Soumettre la recherche
Mettre en ligne
sp2 - sp2 .pdf
•
0 j'aime
•
2 vues
K
karan8801
Suivre
sp2 - sp2 Solution sp2 - sp2.
Lire moins
Lire la suite
Formation
Signaler
Partager
Signaler
Partager
1 sur 1
Télécharger maintenant
Télécharger pour lire hors ligne
Recommandé
1. C. HIV is a virus it can be tansmitted through body fluids liks semen and blood and it causes AIDS. 2. D. 3.C. 4.B. 5.A. 6.B. Solution 1. C. HIV is a virus it can be tansmitted through body fluids liks semen and blood and it causes AIDS. 2. D. 3.C. 4.B. 5.A. 6.B..
1. C.HIV is a virus it can be tansmitted through body fluids liks .pdf
1. C.HIV is a virus it can be tansmitted through body fluids liks .pdf
karan8801
#include #include using namespace std; template class Node{ private: T value; Node *next; public: Node(); Node (T data); T remove(); T getValue(); Node* getNext(); void setValue(T data); void setNext(Node *nextNode); }; template Node::Node(){ T value =NULL; } template Node::Node(T data){ T value=data; } template T Node::getValue(){ return value; } template Node* Node::getNext(){ return next; } template void Node::setValue(T data){ T value=data; } template void Node::setNext(Node *nextNode){ next=nextNode; } template class MyLinkedList{ Node start; //Define a constructor MyLinkedList() { start->setValue(0); start->setNext(NULL); } //Member Functions: // Adds a new element to the list void add (T val) { /* creating a new node to hold value*/ Nodetemp(val,NULL); /* If no start node till now, this node is made as start node*/ if (start->next == NULL) { start.setNext(temp); } /* if start node exists, insert new node at end of list*/ else { s = start->next; /* locating end of list*/ while (s->next != NULL) { s= s->next; } temp->next = NULL; /* attaching new node */ s->next = temp; cout<<\"Element Inserted\"<next->next != NULL) { s = s->next; } //After locating the end node ptr=s->next; s->next=NULL; //deletting end node return ptr; } } //Prints elements of the list void printContent() { struct node *ptr = start; cout<data; ptr=ptr->next; } } } //Prints total number of elements int getSize() { struct node *ptr, *s; int count=0; if (start == NULL) { cout<<\"The List is empty\"<next; count=1; while (ptr != NULL) { count++; ptr=ptr->next; } return count; } //Returns true if there is no element in the list. bool empty() { if(start==NULL) { return empty; } } }; Solution #include #include using namespace std; template class Node{ private: T value; Node *next; public: Node(); Node (T data); T remove(); T getValue(); Node* getNext(); void setValue(T data); void setNext(Node *nextNode); }; template Node::Node(){ T value =NULL; } template Node::Node(T data){ T value=data; } template T Node::getValue(){ return value; } template Node* Node::getNext(){ return next; } template void Node::setValue(T data){ T value=data; } template void Node::setNext(Node *nextNode){ next=nextNode; } template class MyLinkedList{ Node start; //Define a constructor MyLinkedList() { start->setValue(0); start->setNext(NULL); } //Member Functions: // Adds a new element to the list void add (T val) { /* creating a new node to hold value*/ Nodetemp(val,NULL); /* If no start node till now, this node is made as start node*/ if (start->next == NULL) { start.setNext(temp); } /* if start node exists, insert new node at end of list*/ else { s = start->next; /* locating end of list*/ while (s->next != NULL) { s= s->next; } temp->next = NULL; /* attaching new node */ s->next = temp; cout<<\"Element Inserted\"<next->next != NULL) { s = s->next; } //After locating the end node ptr=s->next; s->next=NULL; //deletting end node return ptr; } } //Prints elements of the list void printContent() { struct .
#include iostream #include cstddefusing namespace std;temp.pdf
#include iostream #include cstddefusing namespace std;temp.pdf
karan8801
omparative historical research is a method of social science that examines historical events in order to create explanations that are valid beyond a particular time and place, either by direct comparison to other historical events, theory building, or reference to the present day.[1]Generally, it involves comparisons of social processes across times and places. It overlaps with historical sociology. While the disciplines ofhistory and sociology have always been connected, they have connected in different ways at different times (see \'Major researchers\' below). This form of research may use any of several theoretical orientations. It is distinguished by the types of questions it asks, not the theoretical framework it employs (see \'Illustrations\' below) Some commentators have identified three waves of historical comparative research.[2] The first wave of historical comparative research concerned how societies came to be modern, i.e. based on individual and rational action, with exact definitions varying widely. Some of the major researchers in this mode were Alexis de Tocqueville,[3]Karl Marx,[4]Emile Durkheim,[5]Max Weber,[6] and W.E.B. Du Bois.[7] The second wave reacted to a perceived ahistorical body of theory and sought to show how social systems were not static, but developed over time.[8] Notable authors of this wave include Barrington Moore, Jr.,[9]Theda Skocpol,[10]Charles Tilly,[11]Michael Mann,[12] and Mark Gould.[13] Some have placed the Annales school and Pierre Bourdieu in this general group, despite their stylistic differences.[14] The current wave of historical comparative research sociology is often but not exclusively post-structural in its theoretical orientation. Influential current authors include Julia Adams,[15] Anne Laura Stoler,[16]Philip Gorski,[17] and James Mahoney.[18] There are four major methods that researchers use to collect historical data. These are archival data, secondary sources, running records, and recollections. The archival data, or primary sources, are typically the resources that researchers rely most heavily on. Archival data includes official documents and other items that would be found in archives, museums, etc. Secondary sources are the works of other historians who have written history. Running records are ongoing series of statistical or other sorts of data, such as census data, ship\'s registries, property deeds, etc. Finally recollections include sources such as autobiographies, memoirs or diaries.[19] There are four stages, as discussed by Schutt, to systematic qualitative comparative historical studies: (1) develop the premise of the investigation, identifying events, concepts, etc., that may explain the phenomena; (2) choose the case(s) (location- nation, region) to examine; (3) use what Theda Skocpol has termed as \"interpretive historical sociology\" and examine the similarities and the differences; and (4) based on the information gathered, propose a causal explanation for the pheno.
omparative historical research is a method of social science that e.pdf
omparative historical research is a method of social science that e.pdf
karan8801
1 .D 2.B Solution 1 .D 2.B.
1 .D 2.BSolution 1 .D 2.B.pdf
1 .D 2.BSolution 1 .D 2.B.pdf
karan8801
where is it Solution where is it.
where is it .pdf
where is it .pdf
karan8801
The name of the reaction is aldol condesation (Claisen) (it usesbase to form a enolate and it atacks over the diketone one time andthen it atacks again intramolecular to form the ring), the -OHformed in the condesation are eliminated to form the two doublebonds. The functional group present in the product is a ketone andthe hydroxi groups that were eliminated. Solution The name of the reaction is aldol condesation (Claisen) (it usesbase to form a enolate and it atacks over the diketone one time andthen it atacks again intramolecular to form the ring), the -OHformed in the condesation are eliminated to form the two doublebonds. The functional group present in the product is a ketone andthe hydroxi groups that were eliminated..
The name of the reaction is aldol condesation (Cl.pdf
The name of the reaction is aldol condesation (Cl.pdf
karan8801
Step1 3 moles of H2O react with Cr metal = 2moles Step2 3.17 moles of H2O will react with Cr = 2x3.17/3 =6.14/3 =2.057 Solution Step1 3 moles of H2O react with Cr metal = 2moles Step2 3.17 moles of H2O will react with Cr = 2x3.17/3 =6.14/3 =2.057.
Step1 3 moles of H2O react with Cr metal = 2moles.pdf
Step1 3 moles of H2O react with Cr metal = 2moles.pdf
karan8801
Part A: neutralizes acids - base Part B: produces OH^- in water - base Part C: has a soapy feel - base Part D: turns litmus red - acid Solution Part A: neutralizes acids - base Part B: produces OH^- in water - base Part C: has a soapy feel - base Part D: turns litmus red - acid.
Part A neutralizes acids - base Part B produces.pdf
Part A neutralizes acids - base Part B produces.pdf
karan8801
Recommandé
1. C. HIV is a virus it can be tansmitted through body fluids liks semen and blood and it causes AIDS. 2. D. 3.C. 4.B. 5.A. 6.B. Solution 1. C. HIV is a virus it can be tansmitted through body fluids liks semen and blood and it causes AIDS. 2. D. 3.C. 4.B. 5.A. 6.B..
1. C.HIV is a virus it can be tansmitted through body fluids liks .pdf
1. C.HIV is a virus it can be tansmitted through body fluids liks .pdf
karan8801
#include #include using namespace std; template class Node{ private: T value; Node *next; public: Node(); Node (T data); T remove(); T getValue(); Node* getNext(); void setValue(T data); void setNext(Node *nextNode); }; template Node::Node(){ T value =NULL; } template Node::Node(T data){ T value=data; } template T Node::getValue(){ return value; } template Node* Node::getNext(){ return next; } template void Node::setValue(T data){ T value=data; } template void Node::setNext(Node *nextNode){ next=nextNode; } template class MyLinkedList{ Node start; //Define a constructor MyLinkedList() { start->setValue(0); start->setNext(NULL); } //Member Functions: // Adds a new element to the list void add (T val) { /* creating a new node to hold value*/ Nodetemp(val,NULL); /* If no start node till now, this node is made as start node*/ if (start->next == NULL) { start.setNext(temp); } /* if start node exists, insert new node at end of list*/ else { s = start->next; /* locating end of list*/ while (s->next != NULL) { s= s->next; } temp->next = NULL; /* attaching new node */ s->next = temp; cout<<\"Element Inserted\"<next->next != NULL) { s = s->next; } //After locating the end node ptr=s->next; s->next=NULL; //deletting end node return ptr; } } //Prints elements of the list void printContent() { struct node *ptr = start; cout<data; ptr=ptr->next; } } } //Prints total number of elements int getSize() { struct node *ptr, *s; int count=0; if (start == NULL) { cout<<\"The List is empty\"<next; count=1; while (ptr != NULL) { count++; ptr=ptr->next; } return count; } //Returns true if there is no element in the list. bool empty() { if(start==NULL) { return empty; } } }; Solution #include #include using namespace std; template class Node{ private: T value; Node *next; public: Node(); Node (T data); T remove(); T getValue(); Node* getNext(); void setValue(T data); void setNext(Node *nextNode); }; template Node::Node(){ T value =NULL; } template Node::Node(T data){ T value=data; } template T Node::getValue(){ return value; } template Node* Node::getNext(){ return next; } template void Node::setValue(T data){ T value=data; } template void Node::setNext(Node *nextNode){ next=nextNode; } template class MyLinkedList{ Node start; //Define a constructor MyLinkedList() { start->setValue(0); start->setNext(NULL); } //Member Functions: // Adds a new element to the list void add (T val) { /* creating a new node to hold value*/ Nodetemp(val,NULL); /* If no start node till now, this node is made as start node*/ if (start->next == NULL) { start.setNext(temp); } /* if start node exists, insert new node at end of list*/ else { s = start->next; /* locating end of list*/ while (s->next != NULL) { s= s->next; } temp->next = NULL; /* attaching new node */ s->next = temp; cout<<\"Element Inserted\"<next->next != NULL) { s = s->next; } //After locating the end node ptr=s->next; s->next=NULL; //deletting end node return ptr; } } //Prints elements of the list void printContent() { struct .
#include iostream #include cstddefusing namespace std;temp.pdf
#include iostream #include cstddefusing namespace std;temp.pdf
karan8801
omparative historical research is a method of social science that examines historical events in order to create explanations that are valid beyond a particular time and place, either by direct comparison to other historical events, theory building, or reference to the present day.[1]Generally, it involves comparisons of social processes across times and places. It overlaps with historical sociology. While the disciplines ofhistory and sociology have always been connected, they have connected in different ways at different times (see \'Major researchers\' below). This form of research may use any of several theoretical orientations. It is distinguished by the types of questions it asks, not the theoretical framework it employs (see \'Illustrations\' below) Some commentators have identified three waves of historical comparative research.[2] The first wave of historical comparative research concerned how societies came to be modern, i.e. based on individual and rational action, with exact definitions varying widely. Some of the major researchers in this mode were Alexis de Tocqueville,[3]Karl Marx,[4]Emile Durkheim,[5]Max Weber,[6] and W.E.B. Du Bois.[7] The second wave reacted to a perceived ahistorical body of theory and sought to show how social systems were not static, but developed over time.[8] Notable authors of this wave include Barrington Moore, Jr.,[9]Theda Skocpol,[10]Charles Tilly,[11]Michael Mann,[12] and Mark Gould.[13] Some have placed the Annales school and Pierre Bourdieu in this general group, despite their stylistic differences.[14] The current wave of historical comparative research sociology is often but not exclusively post-structural in its theoretical orientation. Influential current authors include Julia Adams,[15] Anne Laura Stoler,[16]Philip Gorski,[17] and James Mahoney.[18] There are four major methods that researchers use to collect historical data. These are archival data, secondary sources, running records, and recollections. The archival data, or primary sources, are typically the resources that researchers rely most heavily on. Archival data includes official documents and other items that would be found in archives, museums, etc. Secondary sources are the works of other historians who have written history. Running records are ongoing series of statistical or other sorts of data, such as census data, ship\'s registries, property deeds, etc. Finally recollections include sources such as autobiographies, memoirs or diaries.[19] There are four stages, as discussed by Schutt, to systematic qualitative comparative historical studies: (1) develop the premise of the investigation, identifying events, concepts, etc., that may explain the phenomena; (2) choose the case(s) (location- nation, region) to examine; (3) use what Theda Skocpol has termed as \"interpretive historical sociology\" and examine the similarities and the differences; and (4) based on the information gathered, propose a causal explanation for the pheno.
omparative historical research is a method of social science that e.pdf
omparative historical research is a method of social science that e.pdf
karan8801
1 .D 2.B Solution 1 .D 2.B.
1 .D 2.BSolution 1 .D 2.B.pdf
1 .D 2.BSolution 1 .D 2.B.pdf
karan8801
where is it Solution where is it.
where is it .pdf
where is it .pdf
karan8801
The name of the reaction is aldol condesation (Claisen) (it usesbase to form a enolate and it atacks over the diketone one time andthen it atacks again intramolecular to form the ring), the -OHformed in the condesation are eliminated to form the two doublebonds. The functional group present in the product is a ketone andthe hydroxi groups that were eliminated. Solution The name of the reaction is aldol condesation (Claisen) (it usesbase to form a enolate and it atacks over the diketone one time andthen it atacks again intramolecular to form the ring), the -OHformed in the condesation are eliminated to form the two doublebonds. The functional group present in the product is a ketone andthe hydroxi groups that were eliminated..
The name of the reaction is aldol condesation (Cl.pdf
The name of the reaction is aldol condesation (Cl.pdf
karan8801
Step1 3 moles of H2O react with Cr metal = 2moles Step2 3.17 moles of H2O will react with Cr = 2x3.17/3 =6.14/3 =2.057 Solution Step1 3 moles of H2O react with Cr metal = 2moles Step2 3.17 moles of H2O will react with Cr = 2x3.17/3 =6.14/3 =2.057.
Step1 3 moles of H2O react with Cr metal = 2moles.pdf
Step1 3 moles of H2O react with Cr metal = 2moles.pdf
karan8801
Part A: neutralizes acids - base Part B: produces OH^- in water - base Part C: has a soapy feel - base Part D: turns litmus red - acid Solution Part A: neutralizes acids - base Part B: produces OH^- in water - base Part C: has a soapy feel - base Part D: turns litmus red - acid.
Part A neutralizes acids - base Part B produces.pdf
Part A neutralizes acids - base Part B produces.pdf
karan8801
Lewis Acid Base Reaction Solution Lewis Acid Base Reaction.
Lewis Acid Base Reaction .pdf
Lewis Acid Base Reaction .pdf
karan8801
KOH ====> K+ (aq) + OH-(aq) [OH-] =9.7*10^-5 [H+][OH-] = 1.0*10^-14 [H+] = (1.0*10^-14)/(9.7*10^-5) = 1.03*10^-10 pH = -log[H+] = -log(1.03*10^-10) = 9.987 = 10 pH+pOH=14 pOh = 14-pH = 14-10=4 Solution KOH ====> K+ (aq) + OH-(aq) [OH-] =9.7*10^-5 [H+][OH-] = 1.0*10^-14 [H+] = (1.0*10^-14)/(9.7*10^-5) = 1.03*10^-10 pH = -log[H+] = -log(1.03*10^-10) = 9.987 = 10 pH+pOH=14 pOh = 14-pH = 14-10=4.
KOH ==== K+ (aq) + OH-(aq) [OH-] =9.710^-5 [H.pdf
KOH ==== K+ (aq) + OH-(aq) [OH-] =9.710^-5 [H.pdf
karan8801
It is a reaction for secondary amines .Option 2 is correct. Solution It is a reaction for secondary amines .Option 2 is correct..
It is a reaction for secondary amines .Option 2 i.pdf
It is a reaction for secondary amines .Option 2 i.pdf
karan8801
fluorene is mobile phase and fluorenone is stationary phase because : Mostly,it depends on the solvent you are using, but lets assume you use pentane as the solvent for fluorene and diethyl ether as the solvent for fluorenone. Used in this order, fluorene should actaully move slower down the column due to the the polarity of the alumina and the nonpolarity of the solvent (pentane). Since fluorene is less polar so, remember the rule like dissolves like? Well, this kind of applies to chromatography wherein rather than dissolve just replace it with moves with, so like moves with like. That being said the pentane will \"carry\" the fluorene through the alumina slower than the latter (which I will explain). Fluorenone is polar because of its C=O bond, that being said the dielectric constant of diethyl ether is 4.3 which means it has intermediate polarity (remember that pentane has a dielectric constant of 2.1 I think, so it is nonpolar). Since the alumina, diethyl ether, and the fluorenone are all polar, the fluorenone will travel faster through the alumina than would the fluorene, because there is no attraction between all these polar compounds which will allow it to move faster, rather than a nonpolar and polar chemical having an attration towards each other and thus moving more slowly. Solution fluorene is mobile phase and fluorenone is stationary phase because : Mostly,it depends on the solvent you are using, but lets assume you use pentane as the solvent for fluorene and diethyl ether as the solvent for fluorenone. Used in this order, fluorene should actaully move slower down the column due to the the polarity of the alumina and the nonpolarity of the solvent (pentane). Since fluorene is less polar so, remember the rule like dissolves like? Well, this kind of applies to chromatography wherein rather than dissolve just replace it with moves with, so like moves with like. That being said the pentane will \"carry\" the fluorene through the alumina slower than the latter (which I will explain). Fluorenone is polar because of its C=O bond, that being said the dielectric constant of diethyl ether is 4.3 which means it has intermediate polarity (remember that pentane has a dielectric constant of 2.1 I think, so it is nonpolar). Since the alumina, diethyl ether, and the fluorenone are all polar, the fluorenone will travel faster through the alumina than would the fluorene, because there is no attraction between all these polar compounds which will allow it to move faster, rather than a nonpolar and polar chemical having an attration towards each other and thus moving more slowly..
fluorene is mobile phase and fluorenone is statio.pdf
fluorene is mobile phase and fluorenone is statio.pdf
karan8801
equivalent mass is 27.9 . so Molecular mass/moles of electrons per mole = 27.9 looking at the table for atomic masses, the closest we get is for moles of electrons per mole = 2 that is , molecular weight = 55.8 which is for Iron Hence Y is Iron ( Fe ) Solution equivalent mass is 27.9 . so Molecular mass/moles of electrons per mole = 27.9 looking at the table for atomic masses, the closest we get is for moles of electrons per mole = 2 that is , molecular weight = 55.8 which is for Iron Hence Y is Iron ( Fe ).
equivalent mass is 27.9 . so Molecular massmole.pdf
equivalent mass is 27.9 . so Molecular massmole.pdf
karan8801
There is no solution Solution There is no solution.
There is no solutionSolutionThere is no solution.pdf
There is no solutionSolutionThere is no solution.pdf
karan8801
This is a case of allopatric speciation. In this case lice from gorilla moved and solutes into two different species. So this is case of geographical speciation where species become apart from each other to an extent that they cannot interbreed. Solution This is a case of allopatric speciation. In this case lice from gorilla moved and solutes into two different species. So this is case of geographical speciation where species become apart from each other to an extent that they cannot interbreed..
This is a case of allopatric speciation. In this case lice from gori.pdf
This is a case of allopatric speciation. In this case lice from gori.pdf
karan8801
d. 8 The bonding electrons are the electrons taking part in the bonds. C is in the middle and H and O are on either side of C. C shares 2 electrons each with each of the two Hs, so C shares a total of 4 electrons with the Hs. Carbon shares a total of 4 electrons with O. So 4+4= 8 bonding electrons total. Solution d. 8 The bonding electrons are the electrons taking part in the bonds. C is in the middle and H and O are on either side of C. C shares 2 electrons each with each of the two Hs, so C shares a total of 4 electrons with the Hs. Carbon shares a total of 4 electrons with O. So 4+4= 8 bonding electrons total..
d. 8 The bonding electrons are the electrons taki.pdf
d. 8 The bonding electrons are the electrons taki.pdf
karan8801
The genetical changes of single locus are responsible for divergent branches of Y chromosome phylogenetic tree in different geographical locations elicit the human origins. The mitochondrial DNA sequences of Neanderthals and modern humans were compared and analyzed. The Hominins species were found to be 30000 year old and were compared with Neanderthal specimens. The relationships between these populations have drawn with the help of the age of specimens plotted against the average genetic distance. The Hominins type sequences are very rare when compared with the modern European samples. The prehistoric Europeans had shown constant number of differences when compared to today’s Europeans and Neanderthals. The Neanderthals have shown the genetic continuity in modern human’s genealogy. The Neanderthals had shown clear discontinuity with the Hominins from upper Paleolithic period to the similar ages. The Neanderthals and early anatomical modern humans were thought to be regional populations of the same evolving species connected by gene flow and both the archaic and modern humans contribute in different proportions to the present human gene pool. The transition and demographic replacement separated the population as Neanderthals and modern humans are considered just as one population observed at different times. The multi regional models predict that at least some level of genetic continuity can be observed from archaic Neanderthal forms to the contemporary Hominins forms and up to today’s Europeans. The results have not matched with the view that the Neanderthal were genetically related with the anatomically modern ancestors of current Europeans or contributed to the present day human gene pool. Solution The genetical changes of single locus are responsible for divergent branches of Y chromosome phylogenetic tree in different geographical locations elicit the human origins. The mitochondrial DNA sequences of Neanderthals and modern humans were compared and analyzed. The Hominins species were found to be 30000 year old and were compared with Neanderthal specimens. The relationships between these populations have drawn with the help of the age of specimens plotted against the average genetic distance. The Hominins type sequences are very rare when compared with the modern European samples. The prehistoric Europeans had shown constant number of differences when compared to today’s Europeans and Neanderthals. The Neanderthals have shown the genetic continuity in modern human’s genealogy. The Neanderthals had shown clear discontinuity with the Hominins from upper Paleolithic period to the similar ages. The Neanderthals and early anatomical modern humans were thought to be regional populations of the same evolving species connected by gene flow and both the archaic and modern humans contribute in different proportions to the present human gene pool. The transition and demographic replacement separated the population as Neanderthals and modern humans .
The genetical changes of single locus are responsible for divergent .pdf
The genetical changes of single locus are responsible for divergent .pdf
karan8801
The corrects answers are: Thr and Asn are polar amino acids. Isoleucine has more than one stereocenter (chiral center) The Leu side chain does not from hydrogen bonds with other amino acids Solution The corrects answers are: Thr and Asn are polar amino acids. Isoleucine has more than one stereocenter (chiral center) The Leu side chain does not from hydrogen bonds with other amino acids.
The corrects answers areThr and Asn are polar amino acids.Isole.pdf
The corrects answers areThr and Asn are polar amino acids.Isole.pdf
karan8801
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4 i.e Z4 = ( 0,1,2,3) U(8) = { 1,3,5,7} while each element of U(8) has order 1 or 2 So, isomorphic Z4 and U(8) are not isomorphic. Solution Since Z4 has an element of order 4,Z4, its a cyclic group of order 4 i.e Z4 = ( 0,1,2,3) U(8) = { 1,3,5,7} while each element of U(8) has order 1 or 2 So, isomorphic Z4 and U(8) are not isomorphic..
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4.pdf
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4.pdf
karan8801
selected palnts xerophyte verus mesophyte ========================================================= A xerophyte is a types of plant that has adjusted to get by in a domain with minimal fluid water, for example, a forsake or an ice-or snow-canvassed locale in the Alps or the Arctic. The morphology and physiology of xerophytes are differently adjusted to moderate water, and usually likewise to store substantial amounts of water, amid dry periods. Different species might be adjusted to survive long stretches of parching of their tissues, amid which their metabolic action may viably close down. Plants with such morphological and physiological adjustments are xeromorphic. Xerophytic plants may have comparable shapes, structures, and structures and look fundamentally the same as, regardless of the possibility that the plants are not firmly related, through a procedure called concurrent development. For instance, a few types of desert flora (individuals from the family Cactaceae), which advanced just in the Americas, may seem like Euphorbias, which are dispersed around the world. A random types of caudiciforms, plants with swollen bases that are utilized to store water, may likewise show such likenesses. Xerophytic plants can have less general surface territory than different plants, so diminishing the range that is presented to the air and lessening water misfortune by vanishing. Xerophytes can have littler leaves or less branches than different plants. A case of leaf surface decrease are the spines of a desert flora. A case of compaction and diminishment of spreading are the barrel desert flora. Different xerophytes may have their leaves compacted at the base, as in a basal rosette, which might be littler than the plant\'s blossom. This adjustment is displayed by some Agave and Eriogonum species, which can be discovered developing close Death Valley. A few xerophytes have minor hairs on their surface to give a wind break and decrease wind current, along these lines diminishing the rate of dissipation. At the point when a plant surface is secured with minor hairs, it is called tomentose. In a still domain, the regions under the leaves/spines where transpiration is occurring structure a little limited environment that is more soaked than typical with water vapor. In the event that this is not overwhelmed by wind, the water vapor potential angle is diminished as is transpiration. Subsequently, in a windier circumstance, this confinement is not held thus the angle stays high, which helps the loss of water vapor. Spines trap a layer of dampness furthermore moderate air development over tissues. =================== Mesophytes are earthbound plants which are adjusted to neither an especially dry nor especially wet environment. A case of a mesophytic living space would be a country calm glade, which may contain goldenrod, clover, oxeye daisy, and Rosa multiflora. Mesophytic plants have unbending, tough, openly expanded stems and stringy, all around create.
selected palntsxerophyte verus mesophyte========================.pdf
selected palntsxerophyte verus mesophyte========================.pdf
karan8801
Ruth & Associates Income Statement For the Year Ended December 31, 2010 Revenue Consulting Revenue 100000 Interest Revenue 4000 Total Revenue 104000 Expenses Salary Expense 47000 Interest Expense 5000 Total Expenses 52000 Net Income 52000 Ruth & Associates Statement of Changes in Stockholders’ Equity For the Year Ended December 31, 2010 Beginning Common Stock 25000 Plus: Common Stock Issued 15000 Ending Common Stock 40000 Beginning Retained Earnings 18000 Plus: Net Income 52000 Less: Dividends 10000 Ending Retained Earnings 60000 Total Stockholders’ Equity 100000 Ruth & Associates Balance Sheet As of the December 31, 2010 Assets Cash 52000 Accounts Receivable 26000 Supplies 3000 Prepaid Rent 5000 Land 63000 Total Assets 149000 Liabilities Interest Payable 2000 Salaries Payable 7000 Unearned Revenue 8000 Notes Payable 32000 Total Liabilities 49000 Stockholders’ Equity Common Stock 40000 Retained Earnings 60000 Total Stockholders’ Equity 100000 Total Liab. and Stockholders’ Equity 149000 Ruth & Associates Statement of Cash Flows For the Year Ended December 31, 2010 Cash Flow From Operating Activities 52000 Cash Flow From Investing Activities -30000 Cash Flow From Financing Activities 20000 Net Change in Cash Plus: 42000 Beginning Cash Balance 10000 Ending Cash Balance 52000 Ending Cash Balance 52000 Net Change in Cash 42000 = Beginning Cash Balance 10000 Ruth & Associates Income Statement For the Year Ended December 31, 2010 Revenue Consulting Revenue 100000 Interest Revenue 4000 Total Revenue 104000 Expenses Salary Expense 47000 Interest Expense 5000 Total Expenses 52000 Net Income 52000 Ruth & Associates Statement of Changes in Stockholders’ Equity For the Year Ended December 31, 2010 Beginning Common Stock 25000 Plus: Common Stock Issued 15000 Ending Common Stock 40000 Beginning Retained Earnings 18000 Plus: Net Income 52000 Less: Dividends 10000 Ending Retained Earnings 60000 Total Stockholders’ Equity 100000 Ruth & Associates Balance Sheet As of the December 31, 2010 Assets Cash 52000 Accounts Receivable 26000 Supplies 3000 Prepaid Rent 5000 Land 63000 Total Assets 149000 Liabilities Interest Payable 2000 Salaries Payable 7000 Unearned Revenue 8000 Notes Payable 32000 Total Liabilities 49000 Stockholders’ Equity Common Stock 40000 Retained Earnings 60000 Total Stockholders’ Equity 100000 Total Liab. and Stockholders’ Equity 149000 Ruth & Associates Statement of Cash Flows For the Year Ended December 31, 2010 Cash Flow From Operating Activities 52000 Cash Flow From Investing Activities -30000 Cash Flow From Financing Activities 20000 Net Change in Cash Plus: 42000 Beginning Cash Balance 10000 Ending Cash Balance 52000 Ending Cash Balance 52000 Net Change in Cash 42000 = Beginning Cash Balance 10000 Solution Ruth & Associates Income Statement For the Year Ended December 31, 2010 Revenue Consulting Revenue 100000 Interest Revenue 4000 Total Revenue 104000 Expenses Salary Expense 47000 Interest Expense 5000 Total Expenses 52000 Net Income 5200.
Ruth & AssociatesIncome StatementFor the Year Ended December 31,.pdf
Ruth & AssociatesIncome StatementFor the Year Ended December 31,.pdf
karan8801
P(all same) = 6*(1/6)^5 = 1/6^4 = 1/1296 P(all Different) = 1/6 * 1/5 *1/4 * 1/3 *1/2 = 1/720 Solution P(all same) = 6*(1/6)^5 = 1/6^4 = 1/1296 P(all Different) = 1/6 * 1/5 *1/4 * 1/3 *1/2 = 1/720.
P(all same) = 6(16)^5 = 16^4 = 11296P(all Different) = 16 1.pdf
P(all same) = 6(16)^5 = 16^4 = 11296P(all Different) = 16 1.pdf
karan8801
Part A: blood solutes, swell Part B: blood osmotic pressure, blood osmotic pressure Part C: concentrations. Part D: passive Part E: number, increases Part F: Atrial natriuretic peptide (ANP), Anti diuretic Hormone (ADH) Part G: Aldosterone Part H: over hydration,water intoxication. Part I: low, high Part J: swell, increased. Solution Part A: blood solutes, swell Part B: blood osmotic pressure, blood osmotic pressure Part C: concentrations. Part D: passive Part E: number, increases Part F: Atrial natriuretic peptide (ANP), Anti diuretic Hormone (ADH) Part G: Aldosterone Part H: over hydration,water intoxication. Part I: low, high Part J: swell, increased..
Part A blood solutes, swell Part B blood osmotic pressure, blood.pdf
Part A blood solutes, swell Part B blood osmotic pressure, blood.pdf
karan8801
no of units produced no of units of component A required price per unit total= no of units*no of unit of component*price per unit 7- cost of component A 1200 2 1.25 3000 cost of component B 1200 3 0.8 2880 total budgeted cost for may 2018 5880 no of units produced cost per unit of A = 2*1.25 cost per unit of b = 3*.8 total cost =(no of units*cost of A)+(no of units*cost of B) 8- April 1000 2.5 2.4 4900 May 1200 2.5 2.4 5880 June 1250 2.5 2.4 6125 total 16905 9- INVENTORY OF COMPONENT A IN APRIL UNITS PRODUCED 1000 PER UNIT, UNITS OF COMPONENT a REQUIRED 2 UNITS OF COMPONENT A REQUIRED 2000 VALUE OF INVENTORY =2000*1.25 2500 10- total direct labor cost in process department -April to june Month units produced labor cost per unit in process department = labor hour per unit* labor cost per hour total labor cost in process department April 1000 8 8000 May 1200 8 9600 june 1250 8 10000 total direct labor cost in process department -April to june 27600 11- total direct labor cost in june no of units labor cost per unit in process department = labor hour per unit* labor cost per hour labor cost per unit in assembly department = labor hour per unit* labor cost per hour total = no of units produced*labor cost per unit in process department +no of units*labor cost in assembly department 1250 8 6 17500 12- Month units produced direct labor cost per unit = labor cost in process +labor cost in assembly total April 1000 14 14000 May 1200 14 16800 june 1250 14 17500 total direct labor cost in process department -April to june 48300 no of units produced no of units of component A required price per unit total= no of units*no of unit of component*price per unit 7- cost of component A 1200 2 1.25 3000 cost of component B 1200 3 0.8 2880 total budgeted cost for may 2018 5880 no of units produced cost per unit of A = 2*1.25 cost per unit of b = 3*.8 total cost =(no of units*cost of A)+(no of units*cost of B) 8- April 1000 2.5 2.4 4900 May 1200 2.5 2.4 5880 June 1250 2.5 2.4 6125 total 16905 9- INVENTORY OF COMPONENT A IN APRIL UNITS PRODUCED 1000 PER UNIT, UNITS OF COMPONENT a REQUIRED 2 UNITS OF COMPONENT A REQUIRED 2000 VALUE OF INVENTORY =2000*1.25 2500 10- total direct labor cost in process department -April to june Month units produced labor cost per unit in process department = labor hour per unit* labor cost per hour total labor cost in process department April 1000 8 8000 May 1200 8 9600 june 1250 8 10000 total direct labor cost in process department -April to june 27600 11- total direct labor cost in june no of units labor cost per unit in process department = labor hour per unit* labor cost per hour labor cost per unit in assembly department = labor hour per unit* labor cost per hour total = no of units produced*labor cost per unit in process department +no of units*labor cost in assembly department 1250 8 6 17500 12- Month units produced direct labor cost per unit = labor cost in process +labor cost in assembly total April 1000 14 14000 May 1200 14 16.
no of units producedno of units of component A requiredprice per.pdf
no of units producedno of units of component A requiredprice per.pdf
karan8801
benzene is less reactive than toluene towardselectrophilic substitution reactions because in toluene mehyl groupdonates electrons inductively to the benzene ring and activates ittowards electrophilic substitution but in case of nitromethylbenzene the electron donating capacity is decreased because nitrogroup is strongly electron withdrawing and will with drawelectrondensity. Solution benzene is less reactive than toluene towardselectrophilic substitution reactions because in toluene mehyl groupdonates electrons inductively to the benzene ring and activates ittowards electrophilic substitution but in case of nitromethylbenzene the electron donating capacity is decreased because nitrogroup is strongly electron withdrawing and will with drawelectrondensity..
benzene is less reactive than toluene towardselec.pdf
benzene is less reactive than toluene towardselec.pdf
karan8801
Net NonOperating Obligations = Non Operating Liabilities - Non Operating Assets Non Operating Liabilities = Total Liabilities - (Accounts Payable + Accrued Payroll + Accrued Income Taxes + Other Current Liabilites) Non Operating Assets = Marketable securities current + Marketable securities non current + Investments + Goodwill + Intangible assets + Prepaid pension benefits + Other Assets 2008: Non Operating liabilities = 15489 - (1301 + 644 + 350 + 1992) = $11202 Non Operating Assets = 373 + 352 + 111 + 5753 + 1398 + 36 + 1659 = $9682 Net non operating obligations = 11202 - 9682 = $1520 2007: Non operating Liabilities = 12947 - (1505 + 580 + 543 + 1833) = $8486 Net non operating assets = 579 + 480 + 298 + 4589 + 801 + 1378 + 728 = $8853 Net non operating obligations = 8486 - 8853 = -$367 Solution Net NonOperating Obligations = Non Operating Liabilities - Non Operating Assets Non Operating Liabilities = Total Liabilities - (Accounts Payable + Accrued Payroll + Accrued Income Taxes + Other Current Liabilites) Non Operating Assets = Marketable securities current + Marketable securities non current + Investments + Goodwill + Intangible assets + Prepaid pension benefits + Other Assets 2008: Non Operating liabilities = 15489 - (1301 + 644 + 350 + 1992) = $11202 Non Operating Assets = 373 + 352 + 111 + 5753 + 1398 + 36 + 1659 = $9682 Net non operating obligations = 11202 - 9682 = $1520 2007: Non operating Liabilities = 12947 - (1505 + 580 + 543 + 1833) = $8486 Net non operating assets = 579 + 480 + 298 + 4589 + 801 + 1378 + 728 = $8853 Net non operating obligations = 8486 - 8853 = -$367.
Net NonOperating Obligations = Non Operating Liabilities - Non Opera.pdf
Net NonOperating Obligations = Non Operating Liabilities - Non Opera.pdf
karan8801
Malware aimed at mobile devices because: Solution Malware aimed at mobile devices because:.
Malware aimed at mobile devices becauseSolutionMalware aimed .pdf
Malware aimed at mobile devices becauseSolutionMalware aimed .pdf
karan8801
Let´s make a a very short overview of the two stories. Fourth day. A monk, fallen into sin worthy of serious punishment, get rid of the penalty scolding discreetly to his Abbot by that same guilt. The fifth day, the Marchioness of Monferrato restrains, with a treat of chickens and ingenious words, crazy love of the King of France. The themes present are love, eroticism, human intelligence, the virtue, fortune, religious beliefs and the criticism of the customs of the time. Their mean characteristics of two stories are the language is cultured and elegant, but at the same time scam with vulgar meanings. They are true comedy human of the middle ages and represents the complex reality of the world without superstructure and traditional expressive models. Solution Let´s make a a very short overview of the two stories. Fourth day. A monk, fallen into sin worthy of serious punishment, get rid of the penalty scolding discreetly to his Abbot by that same guilt. The fifth day, the Marchioness of Monferrato restrains, with a treat of chickens and ingenious words, crazy love of the King of France. The themes present are love, eroticism, human intelligence, the virtue, fortune, religious beliefs and the criticism of the customs of the time. Their mean characteristics of two stories are the language is cultured and elegant, but at the same time scam with vulgar meanings. They are true comedy human of the middle ages and represents the complex reality of the world without superstructure and traditional expressive models..
Let´s make a a very short overview of the two stories.Fourth day. .pdf
Let´s make a a very short overview of the two stories.Fourth day. .pdf
karan8801
measurement and evaluation in education Tests à référence normative tests à référence critériée Edumétrie Psychométrie
GHASSOUB _Seance 4_ measurement and evaluation in education_-.pptx
GHASSOUB _Seance 4_ measurement and evaluation in education_-.pptx
Abderrahim GHASSOUB
Réunion des directrices et directeurs de la circonscription de Jonzac le 15 mai 2024 au lycée Jean Hyppolite de Jonzac. Enjeux de la protection de l'enfance
Réunion des directeurs de Jonzac - 15 mai 2024
Réunion des directeurs de Jonzac - 15 mai 2024
IEN_Jonzac
Contenu connexe
Plus de karan8801
Lewis Acid Base Reaction Solution Lewis Acid Base Reaction.
Lewis Acid Base Reaction .pdf
Lewis Acid Base Reaction .pdf
karan8801
KOH ====> K+ (aq) + OH-(aq) [OH-] =9.7*10^-5 [H+][OH-] = 1.0*10^-14 [H+] = (1.0*10^-14)/(9.7*10^-5) = 1.03*10^-10 pH = -log[H+] = -log(1.03*10^-10) = 9.987 = 10 pH+pOH=14 pOh = 14-pH = 14-10=4 Solution KOH ====> K+ (aq) + OH-(aq) [OH-] =9.7*10^-5 [H+][OH-] = 1.0*10^-14 [H+] = (1.0*10^-14)/(9.7*10^-5) = 1.03*10^-10 pH = -log[H+] = -log(1.03*10^-10) = 9.987 = 10 pH+pOH=14 pOh = 14-pH = 14-10=4.
KOH ==== K+ (aq) + OH-(aq) [OH-] =9.710^-5 [H.pdf
KOH ==== K+ (aq) + OH-(aq) [OH-] =9.710^-5 [H.pdf
karan8801
It is a reaction for secondary amines .Option 2 is correct. Solution It is a reaction for secondary amines .Option 2 is correct..
It is a reaction for secondary amines .Option 2 i.pdf
It is a reaction for secondary amines .Option 2 i.pdf
karan8801
fluorene is mobile phase and fluorenone is stationary phase because : Mostly,it depends on the solvent you are using, but lets assume you use pentane as the solvent for fluorene and diethyl ether as the solvent for fluorenone. Used in this order, fluorene should actaully move slower down the column due to the the polarity of the alumina and the nonpolarity of the solvent (pentane). Since fluorene is less polar so, remember the rule like dissolves like? Well, this kind of applies to chromatography wherein rather than dissolve just replace it with moves with, so like moves with like. That being said the pentane will \"carry\" the fluorene through the alumina slower than the latter (which I will explain). Fluorenone is polar because of its C=O bond, that being said the dielectric constant of diethyl ether is 4.3 which means it has intermediate polarity (remember that pentane has a dielectric constant of 2.1 I think, so it is nonpolar). Since the alumina, diethyl ether, and the fluorenone are all polar, the fluorenone will travel faster through the alumina than would the fluorene, because there is no attraction between all these polar compounds which will allow it to move faster, rather than a nonpolar and polar chemical having an attration towards each other and thus moving more slowly. Solution fluorene is mobile phase and fluorenone is stationary phase because : Mostly,it depends on the solvent you are using, but lets assume you use pentane as the solvent for fluorene and diethyl ether as the solvent for fluorenone. Used in this order, fluorene should actaully move slower down the column due to the the polarity of the alumina and the nonpolarity of the solvent (pentane). Since fluorene is less polar so, remember the rule like dissolves like? Well, this kind of applies to chromatography wherein rather than dissolve just replace it with moves with, so like moves with like. That being said the pentane will \"carry\" the fluorene through the alumina slower than the latter (which I will explain). Fluorenone is polar because of its C=O bond, that being said the dielectric constant of diethyl ether is 4.3 which means it has intermediate polarity (remember that pentane has a dielectric constant of 2.1 I think, so it is nonpolar). Since the alumina, diethyl ether, and the fluorenone are all polar, the fluorenone will travel faster through the alumina than would the fluorene, because there is no attraction between all these polar compounds which will allow it to move faster, rather than a nonpolar and polar chemical having an attration towards each other and thus moving more slowly..
fluorene is mobile phase and fluorenone is statio.pdf
fluorene is mobile phase and fluorenone is statio.pdf
karan8801
equivalent mass is 27.9 . so Molecular mass/moles of electrons per mole = 27.9 looking at the table for atomic masses, the closest we get is for moles of electrons per mole = 2 that is , molecular weight = 55.8 which is for Iron Hence Y is Iron ( Fe ) Solution equivalent mass is 27.9 . so Molecular mass/moles of electrons per mole = 27.9 looking at the table for atomic masses, the closest we get is for moles of electrons per mole = 2 that is , molecular weight = 55.8 which is for Iron Hence Y is Iron ( Fe ).
equivalent mass is 27.9 . so Molecular massmole.pdf
equivalent mass is 27.9 . so Molecular massmole.pdf
karan8801
There is no solution Solution There is no solution.
There is no solutionSolutionThere is no solution.pdf
There is no solutionSolutionThere is no solution.pdf
karan8801
This is a case of allopatric speciation. In this case lice from gorilla moved and solutes into two different species. So this is case of geographical speciation where species become apart from each other to an extent that they cannot interbreed. Solution This is a case of allopatric speciation. In this case lice from gorilla moved and solutes into two different species. So this is case of geographical speciation where species become apart from each other to an extent that they cannot interbreed..
This is a case of allopatric speciation. In this case lice from gori.pdf
This is a case of allopatric speciation. In this case lice from gori.pdf
karan8801
d. 8 The bonding electrons are the electrons taking part in the bonds. C is in the middle and H and O are on either side of C. C shares 2 electrons each with each of the two Hs, so C shares a total of 4 electrons with the Hs. Carbon shares a total of 4 electrons with O. So 4+4= 8 bonding electrons total. Solution d. 8 The bonding electrons are the electrons taking part in the bonds. C is in the middle and H and O are on either side of C. C shares 2 electrons each with each of the two Hs, so C shares a total of 4 electrons with the Hs. Carbon shares a total of 4 electrons with O. So 4+4= 8 bonding electrons total..
d. 8 The bonding electrons are the electrons taki.pdf
d. 8 The bonding electrons are the electrons taki.pdf
karan8801
The genetical changes of single locus are responsible for divergent branches of Y chromosome phylogenetic tree in different geographical locations elicit the human origins. The mitochondrial DNA sequences of Neanderthals and modern humans were compared and analyzed. The Hominins species were found to be 30000 year old and were compared with Neanderthal specimens. The relationships between these populations have drawn with the help of the age of specimens plotted against the average genetic distance. The Hominins type sequences are very rare when compared with the modern European samples. The prehistoric Europeans had shown constant number of differences when compared to today’s Europeans and Neanderthals. The Neanderthals have shown the genetic continuity in modern human’s genealogy. The Neanderthals had shown clear discontinuity with the Hominins from upper Paleolithic period to the similar ages. The Neanderthals and early anatomical modern humans were thought to be regional populations of the same evolving species connected by gene flow and both the archaic and modern humans contribute in different proportions to the present human gene pool. The transition and demographic replacement separated the population as Neanderthals and modern humans are considered just as one population observed at different times. The multi regional models predict that at least some level of genetic continuity can be observed from archaic Neanderthal forms to the contemporary Hominins forms and up to today’s Europeans. The results have not matched with the view that the Neanderthal were genetically related with the anatomically modern ancestors of current Europeans or contributed to the present day human gene pool. Solution The genetical changes of single locus are responsible for divergent branches of Y chromosome phylogenetic tree in different geographical locations elicit the human origins. The mitochondrial DNA sequences of Neanderthals and modern humans were compared and analyzed. The Hominins species were found to be 30000 year old and were compared with Neanderthal specimens. The relationships between these populations have drawn with the help of the age of specimens plotted against the average genetic distance. The Hominins type sequences are very rare when compared with the modern European samples. The prehistoric Europeans had shown constant number of differences when compared to today’s Europeans and Neanderthals. The Neanderthals have shown the genetic continuity in modern human’s genealogy. The Neanderthals had shown clear discontinuity with the Hominins from upper Paleolithic period to the similar ages. The Neanderthals and early anatomical modern humans were thought to be regional populations of the same evolving species connected by gene flow and both the archaic and modern humans contribute in different proportions to the present human gene pool. The transition and demographic replacement separated the population as Neanderthals and modern humans .
The genetical changes of single locus are responsible for divergent .pdf
The genetical changes of single locus are responsible for divergent .pdf
karan8801
The corrects answers are: Thr and Asn are polar amino acids. Isoleucine has more than one stereocenter (chiral center) The Leu side chain does not from hydrogen bonds with other amino acids Solution The corrects answers are: Thr and Asn are polar amino acids. Isoleucine has more than one stereocenter (chiral center) The Leu side chain does not from hydrogen bonds with other amino acids.
The corrects answers areThr and Asn are polar amino acids.Isole.pdf
The corrects answers areThr and Asn are polar amino acids.Isole.pdf
karan8801
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4 i.e Z4 = ( 0,1,2,3) U(8) = { 1,3,5,7} while each element of U(8) has order 1 or 2 So, isomorphic Z4 and U(8) are not isomorphic. Solution Since Z4 has an element of order 4,Z4, its a cyclic group of order 4 i.e Z4 = ( 0,1,2,3) U(8) = { 1,3,5,7} while each element of U(8) has order 1 or 2 So, isomorphic Z4 and U(8) are not isomorphic..
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4.pdf
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4.pdf
karan8801
selected palnts xerophyte verus mesophyte ========================================================= A xerophyte is a types of plant that has adjusted to get by in a domain with minimal fluid water, for example, a forsake or an ice-or snow-canvassed locale in the Alps or the Arctic. The morphology and physiology of xerophytes are differently adjusted to moderate water, and usually likewise to store substantial amounts of water, amid dry periods. Different species might be adjusted to survive long stretches of parching of their tissues, amid which their metabolic action may viably close down. Plants with such morphological and physiological adjustments are xeromorphic. Xerophytic plants may have comparable shapes, structures, and structures and look fundamentally the same as, regardless of the possibility that the plants are not firmly related, through a procedure called concurrent development. For instance, a few types of desert flora (individuals from the family Cactaceae), which advanced just in the Americas, may seem like Euphorbias, which are dispersed around the world. A random types of caudiciforms, plants with swollen bases that are utilized to store water, may likewise show such likenesses. Xerophytic plants can have less general surface territory than different plants, so diminishing the range that is presented to the air and lessening water misfortune by vanishing. Xerophytes can have littler leaves or less branches than different plants. A case of leaf surface decrease are the spines of a desert flora. A case of compaction and diminishment of spreading are the barrel desert flora. Different xerophytes may have their leaves compacted at the base, as in a basal rosette, which might be littler than the plant\'s blossom. This adjustment is displayed by some Agave and Eriogonum species, which can be discovered developing close Death Valley. A few xerophytes have minor hairs on their surface to give a wind break and decrease wind current, along these lines diminishing the rate of dissipation. At the point when a plant surface is secured with minor hairs, it is called tomentose. In a still domain, the regions under the leaves/spines where transpiration is occurring structure a little limited environment that is more soaked than typical with water vapor. In the event that this is not overwhelmed by wind, the water vapor potential angle is diminished as is transpiration. Subsequently, in a windier circumstance, this confinement is not held thus the angle stays high, which helps the loss of water vapor. Spines trap a layer of dampness furthermore moderate air development over tissues. =================== Mesophytes are earthbound plants which are adjusted to neither an especially dry nor especially wet environment. A case of a mesophytic living space would be a country calm glade, which may contain goldenrod, clover, oxeye daisy, and Rosa multiflora. Mesophytic plants have unbending, tough, openly expanded stems and stringy, all around create.
selected palntsxerophyte verus mesophyte========================.pdf
selected palntsxerophyte verus mesophyte========================.pdf
karan8801
Ruth & Associates Income Statement For the Year Ended December 31, 2010 Revenue Consulting Revenue 100000 Interest Revenue 4000 Total Revenue 104000 Expenses Salary Expense 47000 Interest Expense 5000 Total Expenses 52000 Net Income 52000 Ruth & Associates Statement of Changes in Stockholders’ Equity For the Year Ended December 31, 2010 Beginning Common Stock 25000 Plus: Common Stock Issued 15000 Ending Common Stock 40000 Beginning Retained Earnings 18000 Plus: Net Income 52000 Less: Dividends 10000 Ending Retained Earnings 60000 Total Stockholders’ Equity 100000 Ruth & Associates Balance Sheet As of the December 31, 2010 Assets Cash 52000 Accounts Receivable 26000 Supplies 3000 Prepaid Rent 5000 Land 63000 Total Assets 149000 Liabilities Interest Payable 2000 Salaries Payable 7000 Unearned Revenue 8000 Notes Payable 32000 Total Liabilities 49000 Stockholders’ Equity Common Stock 40000 Retained Earnings 60000 Total Stockholders’ Equity 100000 Total Liab. and Stockholders’ Equity 149000 Ruth & Associates Statement of Cash Flows For the Year Ended December 31, 2010 Cash Flow From Operating Activities 52000 Cash Flow From Investing Activities -30000 Cash Flow From Financing Activities 20000 Net Change in Cash Plus: 42000 Beginning Cash Balance 10000 Ending Cash Balance 52000 Ending Cash Balance 52000 Net Change in Cash 42000 = Beginning Cash Balance 10000 Ruth & Associates Income Statement For the Year Ended December 31, 2010 Revenue Consulting Revenue 100000 Interest Revenue 4000 Total Revenue 104000 Expenses Salary Expense 47000 Interest Expense 5000 Total Expenses 52000 Net Income 52000 Ruth & Associates Statement of Changes in Stockholders’ Equity For the Year Ended December 31, 2010 Beginning Common Stock 25000 Plus: Common Stock Issued 15000 Ending Common Stock 40000 Beginning Retained Earnings 18000 Plus: Net Income 52000 Less: Dividends 10000 Ending Retained Earnings 60000 Total Stockholders’ Equity 100000 Ruth & Associates Balance Sheet As of the December 31, 2010 Assets Cash 52000 Accounts Receivable 26000 Supplies 3000 Prepaid Rent 5000 Land 63000 Total Assets 149000 Liabilities Interest Payable 2000 Salaries Payable 7000 Unearned Revenue 8000 Notes Payable 32000 Total Liabilities 49000 Stockholders’ Equity Common Stock 40000 Retained Earnings 60000 Total Stockholders’ Equity 100000 Total Liab. and Stockholders’ Equity 149000 Ruth & Associates Statement of Cash Flows For the Year Ended December 31, 2010 Cash Flow From Operating Activities 52000 Cash Flow From Investing Activities -30000 Cash Flow From Financing Activities 20000 Net Change in Cash Plus: 42000 Beginning Cash Balance 10000 Ending Cash Balance 52000 Ending Cash Balance 52000 Net Change in Cash 42000 = Beginning Cash Balance 10000 Solution Ruth & Associates Income Statement For the Year Ended December 31, 2010 Revenue Consulting Revenue 100000 Interest Revenue 4000 Total Revenue 104000 Expenses Salary Expense 47000 Interest Expense 5000 Total Expenses 52000 Net Income 5200.
Ruth & AssociatesIncome StatementFor the Year Ended December 31,.pdf
Ruth & AssociatesIncome StatementFor the Year Ended December 31,.pdf
karan8801
P(all same) = 6*(1/6)^5 = 1/6^4 = 1/1296 P(all Different) = 1/6 * 1/5 *1/4 * 1/3 *1/2 = 1/720 Solution P(all same) = 6*(1/6)^5 = 1/6^4 = 1/1296 P(all Different) = 1/6 * 1/5 *1/4 * 1/3 *1/2 = 1/720.
P(all same) = 6(16)^5 = 16^4 = 11296P(all Different) = 16 1.pdf
P(all same) = 6(16)^5 = 16^4 = 11296P(all Different) = 16 1.pdf
karan8801
Part A: blood solutes, swell Part B: blood osmotic pressure, blood osmotic pressure Part C: concentrations. Part D: passive Part E: number, increases Part F: Atrial natriuretic peptide (ANP), Anti diuretic Hormone (ADH) Part G: Aldosterone Part H: over hydration,water intoxication. Part I: low, high Part J: swell, increased. Solution Part A: blood solutes, swell Part B: blood osmotic pressure, blood osmotic pressure Part C: concentrations. Part D: passive Part E: number, increases Part F: Atrial natriuretic peptide (ANP), Anti diuretic Hormone (ADH) Part G: Aldosterone Part H: over hydration,water intoxication. Part I: low, high Part J: swell, increased..
Part A blood solutes, swell Part B blood osmotic pressure, blood.pdf
Part A blood solutes, swell Part B blood osmotic pressure, blood.pdf
karan8801
no of units produced no of units of component A required price per unit total= no of units*no of unit of component*price per unit 7- cost of component A 1200 2 1.25 3000 cost of component B 1200 3 0.8 2880 total budgeted cost for may 2018 5880 no of units produced cost per unit of A = 2*1.25 cost per unit of b = 3*.8 total cost =(no of units*cost of A)+(no of units*cost of B) 8- April 1000 2.5 2.4 4900 May 1200 2.5 2.4 5880 June 1250 2.5 2.4 6125 total 16905 9- INVENTORY OF COMPONENT A IN APRIL UNITS PRODUCED 1000 PER UNIT, UNITS OF COMPONENT a REQUIRED 2 UNITS OF COMPONENT A REQUIRED 2000 VALUE OF INVENTORY =2000*1.25 2500 10- total direct labor cost in process department -April to june Month units produced labor cost per unit in process department = labor hour per unit* labor cost per hour total labor cost in process department April 1000 8 8000 May 1200 8 9600 june 1250 8 10000 total direct labor cost in process department -April to june 27600 11- total direct labor cost in june no of units labor cost per unit in process department = labor hour per unit* labor cost per hour labor cost per unit in assembly department = labor hour per unit* labor cost per hour total = no of units produced*labor cost per unit in process department +no of units*labor cost in assembly department 1250 8 6 17500 12- Month units produced direct labor cost per unit = labor cost in process +labor cost in assembly total April 1000 14 14000 May 1200 14 16800 june 1250 14 17500 total direct labor cost in process department -April to june 48300 no of units produced no of units of component A required price per unit total= no of units*no of unit of component*price per unit 7- cost of component A 1200 2 1.25 3000 cost of component B 1200 3 0.8 2880 total budgeted cost for may 2018 5880 no of units produced cost per unit of A = 2*1.25 cost per unit of b = 3*.8 total cost =(no of units*cost of A)+(no of units*cost of B) 8- April 1000 2.5 2.4 4900 May 1200 2.5 2.4 5880 June 1250 2.5 2.4 6125 total 16905 9- INVENTORY OF COMPONENT A IN APRIL UNITS PRODUCED 1000 PER UNIT, UNITS OF COMPONENT a REQUIRED 2 UNITS OF COMPONENT A REQUIRED 2000 VALUE OF INVENTORY =2000*1.25 2500 10- total direct labor cost in process department -April to june Month units produced labor cost per unit in process department = labor hour per unit* labor cost per hour total labor cost in process department April 1000 8 8000 May 1200 8 9600 june 1250 8 10000 total direct labor cost in process department -April to june 27600 11- total direct labor cost in june no of units labor cost per unit in process department = labor hour per unit* labor cost per hour labor cost per unit in assembly department = labor hour per unit* labor cost per hour total = no of units produced*labor cost per unit in process department +no of units*labor cost in assembly department 1250 8 6 17500 12- Month units produced direct labor cost per unit = labor cost in process +labor cost in assembly total April 1000 14 14000 May 1200 14 16.
no of units producedno of units of component A requiredprice per.pdf
no of units producedno of units of component A requiredprice per.pdf
karan8801
benzene is less reactive than toluene towardselectrophilic substitution reactions because in toluene mehyl groupdonates electrons inductively to the benzene ring and activates ittowards electrophilic substitution but in case of nitromethylbenzene the electron donating capacity is decreased because nitrogroup is strongly electron withdrawing and will with drawelectrondensity. Solution benzene is less reactive than toluene towardselectrophilic substitution reactions because in toluene mehyl groupdonates electrons inductively to the benzene ring and activates ittowards electrophilic substitution but in case of nitromethylbenzene the electron donating capacity is decreased because nitrogroup is strongly electron withdrawing and will with drawelectrondensity..
benzene is less reactive than toluene towardselec.pdf
benzene is less reactive than toluene towardselec.pdf
karan8801
Net NonOperating Obligations = Non Operating Liabilities - Non Operating Assets Non Operating Liabilities = Total Liabilities - (Accounts Payable + Accrued Payroll + Accrued Income Taxes + Other Current Liabilites) Non Operating Assets = Marketable securities current + Marketable securities non current + Investments + Goodwill + Intangible assets + Prepaid pension benefits + Other Assets 2008: Non Operating liabilities = 15489 - (1301 + 644 + 350 + 1992) = $11202 Non Operating Assets = 373 + 352 + 111 + 5753 + 1398 + 36 + 1659 = $9682 Net non operating obligations = 11202 - 9682 = $1520 2007: Non operating Liabilities = 12947 - (1505 + 580 + 543 + 1833) = $8486 Net non operating assets = 579 + 480 + 298 + 4589 + 801 + 1378 + 728 = $8853 Net non operating obligations = 8486 - 8853 = -$367 Solution Net NonOperating Obligations = Non Operating Liabilities - Non Operating Assets Non Operating Liabilities = Total Liabilities - (Accounts Payable + Accrued Payroll + Accrued Income Taxes + Other Current Liabilites) Non Operating Assets = Marketable securities current + Marketable securities non current + Investments + Goodwill + Intangible assets + Prepaid pension benefits + Other Assets 2008: Non Operating liabilities = 15489 - (1301 + 644 + 350 + 1992) = $11202 Non Operating Assets = 373 + 352 + 111 + 5753 + 1398 + 36 + 1659 = $9682 Net non operating obligations = 11202 - 9682 = $1520 2007: Non operating Liabilities = 12947 - (1505 + 580 + 543 + 1833) = $8486 Net non operating assets = 579 + 480 + 298 + 4589 + 801 + 1378 + 728 = $8853 Net non operating obligations = 8486 - 8853 = -$367.
Net NonOperating Obligations = Non Operating Liabilities - Non Opera.pdf
Net NonOperating Obligations = Non Operating Liabilities - Non Opera.pdf
karan8801
Malware aimed at mobile devices because: Solution Malware aimed at mobile devices because:.
Malware aimed at mobile devices becauseSolutionMalware aimed .pdf
Malware aimed at mobile devices becauseSolutionMalware aimed .pdf
karan8801
Let´s make a a very short overview of the two stories. Fourth day. A monk, fallen into sin worthy of serious punishment, get rid of the penalty scolding discreetly to his Abbot by that same guilt. The fifth day, the Marchioness of Monferrato restrains, with a treat of chickens and ingenious words, crazy love of the King of France. The themes present are love, eroticism, human intelligence, the virtue, fortune, religious beliefs and the criticism of the customs of the time. Their mean characteristics of two stories are the language is cultured and elegant, but at the same time scam with vulgar meanings. They are true comedy human of the middle ages and represents the complex reality of the world without superstructure and traditional expressive models. Solution Let´s make a a very short overview of the two stories. Fourth day. A monk, fallen into sin worthy of serious punishment, get rid of the penalty scolding discreetly to his Abbot by that same guilt. The fifth day, the Marchioness of Monferrato restrains, with a treat of chickens and ingenious words, crazy love of the King of France. The themes present are love, eroticism, human intelligence, the virtue, fortune, religious beliefs and the criticism of the customs of the time. Their mean characteristics of two stories are the language is cultured and elegant, but at the same time scam with vulgar meanings. They are true comedy human of the middle ages and represents the complex reality of the world without superstructure and traditional expressive models..
Let´s make a a very short overview of the two stories.Fourth day. .pdf
Let´s make a a very short overview of the two stories.Fourth day. .pdf
karan8801
Plus de karan8801
(20)
Lewis Acid Base Reaction .pdf
Lewis Acid Base Reaction .pdf
KOH ==== K+ (aq) + OH-(aq) [OH-] =9.710^-5 [H.pdf
KOH ==== K+ (aq) + OH-(aq) [OH-] =9.710^-5 [H.pdf
It is a reaction for secondary amines .Option 2 i.pdf
It is a reaction for secondary amines .Option 2 i.pdf
fluorene is mobile phase and fluorenone is statio.pdf
fluorene is mobile phase and fluorenone is statio.pdf
equivalent mass is 27.9 . so Molecular massmole.pdf
equivalent mass is 27.9 . so Molecular massmole.pdf
There is no solutionSolutionThere is no solution.pdf
There is no solutionSolutionThere is no solution.pdf
This is a case of allopatric speciation. In this case lice from gori.pdf
This is a case of allopatric speciation. In this case lice from gori.pdf
d. 8 The bonding electrons are the electrons taki.pdf
d. 8 The bonding electrons are the electrons taki.pdf
The genetical changes of single locus are responsible for divergent .pdf
The genetical changes of single locus are responsible for divergent .pdf
The corrects answers areThr and Asn are polar amino acids.Isole.pdf
The corrects answers areThr and Asn are polar amino acids.Isole.pdf
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4.pdf
Since Z4 has an element of order 4,Z4, its a cyclic group of order 4.pdf
selected palntsxerophyte verus mesophyte========================.pdf
selected palntsxerophyte verus mesophyte========================.pdf
Ruth & AssociatesIncome StatementFor the Year Ended December 31,.pdf
Ruth & AssociatesIncome StatementFor the Year Ended December 31,.pdf
P(all same) = 6(16)^5 = 16^4 = 11296P(all Different) = 16 1.pdf
P(all same) = 6(16)^5 = 16^4 = 11296P(all Different) = 16 1.pdf
Part A blood solutes, swell Part B blood osmotic pressure, blood.pdf
Part A blood solutes, swell Part B blood osmotic pressure, blood.pdf
no of units producedno of units of component A requiredprice per.pdf
no of units producedno of units of component A requiredprice per.pdf
benzene is less reactive than toluene towardselec.pdf
benzene is less reactive than toluene towardselec.pdf
Net NonOperating Obligations = Non Operating Liabilities - Non Opera.pdf
Net NonOperating Obligations = Non Operating Liabilities - Non Opera.pdf
Malware aimed at mobile devices becauseSolutionMalware aimed .pdf
Malware aimed at mobile devices becauseSolutionMalware aimed .pdf
Let´s make a a very short overview of the two stories.Fourth day. .pdf
Let´s make a a very short overview of the two stories.Fourth day. .pdf
Dernier
measurement and evaluation in education Tests à référence normative tests à référence critériée Edumétrie Psychométrie
GHASSOUB _Seance 4_ measurement and evaluation in education_-.pptx
GHASSOUB _Seance 4_ measurement and evaluation in education_-.pptx
Abderrahim GHASSOUB
Réunion des directrices et directeurs de la circonscription de Jonzac le 15 mai 2024 au lycée Jean Hyppolite de Jonzac. Enjeux de la protection de l'enfance
Réunion des directeurs de Jonzac - 15 mai 2024
Réunion des directeurs de Jonzac - 15 mai 2024
IEN_Jonzac
measurement and evaluation in education Psychometry- testing-test design
GHASSOUB _Seance 3_ measurement and evaluation in education.pptx
GHASSOUB _Seance 3_ measurement and evaluation in education.pptx
Abderrahim GHASSOUB
Quitter la nuit est un film réaisé par une femme belge.
Quitter la nuit. pptx
Quitter la nuit. pptx
Txaruka
Film franáis réalisé par Marie Amachoukelli et tourné au Cap Vert et en France.
Àma Gloria.pptx Un film tourné au Cap Vert et en France
Àma Gloria.pptx Un film tourné au Cap Vert et en France
Txaruka
Par Hubert Passot Pur l'association des Amis du Musée de l'Imprimerie Février 1953. Une révolution a lieu en France dans le milieu éditorial avec la parution des premiers titres du Livre de Poche, sous l'impulsion d'Henri Filipacchi. Très vite, le Livre de Poche devient un symbole populaire qui permet l'accès à la culture de nouvelles catégories de population, en particulier les plus jeunes. Les premières années sont modestes, mais dès 1958 plus de 8 millions d'exemplaires ont été édités. C'est la décennie suivante qui voit l'explosion du phénomène avec plus de 28 millions d'exemplaires en 1969. Un succès qui ne s'arrêtera plus puisque depuis sa création il y a 70 ans plus d'un milliard de volumes ont été diffusés! Hubert Passot vous propose un diaporama des cent premiers numéros de la collection, en évoquant le graphisme des couvertures illustrées comme des affiches de cinéma.
Les débuts de la collection "Le livre de poche"
Les débuts de la collection "Le livre de poche"
ArchivesdeLyon
Projet eTwinning de A1 class - 3me lycee de Kastoria
Un petit coin etwinning- Au fil des cultures urbaines
Un petit coin etwinning- Au fil des cultures urbaines
Socratis Vasiopoulos
Peintre française née en Israël qui expose son oeuvre dans le musée d'Orsay
Nathanaëlle Herbelin.pptx Peintre française
Nathanaëlle Herbelin.pptx Peintre française
Txaruka
Mémoire infirmier
PowerPoint-de-Soutenance-de-TFE-infirmier.pdf
PowerPoint-de-Soutenance-de-TFE-infirmier.pdf
DafWafia
Fiche de vocabulaire pour faire une appréciation (niveau A2)
Fiche de vocabulaire pour faire une appréciation
Fiche de vocabulaire pour faire une appréciation
LeBaobabBleu1
assimilé plusieurs informations techniques qui concerne le domaine du gros œuvre en participant à plusieurs tâches dans le bureau ; à savoir l’établissement des plans de coffrages, impression des plans, estimation du béton armé dans un projet de construction…etc.
rapport de stage gros oeuvre_compressed.pdf
rapport de stage gros oeuvre_compressed.pdf
OssamaLachheb
Texte Texte avec différentes critiques positives, négatives ou mitigées à travailler avec le niveau A2
Texte avec différentes critiques positives, négatives ou mitigées
Texte avec différentes critiques positives, négatives ou mitigées
LeBaobabBleu1
Grille d'audit 5S
Exemple de grille d'audit 5S, check liste Audit
Exemple de grille d'audit 5S, check liste Audit
techwinconsulting
Film réalisé par Delphine Girard.
Quitter la nuit. pptx
Quitter la nuit. pptx
Txaruka
Dernier
(14)
GHASSOUB _Seance 4_ measurement and evaluation in education_-.pptx
GHASSOUB _Seance 4_ measurement and evaluation in education_-.pptx
Réunion des directeurs de Jonzac - 15 mai 2024
Réunion des directeurs de Jonzac - 15 mai 2024
GHASSOUB _Seance 3_ measurement and evaluation in education.pptx
GHASSOUB _Seance 3_ measurement and evaluation in education.pptx
Quitter la nuit. pptx
Quitter la nuit. pptx
Àma Gloria.pptx Un film tourné au Cap Vert et en France
Àma Gloria.pptx Un film tourné au Cap Vert et en France
Les débuts de la collection "Le livre de poche"
Les débuts de la collection "Le livre de poche"
Un petit coin etwinning- Au fil des cultures urbaines
Un petit coin etwinning- Au fil des cultures urbaines
Nathanaëlle Herbelin.pptx Peintre française
Nathanaëlle Herbelin.pptx Peintre française
PowerPoint-de-Soutenance-de-TFE-infirmier.pdf
PowerPoint-de-Soutenance-de-TFE-infirmier.pdf
Fiche de vocabulaire pour faire une appréciation
Fiche de vocabulaire pour faire une appréciation
rapport de stage gros oeuvre_compressed.pdf
rapport de stage gros oeuvre_compressed.pdf
Texte avec différentes critiques positives, négatives ou mitigées
Texte avec différentes critiques positives, négatives ou mitigées
Exemple de grille d'audit 5S, check liste Audit
Exemple de grille d'audit 5S, check liste Audit
Quitter la nuit. pptx
Quitter la nuit. pptx
sp2 - sp2 .pdf
1.
sp2 - sp2 Solution sp2
- sp2
Télécharger maintenant