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5 steps: (make sure to press \"See full answer\") - make observation and gathering all available data Solution 5 steps: (make sure to press \"See full answer\") - make observation and gathering all available data.
5 steps (make sure to press See full answer)- make observat.pdf
5 steps (make sure to press See full answer)- make observat.pdf
nareshsonyericcson
2cosx=(cosxtanx+sinx)/(tanx) taking RHS : cosxtanx+sinx)/(tanx) = cosxtanx/tanx + sinx/tanx = cosx + sinx *cosx/sinx = cosx + cosx = 2cosx = LHS hence LHS = RHS Solution 2cosx=(cosxtanx+sinx)/(tanx) taking RHS : cosxtanx+sinx)/(tanx) = cosxtanx/tanx + sinx/tanx = cosx + sinx *cosx/sinx = cosx + cosx = 2cosx = LHS hence LHS = RHS.
2cosx=(cosxtanx+sinx)(tanx)taking RHS cosxtanx+sinx)(tanx)= .pdf
2cosx=(cosxtanx+sinx)(tanx)taking RHS cosxtanx+sinx)(tanx)= .pdf
nareshsonyericcson
1.5 X 10-8 Solution 1.5 X 10-8.
1.5 X 10-8Solution1.5 X 10-8.pdf
1.5 X 10-8Solution1.5 X 10-8.pdf
nareshsonyericcson
#include #include int encrypt(void); int decrypt(void); int encrypt_view(void); int decrypt_view(void); FILE *fp1, *fp2; char ch; int main() { int choice; while(1) { printf(\"Select One of the Following:\ \"); printf(\"\ 1. Encrypt\ \"); printf(\"2. Decrypt\ \"); printf(\"3. View The Encypted File\ \"); printf(\"4. View The Decrypted File\ \"); printf(\"5. Exit\ \"); printf(\"\ Enter Your Choice:\\t\"); scanf(\"%d\", &choice); switch(choice) { case 1: encrypt(); break; case 2: decrypt(); break; case 3: encrypt_view(); break; case 4: decrypt_view(); break; case 5: exit(1); } } printf(\"\ \"); return 0; } int encrypt() { printf(\"\ \"); fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\"); if(fp1 == NULL) { printf(\"Source File Could Not Be Found\ \"); } fp2 = fopen(\"/home/tusharsoni/Desktop/Target\",\"w\"); if(fp2 == NULL) { printf(\"Target File Could Not Be Found\ \"); } while(1) { ch = fgetc(fp1); if(ch == EOF) { printf(\"\ End Of File\ \"); break; } else { ch = ch - (8 * 5 - 3); fputc(ch, fp2); } } fclose(fp1); fclose(fp2); printf(\"\ \"); return 0; } int decrypt() { printf(\"\ \"); fp1 = fopen(\"/home/tusharsoni/Desktop/Target\",\"r\"); if(fp1 == NULL) { printf(\"Source File Could Not Be Found\ \"); } fp2 = fopen(\"/home/tusharsoni/Desktop/Source\",\"w\"); if(fp2 == NULL) { printf(\"Target File Could Not Be Found\ \"); } while(1) { ch = fgetc(fp1); if(ch == EOF) { printf(\"\ End Of File\ \"); break; } else { ch = ch + (8 * 5 - 3); fputc(ch, fp2); } } fclose(fp1); fclose(fp2); printf(\"\ \"); return 0; } int encrypt_view() { printf(\"\ \"); fp1 = fopen(\"/home/tusharsoni/Desktop/Target\",\"r\"); if(fp1 == NULL) { printf(\"No File Found\ \"); exit(1); } else { while(1) { ch = fgetc(fp1); if(ch == EOF) { break; } else { printf(\"%c\", ch); } } printf(\"\ \"); fclose(fp1); } printf(\"\ \"); return 0; } int decrypt_view() { printf(\"\ \"); fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\"); if(fp1 == NULL) { printf(\"No File Found\ \"); exit(1); } else { while(1) { ch = fgetc(fp1); if(ch == EOF) { break; } else { printf(\"%c\", ch); } } printf(\"\ \"); fclose(fp1); } return 0; printf(\"\ \"); } Solution #include #include int encrypt(void); int decrypt(void); int encrypt_view(void); int decrypt_view(void); FILE *fp1, *fp2; char ch; int main() { int choice; while(1) { printf(\"Select One of the Following:\ \"); printf(\"\ 1. Encrypt\ \"); printf(\"2. Decrypt\ \"); printf(\"3. View The Encypted File\ \"); printf(\"4. View The Decrypted File\ \"); printf(\"5. Exit\ \"); printf(\"\ Enter Your Choice:\\t\"); scanf(\"%d\", &choice); switch(choice) { case 1: encrypt(); break; case 2: decrypt(); break; case 3: encrypt_view(); break; case 4: decrypt_view(); break; case 5: exit(1); } } printf(\"\ \"); return 0; } int encrypt() { printf(\"\ \"); fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\"); if(fp1 == NULL) { printf(\"Source File Could Not Be Found\ \"); } fp2 = fopen(\"/home/tusharsoni/Desktop/Target\",\"w\"); if(fp2 == NULL) { printf(\"Ta.
#includestdio.h#includestdlib.hint encrypt(void);int decry.pdf
#includestdio.h#includestdlib.hint encrypt(void);int decry.pdf
nareshsonyericcson
Neuman System Model helps in studying the stressors, find out the defense, gives the feedback and helps in preventing the occurrence through assessment, diagnosis, planning, implementation and evaluation taking the person as a whole. Neuman System Model says that individual person as whole is in interaction with environment. Neuman System Model helps in diagnosis and prevents the stressors. Neuman System Model implements and improves the immunity. Neuman System Model takes the help of social elements such as relatives and neighbors to protect the person. Solution Neuman System Model helps in studying the stressors, find out the defense, gives the feedback and helps in preventing the occurrence through assessment, diagnosis, planning, implementation and evaluation taking the person as a whole. Neuman System Model says that individual person as whole is in interaction with environment. Neuman System Model helps in diagnosis and prevents the stressors. Neuman System Model implements and improves the immunity. Neuman System Model takes the help of social elements such as relatives and neighbors to protect the person..
Neuman System Model helps in studying the stressors, find out the de.pdf
Neuman System Model helps in studying the stressors, find out the de.pdf
nareshsonyericcson
NaOH will completely react will acetic acid and turn it into salt. Conc . of NaOH left = 16.67*.12-2 /36.67*10^-3 =0.0109 pOH = -log(0.0109) =1.963 pH = 14 -1.963 =12.037 Solution NaOH will completely react will acetic acid and turn it into salt. Conc . of NaOH left = 16.67*.12-2 /36.67*10^-3 =0.0109 pOH = -log(0.0109) =1.963 pH = 14 -1.963 =12.037.
NaOH will completely react will acetic acid and t.pdf
NaOH will completely react will acetic acid and t.pdf
nareshsonyericcson
the bonding between the carban and the nitrogen in hydrogen cyanide or hydrocyanic acid is a triple bond, hence the hybrid orbital is sp, due to the linear geometry of the molecule Solution the bonding between the carban and the nitrogen in hydrogen cyanide or hydrocyanic acid is a triple bond, hence the hybrid orbital is sp, due to the linear geometry of the molecule.
the bonding between the carban and the nitrogen i.pdf
the bonding between the carban and the nitrogen i.pdf
nareshsonyericcson
The salts are more ionic, more soluble they are in water since H2O is a polar solvent. LiF and CsI are much less polar than LiI and CSF since the latters are more ionic bonded. Solution The salts are more ionic, more soluble they are in water since H2O is a polar solvent. LiF and CsI are much less polar than LiI and CSF since the latters are more ionic bonded..
The salts are more ionic, more soluble they are i.pdf
The salts are more ionic, more soluble they are i.pdf
nareshsonyericcson
Recommended
5 steps: (make sure to press \"See full answer\") - make observation and gathering all available data Solution 5 steps: (make sure to press \"See full answer\") - make observation and gathering all available data.
5 steps (make sure to press See full answer)- make observat.pdf
5 steps (make sure to press See full answer)- make observat.pdf
nareshsonyericcson
2cosx=(cosxtanx+sinx)/(tanx) taking RHS : cosxtanx+sinx)/(tanx) = cosxtanx/tanx + sinx/tanx = cosx + sinx *cosx/sinx = cosx + cosx = 2cosx = LHS hence LHS = RHS Solution 2cosx=(cosxtanx+sinx)/(tanx) taking RHS : cosxtanx+sinx)/(tanx) = cosxtanx/tanx + sinx/tanx = cosx + sinx *cosx/sinx = cosx + cosx = 2cosx = LHS hence LHS = RHS.
2cosx=(cosxtanx+sinx)(tanx)taking RHS cosxtanx+sinx)(tanx)= .pdf
2cosx=(cosxtanx+sinx)(tanx)taking RHS cosxtanx+sinx)(tanx)= .pdf
nareshsonyericcson
1.5 X 10-8 Solution 1.5 X 10-8.
1.5 X 10-8Solution1.5 X 10-8.pdf
1.5 X 10-8Solution1.5 X 10-8.pdf
nareshsonyericcson
#include #include int encrypt(void); int decrypt(void); int encrypt_view(void); int decrypt_view(void); FILE *fp1, *fp2; char ch; int main() { int choice; while(1) { printf(\"Select One of the Following:\ \"); printf(\"\ 1. Encrypt\ \"); printf(\"2. Decrypt\ \"); printf(\"3. View The Encypted File\ \"); printf(\"4. View The Decrypted File\ \"); printf(\"5. Exit\ \"); printf(\"\ Enter Your Choice:\\t\"); scanf(\"%d\", &choice); switch(choice) { case 1: encrypt(); break; case 2: decrypt(); break; case 3: encrypt_view(); break; case 4: decrypt_view(); break; case 5: exit(1); } } printf(\"\ \"); return 0; } int encrypt() { printf(\"\ \"); fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\"); if(fp1 == NULL) { printf(\"Source File Could Not Be Found\ \"); } fp2 = fopen(\"/home/tusharsoni/Desktop/Target\",\"w\"); if(fp2 == NULL) { printf(\"Target File Could Not Be Found\ \"); } while(1) { ch = fgetc(fp1); if(ch == EOF) { printf(\"\ End Of File\ \"); break; } else { ch = ch - (8 * 5 - 3); fputc(ch, fp2); } } fclose(fp1); fclose(fp2); printf(\"\ \"); return 0; } int decrypt() { printf(\"\ \"); fp1 = fopen(\"/home/tusharsoni/Desktop/Target\",\"r\"); if(fp1 == NULL) { printf(\"Source File Could Not Be Found\ \"); } fp2 = fopen(\"/home/tusharsoni/Desktop/Source\",\"w\"); if(fp2 == NULL) { printf(\"Target File Could Not Be Found\ \"); } while(1) { ch = fgetc(fp1); if(ch == EOF) { printf(\"\ End Of File\ \"); break; } else { ch = ch + (8 * 5 - 3); fputc(ch, fp2); } } fclose(fp1); fclose(fp2); printf(\"\ \"); return 0; } int encrypt_view() { printf(\"\ \"); fp1 = fopen(\"/home/tusharsoni/Desktop/Target\",\"r\"); if(fp1 == NULL) { printf(\"No File Found\ \"); exit(1); } else { while(1) { ch = fgetc(fp1); if(ch == EOF) { break; } else { printf(\"%c\", ch); } } printf(\"\ \"); fclose(fp1); } printf(\"\ \"); return 0; } int decrypt_view() { printf(\"\ \"); fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\"); if(fp1 == NULL) { printf(\"No File Found\ \"); exit(1); } else { while(1) { ch = fgetc(fp1); if(ch == EOF) { break; } else { printf(\"%c\", ch); } } printf(\"\ \"); fclose(fp1); } return 0; printf(\"\ \"); } Solution #include #include int encrypt(void); int decrypt(void); int encrypt_view(void); int decrypt_view(void); FILE *fp1, *fp2; char ch; int main() { int choice; while(1) { printf(\"Select One of the Following:\ \"); printf(\"\ 1. Encrypt\ \"); printf(\"2. Decrypt\ \"); printf(\"3. View The Encypted File\ \"); printf(\"4. View The Decrypted File\ \"); printf(\"5. Exit\ \"); printf(\"\ Enter Your Choice:\\t\"); scanf(\"%d\", &choice); switch(choice) { case 1: encrypt(); break; case 2: decrypt(); break; case 3: encrypt_view(); break; case 4: decrypt_view(); break; case 5: exit(1); } } printf(\"\ \"); return 0; } int encrypt() { printf(\"\ \"); fp1 = fopen(\"/home/tusharsoni/Desktop/Source\",\"r\"); if(fp1 == NULL) { printf(\"Source File Could Not Be Found\ \"); } fp2 = fopen(\"/home/tusharsoni/Desktop/Target\",\"w\"); if(fp2 == NULL) { printf(\"Ta.
#includestdio.h#includestdlib.hint encrypt(void);int decry.pdf
#includestdio.h#includestdlib.hint encrypt(void);int decry.pdf
nareshsonyericcson
Neuman System Model helps in studying the stressors, find out the defense, gives the feedback and helps in preventing the occurrence through assessment, diagnosis, planning, implementation and evaluation taking the person as a whole. Neuman System Model says that individual person as whole is in interaction with environment. Neuman System Model helps in diagnosis and prevents the stressors. Neuman System Model implements and improves the immunity. Neuman System Model takes the help of social elements such as relatives and neighbors to protect the person. Solution Neuman System Model helps in studying the stressors, find out the defense, gives the feedback and helps in preventing the occurrence through assessment, diagnosis, planning, implementation and evaluation taking the person as a whole. Neuman System Model says that individual person as whole is in interaction with environment. Neuman System Model helps in diagnosis and prevents the stressors. Neuman System Model implements and improves the immunity. Neuman System Model takes the help of social elements such as relatives and neighbors to protect the person..
Neuman System Model helps in studying the stressors, find out the de.pdf
Neuman System Model helps in studying the stressors, find out the de.pdf
nareshsonyericcson
NaOH will completely react will acetic acid and turn it into salt. Conc . of NaOH left = 16.67*.12-2 /36.67*10^-3 =0.0109 pOH = -log(0.0109) =1.963 pH = 14 -1.963 =12.037 Solution NaOH will completely react will acetic acid and turn it into salt. Conc . of NaOH left = 16.67*.12-2 /36.67*10^-3 =0.0109 pOH = -log(0.0109) =1.963 pH = 14 -1.963 =12.037.
NaOH will completely react will acetic acid and t.pdf
NaOH will completely react will acetic acid and t.pdf
nareshsonyericcson
the bonding between the carban and the nitrogen in hydrogen cyanide or hydrocyanic acid is a triple bond, hence the hybrid orbital is sp, due to the linear geometry of the molecule Solution the bonding between the carban and the nitrogen in hydrogen cyanide or hydrocyanic acid is a triple bond, hence the hybrid orbital is sp, due to the linear geometry of the molecule.
the bonding between the carban and the nitrogen i.pdf
the bonding between the carban and the nitrogen i.pdf
nareshsonyericcson
The salts are more ionic, more soluble they are in water since H2O is a polar solvent. LiF and CsI are much less polar than LiI and CSF since the latters are more ionic bonded. Solution The salts are more ionic, more soluble they are in water since H2O is a polar solvent. LiF and CsI are much less polar than LiI and CSF since the latters are more ionic bonded..
The salts are more ionic, more soluble they are i.pdf
The salts are more ionic, more soluble they are i.pdf
nareshsonyericcson
They can form sulfides, but the metal sulfides that form will notprecipitate. Solution They can form sulfides, but the metal sulfides that form will notprecipitate..
They can form sulfides, but the metal sulfides th.pdf
They can form sulfides, but the metal sulfides th.pdf
nareshsonyericcson
Look at the position of Pb in the periodic table and the postion of the Alkali Earth metals. Alkali Earth Metals are in noble-gas configuration when they are doubly charged cations. Lead however is on the right hand side, and would like to have more electrons to get into that stablest configuration; making the bonds in lead-compounds more covalent -and stronger- in nature. Insoluble compounds have lattice enthalpies higher than the enthalpy of solvation. Now since Alkali Earth metals like to be in ionic states; their compounds will dissolve in general more easily than the same compounds with lead, since usually the lattice energy of lead compounds is much higher (In case of insoluble compounds higher than the energy that would be released upon solvation, which thus doesn\'t occur). Solution Look at the position of Pb in the periodic table and the postion of the Alkali Earth metals. Alkali Earth Metals are in noble-gas configuration when they are doubly charged cations. Lead however is on the right hand side, and would like to have more electrons to get into that stablest configuration; making the bonds in lead-compounds more covalent -and stronger- in nature. Insoluble compounds have lattice enthalpies higher than the enthalpy of solvation. Now since Alkali Earth metals like to be in ionic states; their compounds will dissolve in general more easily than the same compounds with lead, since usually the lattice energy of lead compounds is much higher (In case of insoluble compounds higher than the energy that would be released upon solvation, which thus doesn\'t occur)..
Look at the position of Pb in the periodic table .pdf
Look at the position of Pb in the periodic table .pdf
nareshsonyericcson
Yes. A tree which has a root and nothing else. Solution Yes. A tree which has a root and nothing else..
Yes. A tree which has a root and nothing else.SolutionYe.pdf
Yes. A tree which has a root and nothing else.SolutionYe.pdf
nareshsonyericcson
First you will want to convert the molarity of each to moles. To do this multiply the molarity by the volume in liters. Next add the moles together and divide that by the total volume of the solution. Molarity = moles/liters Solution First you will want to convert the molarity of each to moles. To do this multiply the molarity by the volume in liters. Next add the moles together and divide that by the total volume of the solution. Molarity = moles/liters.
First you will want to convert the molarity of ea.pdf
First you will want to convert the molarity of ea.pdf
nareshsonyericcson
The process of sperm competition is defined as ‘the competition within a single female between the sperm of two or more males for the fertilization of the ova’. Sperm competition occurs when females mate at least two males, whose sperm must exhibit spatial and temporal overlap. Successful sperm must be good at gaining dominance over those already in store and must be able to resist usurpation by the incoming sperm of later mating males. Effects of sexual selection via sperm competition are much better understood in invertebrates, especially in insects. They often use copulatory plugs that are inserted immediately after a male copulates with a female reducing the possibility of fertilization by subsequent copulations from another male.eg Indian mealmoth (Plodia interpunctella), Bumblebee. In Drosophila melanogaster (fruitflies) males release a toxic seminal fluids, ACPs (accessory gland proteins), from their accessory glands to impede the female from participating in future copulations. Other examples are blue headed wrasse, Thalassoma bifasciatum, mollies (Poecilia), crickets (Teleogryllus oceanicus). There are certain offensive behaviours for suceeding followed by antlers. an example of marine invertebrate showing gamete compatibility and sperm competition is sympatric Asterias sea stars. Solution The process of sperm competition is defined as ‘the competition within a single female between the sperm of two or more males for the fertilization of the ova’. Sperm competition occurs when females mate at least two males, whose sperm must exhibit spatial and temporal overlap. Successful sperm must be good at gaining dominance over those already in store and must be able to resist usurpation by the incoming sperm of later mating males. Effects of sexual selection via sperm competition are much better understood in invertebrates, especially in insects. They often use copulatory plugs that are inserted immediately after a male copulates with a female reducing the possibility of fertilization by subsequent copulations from another male.eg Indian mealmoth (Plodia interpunctella), Bumblebee. In Drosophila melanogaster (fruitflies) males release a toxic seminal fluids, ACPs (accessory gland proteins), from their accessory glands to impede the female from participating in future copulations. Other examples are blue headed wrasse, Thalassoma bifasciatum, mollies (Poecilia), crickets (Teleogryllus oceanicus). There are certain offensive behaviours for suceeding followed by antlers. an example of marine invertebrate showing gamete compatibility and sperm competition is sympatric Asterias sea stars..
The process of sperm competition is defined as ‘the competition with.pdf
The process of sperm competition is defined as ‘the competition with.pdf
nareshsonyericcson
The answer is A. Public, Private The public and private community strings are used within SNMP to read and write respectively. The option C and D are not correct because they do not exist as community strings. Solution The answer is A. Public, Private The public and private community strings are used within SNMP to read and write respectively. The option C and D are not correct because they do not exist as community strings..
The answer isA. Public, PrivateThe public and private community .pdf
The answer isA. Public, PrivateThe public and private community .pdf
nareshsonyericcson
TCO is the assessment of all life time costs from owning ertain kinds of assets.TCO looks three major types of costs initial cost to obtain hardware/software second one is operational expenses for the assets suh as installation and managment of assets third one is indirect cost that affects the bussiness for instance the cost of system downtime from an outage. virtualization can reduce tco in many ways.for example hardware virtualization can reduce the cost associated with owing multiple physical servers due to ability to have them virtualized.Virtualization enables multiple operating systems to run on the same physical platform. It is not unusual to achieve 10:1 virtual to physical machine consolidation. This means that ten server applications can be run on a single machine that had required as many physical computers to provide the unique operating system and technical specification environments in order to operate. Server utilization is optimized and legacy software can maintain old OS configurations while new applications are running in VMs with updated platforms. Use of a VM enables rapid deployment by isolating the application in a known and controlled environment. Unknown factors such as mixed libraries caused by numerous installs can be eliminated. Severe crashes that required hours of reinstallation now take moments by simply copying a virtual image. As server workloads vary, virtualization provides the ability for virtual machines that are over utilizing the resources of a server to be moved to underutilized servers. This dynamic load balancing creates efficient utilization of server resources. Disaster recovery is a critical component for IT, as system crashes can create huge economic losses. Virtualization technology enables a virtual image on a machine to be instantly re-imaged on another server if a machine failure occurs. Multinational flexibility provides seamless transitions between different operating systems on a single machine reducing desktop footprint and hardware expenditure. Virtualization of systems helps prevent system crashes due to memory corruption caused by software like device drivers. VT-d for Directed I/O Architecture provides methods to better control system devices by defining the architecture for DMA and interrupt remapping to ensure improved isolation of I/O resources for greater reliability, security, and availability. Solution TCO is the assessment of all life time costs from owning ertain kinds of assets.TCO looks three major types of costs initial cost to obtain hardware/software second one is operational expenses for the assets suh as installation and managment of assets third one is indirect cost that affects the bussiness for instance the cost of system downtime from an outage. virtualization can reduce tco in many ways.for example hardware virtualization can reduce the cost associated with owing multiple physical servers due to ability to have them virtualized.Virtualization enables multiple operat.
TCO is the assessment of all life time costs from owning ertain kind.pdf
TCO is the assessment of all life time costs from owning ertain kind.pdf
nareshsonyericcson
ROA = Net Income (2015) / [Assets (2014) + Assets (2015)] / 2 = 220 / (500 + 400) / 2 = 48.89% ROE = Net Income (2015) / [Equity (2014) + Equity (2015)]/2 = 220 / (324 + 120) /2 = 99.10% Solution ROA = Net Income (2015) / [Assets (2014) + Assets (2015)] / 2 = 220 / (500 + 400) / 2 = 48.89% ROE = Net Income (2015) / [Equity (2014) + Equity (2015)]/2 = 220 / (324 + 120) /2 = 99.10%.
ROA = Net Income (2015) [Assets (2014) + Assets (2015)] 2= 220.pdf
ROA = Net Income (2015) [Assets (2014) + Assets (2015)] 2= 220.pdf
nareshsonyericcson
Reply if you have any doubts. public class Life { public static final int DEAD = 0; public static final int LIVING = 1; public static final int DYING = 2; public static final int BORNING = 3; private int size; private int[][] cells; /* Constructor is done. */ public Life(int size) { this.size = size; cells = new int[size][size]; } /* getSize is a stub. */ public int getSize() { return this.size; } /* setState is a stub. */ public void setState(int row, int col, int state) { cells[row][col] = state; } /* getState is a stub. */ public int getState(int row, int col) { return this.cells[row][col]; } /* evolve is a stub. */ public void evolve() { //Creating tempArray //only after full cycle cells should evolve int[][] TempCells = new int[size][size]; //looping through all the cells for(int i=0;i 3) { TempCells[i][j] = DEAD; } } else if(CurrentState == DEAD) { if(NeighbourCount == 3) { TempCells[i][j] = LIVING; } } } } //set the cells to nex gen states this.cells = TempCells; } /* getNumberOfNeighbors is a stub; note that it\'s private. */ private int getNumberOfNeighbors(int row, int col) { int neighbourX[] = new int[8]; int neighbourY[] = new int[8]; neighbourX[0] = row - 1; neighbourY[0] = col - 1; neighbourX[1] = row - 1; neighbourY[1] = col; neighbourX[2] = row - 1; neighbourY[2] = col + 1; neighbourX[3] = row; neighbourY[3] = col - 1; neighbourX[4] = row; neighbourY[4] = col + 1; neighbourX[5] = row + 1; neighbourY[5] = col - 1; neighbourX[6] = row + 1; neighbourY[6] = col; neighbourX[7] = row + 1; neighbourY[7] = col + 1; for(int i=0;i<8;i++) { if(neighbourX[i] < 0) neighbourX[i] = size - 1; if(neighbourY[i] < 0) neighbourY[i] = size - 1; if(neighbourX[i] >= size) neighbourX[i] = 0; if(neighbourY[i] >= size) neighbourY[i] = 0; } int count = 0; for(int i=0;i<8;i++) { count += this.cells[neighbourX[i]][neighbourY[i]]; } return count; } /* The main only instantiates the objects and sets them working. */ public static void main(String[] args) { Life life = new Life(25); CellWorld world = new CellWorld(life); } } Solution Reply if you have any doubts. public class Life { public static final int DEAD = 0; public static final int LIVING = 1; public static final int DYING = 2; public static final int BORNING = 3; private int size; private int[][] cells; /* Constructor is done. */ public Life(int size) { this.size = size; cells = new int[size][size]; } /* getSize is a stub. */ public int getSize() { return this.size; } /* setState is a stub. */ public void setState(int row, int col, int state) { cells[row][col] = state; } /* getState is a stub. */ public int getState(int row, int col) { return this.cells[row][col]; } /* evolve is a stub. */ public void evolve() { //Creating tempArray //only after full cycle cells should evolve int[][] TempCells = new int[size][size]; //looping through all the cells for(int i=0;i 3) { TempCells[i][j] = DEAD; } } else if(CurrentState == DEAD) { if(NeighbourCount == 3) { TempCells[i][j] = LIVING; } } } } //set the cells.
Reply if you have any doubts.public class Life { public stati.pdf
Reply if you have any doubts.public class Life { public stati.pdf
nareshsonyericcson
pH + pOH = 14 1. 14-12.2= 1.8pOH pOH = -log(kb) Kb = 10^-1.8 2. -log(6.3*10^-5) = pOH pOH = 4.2 Solution pH + pOH = 14 1. 14-12.2= 1.8pOH pOH = -log(kb) Kb = 10^-1.8 2. -log(6.3*10^-5) = pOH pOH = 4.2.
pH + pOH = 14 1. 14-12.2= 1.8pOH pOH = -log(kb) Kb = 10^-1.pdf
pH + pOH = 14 1. 14-12.2= 1.8pOH pOH = -log(kb) Kb = 10^-1.pdf
nareshsonyericcson
Outear includes the pinna (also called auricle), the ear canal, and the very most superficial layer of the ear drum (also called the tympanic membrane). this portion of the ear is not vital for hearing. The pinna only helps to direct sound through the ear canal to the Middle ear (eardrum). Middle:The middle ear behind the ear drum (tympanic membrane), includes the three ear bones or ossicles: the malleus (or hammer), incus (or anvil), and stapes (or stirrup). The opening of the Eustachian tube is also within the middle ear. The tympanic membrane uses sound energy strikes the tympanic membrane and is concentrated to the smaller footplate. and articulating ear ossicles lead to an increase in the force applied to the stapes footplate then to the malleus. Inner ear: The inner ear includes both the organ of hearing (the cochlea) and a sense organ labyrinth or vestibular apparatus. Function: When sound strikes the ear drum, transferred to the footplate of the stapes, which presses to cochlea. The fluid inside this duct flows against the receptor cells of the Organ of Corti,which stimulate the spiral ganglion, and pass information through the auditory portion of the brain that is eighth cranial nerve. In brief \"The outer ear, collestc osund waves. It then goes to the middle ear, whcih transimits the signal to the ear durm. After that it goes throguh the inner ear. On the other side of your ear drum, is the hammer, the anvil, and the stirrup. These rae the three smallest bones in your body. These bones transimit the signal to the cochlea. The cohclea has nerve hairs that tune differtn sounds. These nerve hairs, send the different souns to the brain, where the sound is interpreted\" Solution Outear includes the pinna (also called auricle), the ear canal, and the very most superficial layer of the ear drum (also called the tympanic membrane). this portion of the ear is not vital for hearing. The pinna only helps to direct sound through the ear canal to the Middle ear (eardrum). Middle:The middle ear behind the ear drum (tympanic membrane), includes the three ear bones or ossicles: the malleus (or hammer), incus (or anvil), and stapes (or stirrup). The opening of the Eustachian tube is also within the middle ear. The tympanic membrane uses sound energy strikes the tympanic membrane and is concentrated to the smaller footplate. and articulating ear ossicles lead to an increase in the force applied to the stapes footplate then to the malleus. Inner ear: The inner ear includes both the organ of hearing (the cochlea) and a sense organ labyrinth or vestibular apparatus. Function: When sound strikes the ear drum, transferred to the footplate of the stapes, which presses to cochlea. The fluid inside this duct flows against the receptor cells of the Organ of Corti,which stimulate the spiral ganglion, and pass information through the auditory portion of the brain that is eighth cranial nerve. In brief \"The outer ear, collestc osund waves. It then goes to the m.
Outear includes the pinna (also called auricle), the ear canal, and .pdf
Outear includes the pinna (also called auricle), the ear canal, and .pdf
nareshsonyericcson
Lewis base: is a donor of an electron pair (i.e., lone pair of electrons) For H2PO4^-, ClO^- and SO3^2-, the O atom and the Cl atom bear lone pair of electrons. Thus, all these three may behavior as Lewis bases. Solution Lewis base: is a donor of an electron pair (i.e., lone pair of electrons) For H2PO4^-, ClO^- and SO3^2-, the O atom and the Cl atom bear lone pair of electrons. Thus, all these three may behavior as Lewis bases..
Lewis base is a donor of an electron pair (i.e., lone pair of ele.pdf
Lewis base is a donor of an electron pair (i.e., lone pair of ele.pdf
nareshsonyericcson
It will be B)- Planning ,directiing and controlling Management is not directly involved in manufacturing and selling Solution It will be B)- Planning ,directiing and controlling Management is not directly involved in manufacturing and selling.
It will be B)- Planning ,directiing and controllingManagement is n.pdf
It will be B)- Planning ,directiing and controllingManagement is n.pdf
nareshsonyericcson
Internet worm is a useless code which effect the code in the software. It mainly infects the code that are going to effect the code with some vulnerabilities, there have been several worms that have given the practice of creating internet worms popularity. Efforts have been made to model the behaviour of worms as they propagate. when the desighner knows about this worms it will be very easy to take it off and run the program very good. The worm spread itself the own vulnerabilities and spread many errors in the code. There are many effects that are cited to avoid the effectiveness of the internet worm. The user must be aware of the internet worms that are cited because there are many tools that are available to find the vulnerabilities thar are occuring through internet. Solution Internet worm is a useless code which effect the code in the software. It mainly infects the code that are going to effect the code with some vulnerabilities, there have been several worms that have given the practice of creating internet worms popularity. Efforts have been made to model the behaviour of worms as they propagate. when the desighner knows about this worms it will be very easy to take it off and run the program very good. The worm spread itself the own vulnerabilities and spread many errors in the code. There are many effects that are cited to avoid the effectiveness of the internet worm. The user must be aware of the internet worms that are cited because there are many tools that are available to find the vulnerabilities thar are occuring through internet..
Internet worm is a useless code which effect the code in the softwar.pdf
Internet worm is a useless code which effect the code in the softwar.pdf
nareshsonyericcson
Here we have the compound V3(PO4)2 Separating it into it\'s ions we have: V2+ and PO43- since vanadium is a transition metal, we denote its oxidation state by roman numerals. Thus, the name of the compound is Vanadium (II) Phosphate. Please don\'t forget to rate, good luck studying! Solution Here we have the compound V3(PO4)2 Separating it into it\'s ions we have: V2+ and PO43- since vanadium is a transition metal, we denote its oxidation state by roman numerals. Thus, the name of the compound is Vanadium (II) Phosphate. Please don\'t forget to rate, good luck studying!.
Here we have the compound V3(PO4)2 Separating it into its ions.pdf
Here we have the compound V3(PO4)2 Separating it into its ions.pdf
nareshsonyericcson
F+ cells have fertility factors while F- cells lack fertility factors F+ is a donor cells while F- is recipient cells F+ cells has circular plasmid separate from chromosome F- cell does not have before receiving plasmid Cell connect through T-pilus Tra K and tra I were required for efficient inter plasmid recombination at ori T whole DNA is transferred to recipient F- Solution F+ cells have fertility factors while F- cells lack fertility factors F+ is a donor cells while F- is recipient cells F+ cells has circular plasmid separate from chromosome F- cell does not have before receiving plasmid Cell connect through T-pilus Tra K and tra I were required for efficient inter plasmid recombination at ori T whole DNA is transferred to recipient F-.
F+ cells have fertility factors while F- cells lack fertility factor.pdf
F+ cells have fertility factors while F- cells lack fertility factor.pdf
nareshsonyericcson
D is correct. D. transport input ssh in vty Line Configuration mode This is the correct command, and must be entered in Line Configuration mode. Solution D is correct. D. transport input ssh in vty Line Configuration mode This is the correct command, and must be entered in Line Configuration mode..
D is correct.D. transport input ssh in vty Line Configuration mode.pdf
D is correct.D. transport input ssh in vty Line Configuration mode.pdf
nareshsonyericcson
Dissociation of 2 protons = dissociation of 2 H+ ions The answer is: C. H3PO4 --> 2H+ + HPO42- Solution Dissociation of 2 protons = dissociation of 2 H+ ions The answer is: C. H3PO4 --> 2H+ + HPO42-.
Dissociation of 2 protons = dissociation of 2 H+ ionsThe answer is.pdf
Dissociation of 2 protons = dissociation of 2 H+ ionsThe answer is.pdf
nareshsonyericcson
CASE ANSWERS: 1.C) Bacteria typically transmitted by airborne droplets because these gram positive bacteria can spread via the droplets expelled from a Sterptococcus pyogenes infected person. 2.D) Ingesting the bacteria and cell debris through phagocytosis because the neutrophils are the white blood cells that are in the first line of defence as an innate immune response that comes to site of infection(Tissues & Blood circulation) throught he chemoattractants produced by the streptococcus pyogenes. 3.The following statements about the antibody productions is TRUE : B) Antibodies are produced by B cells with the help of T helper 2 cell. Solution CASE ANSWERS: 1.C) Bacteria typically transmitted by airborne droplets because these gram positive bacteria can spread via the droplets expelled from a Sterptococcus pyogenes infected person. 2.D) Ingesting the bacteria and cell debris through phagocytosis because the neutrophils are the white blood cells that are in the first line of defence as an innate immune response that comes to site of infection(Tissues & Blood circulation) throught he chemoattractants produced by the streptococcus pyogenes. 3.The following statements about the antibody productions is TRUE : B) Antibodies are produced by B cells with the help of T helper 2 cell..
CASE ANSWERS1.C) Bacteria typically transmitted by airborne dropl.pdf
CASE ANSWERS1.C) Bacteria typically transmitted by airborne dropl.pdf
nareshsonyericcson
Bacteria May have peptidoglycan cell wall Have ester linked lipids Methanogens Methane-producing archaea (methanogens) are strictly anaerobic convert substrate to methane Archae Have RNA polymerase similar to eukaryotes Have ether linked lipids in their cell membranes Solution Bacteria May have peptidoglycan cell wall Have ester linked lipids Methanogens Methane-producing archaea (methanogens) are strictly anaerobic convert substrate to methane Archae Have RNA polymerase similar to eukaryotes Have ether linked lipids in their cell membranes.
Bacteria May have peptidoglycan cell wallHave ester linked lipid.pdf
Bacteria May have peptidoglycan cell wallHave ester linked lipid.pdf
nareshsonyericcson
https://app.box.com/s/h5mhqoyabotgw05s0df0ltw3e39pgnmy
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT TOÁN 2024 - TỪ CÁC TRƯỜNG, TRƯỜNG...
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Transparency, Recognition and the role of eSealing - Ildiko Mazar and Koen No...
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They can form sulfides, but the metal sulfides that form will notprecipitate. Solution They can form sulfides, but the metal sulfides that form will notprecipitate..
They can form sulfides, but the metal sulfides th.pdf
They can form sulfides, but the metal sulfides th.pdf
nareshsonyericcson
Look at the position of Pb in the periodic table and the postion of the Alkali Earth metals. Alkali Earth Metals are in noble-gas configuration when they are doubly charged cations. Lead however is on the right hand side, and would like to have more electrons to get into that stablest configuration; making the bonds in lead-compounds more covalent -and stronger- in nature. Insoluble compounds have lattice enthalpies higher than the enthalpy of solvation. Now since Alkali Earth metals like to be in ionic states; their compounds will dissolve in general more easily than the same compounds with lead, since usually the lattice energy of lead compounds is much higher (In case of insoluble compounds higher than the energy that would be released upon solvation, which thus doesn\'t occur). Solution Look at the position of Pb in the periodic table and the postion of the Alkali Earth metals. Alkali Earth Metals are in noble-gas configuration when they are doubly charged cations. Lead however is on the right hand side, and would like to have more electrons to get into that stablest configuration; making the bonds in lead-compounds more covalent -and stronger- in nature. Insoluble compounds have lattice enthalpies higher than the enthalpy of solvation. Now since Alkali Earth metals like to be in ionic states; their compounds will dissolve in general more easily than the same compounds with lead, since usually the lattice energy of lead compounds is much higher (In case of insoluble compounds higher than the energy that would be released upon solvation, which thus doesn\'t occur)..
Look at the position of Pb in the periodic table .pdf
Look at the position of Pb in the periodic table .pdf
nareshsonyericcson
Yes. A tree which has a root and nothing else. Solution Yes. A tree which has a root and nothing else..
Yes. A tree which has a root and nothing else.SolutionYe.pdf
Yes. A tree which has a root and nothing else.SolutionYe.pdf
nareshsonyericcson
First you will want to convert the molarity of each to moles. To do this multiply the molarity by the volume in liters. Next add the moles together and divide that by the total volume of the solution. Molarity = moles/liters Solution First you will want to convert the molarity of each to moles. To do this multiply the molarity by the volume in liters. Next add the moles together and divide that by the total volume of the solution. Molarity = moles/liters.
First you will want to convert the molarity of ea.pdf
First you will want to convert the molarity of ea.pdf
nareshsonyericcson
The process of sperm competition is defined as ‘the competition within a single female between the sperm of two or more males for the fertilization of the ova’. Sperm competition occurs when females mate at least two males, whose sperm must exhibit spatial and temporal overlap. Successful sperm must be good at gaining dominance over those already in store and must be able to resist usurpation by the incoming sperm of later mating males. Effects of sexual selection via sperm competition are much better understood in invertebrates, especially in insects. They often use copulatory plugs that are inserted immediately after a male copulates with a female reducing the possibility of fertilization by subsequent copulations from another male.eg Indian mealmoth (Plodia interpunctella), Bumblebee. In Drosophila melanogaster (fruitflies) males release a toxic seminal fluids, ACPs (accessory gland proteins), from their accessory glands to impede the female from participating in future copulations. Other examples are blue headed wrasse, Thalassoma bifasciatum, mollies (Poecilia), crickets (Teleogryllus oceanicus). There are certain offensive behaviours for suceeding followed by antlers. an example of marine invertebrate showing gamete compatibility and sperm competition is sympatric Asterias sea stars. Solution The process of sperm competition is defined as ‘the competition within a single female between the sperm of two or more males for the fertilization of the ova’. Sperm competition occurs when females mate at least two males, whose sperm must exhibit spatial and temporal overlap. Successful sperm must be good at gaining dominance over those already in store and must be able to resist usurpation by the incoming sperm of later mating males. Effects of sexual selection via sperm competition are much better understood in invertebrates, especially in insects. They often use copulatory plugs that are inserted immediately after a male copulates with a female reducing the possibility of fertilization by subsequent copulations from another male.eg Indian mealmoth (Plodia interpunctella), Bumblebee. In Drosophila melanogaster (fruitflies) males release a toxic seminal fluids, ACPs (accessory gland proteins), from their accessory glands to impede the female from participating in future copulations. Other examples are blue headed wrasse, Thalassoma bifasciatum, mollies (Poecilia), crickets (Teleogryllus oceanicus). There are certain offensive behaviours for suceeding followed by antlers. an example of marine invertebrate showing gamete compatibility and sperm competition is sympatric Asterias sea stars..
The process of sperm competition is defined as ‘the competition with.pdf
The process of sperm competition is defined as ‘the competition with.pdf
nareshsonyericcson
The answer is A. Public, Private The public and private community strings are used within SNMP to read and write respectively. The option C and D are not correct because they do not exist as community strings. Solution The answer is A. Public, Private The public and private community strings are used within SNMP to read and write respectively. The option C and D are not correct because they do not exist as community strings..
The answer isA. Public, PrivateThe public and private community .pdf
The answer isA. Public, PrivateThe public and private community .pdf
nareshsonyericcson
TCO is the assessment of all life time costs from owning ertain kinds of assets.TCO looks three major types of costs initial cost to obtain hardware/software second one is operational expenses for the assets suh as installation and managment of assets third one is indirect cost that affects the bussiness for instance the cost of system downtime from an outage. virtualization can reduce tco in many ways.for example hardware virtualization can reduce the cost associated with owing multiple physical servers due to ability to have them virtualized.Virtualization enables multiple operating systems to run on the same physical platform. It is not unusual to achieve 10:1 virtual to physical machine consolidation. This means that ten server applications can be run on a single machine that had required as many physical computers to provide the unique operating system and technical specification environments in order to operate. Server utilization is optimized and legacy software can maintain old OS configurations while new applications are running in VMs with updated platforms. Use of a VM enables rapid deployment by isolating the application in a known and controlled environment. Unknown factors such as mixed libraries caused by numerous installs can be eliminated. Severe crashes that required hours of reinstallation now take moments by simply copying a virtual image. As server workloads vary, virtualization provides the ability for virtual machines that are over utilizing the resources of a server to be moved to underutilized servers. This dynamic load balancing creates efficient utilization of server resources. Disaster recovery is a critical component for IT, as system crashes can create huge economic losses. Virtualization technology enables a virtual image on a machine to be instantly re-imaged on another server if a machine failure occurs. Multinational flexibility provides seamless transitions between different operating systems on a single machine reducing desktop footprint and hardware expenditure. Virtualization of systems helps prevent system crashes due to memory corruption caused by software like device drivers. VT-d for Directed I/O Architecture provides methods to better control system devices by defining the architecture for DMA and interrupt remapping to ensure improved isolation of I/O resources for greater reliability, security, and availability. Solution TCO is the assessment of all life time costs from owning ertain kinds of assets.TCO looks three major types of costs initial cost to obtain hardware/software second one is operational expenses for the assets suh as installation and managment of assets third one is indirect cost that affects the bussiness for instance the cost of system downtime from an outage. virtualization can reduce tco in many ways.for example hardware virtualization can reduce the cost associated with owing multiple physical servers due to ability to have them virtualized.Virtualization enables multiple operat.
TCO is the assessment of all life time costs from owning ertain kind.pdf
TCO is the assessment of all life time costs from owning ertain kind.pdf
nareshsonyericcson
ROA = Net Income (2015) / [Assets (2014) + Assets (2015)] / 2 = 220 / (500 + 400) / 2 = 48.89% ROE = Net Income (2015) / [Equity (2014) + Equity (2015)]/2 = 220 / (324 + 120) /2 = 99.10% Solution ROA = Net Income (2015) / [Assets (2014) + Assets (2015)] / 2 = 220 / (500 + 400) / 2 = 48.89% ROE = Net Income (2015) / [Equity (2014) + Equity (2015)]/2 = 220 / (324 + 120) /2 = 99.10%.
ROA = Net Income (2015) [Assets (2014) + Assets (2015)] 2= 220.pdf
ROA = Net Income (2015) [Assets (2014) + Assets (2015)] 2= 220.pdf
nareshsonyericcson
Reply if you have any doubts. public class Life { public static final int DEAD = 0; public static final int LIVING = 1; public static final int DYING = 2; public static final int BORNING = 3; private int size; private int[][] cells; /* Constructor is done. */ public Life(int size) { this.size = size; cells = new int[size][size]; } /* getSize is a stub. */ public int getSize() { return this.size; } /* setState is a stub. */ public void setState(int row, int col, int state) { cells[row][col] = state; } /* getState is a stub. */ public int getState(int row, int col) { return this.cells[row][col]; } /* evolve is a stub. */ public void evolve() { //Creating tempArray //only after full cycle cells should evolve int[][] TempCells = new int[size][size]; //looping through all the cells for(int i=0;i 3) { TempCells[i][j] = DEAD; } } else if(CurrentState == DEAD) { if(NeighbourCount == 3) { TempCells[i][j] = LIVING; } } } } //set the cells to nex gen states this.cells = TempCells; } /* getNumberOfNeighbors is a stub; note that it\'s private. */ private int getNumberOfNeighbors(int row, int col) { int neighbourX[] = new int[8]; int neighbourY[] = new int[8]; neighbourX[0] = row - 1; neighbourY[0] = col - 1; neighbourX[1] = row - 1; neighbourY[1] = col; neighbourX[2] = row - 1; neighbourY[2] = col + 1; neighbourX[3] = row; neighbourY[3] = col - 1; neighbourX[4] = row; neighbourY[4] = col + 1; neighbourX[5] = row + 1; neighbourY[5] = col - 1; neighbourX[6] = row + 1; neighbourY[6] = col; neighbourX[7] = row + 1; neighbourY[7] = col + 1; for(int i=0;i<8;i++) { if(neighbourX[i] < 0) neighbourX[i] = size - 1; if(neighbourY[i] < 0) neighbourY[i] = size - 1; if(neighbourX[i] >= size) neighbourX[i] = 0; if(neighbourY[i] >= size) neighbourY[i] = 0; } int count = 0; for(int i=0;i<8;i++) { count += this.cells[neighbourX[i]][neighbourY[i]]; } return count; } /* The main only instantiates the objects and sets them working. */ public static void main(String[] args) { Life life = new Life(25); CellWorld world = new CellWorld(life); } } Solution Reply if you have any doubts. public class Life { public static final int DEAD = 0; public static final int LIVING = 1; public static final int DYING = 2; public static final int BORNING = 3; private int size; private int[][] cells; /* Constructor is done. */ public Life(int size) { this.size = size; cells = new int[size][size]; } /* getSize is a stub. */ public int getSize() { return this.size; } /* setState is a stub. */ public void setState(int row, int col, int state) { cells[row][col] = state; } /* getState is a stub. */ public int getState(int row, int col) { return this.cells[row][col]; } /* evolve is a stub. */ public void evolve() { //Creating tempArray //only after full cycle cells should evolve int[][] TempCells = new int[size][size]; //looping through all the cells for(int i=0;i 3) { TempCells[i][j] = DEAD; } } else if(CurrentState == DEAD) { if(NeighbourCount == 3) { TempCells[i][j] = LIVING; } } } } //set the cells.
Reply if you have any doubts.public class Life { public stati.pdf
Reply if you have any doubts.public class Life { public stati.pdf
nareshsonyericcson
pH + pOH = 14 1. 14-12.2= 1.8pOH pOH = -log(kb) Kb = 10^-1.8 2. -log(6.3*10^-5) = pOH pOH = 4.2 Solution pH + pOH = 14 1. 14-12.2= 1.8pOH pOH = -log(kb) Kb = 10^-1.8 2. -log(6.3*10^-5) = pOH pOH = 4.2.
pH + pOH = 14 1. 14-12.2= 1.8pOH pOH = -log(kb) Kb = 10^-1.pdf
pH + pOH = 14 1. 14-12.2= 1.8pOH pOH = -log(kb) Kb = 10^-1.pdf
nareshsonyericcson
Outear includes the pinna (also called auricle), the ear canal, and the very most superficial layer of the ear drum (also called the tympanic membrane). this portion of the ear is not vital for hearing. The pinna only helps to direct sound through the ear canal to the Middle ear (eardrum). Middle:The middle ear behind the ear drum (tympanic membrane), includes the three ear bones or ossicles: the malleus (or hammer), incus (or anvil), and stapes (or stirrup). The opening of the Eustachian tube is also within the middle ear. The tympanic membrane uses sound energy strikes the tympanic membrane and is concentrated to the smaller footplate. and articulating ear ossicles lead to an increase in the force applied to the stapes footplate then to the malleus. Inner ear: The inner ear includes both the organ of hearing (the cochlea) and a sense organ labyrinth or vestibular apparatus. Function: When sound strikes the ear drum, transferred to the footplate of the stapes, which presses to cochlea. The fluid inside this duct flows against the receptor cells of the Organ of Corti,which stimulate the spiral ganglion, and pass information through the auditory portion of the brain that is eighth cranial nerve. In brief \"The outer ear, collestc osund waves. It then goes to the middle ear, whcih transimits the signal to the ear durm. After that it goes throguh the inner ear. On the other side of your ear drum, is the hammer, the anvil, and the stirrup. These rae the three smallest bones in your body. These bones transimit the signal to the cochlea. The cohclea has nerve hairs that tune differtn sounds. These nerve hairs, send the different souns to the brain, where the sound is interpreted\" Solution Outear includes the pinna (also called auricle), the ear canal, and the very most superficial layer of the ear drum (also called the tympanic membrane). this portion of the ear is not vital for hearing. The pinna only helps to direct sound through the ear canal to the Middle ear (eardrum). Middle:The middle ear behind the ear drum (tympanic membrane), includes the three ear bones or ossicles: the malleus (or hammer), incus (or anvil), and stapes (or stirrup). The opening of the Eustachian tube is also within the middle ear. The tympanic membrane uses sound energy strikes the tympanic membrane and is concentrated to the smaller footplate. and articulating ear ossicles lead to an increase in the force applied to the stapes footplate then to the malleus. Inner ear: The inner ear includes both the organ of hearing (the cochlea) and a sense organ labyrinth or vestibular apparatus. Function: When sound strikes the ear drum, transferred to the footplate of the stapes, which presses to cochlea. The fluid inside this duct flows against the receptor cells of the Organ of Corti,which stimulate the spiral ganglion, and pass information through the auditory portion of the brain that is eighth cranial nerve. In brief \"The outer ear, collestc osund waves. It then goes to the m.
Outear includes the pinna (also called auricle), the ear canal, and .pdf
Outear includes the pinna (also called auricle), the ear canal, and .pdf
nareshsonyericcson
Lewis base: is a donor of an electron pair (i.e., lone pair of electrons) For H2PO4^-, ClO^- and SO3^2-, the O atom and the Cl atom bear lone pair of electrons. Thus, all these three may behavior as Lewis bases. Solution Lewis base: is a donor of an electron pair (i.e., lone pair of electrons) For H2PO4^-, ClO^- and SO3^2-, the O atom and the Cl atom bear lone pair of electrons. Thus, all these three may behavior as Lewis bases..
Lewis base is a donor of an electron pair (i.e., lone pair of ele.pdf
Lewis base is a donor of an electron pair (i.e., lone pair of ele.pdf
nareshsonyericcson
It will be B)- Planning ,directiing and controlling Management is not directly involved in manufacturing and selling Solution It will be B)- Planning ,directiing and controlling Management is not directly involved in manufacturing and selling.
It will be B)- Planning ,directiing and controllingManagement is n.pdf
It will be B)- Planning ,directiing and controllingManagement is n.pdf
nareshsonyericcson
Internet worm is a useless code which effect the code in the software. It mainly infects the code that are going to effect the code with some vulnerabilities, there have been several worms that have given the practice of creating internet worms popularity. Efforts have been made to model the behaviour of worms as they propagate. when the desighner knows about this worms it will be very easy to take it off and run the program very good. The worm spread itself the own vulnerabilities and spread many errors in the code. There are many effects that are cited to avoid the effectiveness of the internet worm. The user must be aware of the internet worms that are cited because there are many tools that are available to find the vulnerabilities thar are occuring through internet. Solution Internet worm is a useless code which effect the code in the software. It mainly infects the code that are going to effect the code with some vulnerabilities, there have been several worms that have given the practice of creating internet worms popularity. Efforts have been made to model the behaviour of worms as they propagate. when the desighner knows about this worms it will be very easy to take it off and run the program very good. The worm spread itself the own vulnerabilities and spread many errors in the code. There are many effects that are cited to avoid the effectiveness of the internet worm. The user must be aware of the internet worms that are cited because there are many tools that are available to find the vulnerabilities thar are occuring through internet..
Internet worm is a useless code which effect the code in the softwar.pdf
Internet worm is a useless code which effect the code in the softwar.pdf
nareshsonyericcson
Here we have the compound V3(PO4)2 Separating it into it\'s ions we have: V2+ and PO43- since vanadium is a transition metal, we denote its oxidation state by roman numerals. Thus, the name of the compound is Vanadium (II) Phosphate. Please don\'t forget to rate, good luck studying! Solution Here we have the compound V3(PO4)2 Separating it into it\'s ions we have: V2+ and PO43- since vanadium is a transition metal, we denote its oxidation state by roman numerals. Thus, the name of the compound is Vanadium (II) Phosphate. Please don\'t forget to rate, good luck studying!.
Here we have the compound V3(PO4)2 Separating it into its ions.pdf
Here we have the compound V3(PO4)2 Separating it into its ions.pdf
nareshsonyericcson
F+ cells have fertility factors while F- cells lack fertility factors F+ is a donor cells while F- is recipient cells F+ cells has circular plasmid separate from chromosome F- cell does not have before receiving plasmid Cell connect through T-pilus Tra K and tra I were required for efficient inter plasmid recombination at ori T whole DNA is transferred to recipient F- Solution F+ cells have fertility factors while F- cells lack fertility factors F+ is a donor cells while F- is recipient cells F+ cells has circular plasmid separate from chromosome F- cell does not have before receiving plasmid Cell connect through T-pilus Tra K and tra I were required for efficient inter plasmid recombination at ori T whole DNA is transferred to recipient F-.
F+ cells have fertility factors while F- cells lack fertility factor.pdf
F+ cells have fertility factors while F- cells lack fertility factor.pdf
nareshsonyericcson
D is correct. D. transport input ssh in vty Line Configuration mode This is the correct command, and must be entered in Line Configuration mode. Solution D is correct. D. transport input ssh in vty Line Configuration mode This is the correct command, and must be entered in Line Configuration mode..
D is correct.D. transport input ssh in vty Line Configuration mode.pdf
D is correct.D. transport input ssh in vty Line Configuration mode.pdf
nareshsonyericcson
Dissociation of 2 protons = dissociation of 2 H+ ions The answer is: C. H3PO4 --> 2H+ + HPO42- Solution Dissociation of 2 protons = dissociation of 2 H+ ions The answer is: C. H3PO4 --> 2H+ + HPO42-.
Dissociation of 2 protons = dissociation of 2 H+ ionsThe answer is.pdf
Dissociation of 2 protons = dissociation of 2 H+ ionsThe answer is.pdf
nareshsonyericcson
CASE ANSWERS: 1.C) Bacteria typically transmitted by airborne droplets because these gram positive bacteria can spread via the droplets expelled from a Sterptococcus pyogenes infected person. 2.D) Ingesting the bacteria and cell debris through phagocytosis because the neutrophils are the white blood cells that are in the first line of defence as an innate immune response that comes to site of infection(Tissues & Blood circulation) throught he chemoattractants produced by the streptococcus pyogenes. 3.The following statements about the antibody productions is TRUE : B) Antibodies are produced by B cells with the help of T helper 2 cell. Solution CASE ANSWERS: 1.C) Bacteria typically transmitted by airborne droplets because these gram positive bacteria can spread via the droplets expelled from a Sterptococcus pyogenes infected person. 2.D) Ingesting the bacteria and cell debris through phagocytosis because the neutrophils are the white blood cells that are in the first line of defence as an innate immune response that comes to site of infection(Tissues & Blood circulation) throught he chemoattractants produced by the streptococcus pyogenes. 3.The following statements about the antibody productions is TRUE : B) Antibodies are produced by B cells with the help of T helper 2 cell..
CASE ANSWERS1.C) Bacteria typically transmitted by airborne dropl.pdf
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nareshsonyericcson
Bacteria May have peptidoglycan cell wall Have ester linked lipids Methanogens Methane-producing archaea (methanogens) are strictly anaerobic convert substrate to methane Archae Have RNA polymerase similar to eukaryotes Have ether linked lipids in their cell membranes Solution Bacteria May have peptidoglycan cell wall Have ester linked lipids Methanogens Methane-producing archaea (methanogens) are strictly anaerobic convert substrate to methane Archae Have RNA polymerase similar to eukaryotes Have ether linked lipids in their cell membranes.
Bacteria May have peptidoglycan cell wallHave ester linked lipid.pdf
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nareshsonyericcson
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