A Lewis acid is a species that can accept a pair .pdf
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A Lewis acid is a species that can accept a pair of electrons from a Lewis base, which is a donator of electrons. In this reaction, AlCl3 is a strong lewis acid and accepts a lone pair of electrons from the Lewis base Cl-, thereby completing Aluminum\'s octet and forming AlCl4-. Solution A Lewis acid is a species that can accept a pair of electrons from a Lewis base, which is a donator of electrons. In this reaction, AlCl3 is a strong lewis acid and accepts a lone pair of electrons from the Lewis base Cl-, thereby completing Aluminum\'s octet and forming AlCl4-..
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O = O bond is shortest Oxygen electronegativity d.pdf
O = O bond is shortest Oxygen electronegativity difference is higher and also it is
double bond
Solution
O = O bond is shortest Oxygen electronegativity difference is higher and also it is
double bond.
NaCl Na is the symbol for Sodium which bonds to .pdf
NaCl Na is the symbol for Sodium which bonds to Cl which is the symbol for
Chlorine making Sodium Chloride.
Solution
NaCl Na is the symbol for Sodium which bonds to Cl which is the symbol for
Chlorine making Sodium Chloride..
NiCl2 because it not dissociate .pdf
NiCl2 because it not dissociate
Solution
NiCl2 because it not dissociate.
iodide, I-, has the -1 oxidation state. iodine (I.pdf
iodide, I-, has the -1 oxidation state. iodine (I) has 0 oxidation state. I- - e ---> I
Thus, this change is an oxidation reaction.
Solution
iodide, I-, has the -1 oxidation state. iodine (I) has 0 oxidation state. I- - e ---> I
Thus, this change is an oxidation reaction..
Lithium is in group 1 on the peridic table. This .pdf
Lithium is in group 1 on the peridic table. This means that it has only 1 valence
electron. Lithium will tend to lose that electron when it ionizes and become an ion. Remeber, an
ion is any atom, or sometimes molecule, with a charge. When Li loses the electron, it loses one
of its negative charges so the atom becomes an ion with a +1 charge because it now has 3
positively charged protons and only 2 negatively charged electrons.
Solution
Lithium is in group 1 on the peridic table. This means that it has only 1 valence
electron. Lithium will tend to lose that electron when it ionizes and become an ion. Remeber, an
ion is any atom, or sometimes molecule, with a charge. When Li loses the electron, it loses one
of its negative charges so the atom becomes an ion with a +1 charge because it now has 3
positively charged protons and only 2 negatively charged electrons..
Solution1.cpp#include iostreamheader for input output function.pdf
Solution
1.cpp
#include //header for input output function
using namespace std;//it tells the compiler to link std name space
int main()
{//main function
int n=1,m=-1;
if(n<-m)//if else statement
{
cout<.
Recommended
O = O bond is shortest Oxygen electronegativity d.pdf
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double bond
Solution
O = O bond is shortest Oxygen electronegativity difference is higher and also it is
double bond.
NaCl Na is the symbol for Sodium which bonds to .pdf
NaCl Na is the symbol for Sodium which bonds to Cl which is the symbol for
Chlorine making Sodium Chloride.
Solution
NaCl Na is the symbol for Sodium which bonds to Cl which is the symbol for
Chlorine making Sodium Chloride..
NiCl2 because it not dissociate .pdf
NiCl2 because it not dissociate
Solution
NiCl2 because it not dissociate.
iodide, I-, has the -1 oxidation state. iodine (I.pdf
iodide, I-, has the -1 oxidation state. iodine (I) has 0 oxidation state. I- - e ---> I
Thus, this change is an oxidation reaction.
Solution
iodide, I-, has the -1 oxidation state. iodine (I) has 0 oxidation state. I- - e ---> I
Thus, this change is an oxidation reaction..
Lithium is in group 1 on the peridic table. This .pdf
Lithium is in group 1 on the peridic table. This means that it has only 1 valence
electron. Lithium will tend to lose that electron when it ionizes and become an ion. Remeber, an
ion is any atom, or sometimes molecule, with a charge. When Li loses the electron, it loses one
of its negative charges so the atom becomes an ion with a +1 charge because it now has 3
positively charged protons and only 2 negatively charged electrons.
Solution
Lithium is in group 1 on the peridic table. This means that it has only 1 valence
electron. Lithium will tend to lose that electron when it ionizes and become an ion. Remeber, an
ion is any atom, or sometimes molecule, with a charge. When Li loses the electron, it loses one
of its negative charges so the atom becomes an ion with a +1 charge because it now has 3
positively charged protons and only 2 negatively charged electrons..
Solution1.cpp#include iostreamheader for input output function.pdf
Solution
1.cpp
#include //header for input output function
using namespace std;//it tells the compiler to link std name space
int main()
{//main function
int n=1,m=-1;
if(n<-m)//if else statement
{
cout<.
Systematic positionKingdom PlantaeSubkingdom Tracheobionta (Vas.pdf
Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
Anatomy:
A. Root : It shows usually Gymnosperm structure. Cortex, endoderm, inner periderm, vascular
camblum, primary and secondary xylem was present.
B. Stem:
1. Primary structure: Transverse structure shows ridges and grooves. cortex clorenchyma
elongated, endoderm is present in single layer.
2. Secondary structure: Parenchymalous cells present between the vascular bundles to form a
ring of cambium, ray initials and fusiform initial are present.
C. Leaves: Transverse structure of leaves shows epidermis, cuticle, xylem, phloem, air space,
spongy parenchyma tissues. The vascular bundles are collateral closed and primary phloem and
primary xylem are also present.
Adaptations:
Mostly xerophytes scrubs or woody climbers Ephedra Pedunculata. Most species of ephedra are
branched scrubs, commonly climbing over vegetation many species spread by rhizomes Ephedra
Trifura. Under favorable condition it grows tree like habitat.
Ephedra Foliata - scrambling scrub, Ephedra Compacta - Branching type, Ephedra
Camphylophoda - Spendulus branches, Ephedra Intermedia - Shoot, Ephedra Saxatilis - scrub.
Values and functions:
In pharmaceutical industries it works as to access Ephedra drug materials.
Traditionally it is used by indigenous people for medicinal treatments of asthma, hay fever and
cold.
Alkaloids obtained from the species of Ephedra is used in herbal medicine.
Solution
Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
A.
Query to list out all the languages in the table LANGUAGES.SELECT .pdf
Query to list out all the languages in the table LANGUAGES.
SELECT LANGUAGE FROM LANGUAGES
Solution
Query to list out all the languages in the table LANGUAGES.
SELECT LANGUAGE FROM LANGUAGES.
public class LunarLander { double currentFuelFlowRate; d.pdf
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate, double thrust)
{
currentFuelFlowRate=0.0;
verticalSpeed=0.0;
altitude=al;
amountOfFuel=fuel;
massOfTheLander=mass;
maxFuelConsumptionRate=rate;
maxThrust=thrust;
}
/* functions to get the values of instance variables*/
public double getCurrentFuelFlowRate()
{
return currentFuelFlowRate;
}
public double getVerticalSpeed()
{
return verticalSpeed;
}
public double getAltitude()
{
return altitude;
}
public double getAmountOfFuel()
{
return amountOfFuel;
}
public double getMassOfTheLander()
{
return massOfTheLander;
}
public double getMaxFuelConsumptionRate()
{
return maxFuelConsumptionRate;
}
public double getMaxThrust()
{
return maxThrust;
}
/* updates the value of currentFuelFlowRate*/
public void setCurrentFuelFlowRate(double rate)
{
currentFuelFlowRate=rate;
}
/* simulating the passage for a small amount of time t*/
public void setPassage(double t)
{
if(currentFuelFlowRate>0)
{
if(amountOfFuel==0)
{
currentFuelFlowRate=0;
}
}
/* velocity is the verticalSpeed*/
double f=maxThrust*currentFuelFlowRate;
double m=massOfTheLander;
verticalSpeed=t*((f/m)-1.62);
//v is verticalSpeed
altitude=t*verticalSpeed;
//ship is landed
if(altitude<0)
{
altitude=0;
verticalSpeed=0;
}
//r is currentFuelFlowRate
//c is maxFuelConsumptionRate
double changeInRemainFuel=t*currentFuelFlowRate*maxFuelConsumptionRate;
amountOfFuel=amountOfFuel-changeInRemainFuel;
if(amountOfFuel<0)
{
amountOfFuel=0;
}
}
}
////////////////////////////////////////////////////////
public class LunarLanderDemo {
public static void main(String[] args) {
LunarLander L=new LunarLander(1000,1700,900,10,5000);
/*give values between 0.0 , 1.0*/
L.setCurrentFuelFlowRate(0.6);
L.setPassage(0.1);
System.out.println(\"Simulation of Lunar Lander Passage at 0.1 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
L.setCurrentFuelFlowRate(0.3);
L.setPassage(0.09);
System.out.println(\"Simulation of Lunar Lander Passage at 0.09 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
}
}
Solution
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate.
Cooling. The reaction is giving off energy(heat) .pdf
Cooling. The reaction is giving off energy(heat) so you need to control this in some
cases or it can be problematic
Solution
Cooling. The reaction is giving off energy(heat) so you need to control this in some
cases or it can be problematic.
php global $bsize,$playerToken,$myToken,$gameOver,$winArr,$rowAr.pdf
Tic Tac Toe in PHP\';
for($i=0; $i<$size; $i++) {
for($j=0; $j<$size; $j++) {
$count++;
$retVal.=\'\';
}
$retVal.=\'
\';
}
$retVal.=\'\';
echo $retVal;
}
function genBox2($size,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$count=0;
$retVal=\'\';
for($i=0; $i<$size; $i++) {
for($j=0; $j<$size; $j++) {
$count++;
$retVal.=\'\';
}
$retVal.=\'
\';
}
$retVal.=\'\';
echo $retVal;
}
function isEmpty($who) {
if($who==\"\")
return 1;
else
return 0;
}
function move($bsize,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$count=0;
$maxCount=0;
$pos=0;
$retVal=0;
# Build Row Array
for($i=0; $i<$bsize; $i++) {
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$count++;
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
}
$rowArr[$i]=$maxCount;
if($fullCounter==$bsize)
$rowArr[$i]=-1;
}
# Building Column Array
for($i=0; $i<$bsize; $i++) {
$count=$i+1;
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
$count+=$bsize;
}
$colArr[$i]=$maxCount;
if($fullCounter==$bsize)
$colArr[$i]=-1;
}
# Building Diagonal Array
for($i=0; $i<2; $i++) {
if($i==0)
$count=$i+1;
else
$count=$bsize;
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
if($i==0)
$count+=$bsize+1;
else
$count+=$bsize-1;
}
$digArr[$i]=$maxCount;
if($fullCounter==$bsize)
$digArr[$i]=-1;
}
# Finding Max Values
$maxRow=myMax(0,$bsize,\"row\",$rowArr);
$maxCol=myMax(0,$bsize,\"col\",$colArr);
$maxDig=myMax(0,$bsize,\"dig\",$digArr);
$maxArrs[0]=myMax(1,$bsize,\"row\",$rowArr);
$maxArrs[1]=myMax(1,$bsize,\"col\",$colArr);
$maxArrs[2]=myMax(1,$bsize,\"dig\",$digArr);
//$maxArrs=array(max(1,$bsize,\"row\",$rowArr),max(1,$bsize,\"col\",$colArr),max(1,$bsize,\"
dig\",$digArr));
if(myMax(0,$bsize,\"x\",$maxArrs)==0)
$pos=$bsize*($maxRow+1)-$bsize;
if(myMax(0,$bsize,\"x\",$maxArrs)==1)
$pos=$maxCol;
if(myMax(0,$bsize,\"x\",$maxArrs)==2)
if($maxDig==0)
$pos=$maxDig;
else
$pos=$bsize-1;
$retFlag=0;
for($y=0; $y<$bsize; $y++) {
if(!$retFlag) {
if($arr[$pos]==\"\") {
$retVal=$pos;
$retFlag=1;
}
if(myMax(0,$bsize,\"x\",$maxArrs)==0)
$pos++;
if(myMax(0,$bsize,\"x\",$maxArrs)==1)
$pos+=$bsize;
if(myMax(0,$bsize,\"x\",$maxArrs)==2)
if($maxDig==0)
$pos+=$bsize+1;
else
$pos+=$bsize-1;
}
}
return $retVal;
}
function myMax($what,$bsize,$type,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$max=-1;
$maxIndex=-1;
if($type!=\"dig\") {
for($i=0; $i<$bsize; $i++) {
if($arr[$i]>$max) {
$max=$arr[$i];
$maxIndex=$i;
}
}
}
if($type==\"dig\") {
for($i=0; $i<2; $i++) {
if($arr[$i]>$max) {
$ma.
c. cytochrome C It is the enzyme involved in the .pdf
c. cytochrome C It is the enzyme involved in the reaction, so cannot exchange net
electrons
Solution
c. cytochrome C It is the enzyme involved in the reaction, so cannot exchange net
electrons.
Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
Ka=-log(Ka)=-log(5.6*10^-10)=9.25
According to Henderson Hasselbalch Equation
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6)
pH=8.77
(2)
NH3 + HCl =======> NH4Cl
I 0.2 0.10 0.6
C 0.2-0.1 -0.1 +0.1
E 0.1 0 0.7
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7)
pH=8.40
(3)
NH4Cl + NaOH =============> NH3 + NaCl + H2O
I 0.6 0.20 0.2
C -0.20 -0.20 +0.20
E 0.4 0 0.4
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25
Solution
Ka=-log(Ka)=-log(5.6*10^-10)=9.25
According to Henderson Hasselbalch Equation
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6)
pH=8.77
(2)
NH3 + HCl =======> NH4Cl
I 0.2 0.10 0.6
C 0.2-0.1 -0.1 +0.1
E 0.1 0 0.7
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7)
pH=8.40
(3)
NH4Cl + NaOH =============> NH3 + NaCl + H2O
I 0.6 0.20 0.2
C -0.20 -0.20 +0.20
E 0.4 0 0.4
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25.
C is wrong 14 7N contains 7 protons and 7 neutron.pdf
C is wrong 14 7N contains 7 protons and 7 neutrons , N/P = 1 which lies in belt of
stability
Solution
C is wrong 14 7N contains 7 protons and 7 neutrons , N/P = 1 which lies in belt of
stability.
1.The following are the main types of service providers to a mutua.pdf
1.
The following are the main types of service providers to a mutual fund:
Service provider refers to the sponsor of funds. From the public who are ready to invest, the fund
sponsor raises funds. The fund sponsor is one among the following:
The duty of the sponsor is to invest the funds of the investor in stocks, bonds, and shares
according to the preference of the investors.
If the investors demand for 100% safety of the invested funds, the fund sponsor invests in such
shares or bonds. If the investor is ready to meet risks even to 100%, then the fund sponsor will
invest accordingly. For this service, the investor has to pay the fund sponsor up to a certain limit.
2.
Determining the fees of mutual fund advisor:
There are certain restrictions that are imposed by SEBI. An advisor has to follow the same in
order to claim his fees. Mutual fund distributors are generally exempted from being registered as
investment advisors.
For a no-load large-cap fund, the percentage of fees is 0.95 whereas for small cap funds it is
1.20%. Therefore, if the advisor charges more than the specified percentage, the shareholders can
challenge that the fees is excessive.
3.
Differences between mutual funds and closed-end funds:
One of the major differences between the mutual funds and the closed-end funds is in the date of
maturity. A mutual fund can be withdrawn whenever the funds are required. However, it is not
the case in closed-end funds. In closed-end funds, the funds can be taken only at its date of
maturity. Some funds can be taken before the date of maturity. However, the rate of interest will
either be very little or even nil.
The closed-end funds have declined in popularity because of the returns and the maturity date.
Investing in this leads to overtrading and poor returns.
4.
Differences between ETFs and UITs from index mutual funds:
ETF is similar to closed-end funds. They trade closer to NAVs and therefore can be created and
retired whenever required. Mostly ETFs are passively managed with the help of indexes.
However, there are actively managed investments too. They are allowed to sell short and buy on
margin. They are even allowed to purchase a single share.
UITs are sold by investment advisors rather than purchasing from a secondary market. This does
not have a board of directors and is registered with Securities and Exchange Board. This
portfolio has different types of securities too.
Index mutual fund is a type of mutual fund in which a sale and buy index is set. When the rate
falls down to the set index, the funds are sold/bought on its own.
Solution
1.
The following are the main types of service providers to a mutual fund:
Service provider refers to the sponsor of funds. From the public who are ready to invest, the fund
sponsor raises funds. The fund sponsor is one among the following:
The duty of the sponsor is to invest the funds of the investor in stocks, bonds, and shares
according to the preference of the investors.
If the investors demand.
D. planned and directed means to improve the functioning of the clie.pdf
D. planned and directed means to improve the functioning of the client system
Solution
D. planned and directed means to improve the functioning of the client system.
domain - -x-1 0 = x+10 =x-1i.e. (-inf,-1)range-(-in.pdf
domain -
-x-1> 0
=> x+1<0
=>x<-1
i.e. (-inf,-1)
range-
(-inf,+inf)
Solution
domain -
-x-1> 0
=> x+1<0
=>x<-1
i.e. (-inf,-1)
range-
(-inf,+inf).
C. Keystroke loggers are a form of malware that records every key th.pdf
C. Keystroke loggers are a form of malware that records every key the computer user presses to
harvest usernames, passwords, social security numbers, and other confidential information and
then send that data to the person who caused the malware installation.
Solution
C. Keystroke loggers are a form of malware that records every key the computer user presses to
harvest usernames, passwords, social security numbers, and other confidential information and
then send that data to the person who caused the malware installation..
Ans.) Researchers have found much more proof that sexual orientation.pdf
Ans.) Researchers have found much more proof that sexual orientation is to a great extent
controlled by hereditary qualities, not decision. That can undermine a noteworthy contention
against the LBGT people group that claims that these individuals are living \"unnaturally.\"
According to latest research conducted and published in scientific journals the two areas of
human genome - one on the X chromosome and one on chromosome 8.
The relationship amongst science and sexual orientation is a subject of deep research and
analysis. A straightforward and particular determinant for sexual orientation has not been
indisputably illustrated; different studies indicate diverse, notwithstanding clashing positions,
however researchers theorize that a blend of genetics, hormonal and social elements decide
sexual orientation. Biological hypotheses for clarifying the reasons for sexual orientation are
more popular and natural elements may include an intricate interaction of hereditary components
and the early uterine, environment. These variables, which might be identified with the
improvement of a hetero, gay person, promiscuous orientation, incorporate qualities, pre-birth
hormones, and cerebrum structure.
According to a study, every single grown-up twin in Sweden (more than 7,600 twins) found that
same-sex conduct was clarified by both genetic elements and individual-particular natural
sources, (for example, pre-birth environment, involvement with sickness and injury and also peer
bunches and sexual encounters), while impacts of shared-environment factors, for example,
familial environment and social demeanors had a weaker, yet noteworthy impact. Women
demonstrated a factually non-critical pattern to weaker impact of genetic impacts, while men
demonstrated no impact of shared ecological impacts.
Chromosome linkage analysis of sexual orientation has shown the presence of different
contributing hereditary elements all through the genome. In 1993 Dean Hamer and partners
published their finding from a linkage examination of a specimen of 76 homo siblings and their
families. The researcher found that the homo men had more homo male uncles and cousins on
the maternal side of the family than on the fatherly side. Gay siblings who demonstrated this
maternal family were then tried for X chromosome linkage, utilizing twenty-two markers on the
X chromosome to test for comparable alleles. In another discovering, thirty-three of the forty kin
sets tried were found to have comparable alleles in the distal district of Xq28, which was
altogether higher than the normal rates of half for friendly siblings. This was prominently named
the \" homo quality\" in the media, bringing on noteworthy discussion. Sanders et al. in 1998
investigated their comparative study, in which they found that 13% of uncles of homosexual
siblings on the maternal side were homo person, contrasted and 6% on the fatherly side.
Hence, it is concluded that up to some extent homosexuality or sex orient.
Amnion - Protection of embryo (It covers the embryo when first forme.pdf
Amnion - Protection of embryo (It covers the embryo when first formed and filled with amniotic
fluid)
Chorion - Waste removal (The chorion is formed by two layers – trophoblast- outer layer and
mesoderm- inner layer. The mesoderm will be in contact with the amnion and removes waste
produced. The trophoblast provides the nutrients for the fetus during its confinement)
Allantois - gas exchange ( is a sac-like structure below the chorion, which consists of all
embryonic and extra-embryonic tissues)
Yolk - Primary nutrition source.
Albumen - Protein, nutrition.
Solution
Amnion - Protection of embryo (It covers the embryo when first formed and filled with amniotic
fluid)
Chorion - Waste removal (The chorion is formed by two layers – trophoblast- outer layer and
mesoderm- inner layer. The mesoderm will be in contact with the amnion and removes waste
produced. The trophoblast provides the nutrients for the fetus during its confinement)
Allantois - gas exchange ( is a sac-like structure below the chorion, which consists of all
embryonic and extra-embryonic tissues)
Yolk - Primary nutrition source.
Albumen - Protein, nutrition..
A. polar covalent NH3 H2O B. nonpolar covalent CH.pdf
A. polar covalent NH3 H2O B. nonpolar covalent CH4 CO2 C. ionic NaI
Solution
A. polar covalent NH3 H2O B. nonpolar covalent CH4 CO2 C. ionic NaI.
1.) If phi is thrival, kernal will be one to one. And kern = 9.pdf
1.) If phi is thrival, kernal will be one to one.
And |kern | = 9
2.) As phi is one to one, kernal will be trivial.
And |ker| = 9
And for c, kernel is also neither.
Solution
1.) If phi is thrival, kernal will be one to one.
And |kern | = 9
2.) As phi is one to one, kernal will be trivial.
And |ker| = 9
And for c, kernel is also neither..
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Operation Blue Star - Saka Neela Tara
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
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Systematic positionKingdom PlantaeSubkingdom Tracheobionta (Vas.pdf
Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
Anatomy:
A. Root : It shows usually Gymnosperm structure. Cortex, endoderm, inner periderm, vascular
camblum, primary and secondary xylem was present.
B. Stem:
1. Primary structure: Transverse structure shows ridges and grooves. cortex clorenchyma
elongated, endoderm is present in single layer.
2. Secondary structure: Parenchymalous cells present between the vascular bundles to form a
ring of cambium, ray initials and fusiform initial are present.
C. Leaves: Transverse structure of leaves shows epidermis, cuticle, xylem, phloem, air space,
spongy parenchyma tissues. The vascular bundles are collateral closed and primary phloem and
primary xylem are also present.
Adaptations:
Mostly xerophytes scrubs or woody climbers Ephedra Pedunculata. Most species of ephedra are
branched scrubs, commonly climbing over vegetation many species spread by rhizomes Ephedra
Trifura. Under favorable condition it grows tree like habitat.
Ephedra Foliata - scrambling scrub, Ephedra Compacta - Branching type, Ephedra
Camphylophoda - Spendulus branches, Ephedra Intermedia - Shoot, Ephedra Saxatilis - scrub.
Values and functions:
In pharmaceutical industries it works as to access Ephedra drug materials.
Traditionally it is used by indigenous people for medicinal treatments of asthma, hay fever and
cold.
Alkaloids obtained from the species of Ephedra is used in herbal medicine.
Solution
Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
A.
Query to list out all the languages in the table LANGUAGES.SELECT .pdf
Query to list out all the languages in the table LANGUAGES.
SELECT LANGUAGE FROM LANGUAGES
Solution
Query to list out all the languages in the table LANGUAGES.
SELECT LANGUAGE FROM LANGUAGES.
public class LunarLander { double currentFuelFlowRate; d.pdf
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate, double thrust)
{
currentFuelFlowRate=0.0;
verticalSpeed=0.0;
altitude=al;
amountOfFuel=fuel;
massOfTheLander=mass;
maxFuelConsumptionRate=rate;
maxThrust=thrust;
}
/* functions to get the values of instance variables*/
public double getCurrentFuelFlowRate()
{
return currentFuelFlowRate;
}
public double getVerticalSpeed()
{
return verticalSpeed;
}
public double getAltitude()
{
return altitude;
}
public double getAmountOfFuel()
{
return amountOfFuel;
}
public double getMassOfTheLander()
{
return massOfTheLander;
}
public double getMaxFuelConsumptionRate()
{
return maxFuelConsumptionRate;
}
public double getMaxThrust()
{
return maxThrust;
}
/* updates the value of currentFuelFlowRate*/
public void setCurrentFuelFlowRate(double rate)
{
currentFuelFlowRate=rate;
}
/* simulating the passage for a small amount of time t*/
public void setPassage(double t)
{
if(currentFuelFlowRate>0)
{
if(amountOfFuel==0)
{
currentFuelFlowRate=0;
}
}
/* velocity is the verticalSpeed*/
double f=maxThrust*currentFuelFlowRate;
double m=massOfTheLander;
verticalSpeed=t*((f/m)-1.62);
//v is verticalSpeed
altitude=t*verticalSpeed;
//ship is landed
if(altitude<0)
{
altitude=0;
verticalSpeed=0;
}
//r is currentFuelFlowRate
//c is maxFuelConsumptionRate
double changeInRemainFuel=t*currentFuelFlowRate*maxFuelConsumptionRate;
amountOfFuel=amountOfFuel-changeInRemainFuel;
if(amountOfFuel<0)
{
amountOfFuel=0;
}
}
}
////////////////////////////////////////////////////////
public class LunarLanderDemo {
public static void main(String[] args) {
LunarLander L=new LunarLander(1000,1700,900,10,5000);
/*give values between 0.0 , 1.0*/
L.setCurrentFuelFlowRate(0.6);
L.setPassage(0.1);
System.out.println(\"Simulation of Lunar Lander Passage at 0.1 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
L.setCurrentFuelFlowRate(0.3);
L.setPassage(0.09);
System.out.println(\"Simulation of Lunar Lander Passage at 0.09 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
}
}
Solution
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate.
Cooling. The reaction is giving off energy(heat) .pdf
Cooling. The reaction is giving off energy(heat) so you need to control this in some
cases or it can be problematic
Solution
Cooling. The reaction is giving off energy(heat) so you need to control this in some
cases or it can be problematic.
php global $bsize,$playerToken,$myToken,$gameOver,$winArr,$rowAr.pdf
Tic Tac Toe in PHP\';
for($i=0; $i<$size; $i++) {
for($j=0; $j<$size; $j++) {
$count++;
$retVal.=\'\';
}
$retVal.=\'
\';
}
$retVal.=\'\';
echo $retVal;
}
function genBox2($size,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$count=0;
$retVal=\'\';
for($i=0; $i<$size; $i++) {
for($j=0; $j<$size; $j++) {
$count++;
$retVal.=\'\';
}
$retVal.=\'
\';
}
$retVal.=\'\';
echo $retVal;
}
function isEmpty($who) {
if($who==\"\")
return 1;
else
return 0;
}
function move($bsize,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$count=0;
$maxCount=0;
$pos=0;
$retVal=0;
# Build Row Array
for($i=0; $i<$bsize; $i++) {
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$count++;
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
}
$rowArr[$i]=$maxCount;
if($fullCounter==$bsize)
$rowArr[$i]=-1;
}
# Building Column Array
for($i=0; $i<$bsize; $i++) {
$count=$i+1;
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
$count+=$bsize;
}
$colArr[$i]=$maxCount;
if($fullCounter==$bsize)
$colArr[$i]=-1;
}
# Building Diagonal Array
for($i=0; $i<2; $i++) {
if($i==0)
$count=$i+1;
else
$count=$bsize;
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
if($i==0)
$count+=$bsize+1;
else
$count+=$bsize-1;
}
$digArr[$i]=$maxCount;
if($fullCounter==$bsize)
$digArr[$i]=-1;
}
# Finding Max Values
$maxRow=myMax(0,$bsize,\"row\",$rowArr);
$maxCol=myMax(0,$bsize,\"col\",$colArr);
$maxDig=myMax(0,$bsize,\"dig\",$digArr);
$maxArrs[0]=myMax(1,$bsize,\"row\",$rowArr);
$maxArrs[1]=myMax(1,$bsize,\"col\",$colArr);
$maxArrs[2]=myMax(1,$bsize,\"dig\",$digArr);
//$maxArrs=array(max(1,$bsize,\"row\",$rowArr),max(1,$bsize,\"col\",$colArr),max(1,$bsize,\"
dig\",$digArr));
if(myMax(0,$bsize,\"x\",$maxArrs)==0)
$pos=$bsize*($maxRow+1)-$bsize;
if(myMax(0,$bsize,\"x\",$maxArrs)==1)
$pos=$maxCol;
if(myMax(0,$bsize,\"x\",$maxArrs)==2)
if($maxDig==0)
$pos=$maxDig;
else
$pos=$bsize-1;
$retFlag=0;
for($y=0; $y<$bsize; $y++) {
if(!$retFlag) {
if($arr[$pos]==\"\") {
$retVal=$pos;
$retFlag=1;
}
if(myMax(0,$bsize,\"x\",$maxArrs)==0)
$pos++;
if(myMax(0,$bsize,\"x\",$maxArrs)==1)
$pos+=$bsize;
if(myMax(0,$bsize,\"x\",$maxArrs)==2)
if($maxDig==0)
$pos+=$bsize+1;
else
$pos+=$bsize-1;
}
}
return $retVal;
}
function myMax($what,$bsize,$type,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$max=-1;
$maxIndex=-1;
if($type!=\"dig\") {
for($i=0; $i<$bsize; $i++) {
if($arr[$i]>$max) {
$max=$arr[$i];
$maxIndex=$i;
}
}
}
if($type==\"dig\") {
for($i=0; $i<2; $i++) {
if($arr[$i]>$max) {
$ma.
c. cytochrome C It is the enzyme involved in the .pdf
c. cytochrome C It is the enzyme involved in the reaction, so cannot exchange net
electrons
Solution
c. cytochrome C It is the enzyme involved in the reaction, so cannot exchange net
electrons.
Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
Ka=-log(Ka)=-log(5.6*10^-10)=9.25
According to Henderson Hasselbalch Equation
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6)
pH=8.77
(2)
NH3 + HCl =======> NH4Cl
I 0.2 0.10 0.6
C 0.2-0.1 -0.1 +0.1
E 0.1 0 0.7
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7)
pH=8.40
(3)
NH4Cl + NaOH =============> NH3 + NaCl + H2O
I 0.6 0.20 0.2
C -0.20 -0.20 +0.20
E 0.4 0 0.4
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25
Solution
Ka=-log(Ka)=-log(5.6*10^-10)=9.25
According to Henderson Hasselbalch Equation
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6)
pH=8.77
(2)
NH3 + HCl =======> NH4Cl
I 0.2 0.10 0.6
C 0.2-0.1 -0.1 +0.1
E 0.1 0 0.7
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7)
pH=8.40
(3)
NH4Cl + NaOH =============> NH3 + NaCl + H2O
I 0.6 0.20 0.2
C -0.20 -0.20 +0.20
E 0.4 0 0.4
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25.
C is wrong 14 7N contains 7 protons and 7 neutron.pdf
C is wrong 14 7N contains 7 protons and 7 neutrons , N/P = 1 which lies in belt of
stability
Solution
C is wrong 14 7N contains 7 protons and 7 neutrons , N/P = 1 which lies in belt of
stability.
1.The following are the main types of service providers to a mutua.pdf
1.
The following are the main types of service providers to a mutual fund:
Service provider refers to the sponsor of funds. From the public who are ready to invest, the fund
sponsor raises funds. The fund sponsor is one among the following:
The duty of the sponsor is to invest the funds of the investor in stocks, bonds, and shares
according to the preference of the investors.
If the investors demand for 100% safety of the invested funds, the fund sponsor invests in such
shares or bonds. If the investor is ready to meet risks even to 100%, then the fund sponsor will
invest accordingly. For this service, the investor has to pay the fund sponsor up to a certain limit.
2.
Determining the fees of mutual fund advisor:
There are certain restrictions that are imposed by SEBI. An advisor has to follow the same in
order to claim his fees. Mutual fund distributors are generally exempted from being registered as
investment advisors.
For a no-load large-cap fund, the percentage of fees is 0.95 whereas for small cap funds it is
1.20%. Therefore, if the advisor charges more than the specified percentage, the shareholders can
challenge that the fees is excessive.
3.
Differences between mutual funds and closed-end funds:
One of the major differences between the mutual funds and the closed-end funds is in the date of
maturity. A mutual fund can be withdrawn whenever the funds are required. However, it is not
the case in closed-end funds. In closed-end funds, the funds can be taken only at its date of
maturity. Some funds can be taken before the date of maturity. However, the rate of interest will
either be very little or even nil.
The closed-end funds have declined in popularity because of the returns and the maturity date.
Investing in this leads to overtrading and poor returns.
4.
Differences between ETFs and UITs from index mutual funds:
ETF is similar to closed-end funds. They trade closer to NAVs and therefore can be created and
retired whenever required. Mostly ETFs are passively managed with the help of indexes.
However, there are actively managed investments too. They are allowed to sell short and buy on
margin. They are even allowed to purchase a single share.
UITs are sold by investment advisors rather than purchasing from a secondary market. This does
not have a board of directors and is registered with Securities and Exchange Board. This
portfolio has different types of securities too.
Index mutual fund is a type of mutual fund in which a sale and buy index is set. When the rate
falls down to the set index, the funds are sold/bought on its own.
Solution
1.
The following are the main types of service providers to a mutual fund:
Service provider refers to the sponsor of funds. From the public who are ready to invest, the fund
sponsor raises funds. The fund sponsor is one among the following:
The duty of the sponsor is to invest the funds of the investor in stocks, bonds, and shares
according to the preference of the investors.
If the investors demand.
D. planned and directed means to improve the functioning of the clie.pdf
D. planned and directed means to improve the functioning of the client system
Solution
D. planned and directed means to improve the functioning of the client system.
domain - -x-1 0 = x+10 =x-1i.e. (-inf,-1)range-(-in.pdf
domain -
-x-1> 0
=> x+1<0
=>x<-1
i.e. (-inf,-1)
range-
(-inf,+inf)
Solution
domain -
-x-1> 0
=> x+1<0
=>x<-1
i.e. (-inf,-1)
range-
(-inf,+inf).
C. Keystroke loggers are a form of malware that records every key th.pdf
C. Keystroke loggers are a form of malware that records every key the computer user presses to
harvest usernames, passwords, social security numbers, and other confidential information and
then send that data to the person who caused the malware installation.
Solution
C. Keystroke loggers are a form of malware that records every key the computer user presses to
harvest usernames, passwords, social security numbers, and other confidential information and
then send that data to the person who caused the malware installation..
Ans.) Researchers have found much more proof that sexual orientation.pdf
Ans.) Researchers have found much more proof that sexual orientation is to a great extent
controlled by hereditary qualities, not decision. That can undermine a noteworthy contention
against the LBGT people group that claims that these individuals are living \"unnaturally.\"
According to latest research conducted and published in scientific journals the two areas of
human genome - one on the X chromosome and one on chromosome 8.
The relationship amongst science and sexual orientation is a subject of deep research and
analysis. A straightforward and particular determinant for sexual orientation has not been
indisputably illustrated; different studies indicate diverse, notwithstanding clashing positions,
however researchers theorize that a blend of genetics, hormonal and social elements decide
sexual orientation. Biological hypotheses for clarifying the reasons for sexual orientation are
more popular and natural elements may include an intricate interaction of hereditary components
and the early uterine, environment. These variables, which might be identified with the
improvement of a hetero, gay person, promiscuous orientation, incorporate qualities, pre-birth
hormones, and cerebrum structure.
According to a study, every single grown-up twin in Sweden (more than 7,600 twins) found that
same-sex conduct was clarified by both genetic elements and individual-particular natural
sources, (for example, pre-birth environment, involvement with sickness and injury and also peer
bunches and sexual encounters), while impacts of shared-environment factors, for example,
familial environment and social demeanors had a weaker, yet noteworthy impact. Women
demonstrated a factually non-critical pattern to weaker impact of genetic impacts, while men
demonstrated no impact of shared ecological impacts.
Chromosome linkage analysis of sexual orientation has shown the presence of different
contributing hereditary elements all through the genome. In 1993 Dean Hamer and partners
published their finding from a linkage examination of a specimen of 76 homo siblings and their
families. The researcher found that the homo men had more homo male uncles and cousins on
the maternal side of the family than on the fatherly side. Gay siblings who demonstrated this
maternal family were then tried for X chromosome linkage, utilizing twenty-two markers on the
X chromosome to test for comparable alleles. In another discovering, thirty-three of the forty kin
sets tried were found to have comparable alleles in the distal district of Xq28, which was
altogether higher than the normal rates of half for friendly siblings. This was prominently named
the \" homo quality\" in the media, bringing on noteworthy discussion. Sanders et al. in 1998
investigated their comparative study, in which they found that 13% of uncles of homosexual
siblings on the maternal side were homo person, contrasted and 6% on the fatherly side.
Hence, it is concluded that up to some extent homosexuality or sex orient.
Amnion - Protection of embryo (It covers the embryo when first forme.pdf
Amnion - Protection of embryo (It covers the embryo when first formed and filled with amniotic
fluid)
Chorion - Waste removal (The chorion is formed by two layers – trophoblast- outer layer and
mesoderm- inner layer. The mesoderm will be in contact with the amnion and removes waste
produced. The trophoblast provides the nutrients for the fetus during its confinement)
Allantois - gas exchange ( is a sac-like structure below the chorion, which consists of all
embryonic and extra-embryonic tissues)
Yolk - Primary nutrition source.
Albumen - Protein, nutrition.
Solution
Amnion - Protection of embryo (It covers the embryo when first formed and filled with amniotic
fluid)
Chorion - Waste removal (The chorion is formed by two layers – trophoblast- outer layer and
mesoderm- inner layer. The mesoderm will be in contact with the amnion and removes waste
produced. The trophoblast provides the nutrients for the fetus during its confinement)
Allantois - gas exchange ( is a sac-like structure below the chorion, which consists of all
embryonic and extra-embryonic tissues)
Yolk - Primary nutrition source.
Albumen - Protein, nutrition..
A. polar covalent NH3 H2O B. nonpolar covalent CH.pdf
A. polar covalent NH3 H2O B. nonpolar covalent CH4 CO2 C. ionic NaI
Solution
A. polar covalent NH3 H2O B. nonpolar covalent CH4 CO2 C. ionic NaI.
1.) If phi is thrival, kernal will be one to one. And kern = 9.pdf
1.) If phi is thrival, kernal will be one to one.
And |kern | = 9
2.) As phi is one to one, kernal will be trivial.
And |ker| = 9
And for c, kernel is also neither.
Solution
1.) If phi is thrival, kernal will be one to one.
And |kern | = 9
2.) As phi is one to one, kernal will be trivial.
And |ker| = 9
And for c, kernel is also neither..
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Query to list out all the languages in the table LANGUAGES.SELECT .pdf
Query to list out all the languages in the table LANGUAGES.SELECT .pdf
public class LunarLander { double currentFuelFlowRate; d.pdf
public class LunarLander { double currentFuelFlowRate; d.pdf
Cooling. The reaction is giving off energy(heat) .pdf
Cooling. The reaction is giving off energy(heat) .pdf
php global $bsize,$playerToken,$myToken,$gameOver,$winArr,$rowAr.pdf
php global $bsize,$playerToken,$myToken,$gameOver,$winArr,$rowAr.pdf
c. cytochrome C It is the enzyme involved in the .pdf
c. cytochrome C It is the enzyme involved in the .pdf
Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
C is wrong 14 7N contains 7 protons and 7 neutron.pdf
C is wrong 14 7N contains 7 protons and 7 neutron.pdf
1.The following are the main types of service providers to a mutua.pdf
1.The following are the main types of service providers to a mutua.pdf
D. planned and directed means to improve the functioning of the clie.pdf
D. planned and directed means to improve the functioning of the clie.pdf
domain - -x-1 0 = x+10 =x-1i.e. (-inf,-1)range-(-in.pdf
domain - -x-1 0 = x+10 =x-1i.e. (-inf,-1)range-(-in.pdf
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The French Revolution Class 9 Study Material pdf free download
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
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The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
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Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
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June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
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Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
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