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D) II, III, IV, and V only Solution D) II, III, IV, and V only.
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Step1 First compound has 2 double bonds ; Second compound has 5 double bonds; Third compound has 1 double bond. Step2 Second molecule requires maximum energy due to 5 double bonds.This is likely to to have UV/ Visible absorption. Solution Step1 First compound has 2 double bonds ; Second compound has 5 double bonds; Third compound has 1 double bond. Step2 Second molecule requires maximum energy due to 5 double bonds.This is likely to to have UV/ Visible absorption..
Step1 First compound has 2 double bonds ; Second .pdf
Step1 First compound has 2 double bonds ; Second .pdf
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Molarity = moles/volume9liter) Molarity of NH4Cl = 4.50 mol /2.00L=2.25 M Solution Molarity = moles/volume9liter) Molarity of NH4Cl = 4.50 mol /2.00L=2.25 M.
Molarity = molesvolume9liter) Molarity of NH4Cl.pdf
Molarity = molesvolume9liter) Molarity of NH4Cl.pdf
anjanacottonmills
HOCH2CH2OH is more soluble in water. more OH groups the alcohol has, more easily to form hydrogen bond with water molecules. As a result, it is more soluble. Solution HOCH2CH2OH is more soluble in water. more OH groups the alcohol has, more easily to form hydrogen bond with water molecules. As a result, it is more soluble..
HOCH2CH2OH is more soluble in water. more OH grou.pdf
HOCH2CH2OH is more soluble in water. more OH grou.pdf
anjanacottonmills
x = -2 Solution x = -2.
x = -2Solutionx = -2.pdf
x = -2Solutionx = -2.pdf
anjanacottonmills
z=(2n+1)/4 where n=all integers Solution z=(2n+1)/4 where n=all integers.
z=(2n+1)4where n=all integersSolutionz=(2n+1)4where n=al.pdf
z=(2n+1)4where n=all integersSolutionz=(2n+1)4where n=al.pdf
anjanacottonmills
This above plant name is Solidago sempervirens, belong to genus Solidago. Species S.sempervirens commonly know as seaside goldenrod. Authority L. Solidago sempervirens is a succulent, herbaceous perennial that reaches heights of 4–6 feet (120- 180 cm). It is unusual in the genus in having toothless, hair-less leaves, thicker than those of most other Solidago species. Flower heads are found in a large paniculiforminflorescence at the top of the plant, often with branches that bend backwards towards the base. This species blooms in late summer and well into the fall, later in the season than most of its relatives. Its fruits are wind-dispersed achenes. They are yellow often, and have sprouts of buds at the end of the short branchesKingdomPlantaePlants, but not fungi, lichens, or algae (from Stearn’s Botanical Latin)SubkingdomTracheobiontaVascular plants—plants with a “circulatory system” for delivering water and nutrientsDivisionMagnoliophytaFlowering plants, also known as angiospermsClassMagnoliopsidaDicotyledons—plants with two initial seed leavesSubclassAsteridaeA large class that encompasses astersOrderAsteralesFlowering plants with a central disk flower and surrounding petals, like daisiesFamilyAsteraceaeThe aster family, which also includes daisies and sunflowers; from the Greek , “star,” for the star-shaped flowersGenusSolidagoFrom Latin solido, “to make whole or heal,” because it was believed these species had healing propertiesSpeciessempervirens\"Evergreen\" Solution This above plant name is Solidago sempervirens, belong to genus Solidago. Species S.sempervirens commonly know as seaside goldenrod. Authority L. Solidago sempervirens is a succulent, herbaceous perennial that reaches heights of 4–6 feet (120- 180 cm). It is unusual in the genus in having toothless, hair-less leaves, thicker than those of most other Solidago species. Flower heads are found in a large paniculiforminflorescence at the top of the plant, often with branches that bend backwards towards the base. This species blooms in late summer and well into the fall, later in the season than most of its relatives. Its fruits are wind-dispersed achenes. They are yellow often, and have sprouts of buds at the end of the short branchesKingdomPlantaePlants, but not fungi, lichens, or algae (from Stearn’s Botanical Latin)SubkingdomTracheobiontaVascular plants—plants with a “circulatory system” for delivering water and nutrientsDivisionMagnoliophytaFlowering plants, also known as angiospermsClassMagnoliopsidaDicotyledons—plants with two initial seed leavesSubclassAsteridaeA large class that encompasses astersOrderAsteralesFlowering plants with a central disk flower and surrounding petals, like daisiesFamilyAsteraceaeThe aster family, which also includes daisies and sunflowers; from the Greek , “star,” for the star-shaped flowersGenusSolidagoFrom Latin solido, “to make whole or heal,” because it was believed these species had healing propertiesSpeciessempervirens\"Evergreen\".
This above plant name is Solidago sempervirens, belong to genus Soli.pdf
This above plant name is Solidago sempervirens, belong to genus Soli.pdf
anjanacottonmills
The Java Program for the above given is: import java.io.File; import java.io.FileNotFoundException; import java.io.IOException; import java.util.Scanner; import java.util.NoSuchElementException; import java.io.PrintWriter; public class InputPrototype { public static void main(String[] args) throws FileNotFoundException { Scanner in = new Scanner(System.in); System.out.print(\"Input File Name: \"); String inputFileName = in.next(); System.out.print(\"Output File Name: \"); String outputFileName = in.next(); File inputFile = new File(inputFileName); Scanner input = new Scanner(inputFile); PrintWriter out = new PrintWriter(outputFileName); double total= 0; while (in.hasNextDouble()) { double value = in.nextDouble(); out.printf(\"%15.2f\ \", value); total = total + value; } out.printf(\"Total: %8.2f\ \", total); in.close(); out.close(); } } } Solution The Java Program for the above given is: import java.io.File; import java.io.FileNotFoundException; import java.io.IOException; import java.util.Scanner; import java.util.NoSuchElementException; import java.io.PrintWriter; public class InputPrototype { public static void main(String[] args) throws FileNotFoundException { Scanner in = new Scanner(System.in); System.out.print(\"Input File Name: \"); String inputFileName = in.next(); System.out.print(\"Output File Name: \"); String outputFileName = in.next(); File inputFile = new File(inputFileName); Scanner input = new Scanner(inputFile); PrintWriter out = new PrintWriter(outputFileName); double total= 0; while (in.hasNextDouble()) { double value = in.nextDouble(); out.printf(\"%15.2f\ \", value); total = total + value; } out.printf(\"Total: %8.2f\ \", total); in.close(); out.close(); } } }.
The Java Program for the above given isimport java.io.File;impo.pdf
The Java Program for the above given isimport java.io.File;impo.pdf
anjanacottonmills
The fidelity of DNA replication determines the genome stability and it plays a main role in evolution of species. The DNA replication fidelity is a key factor of mutations which in turn leads to many human diseases. The high fidelity of DNA replication in E.coli is achieved by DNA polymerase III holoenzyme, which is a complex of 17 proteins. This is responsible for simultaneous replication of both leading and lagging DNA strand with high speed and high fidelity. There are also four other accessory DNA polymerases present in E.coli such as Pol I, II, IV and V. Pol II plays an important role in proofreading of DNA intra-strand crosslinks, DNA strand exposed to UV irradiation, oxidation, stress-induced mutagenesis. This enzyme is responsible for translesion synthesis. Pol IV and V are specific to lagging stand. Pol IV is responsible for proceeding the DNA replication even when there is a mismatch. The Pol IV has a high constitutive intracellular concentration. Pol V is induced constitutively in the absence of exogenous DNA damage. The primary role of Pol I in DNA replication is to remove RNA primers and fill the resulting gap during maturation of Okazaki fragments in the lagging strand. In eukaryotes, the replication of leading and lagging strands takes place by Pol and Pol respectively, which leads to fidelity differences. Whereas in E.coli the replication of both leading and lagging strands takes place by same enzyme polymerase III. This is also one of the reason why E.coli has high fidelity of DNA replication. The tau subunit of Pol III plays a central role in DNA replication fidelity, as it interacts with both leading and lagging strand. This subunit also contributes for polymerase switching during replication. Solution The fidelity of DNA replication determines the genome stability and it plays a main role in evolution of species. The DNA replication fidelity is a key factor of mutations which in turn leads to many human diseases. The high fidelity of DNA replication in E.coli is achieved by DNA polymerase III holoenzyme, which is a complex of 17 proteins. This is responsible for simultaneous replication of both leading and lagging DNA strand with high speed and high fidelity. There are also four other accessory DNA polymerases present in E.coli such as Pol I, II, IV and V. Pol II plays an important role in proofreading of DNA intra-strand crosslinks, DNA strand exposed to UV irradiation, oxidation, stress-induced mutagenesis. This enzyme is responsible for translesion synthesis. Pol IV and V are specific to lagging stand. Pol IV is responsible for proceeding the DNA replication even when there is a mismatch. The Pol IV has a high constitutive intracellular concentration. Pol V is induced constitutively in the absence of exogenous DNA damage. The primary role of Pol I in DNA replication is to remove RNA primers and fill the resulting gap during maturation of Okazaki fragments in the lagging strand. In eukaryotes, the replication of leading .
The fidelity of DNA replication determines the genome stability and .pdf
The fidelity of DNA replication determines the genome stability and .pdf
anjanacottonmills
Recommended
Step1 First compound has 2 double bonds ; Second compound has 5 double bonds; Third compound has 1 double bond. Step2 Second molecule requires maximum energy due to 5 double bonds.This is likely to to have UV/ Visible absorption. Solution Step1 First compound has 2 double bonds ; Second compound has 5 double bonds; Third compound has 1 double bond. Step2 Second molecule requires maximum energy due to 5 double bonds.This is likely to to have UV/ Visible absorption..
Step1 First compound has 2 double bonds ; Second .pdf
Step1 First compound has 2 double bonds ; Second .pdf
anjanacottonmills
Molarity = moles/volume9liter) Molarity of NH4Cl = 4.50 mol /2.00L=2.25 M Solution Molarity = moles/volume9liter) Molarity of NH4Cl = 4.50 mol /2.00L=2.25 M.
Molarity = molesvolume9liter) Molarity of NH4Cl.pdf
Molarity = molesvolume9liter) Molarity of NH4Cl.pdf
anjanacottonmills
HOCH2CH2OH is more soluble in water. more OH groups the alcohol has, more easily to form hydrogen bond with water molecules. As a result, it is more soluble. Solution HOCH2CH2OH is more soluble in water. more OH groups the alcohol has, more easily to form hydrogen bond with water molecules. As a result, it is more soluble..
HOCH2CH2OH is more soluble in water. more OH grou.pdf
HOCH2CH2OH is more soluble in water. more OH grou.pdf
anjanacottonmills
x = -2 Solution x = -2.
x = -2Solutionx = -2.pdf
x = -2Solutionx = -2.pdf
anjanacottonmills
z=(2n+1)/4 where n=all integers Solution z=(2n+1)/4 where n=all integers.
z=(2n+1)4where n=all integersSolutionz=(2n+1)4where n=al.pdf
z=(2n+1)4where n=all integersSolutionz=(2n+1)4where n=al.pdf
anjanacottonmills
This above plant name is Solidago sempervirens, belong to genus Solidago. Species S.sempervirens commonly know as seaside goldenrod. Authority L. Solidago sempervirens is a succulent, herbaceous perennial that reaches heights of 4–6 feet (120- 180 cm). It is unusual in the genus in having toothless, hair-less leaves, thicker than those of most other Solidago species. Flower heads are found in a large paniculiforminflorescence at the top of the plant, often with branches that bend backwards towards the base. This species blooms in late summer and well into the fall, later in the season than most of its relatives. Its fruits are wind-dispersed achenes. They are yellow often, and have sprouts of buds at the end of the short branchesKingdomPlantaePlants, but not fungi, lichens, or algae (from Stearn’s Botanical Latin)SubkingdomTracheobiontaVascular plants—plants with a “circulatory system” for delivering water and nutrientsDivisionMagnoliophytaFlowering plants, also known as angiospermsClassMagnoliopsidaDicotyledons—plants with two initial seed leavesSubclassAsteridaeA large class that encompasses astersOrderAsteralesFlowering plants with a central disk flower and surrounding petals, like daisiesFamilyAsteraceaeThe aster family, which also includes daisies and sunflowers; from the Greek , “star,” for the star-shaped flowersGenusSolidagoFrom Latin solido, “to make whole or heal,” because it was believed these species had healing propertiesSpeciessempervirens\"Evergreen\" Solution This above plant name is Solidago sempervirens, belong to genus Solidago. Species S.sempervirens commonly know as seaside goldenrod. Authority L. Solidago sempervirens is a succulent, herbaceous perennial that reaches heights of 4–6 feet (120- 180 cm). It is unusual in the genus in having toothless, hair-less leaves, thicker than those of most other Solidago species. Flower heads are found in a large paniculiforminflorescence at the top of the plant, often with branches that bend backwards towards the base. This species blooms in late summer and well into the fall, later in the season than most of its relatives. Its fruits are wind-dispersed achenes. They are yellow often, and have sprouts of buds at the end of the short branchesKingdomPlantaePlants, but not fungi, lichens, or algae (from Stearn’s Botanical Latin)SubkingdomTracheobiontaVascular plants—plants with a “circulatory system” for delivering water and nutrientsDivisionMagnoliophytaFlowering plants, also known as angiospermsClassMagnoliopsidaDicotyledons—plants with two initial seed leavesSubclassAsteridaeA large class that encompasses astersOrderAsteralesFlowering plants with a central disk flower and surrounding petals, like daisiesFamilyAsteraceaeThe aster family, which also includes daisies and sunflowers; from the Greek , “star,” for the star-shaped flowersGenusSolidagoFrom Latin solido, “to make whole or heal,” because it was believed these species had healing propertiesSpeciessempervirens\"Evergreen\".
This above plant name is Solidago sempervirens, belong to genus Soli.pdf
This above plant name is Solidago sempervirens, belong to genus Soli.pdf
anjanacottonmills
The Java Program for the above given is: import java.io.File; import java.io.FileNotFoundException; import java.io.IOException; import java.util.Scanner; import java.util.NoSuchElementException; import java.io.PrintWriter; public class InputPrototype { public static void main(String[] args) throws FileNotFoundException { Scanner in = new Scanner(System.in); System.out.print(\"Input File Name: \"); String inputFileName = in.next(); System.out.print(\"Output File Name: \"); String outputFileName = in.next(); File inputFile = new File(inputFileName); Scanner input = new Scanner(inputFile); PrintWriter out = new PrintWriter(outputFileName); double total= 0; while (in.hasNextDouble()) { double value = in.nextDouble(); out.printf(\"%15.2f\ \", value); total = total + value; } out.printf(\"Total: %8.2f\ \", total); in.close(); out.close(); } } } Solution The Java Program for the above given is: import java.io.File; import java.io.FileNotFoundException; import java.io.IOException; import java.util.Scanner; import java.util.NoSuchElementException; import java.io.PrintWriter; public class InputPrototype { public static void main(String[] args) throws FileNotFoundException { Scanner in = new Scanner(System.in); System.out.print(\"Input File Name: \"); String inputFileName = in.next(); System.out.print(\"Output File Name: \"); String outputFileName = in.next(); File inputFile = new File(inputFileName); Scanner input = new Scanner(inputFile); PrintWriter out = new PrintWriter(outputFileName); double total= 0; while (in.hasNextDouble()) { double value = in.nextDouble(); out.printf(\"%15.2f\ \", value); total = total + value; } out.printf(\"Total: %8.2f\ \", total); in.close(); out.close(); } } }.
The Java Program for the above given isimport java.io.File;impo.pdf
The Java Program for the above given isimport java.io.File;impo.pdf
anjanacottonmills
The fidelity of DNA replication determines the genome stability and it plays a main role in evolution of species. The DNA replication fidelity is a key factor of mutations which in turn leads to many human diseases. The high fidelity of DNA replication in E.coli is achieved by DNA polymerase III holoenzyme, which is a complex of 17 proteins. This is responsible for simultaneous replication of both leading and lagging DNA strand with high speed and high fidelity. There are also four other accessory DNA polymerases present in E.coli such as Pol I, II, IV and V. Pol II plays an important role in proofreading of DNA intra-strand crosslinks, DNA strand exposed to UV irradiation, oxidation, stress-induced mutagenesis. This enzyme is responsible for translesion synthesis. Pol IV and V are specific to lagging stand. Pol IV is responsible for proceeding the DNA replication even when there is a mismatch. The Pol IV has a high constitutive intracellular concentration. Pol V is induced constitutively in the absence of exogenous DNA damage. The primary role of Pol I in DNA replication is to remove RNA primers and fill the resulting gap during maturation of Okazaki fragments in the lagging strand. In eukaryotes, the replication of leading and lagging strands takes place by Pol and Pol respectively, which leads to fidelity differences. Whereas in E.coli the replication of both leading and lagging strands takes place by same enzyme polymerase III. This is also one of the reason why E.coli has high fidelity of DNA replication. The tau subunit of Pol III plays a central role in DNA replication fidelity, as it interacts with both leading and lagging strand. This subunit also contributes for polymerase switching during replication. Solution The fidelity of DNA replication determines the genome stability and it plays a main role in evolution of species. The DNA replication fidelity is a key factor of mutations which in turn leads to many human diseases. The high fidelity of DNA replication in E.coli is achieved by DNA polymerase III holoenzyme, which is a complex of 17 proteins. This is responsible for simultaneous replication of both leading and lagging DNA strand with high speed and high fidelity. There are also four other accessory DNA polymerases present in E.coli such as Pol I, II, IV and V. Pol II plays an important role in proofreading of DNA intra-strand crosslinks, DNA strand exposed to UV irradiation, oxidation, stress-induced mutagenesis. This enzyme is responsible for translesion synthesis. Pol IV and V are specific to lagging stand. Pol IV is responsible for proceeding the DNA replication even when there is a mismatch. The Pol IV has a high constitutive intracellular concentration. Pol V is induced constitutively in the absence of exogenous DNA damage. The primary role of Pol I in DNA replication is to remove RNA primers and fill the resulting gap during maturation of Okazaki fragments in the lagging strand. In eukaryotes, the replication of leading .
The fidelity of DNA replication determines the genome stability and .pdf
The fidelity of DNA replication determines the genome stability and .pdf
anjanacottonmills
E= 0.28V reactants are favored Solution E= 0.28V reactants are favored.
E= 0.28V reactants are favored .pdf
E= 0.28V reactants are favored .pdf
anjanacottonmills
The differences between ipv4 and ipv6 are as below 1. Header: 1.Checksum and options feild are present and the length of header with 20 bytes in ipv4. 2.extension header is present with length 40 bytes in ipv6. 2.Addressing scheme: 1. A 32-bit permiting a amont of 2^32 in ipv4. 2. The adress of ipv6 is 128 bit. 3.Translation: Ipv4 is using 4 bytes and ipv6 is is utilizing 16 bytes to transit from ipv4 to ipv6. 4.Data flow: The ipv4 is unable to recognise packet flow for QOS handling.In ipv6 flow label is present to describe packet flow for QOS handling. 5.Duplicate adress detection: in ipv4 ARP is used to detect the duplicate adress and in ipv6 SLAAC +DHPCPV6 is used. 6.Multicast and routing: ipv6 is healthier in multicast routing when compared to ipv6.in ipv4 with TCP protocol is about 576 bytes.where as in ipv6 with 120 bytes. 7.current status of ipv6 adoption: Gooogle dignified ipv6 adoption at arriving 10% positioning spot. Solution The differences between ipv4 and ipv6 are as below 1. Header: 1.Checksum and options feild are present and the length of header with 20 bytes in ipv4. 2.extension header is present with length 40 bytes in ipv6. 2.Addressing scheme: 1. A 32-bit permiting a amont of 2^32 in ipv4. 2. The adress of ipv6 is 128 bit. 3.Translation: Ipv4 is using 4 bytes and ipv6 is is utilizing 16 bytes to transit from ipv4 to ipv6. 4.Data flow: The ipv4 is unable to recognise packet flow for QOS handling.In ipv6 flow label is present to describe packet flow for QOS handling. 5.Duplicate adress detection: in ipv4 ARP is used to detect the duplicate adress and in ipv6 SLAAC +DHPCPV6 is used. 6.Multicast and routing: ipv6 is healthier in multicast routing when compared to ipv6.in ipv4 with TCP protocol is about 576 bytes.where as in ipv6 with 120 bytes. 7.current status of ipv6 adoption: Gooogle dignified ipv6 adoption at arriving 10% positioning spot..
The differences between ipv4 and ipv6 are as below1. Header1.Ch.pdf
The differences between ipv4 and ipv6 are as below1. Header1.Ch.pdf
anjanacottonmills
TCP - TCP breaks data into manageable packets and tracks information such as source and destination of packets. It is able to reroute packets and is responsible for guaranteed delivery of the data. IP - This is a connectionless protocol, which means that a session is not created before sending data. IP is responsible for addressing and routing of packets between computers. It does not guarantee delivery and does not give acknowledgement of packets that are lost or sent out of order as this is the responsibility of higher layer protocols such as TCP. UDP - A connectionless, datagram service that provides an unreliable, best-effort delivery. ICMP - Internet Control Message Protocol enables systems on a TCP/IP network to share status and error information such as with the use of PING and TRACERT utilities. SMTP - Used to reliably send and receive mail over the Internet. FTP - File transfer protocol is used for transferring files between remote systems. Must resolve host name to IP address to establish communication. It is connection oriented (i.e. verifies that packets reach destination). TFTP - Same as FTP but not connection oriented. ARP - provides IP-address to MAC address resolution for IP packets. A MAC address is your computer\'s unique hardware number and appears in the form 00-A0-F1-27-64-E1 (for example). Each computer stores an ARP cache of other computers ARP-IP combinations. POP3 - Post Office Protocol. A POP3 mail server holds mail until the workstation is ready to receive it. IMAP - Like POP3, Internet Message Access Protocol is a standard protocol for accessing e- mail from your local server. IMAP (the latest version is IMAP4) is a client/server protocol in which e-mail is received and held for you by your Internet server. TELNET - Provides a virtual terminal or remote login across the network that is connection- based. The remote server must be running a Telnet service for clients to connect. HTTP - The Hypertext Transfer Protocol is the set of rules for exchanging files (text, graphic images, sound, video, and other multimedia files) on the World Wide Web. It is the protocol controlling the transfer and addressing of HTTP requests and responses. HTTPS - Signifies that a web page is using the Secure Sockets Layer (SSL) protocol and is providing a secure connection. This is used for secure internet business transactions. NTP - Network Time Protocol is a protocol that is used to synchronize computer clock times in a network of computers. SNMP - Stands for Simple Network Management Protocol and is used for monitoring and status information on a network. SNMP can be used to monitor any device that is SNMP capable and this can include computers, printers, routers, servers, gateways and many more using agents on the target systems. The agents report information back to the management systems by the use of “traps” which capture snapshot data of the system. This trap information could be system errors, resource information, or other info.
TCP - TCP breaks data into manageable packets and tracks information.pdf
TCP - TCP breaks data into manageable packets and tracks information.pdf
anjanacottonmills
Static arrays are structures whose size is fixed at compile time and therefore cannot be extended or reduced to fit the data set. A dynamic array can be extended by doubling the size but there is overhead associated with the operation of copying old data and freeing the memory associated with the old data structure. One potential problem of using arrays for storing data is that arrays require a contiguous block of memory which may not be available, if the requested contiguous block is too large. However the advantages of using arrays are that each element in the array can be accessed very efficiently using an index. However, for applications that can be better managed without using contiguous memory we define a concept called “linked lists”. A linked list is a collection of objects linked together by references from one object to another object. By convention these objects are named as nodes. So the basic linked list is collection of nodes where each node contains one or more data fields AND a reference to the next node. The last node points to a NULL reference to indicate the end of the list. Types of Linked Lists Linked lists are widely used in many applications because of the flexibility it provides. Unlike arrays that are dynamically assigned, linked lists do not require memory from a contiguous block. This makes it very appealing to store data in a linked list, when the data set is large or device (eg: PDA) has limited memory. One of the disadvantages of linked lists is that they are not random accessed like arrays. To find information in a linked list one must start from the head of the list and traverse the list sequentially until it finds (or not find) the node. Another advantage of linked lists over arrays is that when a node is inserted or deleted, there is no need to “adjust” the array. There are few different types of linked lists. A singly linked list as described above provides access to the list from the head node. Traversal is allowed only one way and there is no going back. A doubly linked list is a list that has two references, one to the next node and another to previous node. Doubly linked list also starts from head node, but provide access both ways. That is one can traverse forward or backward from any node. A multilinked list (see figures 1 & 2) is a more general linked list with multiple links from nodes. For examples, we can define a Node that has two references, age pointer and a name pointer. With this structure it is possible to maintain a single list, where if we follow the name pointer we can traverse the list in alphabetical order of names and if we traverse the age pointer, we can traverse the list sorted by ages. This type of node organization may be useful for maintaining a customer list in a bank where same list can be traversed in any order (name, age, or any other criteria) based on the need. Designing the Node of a Linked List Linked list is a collection of linked nodes. A node is a struct with at least a.
Static arrays are structures whose size is fixed at compile time and.pdf
Static arrays are structures whose size is fixed at compile time and.pdf
anjanacottonmills
NumberList.java (implements the linked list) public class NumberList{ Node first; Node last; public NumberList(){ first = null; last = null; } public NumberList(Node node){ this.first = node; this.last = node; } public boolean isEmpty(){ return first == null; } public void setLast(Node node){ this.last = node; } public void insert(Node node){ if (first==null){ this.first = node; this.last = node; } else{ node.previous = this.last; node.previous.next = node; node.next = null; this.last = node; } } public boolean inList(String num){ Node l = first; while (l.next != null){ if (l.number.equals(num)){ return true; } else{ l = l.next; } } return false; } public void printList(){ Node l = first; while(l.next!=null){ System.out.print(l.number+\" \"); l = l.next; } System.out.println(); } } Node.java (implements a single node in the linked list; stores number and pointers) public class Node{ String number; Node previous; Node next; public Node(String num){ number = num; previous = null; next = null; } public Node(String num, Node p, Node n){ number = num; previous = p; next = n; } } Main.java (contains the main method and the helper methods to solve the questions given) import java.util.Scanner; import java.util.List; import java.util.ArrayList; public class Main{ static boolean checkHappiness(String num){ Node current = new Node(num, null, null); NumberList numberList = new NumberList(current); int len = num.length(); int resultant = 0; for (int i=0; i happyNumbersfrom1to10000(){ ArrayList numbers = new ArrayList(); for (int j=0; j<10000; j++){ if (checkHappiness(String.valueOf(j+1))){ numbers.add(String.valueOf(j+1)); } } return numbers; } static void happyNumbersfrom9001to10000(){ for (int j=9001; j<=10000; j++){ //System.out.println(j); if (checkHappiness(String.valueOf(j))){ System.out.println(String.valueOf(j)); } } } static String getLargeHappyNumber(){ for (long i=10000000000000000000L; i<1000000000000000000000L; i++){ if (checkHappiness(String.valueOf(i))){ return String.valueOf(i); } return \"-1\"; } } static String getLargeUnhappyNumber(){ for (long i=10000000000000000000L; i<1000000000000000000000L; i++){ if (!checkHappiness(String.valueOf(i))){ return String.valueOf(i); } return \"-1\"; } } public static void main(String[] args){ happyNumbersfrom9001to10000(); System.out.println(happyNumbersfrom1to10000()); } } Solution NumberList.java (implements the linked list) public class NumberList{ Node first; Node last; public NumberList(){ first = null; last = null; } public NumberList(Node node){ this.first = node; this.last = node; } public boolean isEmpty(){ return first == null; } public void setLast(Node node){ this.last = node; } public void insert(Node node){ if (first==null){ this.first = node; this.last = node; } else{ node.previous = this.last; node.previous.next = node; node.next = null; this.last = node; } } public boolean inList(String num){ Node l = first; while (l.next != null){ if (l.number.equals(num)){ return true; } else{ l = l.
NumberList.java (implements the linked list)public class NumberLis.pdf
NumberList.java (implements the linked list)public class NumberLis.pdf
anjanacottonmills
c) H2(g) + 1/2 O2(g) ->H2O (l) as all are in there natural state. therefore it will represent the enthalpy of formation Solution c) H2(g) + 1/2 O2(g) ->H2O (l) as all are in there natural state. therefore it will represent the enthalpy of formation.
c) H2(g) + 12 O2(g) -H2O (l) as all are in the.pdf
c) H2(g) + 12 O2(g) -H2O (l) as all are in the.pdf
anjanacottonmills
MagicSquareTest.java import java.util.Scanner; public class MagicSquareTest { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner scan = new Scanner(System.in); System.out.println(\"Enter Number of rows:\"); int rows = scan.nextInt(); System.out.println(\"Enter Nummber of columns:\"); int cols = scan.nextInt(); int a[][] = new int[rows][cols]; System.out.println(\"Enter matrix elements;\"); for(int i=0; i Solution MagicSquareTest.java import java.util.Scanner; public class MagicSquareTest { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner scan = new Scanner(System.in); System.out.println(\"Enter Number of rows:\"); int rows = scan.nextInt(); System.out.println(\"Enter Nummber of columns:\"); int cols = scan.nextInt(); int a[][] = new int[rows][cols]; System.out.println(\"Enter matrix elements;\"); for(int i=0; i.
MagicSquareTest.java import java.util.Scanner;public class Mag.pdf
MagicSquareTest.java import java.util.Scanner;public class Mag.pdf
anjanacottonmills
Interphase. This is the phase where the cell prepares for the next round of division. It has three phases further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in size. The checkpoints ensure that the cell is normal and all the things necessary for DNA synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only very short period of the cell cycle. Solution Interphase. This is the phase where the cell prepares for the next round of division. It has three phases further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in size. The checkpoints ensure that the cell is normal and all the things necessary for DNA synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only very short period of the cell cycle..
Interphase.This is the phase where the cell prepares for the next .pdf
Interphase.This is the phase where the cell prepares for the next .pdf
anjanacottonmills
B.the rotation will be zero as L and D will cancel out each other.. Solution B.the rotation will be zero as L and D will cancel out each other...
B.the rotation will be zero as L and D will cance.pdf
B.the rotation will be zero as L and D will cance.pdf
anjanacottonmills
In the above conversation it is belonging to stereotypes. Stereotypes are a form of generalization. When we generalize, we group or classify people, places, or things according to the traits they have in common. For example, we may say most Masai tribesmen are unusually tall or Scandinavians are usually fair-skinned. If our observations are careless or too limited, the generalization may be faulty, as when someone says, \"Hollywood hasn\'t produced any quality movies in the past fifteen years.\" But stereotypes are more serious than mere faulty generalizations. They are fixed, unbending generalizations about people, places, or things. When a stereotype is challenged, the person who holds it is unlikely to modify or discard it, because it is based on a distortion of perception. As Walter Lippmann explains, when we stereotype,\" we do not so much see this man and that sunset; rather we notice that the thing is man or sunset, and then see chiefly what our mind is full of on those subjects.\"75 The most common kinds of stereotyping are ethnic and religious. Jews are shrewd and cunning, clannish, have a financial genius matched only by their greed. Italians are hot-tempered, coarse, and sensual. The Irish, like the Poles, are big and stupid; in addition, they brawl, lust after heavy liquor and light conversation. Blacks are primitive and slowwitted. (Often each of these stereotypes includes a virtue or two – Jews are good family members, Italians artistic, Poles brave, the Irish devout, blacks athletic.) Beyond these stereotypes are numerous other, less common ones; Swedish women, foreign film directors, Southern senators, physical education instructors, fundamentalist clergymen, agnostics, atheists, democrats, republicans, Mexicans, scientist, prostitutes, politicians, English teachers, psychiatrists, construction workers, black militants, college dropouts, homosexuals, and society matrons. There are stereotypes of institutions as well: marriage, the church, government, the military, the Founding Fathers, Western culture, the Judeo-Christian tradition. (A full list would include even God and mother.) FACTS DON\'T MATTER It is pleasant to assume that when the facts are known, stereotypes disappear. However, that is seldom the case. The late Harvard psychologist Gordon Allport, in The nature of Prejudice, pointed out that \"it is possible for a stereotype to grow in defiance of all evidence…\"76 People who stereotype don\'t dust accept the facts that are offered to them. They measure those facts against what they already \"know.\" That is, they measure them against the stereotype itself. Instead of seeing that the stereotype is false and therefore dismissing it, they reject the unfamiliar facts. People who think in terms of stereotypes tend to be selective in their perceptions. They reject conditions that challenge their preformed judgment and retain those that reinforce it. Thus a person can notice the Jewish employer promoting another Jew but ign.
In the above conversation it is belonging to stereotypes.Stereotyp.pdf
In the above conversation it is belonging to stereotypes.Stereotyp.pdf
anjanacottonmills
In general, the objective of an audit is to assess the risk of material misstatements in the financial statements. Material misstatements can arise from inadequacies in internal controls and from inaccurate management assertions. Thus, testing the validity of the various implicit managerial assertions is a key objective of an auditor. Existence and Completeness Auditing standards require that auditors test basic underlying management assertions implicit in the financial statements. Key among these various assertions are existence or occurrence, which describe a singular concept: Journal entries are not fiction. As the name implies, an auditor will conduct various procedures to verify that assets do in fact exist and that recorded transactions did in fact occur. Additionally, an auditor will seek evidence of completeness, so that the financial statements include all material transactions that occurred, and so the records do not omit material transactions for any reason. Rights and Obligations The various rights and obligations of the company are important management assertions inherent in the financial statements. Thus, an auditor will obtain evidence regarding a company\'s rights, such as proper title to assets and status of intellectual property. An auditor will be concerned with assertions relating to the company\'s obligations, such as accounts payable balances, long- term debts and tax liabilities. Thus, the audit objectives will be fulfilled upon validating these specific assertions. Valuation or Allocation Valuation or allocation are managerial assertions which are often material to the financial statements; thus, an auditor will diligently conduct audit procedures relating to these objectives. Generally accepted accounting principles, or GAAP, require that certain balance sheet items be presented using different valuation methodologies. Meeting these standards is a key audit objective, as the risk of material misstatement is low in probability, but high in magnitude. Thus, among other things, the historical cost of assets is verified, depreciation methods are scrutinized and the fair value of investments are calculated to satisfy this objective. Presentation and Disclosure Another specific audit objective is validating the presentation of the financial statements and the adequacy of the disclosures therein. Financial statements should conform to certain requirements and expectations, and should include the balance sheet, income statement, statement of cash flows and the statement of owner\'s equity. Relating to disclosure, the auditor will consider the sufficiency and clarity of footnotes and the transparency in management discussion and analysis, so he can assess the risk of material misstatement and fulfill the audit objective. Specific audit objectives are the application of the general audit objectives to a given class of transactions, account balance, or presentation and disclosure. There must be at least one specific audit object.
In general, the objective of an audit is to assess the risk of mater.pdf
In general, the objective of an audit is to assess the risk of mater.pdf
anjanacottonmills
if one can understand a few things it is easier to solve these kind of questions. first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and make Z and Y products in the presence of lactose only. I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose. Is is a dominant mutation will stop both the products Z and Y. if you have atlease one Is in the question neither the products will from no matter whether the lactose is presence or not. then comes to Operator (O) ,O+ is a wild type and can produce both the products of Z and Y. in the presence of lactose.and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. next comes to Z+ represents production of -galactosidase and Z- represents no production of - galactosidase. 1.Ans. I+P-O+Z+Y+ In this haploid genotype the transcription of operon genes totally absent. Transcription of structural genes is not occured. because here P- indicates the mutation in the promoter site, the RNA polymerase unable to bind to the Promoter and initiate transcription of structural genes. so these cells cannot able to grow on media which contain lactose as sole carbon source. the Strain is lac- 2.Ans.I+P+OcZ+Y- here Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose means constitutively. but here only the product beta galctosidase produced constitutively and the permease is not produced because Permease gene is is mutated (Y-). these cells also cannot able to grow on media which contain lactose as sole carbon source.because the absence of functional permease enzyme so the Strain is lac- 3.Ans. I-P+OcZ+Y+ here I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. so this haploid genotype transcribes operon genes constitutively. and both the products of Z and Y expressed continuously, so these cells able to grow on the media which contain lactose as sole carbon source. the strain is lac+ b. partially diploid cap-I+P+O+Z-Y+/Cap+I-P+O+Z+Y- these cells can able to grow on lactose medium because these cells have presence of wild promoter which can able to active the transcription of structural genes. so these cells can able to grow on the media which contain lactose as sole carbon source. ii. Transcription of beta-galactosidase and permease is Inducible though diploid cell have mutated inducer region which can able to activate constitutive expression but Wild type (I+) is dominant over I- . so it is inducible operon. Solution if one can understand a few things it is easier to solve these kind of questions. first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and make Z and Y products in the presence of lactose only. I- is a recessive mutation .
if one can understand a few things it is easier to solve these kind .pdf
if one can understand a few things it is easier to solve these kind .pdf
anjanacottonmills
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancy Balanced MolecularEquation is : CuCl2 +H2SiO4 ---> CuSiO4 (s) +2HCl Total ionic Equation is : Cu2+ +2 Cl - + 2H+ +SO42- ---> CuSiO4 +2H+ + 2Cl- Spectator ions are : Cl - &H+ Spectator ions are : Cl - &H+ Net ionic Equation is : Cu2+ + SO42- --->CuSiO4 Solution I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancy Balanced MolecularEquation is : CuCl2 +H2SiO4 ---> CuSiO4 (s) +2HCl Total ionic Equation is : Cu2+ +2 Cl - + 2H+ +SO42- ---> CuSiO4 +2H+ + 2Cl- Spectator ions are : Cl - &H+ Spectator ions are : Cl - &H+ Net ionic Equation is : Cu2+ + SO42- --->CuSiO4.
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancyBalanc.pdf
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancyBalanc.pdf
anjanacottonmills
fixed 3 issues. I am working on remaining. package chegg; public class HugeIntegerApp { public static void main(String args[]) { String[][] testInputs = { { \"987654321\", \"234567890\" }, { \"987654321\", \"-234567890\" }, { \"-987654321\", \"234567890\" }, { \"-987654321\", \"-234567890\" }, { \"234567890\", \"987654321\" }, { \"234567890\", \"-987654321\" }, { \"-234567890\", \"987654321\" }, { \"-234567890\", \"-987654321\" } }; for (String[] ints : testInputs) { HugeInteger h1 = new HugeInteger(ints[0]); HugeInteger h2 = new HugeInteger(ints[1]); HugeInteger hcopy=new HugeInteger(ints[0]); HugeInteger hcopy1=new HugeInteger(ints[0]); HugeInteger hcopy2=new HugeInteger(ints[0]); System.out.println(\"h1=\" + h1); System.out.println(\"h2=\" + h2); if (h1.isEqualTo(h2)) { System.out.println(\"h1 is equal to h2.\"); } if (h1.isNotEqualTo(h2)) { System.out.println(\"h1 is not equal to h2.\"); } if (h1.isGreaterThan(h2)) { System.out.println(\"h1 is greater than h2.\"); } if (h1.isLessThan(h2)) { System.out.println(\"h1 is less than to h2.\"); } if (h1.isGreaterThanOrEqualTo(h2)) { System.out.println(\"h1 is greater than or equal to h2.\"); } if (h1.isLessThanOrEqualTo(h2)) { System.out.println(\"h1 is less than or equal to h2.\"); } h1.add(h2); // h1 += h2 System.out.println(\"h1.add(h2);\"); System.out.printf(\"h1=%s\ \", h1); hcopy.subtract(h2); // h1 -= h2 System.out.println(\"h1.subtract(h2);\"); System.out.println(\"h1.subtract(h2);\"); System.out.printf(\"h1=%s\ \", hcopy); //h1.subtract(h2); // h1 -= h2 hcopy1.add(h2); // h1 += h2 System.out.println(\"h1.add(h2);\"); hcopy2.multiply(h2); // h1 *= h2 System.out.println(\"h1.multiply(h2);\"); System.out.printf(\"h1=%s\ \ \", h1); } } } package chegg; public class HugeInteger { private static final int NUM_DIGITS = 40; private int digits[] = new int[NUM_DIGITS]; private boolean positive; public HugeInteger(String num) { int len = num.length(); if (num.indexOf(0) == \'-\') { len--; positive = false; } else { positive = true; } for (int i = 0; i < len; i++) { digits[NUM_DIGITS - 1 - i] = num.charAt(len - 1 - i) - (int) \'0\'; digits.toString(); } } public void multiply(HugeInteger hi) { this.digits = multiplyArraydigits(this.digits, hi.digits); } private int[] multiplyArraydigits(int[] array1, int[] array2) { int[] result = new int[array1.length + array2.length]; int carry = 0; int set = 0; for (int i = array1.length - 1; i >= 0; i--) { int temp = result.length - 1 - set; for (int j = array2.length - 1; j >= 0; j--) { int sum = result[temp] + (array1[i] * array2[j]) + carry; result[temp] = sum % 10; carry = sum / 10; temp--; } if (carry > 0) { result[temp] += carry; while (result[temp] >= 10) { carry = result[temp] / 10; result[temp] = result[temp] % 10; result[temp--] += carry; } carry = 0; } set++; } return result; } public void subtract(HugeInteger hi) { // if the signs of the two numbers are not the same if (this.positive != hi.positive) { if (this.positive) { // \'this\' is positive... .
fixed 3 issues. I am working on remaining.package chegg;public c.pdf
fixed 3 issues. I am working on remaining.package chegg;public c.pdf
anjanacottonmills
class Mesh{ public: /*Mesh data*? vectorvertices; vector indices; vectortextures; ?*functions*? Mesh(vectorvertices,vectorindices,vector textures); void draw(shader shader); private: /* Render data*/ gluint vao,vbo,ebo; /* functions */ void setupmesh(); }; Solution class Mesh{ public: /*Mesh data*? vectorvertices; vector indices; vectortextures; ?*functions*? Mesh(vectorvertices,vectorindices,vector textures); void draw(shader shader); private: /* Render data*/ gluint vao,vbo,ebo; /* functions */ void setupmesh(); };.
class Mesh{publicMesh datavectorvertexvertices;vector.pdf
class Mesh{publicMesh datavectorvertexvertices;vector.pdf
anjanacottonmills
Bacteria was used in transformation experiment, to prove Dna was the genetic material and replicated semi conservatively, used in gene expression and regulation, preferred host for cloning genes, and in manipulation of DNA Solution Bacteria was used in transformation experiment, to prove Dna was the genetic material and replicated semi conservatively, used in gene expression and regulation, preferred host for cloning genes, and in manipulation of DNA.
Bacteria was used in transformation experiment, to prove Dna was the.pdf
Bacteria was used in transformation experiment, to prove Dna was the.pdf
anjanacottonmills
c) 11 Solution c) 11.
c) 11Solutionc) 11.pdf
c) 11Solutionc) 11.pdf
anjanacottonmills
assembly mips subroutine Solution assembly mips subroutine.
assembly mips subroutineSolutionassembly mips subroutine.pdf
assembly mips subroutineSolutionassembly mips subroutine.pdf
anjanacottonmills
1. Tyrosinase is involved in the synthesis of Melanin pigment in mammalian cells by using tyrosine as a substrate. In general phenolic compounds are best substrates for tyrosinase as it is responsible for browning of fruits and cut vegetables. 2. No, tyrosinase does not exhibit high level of specificity because it has broad substrate specificity for phenolic compounds. 3. Tyrosinase exhibit group specificity that is the presence of functional group which is phenol group. 4. Inhibition can be competitive or non-competitive. Competitive inhibitors are substrate analogs that is it resemble the substrate. Non-competitive inhibitors bind to the region other than active site for substrate binding. Benzoic acid is the substrate analog so it is competitive inhibitor. Phenylthiourea and resorcinol is a non-competitive inhibitor. 6. 0 C is freezing temperature and 70 C is denaturing temperature for enzyme. Enzyme reaction is best at the optimum temperature or physiological temperature. Solution 1. Tyrosinase is involved in the synthesis of Melanin pigment in mammalian cells by using tyrosine as a substrate. In general phenolic compounds are best substrates for tyrosinase as it is responsible for browning of fruits and cut vegetables. 2. No, tyrosinase does not exhibit high level of specificity because it has broad substrate specificity for phenolic compounds. 3. Tyrosinase exhibit group specificity that is the presence of functional group which is phenol group. 4. Inhibition can be competitive or non-competitive. Competitive inhibitors are substrate analogs that is it resemble the substrate. Non-competitive inhibitors bind to the region other than active site for substrate binding. Benzoic acid is the substrate analog so it is competitive inhibitor. Phenylthiourea and resorcinol is a non-competitive inhibitor. 6. 0 C is freezing temperature and 70 C is denaturing temperature for enzyme. Enzyme reaction is best at the optimum temperature or physiological temperature..
1. Tyrosinase is involved in the synthesis of Melanin pigment in mam.pdf
1. Tyrosinase is involved in the synthesis of Melanin pigment in mam.pdf
anjanacottonmills
1) A covalent crystal differs from an ionic crystal in that a)only one of them is a poor electrical conductor. b)only one of them has bonding that involves overlap of electronic orbitals. c)only one of them possesses atomic order. d)only one of them has high melting points. e)only one of them is hard. The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic crystals involve electrostatic bonding between a positive and negative ion. 2)What are the important intermolecular forces acting in CHF3? a)London dispersion forces only b)only dipole-dipole interactions and H-bonding c)dipole-dipole interactions and London dispersion forces d)H-bonding, dipole-dipole interactions and London dispersion forces e)ionic bonding In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and Fluorine are all capable of hydrogen bonding.) Solution 1) A covalent crystal differs from an ionic crystal in that a)only one of them is a poor electrical conductor. b)only one of them has bonding that involves overlap of electronic orbitals. c)only one of them possesses atomic order. d)only one of them has high melting points. e)only one of them is hard. The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic crystals involve electrostatic bonding between a positive and negative ion. 2)What are the important intermolecular forces acting in CHF3? a)London dispersion forces only b)only dipole-dipole interactions and H-bonding c)dipole-dipole interactions and London dispersion forces d)H-bonding, dipole-dipole interactions and London dispersion forces e)ionic bonding In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and Fluorine are all capable of hydrogen bonding.).
1) A covalent crystal differs from an ionic crystal in that a)on.pdf
1) A covalent crystal differs from an ionic crystal in that a)on.pdf
anjanacottonmills
https://medicaleducationelearning.blogspot.com/2024/02/using-micro-scholarship-to-incentivize.html
Micro-Scholarship, What it is, How can it help me.pdf
Micro-Scholarship, What it is, How can it help me.pdf
Poh-Sun Goh
Wizards are very useful for creating a good user experience. In all businesses, interactive sessions are most beneficial. To improve the user experience, wizards in Odoo provide an interactive session. For creating wizards, we can use transient models or abstract models. This gives features of a model class except the data storing. Transient and abstract models have permanent database persistence. For them, database tables are made, and the records in such tables are kept until they are specifically erased.
How to Create and Manage Wizard in Odoo 17
How to Create and Manage Wizard in Odoo 17
Celine George
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E= 0.28V reactants are favored Solution E= 0.28V reactants are favored.
E= 0.28V reactants are favored .pdf
E= 0.28V reactants are favored .pdf
anjanacottonmills
The differences between ipv4 and ipv6 are as below 1. Header: 1.Checksum and options feild are present and the length of header with 20 bytes in ipv4. 2.extension header is present with length 40 bytes in ipv6. 2.Addressing scheme: 1. A 32-bit permiting a amont of 2^32 in ipv4. 2. The adress of ipv6 is 128 bit. 3.Translation: Ipv4 is using 4 bytes and ipv6 is is utilizing 16 bytes to transit from ipv4 to ipv6. 4.Data flow: The ipv4 is unable to recognise packet flow for QOS handling.In ipv6 flow label is present to describe packet flow for QOS handling. 5.Duplicate adress detection: in ipv4 ARP is used to detect the duplicate adress and in ipv6 SLAAC +DHPCPV6 is used. 6.Multicast and routing: ipv6 is healthier in multicast routing when compared to ipv6.in ipv4 with TCP protocol is about 576 bytes.where as in ipv6 with 120 bytes. 7.current status of ipv6 adoption: Gooogle dignified ipv6 adoption at arriving 10% positioning spot. Solution The differences between ipv4 and ipv6 are as below 1. Header: 1.Checksum and options feild are present and the length of header with 20 bytes in ipv4. 2.extension header is present with length 40 bytes in ipv6. 2.Addressing scheme: 1. A 32-bit permiting a amont of 2^32 in ipv4. 2. The adress of ipv6 is 128 bit. 3.Translation: Ipv4 is using 4 bytes and ipv6 is is utilizing 16 bytes to transit from ipv4 to ipv6. 4.Data flow: The ipv4 is unable to recognise packet flow for QOS handling.In ipv6 flow label is present to describe packet flow for QOS handling. 5.Duplicate adress detection: in ipv4 ARP is used to detect the duplicate adress and in ipv6 SLAAC +DHPCPV6 is used. 6.Multicast and routing: ipv6 is healthier in multicast routing when compared to ipv6.in ipv4 with TCP protocol is about 576 bytes.where as in ipv6 with 120 bytes. 7.current status of ipv6 adoption: Gooogle dignified ipv6 adoption at arriving 10% positioning spot..
The differences between ipv4 and ipv6 are as below1. Header1.Ch.pdf
The differences between ipv4 and ipv6 are as below1. Header1.Ch.pdf
anjanacottonmills
TCP - TCP breaks data into manageable packets and tracks information such as source and destination of packets. It is able to reroute packets and is responsible for guaranteed delivery of the data. IP - This is a connectionless protocol, which means that a session is not created before sending data. IP is responsible for addressing and routing of packets between computers. It does not guarantee delivery and does not give acknowledgement of packets that are lost or sent out of order as this is the responsibility of higher layer protocols such as TCP. UDP - A connectionless, datagram service that provides an unreliable, best-effort delivery. ICMP - Internet Control Message Protocol enables systems on a TCP/IP network to share status and error information such as with the use of PING and TRACERT utilities. SMTP - Used to reliably send and receive mail over the Internet. FTP - File transfer protocol is used for transferring files between remote systems. Must resolve host name to IP address to establish communication. It is connection oriented (i.e. verifies that packets reach destination). TFTP - Same as FTP but not connection oriented. ARP - provides IP-address to MAC address resolution for IP packets. A MAC address is your computer\'s unique hardware number and appears in the form 00-A0-F1-27-64-E1 (for example). Each computer stores an ARP cache of other computers ARP-IP combinations. POP3 - Post Office Protocol. A POP3 mail server holds mail until the workstation is ready to receive it. IMAP - Like POP3, Internet Message Access Protocol is a standard protocol for accessing e- mail from your local server. IMAP (the latest version is IMAP4) is a client/server protocol in which e-mail is received and held for you by your Internet server. TELNET - Provides a virtual terminal or remote login across the network that is connection- based. The remote server must be running a Telnet service for clients to connect. HTTP - The Hypertext Transfer Protocol is the set of rules for exchanging files (text, graphic images, sound, video, and other multimedia files) on the World Wide Web. It is the protocol controlling the transfer and addressing of HTTP requests and responses. HTTPS - Signifies that a web page is using the Secure Sockets Layer (SSL) protocol and is providing a secure connection. This is used for secure internet business transactions. NTP - Network Time Protocol is a protocol that is used to synchronize computer clock times in a network of computers. SNMP - Stands for Simple Network Management Protocol and is used for monitoring and status information on a network. SNMP can be used to monitor any device that is SNMP capable and this can include computers, printers, routers, servers, gateways and many more using agents on the target systems. The agents report information back to the management systems by the use of “traps” which capture snapshot data of the system. This trap information could be system errors, resource information, or other info.
TCP - TCP breaks data into manageable packets and tracks information.pdf
TCP - TCP breaks data into manageable packets and tracks information.pdf
anjanacottonmills
Static arrays are structures whose size is fixed at compile time and therefore cannot be extended or reduced to fit the data set. A dynamic array can be extended by doubling the size but there is overhead associated with the operation of copying old data and freeing the memory associated with the old data structure. One potential problem of using arrays for storing data is that arrays require a contiguous block of memory which may not be available, if the requested contiguous block is too large. However the advantages of using arrays are that each element in the array can be accessed very efficiently using an index. However, for applications that can be better managed without using contiguous memory we define a concept called “linked lists”. A linked list is a collection of objects linked together by references from one object to another object. By convention these objects are named as nodes. So the basic linked list is collection of nodes where each node contains one or more data fields AND a reference to the next node. The last node points to a NULL reference to indicate the end of the list. Types of Linked Lists Linked lists are widely used in many applications because of the flexibility it provides. Unlike arrays that are dynamically assigned, linked lists do not require memory from a contiguous block. This makes it very appealing to store data in a linked list, when the data set is large or device (eg: PDA) has limited memory. One of the disadvantages of linked lists is that they are not random accessed like arrays. To find information in a linked list one must start from the head of the list and traverse the list sequentially until it finds (or not find) the node. Another advantage of linked lists over arrays is that when a node is inserted or deleted, there is no need to “adjust” the array. There are few different types of linked lists. A singly linked list as described above provides access to the list from the head node. Traversal is allowed only one way and there is no going back. A doubly linked list is a list that has two references, one to the next node and another to previous node. Doubly linked list also starts from head node, but provide access both ways. That is one can traverse forward or backward from any node. A multilinked list (see figures 1 & 2) is a more general linked list with multiple links from nodes. For examples, we can define a Node that has two references, age pointer and a name pointer. With this structure it is possible to maintain a single list, where if we follow the name pointer we can traverse the list in alphabetical order of names and if we traverse the age pointer, we can traverse the list sorted by ages. This type of node organization may be useful for maintaining a customer list in a bank where same list can be traversed in any order (name, age, or any other criteria) based on the need. Designing the Node of a Linked List Linked list is a collection of linked nodes. A node is a struct with at least a.
Static arrays are structures whose size is fixed at compile time and.pdf
Static arrays are structures whose size is fixed at compile time and.pdf
anjanacottonmills
NumberList.java (implements the linked list) public class NumberList{ Node first; Node last; public NumberList(){ first = null; last = null; } public NumberList(Node node){ this.first = node; this.last = node; } public boolean isEmpty(){ return first == null; } public void setLast(Node node){ this.last = node; } public void insert(Node node){ if (first==null){ this.first = node; this.last = node; } else{ node.previous = this.last; node.previous.next = node; node.next = null; this.last = node; } } public boolean inList(String num){ Node l = first; while (l.next != null){ if (l.number.equals(num)){ return true; } else{ l = l.next; } } return false; } public void printList(){ Node l = first; while(l.next!=null){ System.out.print(l.number+\" \"); l = l.next; } System.out.println(); } } Node.java (implements a single node in the linked list; stores number and pointers) public class Node{ String number; Node previous; Node next; public Node(String num){ number = num; previous = null; next = null; } public Node(String num, Node p, Node n){ number = num; previous = p; next = n; } } Main.java (contains the main method and the helper methods to solve the questions given) import java.util.Scanner; import java.util.List; import java.util.ArrayList; public class Main{ static boolean checkHappiness(String num){ Node current = new Node(num, null, null); NumberList numberList = new NumberList(current); int len = num.length(); int resultant = 0; for (int i=0; i happyNumbersfrom1to10000(){ ArrayList numbers = new ArrayList(); for (int j=0; j<10000; j++){ if (checkHappiness(String.valueOf(j+1))){ numbers.add(String.valueOf(j+1)); } } return numbers; } static void happyNumbersfrom9001to10000(){ for (int j=9001; j<=10000; j++){ //System.out.println(j); if (checkHappiness(String.valueOf(j))){ System.out.println(String.valueOf(j)); } } } static String getLargeHappyNumber(){ for (long i=10000000000000000000L; i<1000000000000000000000L; i++){ if (checkHappiness(String.valueOf(i))){ return String.valueOf(i); } return \"-1\"; } } static String getLargeUnhappyNumber(){ for (long i=10000000000000000000L; i<1000000000000000000000L; i++){ if (!checkHappiness(String.valueOf(i))){ return String.valueOf(i); } return \"-1\"; } } public static void main(String[] args){ happyNumbersfrom9001to10000(); System.out.println(happyNumbersfrom1to10000()); } } Solution NumberList.java (implements the linked list) public class NumberList{ Node first; Node last; public NumberList(){ first = null; last = null; } public NumberList(Node node){ this.first = node; this.last = node; } public boolean isEmpty(){ return first == null; } public void setLast(Node node){ this.last = node; } public void insert(Node node){ if (first==null){ this.first = node; this.last = node; } else{ node.previous = this.last; node.previous.next = node; node.next = null; this.last = node; } } public boolean inList(String num){ Node l = first; while (l.next != null){ if (l.number.equals(num)){ return true; } else{ l = l.
NumberList.java (implements the linked list)public class NumberLis.pdf
NumberList.java (implements the linked list)public class NumberLis.pdf
anjanacottonmills
c) H2(g) + 1/2 O2(g) ->H2O (l) as all are in there natural state. therefore it will represent the enthalpy of formation Solution c) H2(g) + 1/2 O2(g) ->H2O (l) as all are in there natural state. therefore it will represent the enthalpy of formation.
c) H2(g) + 12 O2(g) -H2O (l) as all are in the.pdf
c) H2(g) + 12 O2(g) -H2O (l) as all are in the.pdf
anjanacottonmills
MagicSquareTest.java import java.util.Scanner; public class MagicSquareTest { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner scan = new Scanner(System.in); System.out.println(\"Enter Number of rows:\"); int rows = scan.nextInt(); System.out.println(\"Enter Nummber of columns:\"); int cols = scan.nextInt(); int a[][] = new int[rows][cols]; System.out.println(\"Enter matrix elements;\"); for(int i=0; i Solution MagicSquareTest.java import java.util.Scanner; public class MagicSquareTest { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner scan = new Scanner(System.in); System.out.println(\"Enter Number of rows:\"); int rows = scan.nextInt(); System.out.println(\"Enter Nummber of columns:\"); int cols = scan.nextInt(); int a[][] = new int[rows][cols]; System.out.println(\"Enter matrix elements;\"); for(int i=0; i.
MagicSquareTest.java import java.util.Scanner;public class Mag.pdf
MagicSquareTest.java import java.util.Scanner;public class Mag.pdf
anjanacottonmills
Interphase. This is the phase where the cell prepares for the next round of division. It has three phases further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in size. The checkpoints ensure that the cell is normal and all the things necessary for DNA synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only very short period of the cell cycle. Solution Interphase. This is the phase where the cell prepares for the next round of division. It has three phases further, Gap1, synthesis, and gap 2. During Gap1, the cell acquired nutrients and increases in size. The checkpoints ensure that the cell is normal and all the things necessary for DNA synthesis are prepared. During synthesis phase, DNA replicates. During the Gap 2 phase, the cell grow further and the checkpoints ensure that the cell is correctly prepared for mitosis phase. The cell spends most of its time in G1 phase, followed by S phase. Remaining phases comprise only very short period of the cell cycle..
Interphase.This is the phase where the cell prepares for the next .pdf
Interphase.This is the phase where the cell prepares for the next .pdf
anjanacottonmills
B.the rotation will be zero as L and D will cancel out each other.. Solution B.the rotation will be zero as L and D will cancel out each other...
B.the rotation will be zero as L and D will cance.pdf
B.the rotation will be zero as L and D will cance.pdf
anjanacottonmills
In the above conversation it is belonging to stereotypes. Stereotypes are a form of generalization. When we generalize, we group or classify people, places, or things according to the traits they have in common. For example, we may say most Masai tribesmen are unusually tall or Scandinavians are usually fair-skinned. If our observations are careless or too limited, the generalization may be faulty, as when someone says, \"Hollywood hasn\'t produced any quality movies in the past fifteen years.\" But stereotypes are more serious than mere faulty generalizations. They are fixed, unbending generalizations about people, places, or things. When a stereotype is challenged, the person who holds it is unlikely to modify or discard it, because it is based on a distortion of perception. As Walter Lippmann explains, when we stereotype,\" we do not so much see this man and that sunset; rather we notice that the thing is man or sunset, and then see chiefly what our mind is full of on those subjects.\"75 The most common kinds of stereotyping are ethnic and religious. Jews are shrewd and cunning, clannish, have a financial genius matched only by their greed. Italians are hot-tempered, coarse, and sensual. The Irish, like the Poles, are big and stupid; in addition, they brawl, lust after heavy liquor and light conversation. Blacks are primitive and slowwitted. (Often each of these stereotypes includes a virtue or two – Jews are good family members, Italians artistic, Poles brave, the Irish devout, blacks athletic.) Beyond these stereotypes are numerous other, less common ones; Swedish women, foreign film directors, Southern senators, physical education instructors, fundamentalist clergymen, agnostics, atheists, democrats, republicans, Mexicans, scientist, prostitutes, politicians, English teachers, psychiatrists, construction workers, black militants, college dropouts, homosexuals, and society matrons. There are stereotypes of institutions as well: marriage, the church, government, the military, the Founding Fathers, Western culture, the Judeo-Christian tradition. (A full list would include even God and mother.) FACTS DON\'T MATTER It is pleasant to assume that when the facts are known, stereotypes disappear. However, that is seldom the case. The late Harvard psychologist Gordon Allport, in The nature of Prejudice, pointed out that \"it is possible for a stereotype to grow in defiance of all evidence…\"76 People who stereotype don\'t dust accept the facts that are offered to them. They measure those facts against what they already \"know.\" That is, they measure them against the stereotype itself. Instead of seeing that the stereotype is false and therefore dismissing it, they reject the unfamiliar facts. People who think in terms of stereotypes tend to be selective in their perceptions. They reject conditions that challenge their preformed judgment and retain those that reinforce it. Thus a person can notice the Jewish employer promoting another Jew but ign.
In the above conversation it is belonging to stereotypes.Stereotyp.pdf
In the above conversation it is belonging to stereotypes.Stereotyp.pdf
anjanacottonmills
In general, the objective of an audit is to assess the risk of material misstatements in the financial statements. Material misstatements can arise from inadequacies in internal controls and from inaccurate management assertions. Thus, testing the validity of the various implicit managerial assertions is a key objective of an auditor. Existence and Completeness Auditing standards require that auditors test basic underlying management assertions implicit in the financial statements. Key among these various assertions are existence or occurrence, which describe a singular concept: Journal entries are not fiction. As the name implies, an auditor will conduct various procedures to verify that assets do in fact exist and that recorded transactions did in fact occur. Additionally, an auditor will seek evidence of completeness, so that the financial statements include all material transactions that occurred, and so the records do not omit material transactions for any reason. Rights and Obligations The various rights and obligations of the company are important management assertions inherent in the financial statements. Thus, an auditor will obtain evidence regarding a company\'s rights, such as proper title to assets and status of intellectual property. An auditor will be concerned with assertions relating to the company\'s obligations, such as accounts payable balances, long- term debts and tax liabilities. Thus, the audit objectives will be fulfilled upon validating these specific assertions. Valuation or Allocation Valuation or allocation are managerial assertions which are often material to the financial statements; thus, an auditor will diligently conduct audit procedures relating to these objectives. Generally accepted accounting principles, or GAAP, require that certain balance sheet items be presented using different valuation methodologies. Meeting these standards is a key audit objective, as the risk of material misstatement is low in probability, but high in magnitude. Thus, among other things, the historical cost of assets is verified, depreciation methods are scrutinized and the fair value of investments are calculated to satisfy this objective. Presentation and Disclosure Another specific audit objective is validating the presentation of the financial statements and the adequacy of the disclosures therein. Financial statements should conform to certain requirements and expectations, and should include the balance sheet, income statement, statement of cash flows and the statement of owner\'s equity. Relating to disclosure, the auditor will consider the sufficiency and clarity of footnotes and the transparency in management discussion and analysis, so he can assess the risk of material misstatement and fulfill the audit objective. Specific audit objectives are the application of the general audit objectives to a given class of transactions, account balance, or presentation and disclosure. There must be at least one specific audit object.
In general, the objective of an audit is to assess the risk of mater.pdf
In general, the objective of an audit is to assess the risk of mater.pdf
anjanacottonmills
if one can understand a few things it is easier to solve these kind of questions. first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and make Z and Y products in the presence of lactose only. I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose. Is is a dominant mutation will stop both the products Z and Y. if you have atlease one Is in the question neither the products will from no matter whether the lactose is presence or not. then comes to Operator (O) ,O+ is a wild type and can produce both the products of Z and Y. in the presence of lactose.and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. next comes to Z+ represents production of -galactosidase and Z- represents no production of - galactosidase. 1.Ans. I+P-O+Z+Y+ In this haploid genotype the transcription of operon genes totally absent. Transcription of structural genes is not occured. because here P- indicates the mutation in the promoter site, the RNA polymerase unable to bind to the Promoter and initiate transcription of structural genes. so these cells cannot able to grow on media which contain lactose as sole carbon source. the Strain is lac- 2.Ans.I+P+OcZ+Y- here Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose means constitutively. but here only the product beta galctosidase produced constitutively and the permease is not produced because Permease gene is is mutated (Y-). these cells also cannot able to grow on media which contain lactose as sole carbon source.because the absence of functional permease enzyme so the Strain is lac- 3.Ans. I-P+OcZ+Y+ here I- is a recessive mutation will make products of Z and Y in the presence and the absence of lactose and Oc is the mutation and can produce both Z and Y products in the Presence and even in the absence of lactose. so this haploid genotype transcribes operon genes constitutively. and both the products of Z and Y expressed continuously, so these cells able to grow on the media which contain lactose as sole carbon source. the strain is lac+ b. partially diploid cap-I+P+O+Z-Y+/Cap+I-P+O+Z+Y- these cells can able to grow on lactose medium because these cells have presence of wild promoter which can able to active the transcription of structural genes. so these cells can able to grow on the media which contain lactose as sole carbon source. ii. Transcription of beta-galactosidase and permease is Inducible though diploid cell have mutated inducer region which can able to activate constitutive expression but Wild type (I+) is dominant over I- . so it is inducible operon. Solution if one can understand a few things it is easier to solve these kind of questions. first comes to I which codes for repressor. there are 3 types - I+ Wild type and is normal and make Z and Y products in the presence of lactose only. I- is a recessive mutation .
if one can understand a few things it is easier to solve these kind .pdf
if one can understand a few things it is easier to solve these kind .pdf
anjanacottonmills
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancy Balanced MolecularEquation is : CuCl2 +H2SiO4 ---> CuSiO4 (s) +2HCl Total ionic Equation is : Cu2+ +2 Cl - + 2H+ +SO42- ---> CuSiO4 +2H+ + 2Cl- Spectator ions are : Cl - &H+ Spectator ions are : Cl - &H+ Net ionic Equation is : Cu2+ + SO42- --->CuSiO4 Solution I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancy Balanced MolecularEquation is : CuCl2 +H2SiO4 ---> CuSiO4 (s) +2HCl Total ionic Equation is : Cu2+ +2 Cl - + 2H+ +SO42- ---> CuSiO4 +2H+ + 2Cl- Spectator ions are : Cl - &H+ Spectator ions are : Cl - &H+ Net ionic Equation is : Cu2+ + SO42- --->CuSiO4.
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancyBalanc.pdf
I think it is H2SiO4 not H4SiO4, since it exceeds itsvalancyBalanc.pdf
anjanacottonmills
fixed 3 issues. I am working on remaining. package chegg; public class HugeIntegerApp { public static void main(String args[]) { String[][] testInputs = { { \"987654321\", \"234567890\" }, { \"987654321\", \"-234567890\" }, { \"-987654321\", \"234567890\" }, { \"-987654321\", \"-234567890\" }, { \"234567890\", \"987654321\" }, { \"234567890\", \"-987654321\" }, { \"-234567890\", \"987654321\" }, { \"-234567890\", \"-987654321\" } }; for (String[] ints : testInputs) { HugeInteger h1 = new HugeInteger(ints[0]); HugeInteger h2 = new HugeInteger(ints[1]); HugeInteger hcopy=new HugeInteger(ints[0]); HugeInteger hcopy1=new HugeInteger(ints[0]); HugeInteger hcopy2=new HugeInteger(ints[0]); System.out.println(\"h1=\" + h1); System.out.println(\"h2=\" + h2); if (h1.isEqualTo(h2)) { System.out.println(\"h1 is equal to h2.\"); } if (h1.isNotEqualTo(h2)) { System.out.println(\"h1 is not equal to h2.\"); } if (h1.isGreaterThan(h2)) { System.out.println(\"h1 is greater than h2.\"); } if (h1.isLessThan(h2)) { System.out.println(\"h1 is less than to h2.\"); } if (h1.isGreaterThanOrEqualTo(h2)) { System.out.println(\"h1 is greater than or equal to h2.\"); } if (h1.isLessThanOrEqualTo(h2)) { System.out.println(\"h1 is less than or equal to h2.\"); } h1.add(h2); // h1 += h2 System.out.println(\"h1.add(h2);\"); System.out.printf(\"h1=%s\ \", h1); hcopy.subtract(h2); // h1 -= h2 System.out.println(\"h1.subtract(h2);\"); System.out.println(\"h1.subtract(h2);\"); System.out.printf(\"h1=%s\ \", hcopy); //h1.subtract(h2); // h1 -= h2 hcopy1.add(h2); // h1 += h2 System.out.println(\"h1.add(h2);\"); hcopy2.multiply(h2); // h1 *= h2 System.out.println(\"h1.multiply(h2);\"); System.out.printf(\"h1=%s\ \ \", h1); } } } package chegg; public class HugeInteger { private static final int NUM_DIGITS = 40; private int digits[] = new int[NUM_DIGITS]; private boolean positive; public HugeInteger(String num) { int len = num.length(); if (num.indexOf(0) == \'-\') { len--; positive = false; } else { positive = true; } for (int i = 0; i < len; i++) { digits[NUM_DIGITS - 1 - i] = num.charAt(len - 1 - i) - (int) \'0\'; digits.toString(); } } public void multiply(HugeInteger hi) { this.digits = multiplyArraydigits(this.digits, hi.digits); } private int[] multiplyArraydigits(int[] array1, int[] array2) { int[] result = new int[array1.length + array2.length]; int carry = 0; int set = 0; for (int i = array1.length - 1; i >= 0; i--) { int temp = result.length - 1 - set; for (int j = array2.length - 1; j >= 0; j--) { int sum = result[temp] + (array1[i] * array2[j]) + carry; result[temp] = sum % 10; carry = sum / 10; temp--; } if (carry > 0) { result[temp] += carry; while (result[temp] >= 10) { carry = result[temp] / 10; result[temp] = result[temp] % 10; result[temp--] += carry; } carry = 0; } set++; } return result; } public void subtract(HugeInteger hi) { // if the signs of the two numbers are not the same if (this.positive != hi.positive) { if (this.positive) { // \'this\' is positive... .
fixed 3 issues. I am working on remaining.package chegg;public c.pdf
fixed 3 issues. I am working on remaining.package chegg;public c.pdf
anjanacottonmills
class Mesh{ public: /*Mesh data*? vectorvertices; vector indices; vectortextures; ?*functions*? Mesh(vectorvertices,vectorindices,vector textures); void draw(shader shader); private: /* Render data*/ gluint vao,vbo,ebo; /* functions */ void setupmesh(); }; Solution class Mesh{ public: /*Mesh data*? vectorvertices; vector indices; vectortextures; ?*functions*? Mesh(vectorvertices,vectorindices,vector textures); void draw(shader shader); private: /* Render data*/ gluint vao,vbo,ebo; /* functions */ void setupmesh(); };.
class Mesh{publicMesh datavectorvertexvertices;vector.pdf
class Mesh{publicMesh datavectorvertexvertices;vector.pdf
anjanacottonmills
Bacteria was used in transformation experiment, to prove Dna was the genetic material and replicated semi conservatively, used in gene expression and regulation, preferred host for cloning genes, and in manipulation of DNA Solution Bacteria was used in transformation experiment, to prove Dna was the genetic material and replicated semi conservatively, used in gene expression and regulation, preferred host for cloning genes, and in manipulation of DNA.
Bacteria was used in transformation experiment, to prove Dna was the.pdf
Bacteria was used in transformation experiment, to prove Dna was the.pdf
anjanacottonmills
c) 11 Solution c) 11.
c) 11Solutionc) 11.pdf
c) 11Solutionc) 11.pdf
anjanacottonmills
assembly mips subroutine Solution assembly mips subroutine.
assembly mips subroutineSolutionassembly mips subroutine.pdf
assembly mips subroutineSolutionassembly mips subroutine.pdf
anjanacottonmills
1. Tyrosinase is involved in the synthesis of Melanin pigment in mammalian cells by using tyrosine as a substrate. In general phenolic compounds are best substrates for tyrosinase as it is responsible for browning of fruits and cut vegetables. 2. No, tyrosinase does not exhibit high level of specificity because it has broad substrate specificity for phenolic compounds. 3. Tyrosinase exhibit group specificity that is the presence of functional group which is phenol group. 4. Inhibition can be competitive or non-competitive. Competitive inhibitors are substrate analogs that is it resemble the substrate. Non-competitive inhibitors bind to the region other than active site for substrate binding. Benzoic acid is the substrate analog so it is competitive inhibitor. Phenylthiourea and resorcinol is a non-competitive inhibitor. 6. 0 C is freezing temperature and 70 C is denaturing temperature for enzyme. Enzyme reaction is best at the optimum temperature or physiological temperature. Solution 1. Tyrosinase is involved in the synthesis of Melanin pigment in mammalian cells by using tyrosine as a substrate. In general phenolic compounds are best substrates for tyrosinase as it is responsible for browning of fruits and cut vegetables. 2. No, tyrosinase does not exhibit high level of specificity because it has broad substrate specificity for phenolic compounds. 3. Tyrosinase exhibit group specificity that is the presence of functional group which is phenol group. 4. Inhibition can be competitive or non-competitive. Competitive inhibitors are substrate analogs that is it resemble the substrate. Non-competitive inhibitors bind to the region other than active site for substrate binding. Benzoic acid is the substrate analog so it is competitive inhibitor. Phenylthiourea and resorcinol is a non-competitive inhibitor. 6. 0 C is freezing temperature and 70 C is denaturing temperature for enzyme. Enzyme reaction is best at the optimum temperature or physiological temperature..
1. Tyrosinase is involved in the synthesis of Melanin pigment in mam.pdf
1. Tyrosinase is involved in the synthesis of Melanin pigment in mam.pdf
anjanacottonmills
1) A covalent crystal differs from an ionic crystal in that a)only one of them is a poor electrical conductor. b)only one of them has bonding that involves overlap of electronic orbitals. c)only one of them possesses atomic order. d)only one of them has high melting points. e)only one of them is hard. The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic crystals involve electrostatic bonding between a positive and negative ion. 2)What are the important intermolecular forces acting in CHF3? a)London dispersion forces only b)only dipole-dipole interactions and H-bonding c)dipole-dipole interactions and London dispersion forces d)H-bonding, dipole-dipole interactions and London dispersion forces e)ionic bonding In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and Fluorine are all capable of hydrogen bonding.) Solution 1) A covalent crystal differs from an ionic crystal in that a)only one of them is a poor electrical conductor. b)only one of them has bonding that involves overlap of electronic orbitals. c)only one of them possesses atomic order. d)only one of them has high melting points. e)only one of them is hard. The covalent crystal involves formal, covalent bonds (overlapping orbitals) whereas ionic crystals involve electrostatic bonding between a positive and negative ion. 2)What are the important intermolecular forces acting in CHF3? a)London dispersion forces only b)only dipole-dipole interactions and H-bonding c)dipole-dipole interactions and London dispersion forces d)H-bonding, dipole-dipole interactions and London dispersion forces e)ionic bonding In all covalent compounds, london dispersion and dipole-dipole interactions are involved. When Fluorine and Hydrogen are present, Hydrogen bonding is possible. (Oxygen, Nitrogen, and Fluorine are all capable of hydrogen bonding.).
1) A covalent crystal differs from an ionic crystal in that a)on.pdf
1) A covalent crystal differs from an ionic crystal in that a)on.pdf
anjanacottonmills
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1.
D) II, III,
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