import java.util.Scanner; import java.math.BigInteger; public cl.pdf

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annucommunication1

import java.util.Scanner; import java.math.BigInteger; public class Money{ private int dollars; private int cents; public Money(){ dollars=0; cents=0; } public Money(int d,int c){ dollars=d+c/100; cents=c%100; } public Money(int c){ dollars=c/100; cents=c%100; } public void add(int c){ cents+=c; dollars+=cents/100; cents=cents%100; } public void add(int d,int c){ cents+=c; dollars=dollars+d+cents/100; cents=cents%100; } public void remove(int c){ cents=dollars*100+cents; if(c>cents){ cents=0; } else{ cents-=c; } dollars=cents/100; cents=cents%100; } public String getMoney(){ return dollars+\" dollars and \"+cents+\" cents\"; } public static void main(String []args){ Money h=new Money(1,75); h.add(190); System.out.println(h.getMoney()); } } Solution import java.util.Scanner; import java.math.BigInteger; public class Money{ private int dollars; private int cents; public Money(){ dollars=0; cents=0; } public Money(int d,int c){ dollars=d+c/100; cents=c%100; } public Money(int c){ dollars=c/100; cents=c%100; } public void add(int c){ cents+=c; dollars+=cents/100; cents=cents%100; } public void add(int d,int c){ cents+=c; dollars=dollars+d+cents/100; cents=cents%100; } public void remove(int c){ cents=dollars*100+cents; if(c>cents){ cents=0; } else{ cents-=c; } dollars=cents/100; cents=cents%100; } public String getMoney(){ return dollars+\" dollars and \"+cents+\" cents\"; } public static void main(String []args){ Money h=new Money(1,75); h.add(190); System.out.println(h.getMoney()); } }.

$import java.util.Scanner; import java.math.BigInteger; public class Money{ private int dollars; private int cents; public Money(){ dollars=0; cents=0; } public Money(int d,int c){ dollars=d+c/100; cents=c%100; } public Money(int c){ dollars=c/100; cents=c%100; } public void add(int c){ cents+=c; dollars+=cents/100; cents=cents%100; } public void add(int d,int c){ cents+=c; dollars=dollars+d+cents/100; cents=cents%100; } public void remove(int c){ cents=dollars*100+cents;$

if(c>cents){
cents=0;
}
else{
cents-=c;
}
dollars=cents/100;
cents=cents%100;
}
public String getMoney(){
return dollars+" dollars and "+cents+" cents";
}
public static void main(String []args){
Money h=new Money(1,75);
h.add(190);
System.out.println(h.getMoney());
}
}
Solution
import java.util.Scanner;
import java.math.BigInteger;
public class Money{
private int dollars;
private int cents;
public Money(){
dollars=0;
cents=0;
}
public Money(int d,int c){
dollars=d+c/100;
cents=c%100;

}
public Money(int c){
dollars=c/100;
cents=c%100;
}
public void add(int c){
cents+=c;
dollars+=cents/100;
cents=cents%100;
}
public void add(int d,int c){
cents+=c;
dollars=dollars+d+cents/100;
cents=cents%100;
}
public void remove(int c){
cents=dollars*100+cents;
if(c>cents){
cents=0;
}
else{
cents-=c;
}
dollars=cents/100;
cents=cents%100;
}
public String getMoney(){
return dollars+" dollars and "+cents+" cents";
}
public static void main(String []args){
Money h=new Money(1,75);

h.add(190);
System.out.println(h.getMoney());
}
}

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Eukaryotic Chromosome Structure The length of DNA in the nucleus is far greater than the size of the compartment in which it is contained. To fit into this compartment the DNA has to be condensed in some manner. The degree to which DNA is condensed is expressed as its packing ratio. Packing ratio - the length of DNA divided by the length into which it is packaged For example, the shortest human chromosome contains 4.6 x 107 bp of DNA (about 10 times the genome size of E. coli). This is equivalent to 14,000 µm of extended DNA. In its most condensed state during mitosis, the chromosome is about 2 µm long. This gives a packing ratio of 7000 (14,000/2). To achieve the overall packing ratio, DNA is not packaged directly into final structure of chromatin. Instead, it contains several hierarchies of organization. The first level of packing is achieved by the winding of DNA around a protein core to produce a \"bead-like\" structure called a nucleosome. This gives a packing ratio of about 6. This structure is invariant in both the euchromatin and heterochromatin of all chromosomes. The second level of packing is the coiling of beads in a helical structure called the 30 nm fiber that is found in both interphase chromatin and mitotic chromosomes. This structure increases the packing ratio to about 40. The final packaging occurs when the fiber is organized in loops, scaffolds and domains that give a final packing ratio of about 1000 in interphase chromosomes and about 10,000 in mitotic chromosomes. Eukaryotic chromosomes consist of a DNA-protein complex that is organized in a compact manner which permits the large amount of DNA to be stored in the nucleus of the cell. The subunit designation of the chromosome is chromatin. The fundamental unit of chromatin is the nucleosome. Chromatin - the unit of analysis of the chromosome; chromatin reflects the general structure of the chromosome but is not unique to any particular chromosome Nucleosome - simplest packaging structure of DNA that is found in all eukaryotic chromosomes; DNA is wrapped around an octamer of small basic proteins called histones; 146 bp is wrapped around the core and the remaining bases link to the next nucleosome; this structure causes negative supercoiling The nucleosome consists of about 200 bp wrapped around a histone octamer that contains two copies of histone proteins H2A, H2B, H3 and H4. These are known as the core histones. Histones are basic proteins that have an affinity for DNA and are the most abundant proteins associated with DNA. The amino acid sequence of these four histones is conserved suggesting a similar function for all. The length of DNA that is associated with the nucleosome unit varies between species. But regardless of the size, two DNA components are involved. Core DNA is the DNA that is actually associated with the histone octamer. This value is invariant and is 146 base pairs. The core DNA forms two loops around the octamer, and this permits two regions that are .

Eukaryotic Chromosome Structure The length of DN.pdf

Eukaryotic Chromosome Structure The length of DN.pdf

Eukaryotic Chromosome Structure The length of DN.pdf

annucommunication1

Fe + S ----> FeS moles of Fe = 9.42/56 = 0.1682 moles of S = 8.5/32 = 0.265625 here Fe is limiting so, all the Fe will react to form same amount of FeS as Fe. S0, amount of FeS = 0.1682moles mass of FeS = 0.1682*88 = 14.8g Solution Fe + S ----> FeS moles of Fe = 9.42/56 = 0.1682 moles of S = 8.5/32 = 0.265625 here Fe is limiting so, all the Fe will react to form same amount of FeS as Fe. S0, amount of FeS = 0.1682moles mass of FeS = 0.1682*88 = 14.8g.

Fe + S ---- FeS moles of Fe = 9.4256 = 0.1682 m.pdf

Fe + S ---- FeS moles of Fe = 9.4256 = 0.1682 m.pdf

Fe + S ---- FeS moles of Fe = 9.4256 = 0.1682 m.pdf

annucommunication1

D. is correct. Solution .pdf

D. is correct. Solution .pdf

D. is correct. Solution .pdf

annucommunication1

The various ways to finance a project are: 1)Debt financing-One can take out a loan and finance the project costs Pros: Ownership is maintained Lower interest rate Lower cost of capital Cons: Repayment needed even if business goes bankrupt High rates for old debt even if macroeconomic conditions change Collateral needed 2) Equity financing is when project financing provides share in a firm in lieu of financing Pros- Easily availaible than debt Not to be repaid in case of bankrupcy Periodic payment as in case of debt is not required. Dividends can be paid as when the owner feels so Cons- have to share profit High return required by investor May lose control of company Private equity funding is when a large investor invests in company and his share cannot be publicly tarded: Pros- Capital is easily availaible and all capital can be raised at once The private equity investor provides his insights No return requirement and no cash flow has to be paid out as in case of equity Cons- More interference in day to day matters by investor investor may want to recoup his investment when he feels so but company may still have better profits Solution The various ways to finance a project are: 1)Debt financing-One can take out a loan and finance the project costs Pros: Ownership is maintained Lower interest rate Lower cost of capital Cons: Repayment needed even if business goes bankrupt High rates for old debt even if macroeconomic conditions change Collateral needed 2) Equity financing is when project financing provides share in a firm in lieu of financing Pros- Easily availaible than debt Not to be repaid in case of bankrupcy Periodic payment as in case of debt is not required. Dividends can be paid as when the owner feels so Cons- have to share profit High return required by investor May lose control of company Private equity funding is when a large investor invests in company and his share cannot be publicly tarded: Pros- Capital is easily availaible and all capital can be raised at once The private equity investor provides his insights No return requirement and no cash flow has to be paid out as in case of equity Cons- More interference in day to day matters by investor investor may want to recoup his investment when he feels so but company may still have better profits.

The various ways to finance a project are 1)Debt financing-One ca.pdf

The various ways to finance a project are 1)Debt financing-One ca.pdf

The various ways to finance a project are 1)Debt financing-One ca.pdf

annucommunication1

boiling point: 1-propanol > propanone note: there exists intermolecular hydrogen bonding interaction between 1-propanol (i.e., H-O...H bond), whereas no such interaction between propanone molecules. Thus, bp of 1-propanol is higher than propanone. Solution boiling point: 1-propanol > propanone note: there exists intermolecular hydrogen bonding interaction between 1-propanol (i.e., H-O...H bond), whereas no such interaction between propanone molecules. Thus, bp of 1-propanol is higher than propanone..

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annucommunication1

The organs of urinary system includesPaired kidneys, paired ureter.pdf

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annucommunication1

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The incorrect statement ise. Glucose is a stereoisomer of ribose..pdf

annucommunication1

The answer is: a. The intermolecular forces are much greater in water than in hydrogen sulfide and hydrogen selenide. Boiling point increases with strength of intermolecular forces as these intermolecular bonds need to be broken for the molecules to escape into the gas phase. Water has strong intermolecular hydrogen bonds not present in hydrogen sulfide and hydrogen selenide. Thus water has a much higher boiling point. Solution The answer is: a. The intermolecular forces are much greater in water than in hydrogen sulfide and hydrogen selenide. Boiling point increases with strength of intermolecular forces as these intermolecular bonds need to be broken for the molecules to escape into the gas phase. Water has strong intermolecular hydrogen bonds not present in hydrogen sulfide and hydrogen selenide. Thus water has a much higher boiling point..

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The answer is a. The intermolecular forces are much greater in wate.pdf

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import java.util.Scanner; import java.math.BigInteger; public cl.pdf

1. import java.util.Scanner; import java.math.BigInteger; public class Money{ private int dollars; private int cents; public Money(){ dollars=0; cents=0; } public Money(int d,int c){ dollars=d+c/100; cents=c%100; } public Money(int c){ dollars=c/100; cents=c%100; } public void add(int c){ cents+=c; dollars+=cents/100; cents=cents%100; } public void add(int d,int c){ cents+=c; dollars=dollars+d+cents/100; cents=cents%100; } public void remove(int c){ cents=dollars*100+cents;

2. if(c>cents){ cents=0; } else{ cents-=c; } dollars=cents/100; cents=cents%100; } public String getMoney(){ return dollars+" dollars and "+cents+" cents"; } public static void main(String []args){ Money h=new Money(1,75); h.add(190); System.out.println(h.getMoney()); } } Solution import java.util.Scanner; import java.math.BigInteger; public class Money{ private int dollars; private int cents; public Money(){ dollars=0; cents=0; } public Money(int d,int c){ dollars=d+c/100; cents=c%100;

3. } public Money(int c){ dollars=c/100; cents=c%100; } public void add(int c){ cents+=c; dollars+=cents/100; cents=cents%100; } public void add(int d,int c){ cents+=c; dollars=dollars+d+cents/100; cents=cents%100; } public void remove(int c){ cents=dollars*100+cents; if(c>cents){ cents=0; } else{ cents-=c; } dollars=cents/100; cents=cents%100; } public String getMoney(){ return dollars+" dollars and "+cents+" cents"; } public static void main(String []args){ Money h=new Money(1,75);

4. h.add(190); System.out.println(h.getMoney()); } }