theref.pdf
C gained 2 hydrogen. It was reduced. It is therefore an oxidizing agent.
Solution
C gained 2 hydrogen. It was reduced. It is therefore an oxidizing agent..
A is stronger acid because it give more H+ to sol.pdf
A is stronger acid because it give more H+ to solution
Solution
A is stronger acid because it give more H+ to solution.
Error message number = int(input(Enter a number )) Name.pdf
Error message
number = int(input(\"Enter a number: \")) NameError: name \'C\' is not defined
Solution
Error message
number = int(input(\"Enter a number: \")) NameError: name \'C\' is not defined.
Dear,The answer is .7.SolutionDear,The answer is .7..pdf
The document is a short note stating that the answer to an unknown problem or question is 0.7. It provides no context or explanation for how the answer was determined. In just two sentences, it directly states that the answer is 0.7 without any additional supporting details or calculations.
Crayfish contain an exoskeleton which means it needs to get rid of i.pdf
Crayfish contain an exoskeleton which means it needs to get rid of it in order to grow in size.
This is called molting. If the food is abundant they will molt several times quite often early in
life. The crayfish hides and crawl out of its exoskeleton through a slit along its dorsal surface. It
will and wait for a while until the new skeleton hardens.
In preparation for molting the crayfish withdraws most of the calcium from its shell, and stores
it in two white \"tablets\"in the sides of its head.
An incomplete cycle of molting can be caused due to mutation in a certain transcription factor
such that it could not bind to the DNA and transcribed the m RNA that encodes ecdysone. So,
the answer of the above qujestion wil be Ans. C.
Solution
Crayfish contain an exoskeleton which means it needs to get rid of it in order to grow in size.
This is called molting. If the food is abundant they will molt several times quite often early in
life. The crayfish hides and crawl out of its exoskeleton through a slit along its dorsal surface. It
will and wait for a while until the new skeleton hardens.
In preparation for molting the crayfish withdraws most of the calcium from its shell, and stores
it in two white \"tablets\"in the sides of its head.
An incomplete cycle of molting can be caused due to mutation in a certain transcription factor
such that it could not bind to the DNA and transcribed the m RNA that encodes ecdysone. So,
the answer of the above qujestion wil be Ans. C..
BasicPizza.javapublic class BasicPizza { String type; String c.pdf
BasicPizza.java
public class BasicPizza {
String type;
String crust;
String ingredients;
double cost;
public BasicPizza(String type) {
super();
this.type = type;
}
public String getType() {
return type;
}
public void setType(String type) {
this.type = type;
}
public String getCrust() {
return crust;
}
public void setCrust(String crust) {
this.crust = crust;
}
public BasicPizza() {
super();
}
public String getIngredients() {
return ingredients;
}
public void setIngredients(String ingredients) {
this.ingredients = ingredients;
}
public double getCost() {
return cost;
}
public void setCost() {
this.cost = 5;
}
@Override
public String toString() {
return \"BasicPizza [type=\" + type + \", crust=\" + crust + \", ingredients=\"
+ ingredients + \", cost=\" + cost + \"]\";
}
}
________________________________________________________________________
LiFiCheese.java
public class LiFiCheese extends BasicPizza{
private String crust;
private double cost;
private String ingredients;
public LiFiCheese() {
super(\"Cheese\");
this.cost=5;
}
public void setCrust()
{
this.crust=\"thin\";
}
public String getCrust()
{
return crust;
}
public double getCost() {
return cost;
}
public void setCost(double cost) {
}
public String getIngredients() {
return ingredients;
}
public void setIngredients(String ingredients) {
this.ingredients = ingredients;
}
@Override
public String toString() {
System.out.println(\"You Ordered :\");
System.out.println(getType());
setCrust();
System.out.println(getCrust());
System.out.println(\"Total Cost of :\"+getCost());
return \"\";
}
}
_______________________________________________________________________
LiFiPizza.java
import java.util.Scanner;
public class LiFiPizza extends BasicPizza {
private String type;
private double cost;
private String crust;
private String ingredients;
public LiFiPizza()
{
super();
this.type = \"Meat\";
this.cost=5;
}
public String getType() {
return type;
}
public void setType(String type) {
this.type = type;
}
public double getCost() {
return cost;
}
public void setCost() {
this.cost = this.cost+2;
}
public String getCrust() {
return crust;
}
public void setCrust(String crust) {
if(crust.equals(\"Thin\"))
{
this.crust=\"Thin\";
}
else if(crust.equals(\"Thick\"))
{
this.crust=\"Thick\";
}
}
public String getIngredients() {
return ingredients;
}
public void setIngredients(String ingredients) {
this.ingredients = ingredients;
}
@Override
public String toString() {
System.out.println(\"You Ordered :\");
System.out.println(getType());
System.out.println(getIngredients()+\"<+$2.00>\");
System.out.println(getCrust());
System.out.println(\"Total Cost of :\"+getCost());
return \"\";
}
}
_____________________________________________________________________
LiFiUnit5Ch14.java
import java.util.Scanner;
public class LiFiUnit5Ch14 {
public static void main(String[] args) {
BasicPizza bp=null;
String typeOfPizza;
Scanner sc=new Scanner(System.in);
System.out.print(\"What type of pizza would you like :\");
typeOfPizza=sc.nex.
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