What do the following cities have in common Weimar, Cothen, Leipzig.pdf

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What do the following cities have in common: Weimar, Cothen, Leipzig? a. all are homes of famous violin makers b. all are sites of the premieres of Handel\'s oratorios c. all are cities where Bach was employed d. all of the above Solution Correct Option is All of the above. Bach were employed in these cities: Weimar = 1709 - 1717 Cothen = 1717-1723 Leipzig = 1723 - 1750 Bach and Handel was famous violin player.

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Describe the attributes that make a vision meaningful to stakeholder.pdf

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Create a C program which auto-completes suggestions when beginning to type a command and pressing tab. The program needs to read all the files names in a directory given by the user and store them in a data structure made up of a 26-element array of linked lists. A linked list implementation in C is required. Create a linked list for each letter of the alphabet in which you\'ll store the filenames which begin with that letter in some order. After reading the filenames, the user provides the beginning of a filename they\'ll like to search. The output should be all the filenames which match that prefix. The program should continue to ask for beginnings of filenames until an empty string is entered. Solution Code ----------- #include #include #include struct node { char data[10] ; struct node *next; }*head; node* arrNode[26]; char lettersPosition[26] = {\'a\',\'b\',\'c\',\'d\',\'e\',\'f\',\'g\',\'h\',\'i\',\'j\',\'k\',\'l\',\'m\',\'n\',\'o\',\'p\',\'q\',\'r\',\'s\',\'t\',\'u\',\'v\',\'w\',\'x \',\'y\',\'z\'}; /* * Creating Node */ node* create_node(char fileName[]) { node* newnode = (node *)malloc(sizeof(node)); if (newnode == NULL) { printf(\"\ Memory was not allocated\"); return 0; } else { strcpy(newnode->data,fileName); newnode->next = NULL; return newnode; } } void display(struct node *r) { r=head; if(r==NULL) { return; } while(r!=NULL) { printf(\"%s \",r->data); r=r->next; } printf(\"\ \"); } void search(char beginningFilePrefix[]) { node* newnode = NULL; char str[10]; strcpy(str,beginningFilePrefix); char dirPos = str[0]; node* r = arrNode[lettersPosition[dirPos]]; if(r==NULL){ printf( \"No file is avialable in directory\"); return; }else{ while(r!=NULL){ char tempFile[10]; bool isMathched = true; strcpy(tempFile,r->data); for( int i = 0; i < 1 ; i++){ printf(\"%c %c\",str[i],tempFile[i]); if(!(str[i] == tempFile[i])){ isMathched = false; } } if(isMathched){ if(newnode == NULL){ newnode = (node *)malloc(sizeof(node)); strcpy(newnode->data,r->data); }else{ struct node *temp,*right; temp= (struct node *)malloc(sizeof(struct node)); strcpy(temp->data,r->data); right=(struct node *)newnode; while(right->next != NULL){ right=right->next; } right->next =temp; right=temp; right->next=NULL; } } r=r->next; } } if(newnode==NULL){ printf(\"File not found with given search critieria\"); }else{ printf(\" Below are the output or file list which are matched with search criteria \ \"); while(newnode!=NULL){ printf(\"%s \ \",newnode->data); newnode=newnode->next; } } printf(\"\ \"); } void displayFromArray(int dirPosition) { node* r= arrNode[dirPosition]; if(r==NULL){ printf( \"r is null\"); return; } while(r!=NULL){ printf(\"%s \",r->data); r=r->next; } printf(\"\ \"); } void insertIntoArray(char fileName[],int dirPosition){ struct node *temp; temp=(struct node *)malloc(sizeof(struct node)); strcpy(temp->data,fileName); node* r = arrNode[dirPosition]; if(lettersPosition[dirPosition] == fileName[0]){ printf(\"directory starts with letter %c cannot hold fi.

Create a C program which auto-completes suggestions when beginning t.pdf

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Compare the DHCP and DHCPv6 Solution The comparison between the DHCP and DHCPv6 is highlighted as below 1) DHCP packets are send over IPv4 while as DHCPv6 packets are sent over IPv6. 2) DHCPV6 used IPV6 multicast messages while as DHCP used IPV4 broadcast messsages over the LAN. 3) DHCP has a default IP gateway address for the client while as DHCPv6 Does not have it. 4) DHCP can be configures over a router while as DHCPv6 does not have this option. 5) Scopes and leases is the concept used by both protocols. 6) DHCP Client, Relay and Server is the concept used by both the protocols. 7) Both messages types are different but perform olmost the same funtion. 8) Both Protocols use the DHCP end node for the additional information..

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Consider in a given economy: For a given time period the following information indicated numbers of goods and services needed (Items) Unit Price for these items for selected time are shown in the table below. Assume base year is 1981: calculate CPI for 1971, 1981, 1992 and 2012, Make sure show all your calculations Consider the following information, Use the CPI from part one, and assume base is 1992. Complete the table Calculate percentage change of Nominal GDP from 1971 to 2012. Calculate percentage change of Real GDP from 19711 to 2012. Explain the difference between nominal percentage change and real percentage change. Solution CPI=current year expenditure * 100/base year expenditure.

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Background: Many cities such as Detroit, MI have been in the news recently regarding underfunded pension funds. This exercise addresses a simplified version of some of the issues involved (doing this as a starting point is common: the key generally is to address the core work, adding more complexity later to make the analysis more accurate). Solution We have to analysis the background by the background data.

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Can someone put this code in a zip file. I tried running it last tim.pdf

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Assignment 1. List the three visual elements Brieny descrbe each. 2 What is the PCL? Give an example of how it is used 3. What is the file ending lor an Android project? Windows Project? Windows Phone Project? Mac project? A. Explain where Form int can be found for Windows, Windows phone, Android, and Mac applicatons. 5 Each platform must instantiate the APP class (wihich is located in the PCL This occurs in a can to the oadApplication Explain where this call occurs in the Windows, Windows Phone, Android, and Mac platforms. 6 What is the difference between PCL and SAP? 7 Explain how to create a label 8 Explain how padding works Make sure to include a differences that exist with damerent devices 9 How do you set a margin? 10 How do you center a label? How do you center text within a label? Solution Form: The form of a piece is its form, including its volume or perceived volume. A three-dimensional artwork has depth as well as breadth and height. Three-dimensional form is the basis of sculpture.However, two-dimensional artwork will win the illusion of type with the employment of perspective and/or shading or modelling techniques. Formalism is the analysis of works by their form or shapes in humanities or anthropology. Line Lines and curves are marks that span a distance between 2 points . As an component of visual art, line is the use of assorted marks, outlines, and implied lines in design and style. A line has a width, direction, and length. A line\'s width is typically referred to as its \"thickness\". Lines are typically referred to as \"strokes\", especially once referring to lines in digital design. Color Color is the element of art that\'s made once light-weight, striking AN object, is reflected back to the eye. There are 3 properties to color. The first is hue, which merely means that the name we have a tendency to offer to a color . The second element is intensity, which refers to the vividness of the color. A color\'s intensity is sometimes stated as its \"colorfulness\", its \"saturation\", its \"purity\" or its \"strength\".The third and final property of color is its value, meaning however light-weight or dark it is. The terms shade and tint refer to value changes in colours. In painting, shades are created by adding black to a color, while tints ar created by adding white to a color..

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A. Intergenic sequences make up >60% of the human genome. Where do these intergenic sequences come from? B. What are some of their functions? a. May encode mRNAs b. May encode miRNAs c. May be nonfunctional sequences d. May serve as regulatory sequences for transcription Solution A)Intergenic regions are short -coding regions that are found with in genes iin eukaryotic organisms. It is one of the DNA sequences sometimes referred to as junk DNA, Intergenic regions are a subset of noncoding DNA. these intergenic dna consists of parasitic dna regions of transposable elements.An Intergenic sequence is a stretch of DNA sequences located between genes.Intergenic regions are a subset of noncoding DNA. some intergenicDNA acts to control genes. B) d)May serve as regulatory sequences for transcription.

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a) How can habitat selection and sexual selection drive sympatric speciation? b) What are hybrid zones and why can they be viewed as “natural laboratories” in which to study speciation? Solution a)sympatric speciation is when a small population becomes a new species without geographic separation due to reduced gene flow by factors like polyploidy, habitat differentiation, and sexual selection.This type of speciation takes place in geographically overlapping populations.There is no physical barrier, but still interruption of gene flow is there. For example in fruit flies, food source (habitat selection) and apple maget fruit flies Ex. Cichlids(Differ in coloration of breeding males). Mate choice based on male breeding coloration is the main reproductive barrier that normally keeps these gene pools of these two species separate. b)Hybrid zones are regions in which members of different species meet and mate, producing some offspring of mixed ancestry. Such regions are \"natural laboratories\" in which to study speciation because scientists can directly observe factors that cause (or fail to cause) reproductive isolation..

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50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, Ka=1.9*10^-5) is titrated with a 0.1000M solution of sodium Hydroxide. Calculate the pH at the following points a. Before Titration b. After the addition of 5.00mL of NaOH c. After the addition of 25.00mL of NaOH d. After the addition of 50.00mL of NaOH e. After the addition of 5.00mL of NaOH Solution HN3+ NaOH--> NaN3 +H2O mini moles of HN3=5 ka=Ka=1.9*10^-5 pka=4.72 a) ph before addition before addition only weak acid is present alpha=sqrt(ka/c) =1.37*10^-2 c.alpha= [H+]= 1.37*10^-3 ph=2.86 b) after addition of 5 ml mini moles of naoh added=0.5 ph=4.72+log(0.5/(5-0.5)) ph=3.76 c) after addition of 25 ml Naoh ph=4.72+log(2.5/(5-2.5)) ph=4.72 d)after addition of 50 ml now this is equivalence point. Here N1v1=N2v2 here salt hydrolysis will take place N3- +H2O <=> HN3+OH- pOH=1/2(pKw-pKa-logC) C= 5/(50+50) =1/20M pOH=1/2(14-4.72+1.30) =5.29 ph=8.71.

50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, Ka=1.91.pdf

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4. For each of the following electrochemical cells: i. Identify the anode, the cathode, the reduced species and the oxidized species. ii. Write the half reactions for the anode and cathode, and the net reaction. ii. Calculate the standard cell potential (E°). *Note- For learning purposes, the cathodes are not always on the left! e. C)SO (),H\'(H2S03(a)Co\"(a)Co) Solution a. Cathode : Cr2+/Cr Anode : Sn/Sn2+ Oxidized : Sn Reduced : Cr2+ half-reactions, Sn --> Sn2+ + 2e- Cr2+ + 2e- --> Cr net reaction : Sn + Cr2+ --> Sn2+ + Cr Eo = Ecathode - Eanode = -0.14 - (-0.91) = 0.77 V b. Cathode : Ni2+/Ni Anode : Fe/Fe2+ Oxidized : Fe Reduced : Ni2+ half-reactions, Fe --> Fe2+ + 2e- Ni2+ + 2e- --> Ni net reaction : Fe + Ni2+ --> Fe2+ + Ni Eo = Ecathode - Eanode = -0.25 - (-0.44) = 0.19 V c. Cathode : Cu2+/Cu Anode : Pb/Pb2+ Oxidized : Pb Reduced : Cu2+ half-reactions, Pb --> Pb2+ + 2e- Cu2+ + 2e- --> Cu net reaction : Pb + Cu2+ --> Pb2+ + Cu Eo = Ecathode - Eanode = 0.337 - 0.126 = 0.211 V d. Cathode : Cl2/Cl- Anode : Zn/Zn2+ Oxidized : Zn Reduced : Cl2 half-reactions, Zn --> Zn2+ + 2e- Cl2 + 2e- --> 2Cl- net reaction : Zn + Cl2 --> Zn2+ + 2Cl- Eo = Ecathode - Eanode = 1.358 - (-0.763) = 2.121 V e. Cathode : SO4^2-/H2SO3 Anode : Co/Co2+ Oxidized : Co Reduced : SO4^2- half-reactions, Co --> Co2+ + 2e- SO4^2- + 4H+ + 2e- --> H2SO3 + H2O net reaction : 2Co + SO4^2- + 4H+ --> 2Co2+ + H2SO3 + H2O Eo = Ecathode - Eanode = 0.17 - (-0.28) = 0.45 V f. Cathode : O2/H2O2 Anode : Cu2+/Cu Oxidized : Cu Reduced : O2 half-reactions, Cu --> Cu2+ + 2e- O2 + 2H+ + 2e- --> H2O2 net reaction : Cu + O2 + 2H+ --> Cu2+ + H2O2 Eo = Ecathode - Eanode = 0.682 - 0.337 = 0.345 V g. Cathode : Cr2O7^2-/Cr3+ Anode : Fe2+/Fe3+ Oxidized : Fe2+ Reduced : Cr2O7^2- half-reactions, Fe --> Fe2+ + 2e- Cr2O7^2- + 14H+ 6e- --> 2Cr3+ + 7H2O net reaction : 6Fe + Cr2O7^2- + 14H+ 6e- --> 6Fe2+ + 2Cr3+ + 7H2O Eo = Ecathode - Eanode = 1.33 - 0.771 = 0.559 V.

4. For each of the following electrochemical cells i. Identify the a.pdf

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What are some advantages and disadvantages of using a doubly linked list versus a linked list? Solution The advantages and disadvantages of using a doubly linked list versus a linked list: -> It is easy to operate on many operations because double linked list offers easy implementation. whereas the linked list has to be required more data for the same operation. -> A double linked list can bisect in both directions whereas linked list bisect only in one direction. -> One of the disadvantage of double linked list is it occupies more space so that more operations are required when compared to a linked list of the same operation. -> Insertion and deletion takes more time in double linked list when compared to linked list because in the double linked list more pointer operations are required..

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Use the References to access important values if ne To answer the questions, interpret the following Lewis structure for PO3 3- O: #0: 1. For the central phosphorus atom: The number of lone pairs The number of single bonds The number of double bonds 2. The central phosphorus atom Submit Answer Retry Entire r obeys the octet rule pts remainin has an incomplete octet has an expanded octet Solution I am solving the Q1 as per Chegg guidelines, post multiple question to get the remaining answers Q) Number of lone pairs on Phosphorus = 0 (there are no electron pairs on the Phosphorus atom) Number of single bonds = 4 (Phosphours forms four single bonds with each of Phosphorus atoms) Number of double bonds = 0 There is no double bond in the structure of PO4(3-) The central phosphorus atom has a charge of +1 in the structure, hence it will be having an incomplete octet.

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What is the difference between a personal exemption and a dependency exemption? Are all taxpayers allowed a personal exemption? Solution the effect of this provision is allowed to only 1exemption per individual. if there is a tax payment, then total number of exemptions are equal to total US populationExplanationpersonaldependencyApplicabilityapplies to all tax payers (never greater than twoapplies to those individuals who qualifies taxpayer dependententitlementnot all taxpayers entitled to personal exemptiondependent of another not allowed to claim personal exemption.

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Why should a system administrator log in to a computer running a GUI environment as his/herself and then switch to the administrator account instead of logging in directly as the administrator? Solution In most of the cases system administrator must log in to a computers mostly with windows because it os required so that he/she can log in in to the GUI as root. other than the windows means unlike windows you cannot reboot this point to start using the system. From there you will be taken to the administrator account by using the existing account by using the administrator..

Why should a system administrator log in to a computer running a GUI.pdf

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What motivated so many Americans to fight and kill one another in the long and bloody American Civil War? What did they believe they were fighting for? Solution There were several reasons for the civil war: 1. Abolition of slavery - The people in the south were mostly plantation farm owners which is a labour intensive work. So, when Abrahim Lincoln decided to abolish slavery, it was seen as a direct attack on their way of life. They did not want to be dictated terms by the people from the north who were mostly business owners and the northerners did not support slavery. 2. Union over State powers - In the decision to create policies such as abolition of slavery, the states had no individual power to decide if it supported or opposed the idea. This made the southern states feel threatened by the power of the union over them which they were not accustomed to prior to 1789. 3. Denial of request of secession - Some states in the south wanted to secede from the United States but the then president Abrahim Lincoln refused to entertain this idea. This riled up the people in the south even further. The Northerners believed to be fighting for the abolition of slavery as well as uniting the nation under a single rule of the union. On the other hand, the southerners believed to be fighting for their way of life and seceding from the rest of the United States..

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Using examples, discuss the main determinants of migration. Solution Broadly migration factors can be divided into– 1. Economic factor: The push factor for these citizens can be poverty, low wages and unemployment. For e.g. Often those who migrate from poorer countries like African and South Asian countries face severe economic hardships. They’re pulled by factors like higher wages and better standard of living. 2. Political factor: Civil unrest and war can lead to many citizens becoming refugees as it leads to massive loss of life and property in a country. This becomes a natural factor due to which people migrate to politically stable countries. For e.g. The recent war in Syria has pushed Syrian to migrate to Europe and other developed countries. 3. Socio-cultural factors: Often discrimination based on ethnicity, religion or color can also result in migration to other countries. Ethnic cleansing in African countries has been one of the reasons why Africans moved to nearby countries where they felt safe..

Using examples, discuss the main determinants of migration.Solut.pdf

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The ventricles are not reaching systole in every cardiac cycle.The heart would pump with too much pressure (hypertension).The heart is not pumping any blood at all.The atria are not pumping blood. Solution Answer: (a) The ventricles are not reaching systole in every cardiac cycle. The ECG in C represents a case of second degree blockage. The third QRS complex is missing. Second degree heart block involves a greater level of signal slowing, to the extent where some signals don’t reach the ventricles at all..

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