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The equation you should use is just a variation of the PV formula: r = (FV/PV)1/t -1 r = ($3.20/$1.50)1/5 - 1 = Solution The equation you should use is just a variation of the PV formula: r = (FV/PV)1/t -1 r = ($3.20/$1.50)1/5 - 1 =.
The equation you should use is just a variation of the PV formula.pdf
The equation you should use is just a variation of the PV formula.pdf
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True: short double stranded mRNA are involved in initiating the interference Answer: C: genes are reshuffled in the mRNA segments Answer B: Enhancers can function in either orientation whilst the UAS only functions in one. False: eukaryotes utilise enhancers Solution True: short double stranded mRNA are involved in initiating the interference Answer: C: genes are reshuffled in the mRNA segments Answer B: Enhancers can function in either orientation whilst the UAS only functions in one. False: eukaryotes utilise enhancers.
True short double stranded mRNA are involved in initiating the inte.pdf
True short double stranded mRNA are involved in initiating the inte.pdf
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Yes, it would react similarly, since boric oxide is similar to carbon dioxide in dissolving water and will produce boric acid as result Solution Yes, it would react similarly, since boric oxide is similar to carbon dioxide in dissolving water and will produce boric acid as result.
Yes, it would react similarly, since boric oxide .pdf
Yes, it would react similarly, since boric oxide .pdf
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sol: Solution A,D,E: Crenation.
sol Solution A,D,E Crenation Solution B Hemol.pdf
sol Solution A,D,E Crenation Solution B Hemol.pdf
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please post the figure.... Solution please post the figure.....
please post the figure.... .pdf
please post the figure.... .pdf
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Intermediate species means species that have 2 characteristics 1. they are a product in a previous step 2. they are consumed (they are reactants) in the following step Note that they must have both of these characteristics to be considered intermediates. If they have one or the other they are not counted as intermediates. Notice how NO is being produced in step I and being used up in step II. That is why it is an intermediate. Solution Intermediate species means species that have 2 characteristics 1. they are a product in a previous step 2. they are consumed (they are reactants) in the following step Note that they must have both of these characteristics to be considered intermediates. If they have one or the other they are not counted as intermediates. Notice how NO is being produced in step I and being used up in step II. That is why it is an intermediate..
Intermediate species means species that have 2 ch.pdf
Intermediate species means species that have 2 ch.pdf
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In primitive cells atoms are present at the corners only. Among the given unit cells unit cell 1 is primitive cell. In remaining two unit cells there are atoms at center also. Solution In primitive cells atoms are present at the corners only. Among the given unit cells unit cell 1 is primitive cell. In remaining two unit cells there are atoms at center also..
In primitive cells atoms are present at the corne.pdf
In primitive cells atoms are present at the corne.pdf
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Hg2+ is more water soluble, methyl mercury HgCH3+ is more lipid soluble. CH3Hg, polar molecules would combine to form a substance similar to water but would be more acidic as a result of mixing two non-like particles in a solution Solution Hg2+ is more water soluble, methyl mercury HgCH3+ is more lipid soluble. CH3Hg, polar molecules would combine to form a substance similar to water but would be more acidic as a result of mixing two non-like particles in a solution.
Hg2+ is more water soluble, methyl mercury HgCH3+.pdf
Hg2+ is more water soluble, methyl mercury HgCH3+.pdf
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The equation you should use is just a variation of the PV formula: r = (FV/PV)1/t -1 r = ($3.20/$1.50)1/5 - 1 = Solution The equation you should use is just a variation of the PV formula: r = (FV/PV)1/t -1 r = ($3.20/$1.50)1/5 - 1 =.
The equation you should use is just a variation of the PV formula.pdf
The equation you should use is just a variation of the PV formula.pdf
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True: short double stranded mRNA are involved in initiating the interference Answer: C: genes are reshuffled in the mRNA segments Answer B: Enhancers can function in either orientation whilst the UAS only functions in one. False: eukaryotes utilise enhancers Solution True: short double stranded mRNA are involved in initiating the interference Answer: C: genes are reshuffled in the mRNA segments Answer B: Enhancers can function in either orientation whilst the UAS only functions in one. False: eukaryotes utilise enhancers.
True short double stranded mRNA are involved in initiating the inte.pdf
True short double stranded mRNA are involved in initiating the inte.pdf
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Yes, it would react similarly, since boric oxide is similar to carbon dioxide in dissolving water and will produce boric acid as result Solution Yes, it would react similarly, since boric oxide is similar to carbon dioxide in dissolving water and will produce boric acid as result.
Yes, it would react similarly, since boric oxide .pdf
Yes, it would react similarly, since boric oxide .pdf
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sol: Solution A,D,E: Crenation.
sol Solution A,D,E Crenation Solution B Hemol.pdf
sol Solution A,D,E Crenation Solution B Hemol.pdf
annaimobiles
please post the figure.... Solution please post the figure.....
please post the figure.... .pdf
please post the figure.... .pdf
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Intermediate species means species that have 2 characteristics 1. they are a product in a previous step 2. they are consumed (they are reactants) in the following step Note that they must have both of these characteristics to be considered intermediates. If they have one or the other they are not counted as intermediates. Notice how NO is being produced in step I and being used up in step II. That is why it is an intermediate. Solution Intermediate species means species that have 2 characteristics 1. they are a product in a previous step 2. they are consumed (they are reactants) in the following step Note that they must have both of these characteristics to be considered intermediates. If they have one or the other they are not counted as intermediates. Notice how NO is being produced in step I and being used up in step II. That is why it is an intermediate..
Intermediate species means species that have 2 ch.pdf
Intermediate species means species that have 2 ch.pdf
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In primitive cells atoms are present at the corners only. Among the given unit cells unit cell 1 is primitive cell. In remaining two unit cells there are atoms at center also. Solution In primitive cells atoms are present at the corners only. Among the given unit cells unit cell 1 is primitive cell. In remaining two unit cells there are atoms at center also..
In primitive cells atoms are present at the corne.pdf
In primitive cells atoms are present at the corne.pdf
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Hg2+ is more water soluble, methyl mercury HgCH3+ is more lipid soluble. CH3Hg, polar molecules would combine to form a substance similar to water but would be more acidic as a result of mixing two non-like particles in a solution Solution Hg2+ is more water soluble, methyl mercury HgCH3+ is more lipid soluble. CH3Hg, polar molecules would combine to form a substance similar to water but would be more acidic as a result of mixing two non-like particles in a solution.
Hg2+ is more water soluble, methyl mercury HgCH3+.pdf
Hg2+ is more water soluble, methyl mercury HgCH3+.pdf
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HCl is a polar-covalent compound and also Acetonitrile they are soluble dueto hydrogen - bonding between them. So they have London-dispersion forces between them. Solution HCl is a polar-covalent compound and also Acetonitrile they are soluble dueto hydrogen - bonding between them. So they have London-dispersion forces between them..
HCl is a polar-covalent compound and also Acetoni.pdf
HCl is a polar-covalent compound and also Acetoni.pdf
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for a compound to be soluble in water 1. it should have a large dipolemoment or 2. it should be ionic or 3. it should form H bond with water in first pair CH3CH2CH3 is non polar and is not soluble but CH2OHCHOHCH2OH is soluble in water by making H bonds. in second pair Though CH3Cl has little dipolemoment HCl has very higher dipolemomet because of charge separation and is readily soluble in water Solution for a compound to be soluble in water 1. it should have a large dipolemoment or 2. it should be ionic or 3. it should form H bond with water in first pair CH3CH2CH3 is non polar and is not soluble but CH2OHCHOHCH2OH is soluble in water by making H bonds. in second pair Though CH3Cl has little dipolemoment HCl has very higher dipolemomet because of charge separation and is readily soluble in water.
for a compound to be soluble in water 1. it shou.pdf
for a compound to be soluble in water 1. it shou.pdf
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e) II and III have the loest, equal stability Solution e) II and III have the loest, equal stability.
e) II and III have the loest, equal stability .pdf
e) II and III have the loest, equal stability .pdf
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wait doing Solution wait doing.
wait doingSolutionwait doing.pdf
wait doingSolutionwait doing.pdf
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TRUE The Reynolds number of falling 2.5 micrometer flyash particles in air exceed that of same size of soot particle by more than 50% because the soot particle is a outcome produced due to in complete combustion of carbon particles and very lighter in density compared to the flyash. Solution TRUE The Reynolds number of falling 2.5 micrometer flyash particles in air exceed that of same size of soot particle by more than 50% because the soot particle is a outcome produced due to in complete combustion of carbon particles and very lighter in density compared to the flyash..
TRUE The Reynolds number of falling 2.5 micrometer flyash particles .pdf
TRUE The Reynolds number of falling 2.5 micrometer flyash particles .pdf
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the electron rich molecule is a central Xe with 4 F singlebonded to it, 1 lone pair, and 1 oxygen molecule double bonded toit (if this is confusing the molecule is called xenonoxytetrafluoride so you can google it or whatever) XeF6+H2O XeOF4 +2HF Solution the electron rich molecule is a central Xe with 4 F singlebonded to it, 1 lone pair, and 1 oxygen molecule double bonded toit (if this is confusing the molecule is called xenonoxytetrafluoride so you can google it or whatever) XeF6+H2O XeOF4 +2HF.
the electron rich molecule is a central Xe with 4 F singlebonded.pdf
the electron rich molecule is a central Xe with 4 F singlebonded.pdf
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The given formulae themselves are explicit. Explicit means direct i.e., the ones having a clear expression of one variable in terms of several others which is the case here, as an is expressed as a direct function of n and several lower order terms. By direct or clear we mean that expression which represents, the variable (on the LHS of the equation) that is expressed, as a function of other variables does not involve the expressed variable (in the RHS of the equation). Solution The given formulae themselves are explicit. Explicit means direct i.e., the ones having a clear expression of one variable in terms of several others which is the case here, as an is expressed as a direct function of n and several lower order terms. By direct or clear we mean that expression which represents, the variable (on the LHS of the equation) that is expressed, as a function of other variables does not involve the expressed variable (in the RHS of the equation)..
The given formulae themselves are explicit. Explicit means direct i..pdf
The given formulae themselves are explicit. Explicit means direct i..pdf
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The answer is Meiosis, It occurs in human body at reporductive organs only. For healing and growth Mitosis plays a crucial role. Solution The answer is Meiosis, It occurs in human body at reporductive organs only. For healing and growth Mitosis plays a crucial role..
The answer is Meiosis, It occurs in human body at reporductive organ.pdf
The answer is Meiosis, It occurs in human body at reporductive organ.pdf
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Structure of Hemoglobin Hemoglobin is a chromo protein and is found in red blood cells, it’s a conjugated protein (heme as prosthetic group + globin as the protein part apoprotein). Adults have 14.0 to 16.0 gm% of Hb and 90 mg/kg of Hb will be produced and destroyed in the body on daily basis. Molecular weight of Hb is 67,000 and 3.4 mg of iron present in each gram of Hb. The combination of iron with a porphyrin ring produces the heme. Structure of Heme Heme is derived from porphyrin and porphyrins are cyclic compounds and are formed when 4 pyrrole rings fuse and are linked by methenyl bridges and the four rings are named as I,II,III, IV and Alpha, beta, gamma and delta are the bridges. To the side chain of Porphyrins, four pyrrole rings are attached. We can find one ferrous atom (Fe++) co-ordinated at the centre of the of protoporphyrin IX tetra pyrrole ring. Structure of Globin We can find the tetramer of globin polypeptide chains and each Hb molecule will have 4 Heme units and the subunits of hemoglobin are found to be arranged in a tetrahedral array. And this arrangement will give tight spherical overall appearance (which allows the polar residues being on the exposed surface and keeps the non-polar interactions internal). A molecule of hemoglobin is known to transport up to four oxygen molecules and here iron ion interacts with oxygen molecule to form oxyhemoglobin. Oxyhemoglobin blood is bright red and interactions between the iron–oxygen are very weak and thus can easily be separated without disturbing the heme unit/ the oxygen molecule (completely reversible binding). Deoxyhemoglobin is the hemoglobin molecule without oxygen and dark red.. Primary structure of hemoglobin 141 AA residues will be present in linear sequence of alpha chain contains and non- (, and ) chains will be of 146 amino acids in length (here the beta chain will have valine and histidine as their first residues and Tyr b145 and His b146 found at C-terminal residues). Only 10 residue difference between the delta chain and the beta chain. Secondary structure of hemoglobin We can find nearly 75 percent of the amino acids in or chains in a helical arrangement and 8 helical areas will be found in the chains. Tertiary structure of or chains Sphere type o structure will result during the tertiary folding of each globin chain and this folding brings the Polar or charged side chains directed towards the outer surface of the subunit and non-polar structures directed inwards making Hb water soluble, Heme pocket will be created and is open-toped cleft in the surface and this folding will bring Hb in correct orientation to allow these bonds to form. Quaternary structure of hemoglobin Finally the Hb tetramer will be formed composed of two identical dimers ()1and ()2. These two polypeptide chains are held together tightly (though hydrophobic and ionic interactions andy hydrogen bonding). The two dimers can move with respect to each other. T and R forms of Hb T form (taut structure.
Structure of HemoglobinHemoglobin is a chromo protein and is found.pdf
Structure of HemoglobinHemoglobin is a chromo protein and is found.pdf
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Solution : Selection Sort after two iterations: 1 14 8 9 5 16 2 1 2 14 9 8 16 5 program: class Sorting { public static void main(String args[]) { int a[]=new int[]{8,14,2,9,5,16,1}; int i,j,k; for(i=0;i<2;i++) { for(j=i+1;j<7;j++) { if(a[i]>a[j]) { int temp=a[i]; a[i]=a[j]; a[j]=temp; } } for(k=0;k<7;k++) { System.out.print(a[k]+\" \"); } System.out.println(); } } } INsertion Sort: after 2 iterations 8 14 2 9 5 16 1 2 8 14 9 5 16 1 class Sorting { public static void main(String args[]) { int a[]=new int[]{14,8,2,9,5,16,1}; int i,j,k; for (j = 1; j < 7; j++) { int temp = a[j]; i = j-1; while ( (i > -1) && ( a [i] > temp ) ) { a [i+1] = a [i]; i--; } a[i+1] = temp; for(k=0;k<7;k++) { System.out.print(a[k]+\" \"); } System.out.println(); } } } Bubble Sort: after 1 Iteration: 8 2 9 5 14 1 16 class Sorting { public static void main(String args[]) { int a[]=new int[]{14,8,2,9,5,16,1}; int i,j,k,temp; for(i=0; i < 7; i++){ for(j=1; j < (7-i); j++){ if(a[j-1] > a[j]){ //swap the elements! temp = a[j-1]; a[j-1] = a[j]; a[j] = temp; } } for(k=0;k<7;k++) { System.out.print(a[k]+\" \"); } System.out.println(); } } }.
SolutionSelection Sort after two iterations1 14 8 9 5 16 2 1.pdf
SolutionSelection Sort after two iterations1 14 8 9 5 16 2 1.pdf
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Starch agar consists of starch. The secretion of amylase enzyme by some microbes cause hydrolysis of starch around the colony. The plate flooded with iodine solution after incubation period. The amylase prodding colonies appear unstained and form a color zone because they’ve hydrolyzed starch. The colonies that don’t produce amylase, still have starch in their surrounding and appear dark-blue or black (color of starch-iodine complex) by reaction of starch with iodine. There is no clearance zone around the amylase-negative colonies. Thus, the media differentiates amylase producers from non-amylase produces and can be treated as a differential media. Solution Starch agar consists of starch. The secretion of amylase enzyme by some microbes cause hydrolysis of starch around the colony. The plate flooded with iodine solution after incubation period. The amylase prodding colonies appear unstained and form a color zone because they’ve hydrolyzed starch. The colonies that don’t produce amylase, still have starch in their surrounding and appear dark-blue or black (color of starch-iodine complex) by reaction of starch with iodine. There is no clearance zone around the amylase-negative colonies. Thus, the media differentiates amylase producers from non-amylase produces and can be treated as a differential media..
Starch agar consists of starch. The secretion of amylase enzyme by s.pdf
Starch agar consists of starch. The secretion of amylase enzyme by s.pdf
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Solution : find the solution below and please give the feedback and please let me know if you have any query regarding this question. Time (hrs) Intensity(in/hr) Intensity (cm/hr) F value Infiltration value(f) 0.25 1.2 3.048 1.137 2.503671944 0.5 2.4 6.096 1.6759 1.907607852 0.75 4.8 12.192 2.1167 1.645712666 1 3.6 9.144 2.5073 1.490595461 1.25 1.4 3.556 2.866 1.385389044 1.5 0.6 1.524 3.2022 1.308180314 Total rainfall 3.2 8.128 2.56028932 (Total Infiltration) Type of soil silty loam Features: Porosity 0.501 residual porosity 0.015 effective porosity 0.486 suction head(cm) 16.68 Conductivity(k) (cm/hr) 0.65 Degree os saturation 0.6 delta =(effective porosity(1-degree of saturation)) 0.1944 si 3.242592 F (F-si*LN((1+(F/si))-k*t)=0 Time (hrs) Intensity(in/hr) Intensity (cm/hr) F value Infiltration value(f) 0.25 1.2 3.048 1.137 2.503671944 0.5 2.4 6.096 1.6759 1.907607852 0.75 4.8 12.192 2.1167 1.645712666 1 3.6 9.144 2.5073 1.490595461 1.25 1.4 3.556 2.866 1.385389044 1.5 0.6 1.524 3.2022 1.308180314 Total rainfall 3.2 8.128 2.56028932 (Total Infiltration) Type of soil silty loam Features: Porosity 0.501 residual porosity 0.015 effective porosity 0.486 suction head(cm) 16.68 Conductivity(k) (cm/hr) 0.65 Degree os saturation 0.6 delta =(effective porosity(1-degree of saturation)) 0.1944 si 3.242592 F (F-si*LN((1+(F/si))-k*t)=0.
Solution find the solution below and please give the feedback and p.pdf
Solution find the solution below and please give the feedback and p.pdf
annaimobiles
C BOTH of them Solution C BOTH of them.
C BOTH of them .pdf
C BOTH of them .pdf
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public class DoubleArraySeq implements Cloneable { // Private Instance Variables private double[] data; private int manyItems; private int currentIndex; //Constructor Methods /** * Initialize an empty sequence with an initial capacity of 10. **/ public DoubleArraySeq() { try{ //Set a default capacity for new DoubleArraySeq\'s. int INITAL_CAPACITY = 10; //Set each instance variable to its initial value. data = new double[INITAL_CAPACITY]; manyItems = 0; currentIndex = 0; }//end try catch (OutOfMemoryError e){ throw new OutOfMemoryError (\"There is not enough memory to create a new sequence!\"); }//end catch }//end DoubleArraySeq() method /** * Initialize an empty sequence with a specified initial capacity. Note that the addAfter and addBefore methods work * efficiently (without needing more memory) until this capacity is reached. **/ public DoubleArraySeq(int initialCapacity) { try{ //Set each instance variable to its initial value. data = new double[initialCapacity]; currentIndex = 0; manyItems = 0; }//end try catch (OutOfMemoryError e){ throw new OutOfMemoryError (\"There is not enough memory to create a new sequence of capacity \" + initialCapacity + \"!\"); }//end catch }//end DoubleArraySeq(int initialCapacity) method // Accessor Methods /** **/ public boolean isCurrent() { return (currentIndex < manyItems); }//end isCurrent() method /** * Accessor method to get the current element of this sequence. **/ public double getCurrent() { //Confirm that there is a current element first. if (isCurrent()) return data[currentIndex]; else throw new IllegalStateException(\"There is no current element! Please specify a current element first.\"); }//end getCurrent() method /** * Accessor method to get the current capacity of this sequence. **/ public int getCapacity() { //Returns the number of indexes in the array. return data.length; }//end getCapacity() method /** * Accessor method to get the available capacity (number of empty indexes) of this sequence. * The available capacity (number of empty indexes) of this sequence. **/ public int getAvailCapacity() { //Returns the number of empty indexes in the array. return data.length - manyItems; }//end getAvailCapacity() method /** **/ public int size() { //Returns the number of elements in the sequence. return manyItems; }//end size() method // Setter Methods /** * A method to move forward, so the current element is now the next element in this sequence. **/ public void advance() { if (isCurrent()) currentIndex++; else throw new IllegalStateException (\"There is no current element! Advance may not be called.\"); }//end advance() method /** * A method to set the current element at the front of this sequence. **/ public void start() { if (manyItems > 0) currentIndex = 0; else throw new IllegalStateException(\"This sequence is empty!\"); }//end start() method /** * A method that makes the last element of the sequence the current element. **/ public void setCurrentLast() { if (manyItems > 0) currentIndex = ma.
public class DoubleArraySeq implements Cloneable { Priva.pdf
public class DoubleArraySeq implements Cloneable { Priva.pdf
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plese post the question clearly not able to understand Solution plese post the question clearly not able to understand.
plese post the question clearly not able to understandSolution.pdf
plese post the question clearly not able to understandSolution.pdf
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Lymphatic system function is to produce T- helper cells and B- helper cells,it is majorly concerned about productions of Antibodies that fights aganish infections.T he B& T cells produce memory cells which can recognise the germ if it affects second time .It plays a important role in immune response that protects body from infections Both cardiovascular and lymphatic systems are considered about transport of fluids entire body The difference is cardiovascular system considered about transporting materials like oxygen,hormones,salts etc....whereas lymphatic system considered about removal of wastes from the body fluids Solution Lymphatic system function is to produce T- helper cells and B- helper cells,it is majorly concerned about productions of Antibodies that fights aganish infections.T he B& T cells produce memory cells which can recognise the germ if it affects second time .It plays a important role in immune response that protects body from infections Both cardiovascular and lymphatic systems are considered about transport of fluids entire body The difference is cardiovascular system considered about transporting materials like oxygen,hormones,salts etc....whereas lymphatic system considered about removal of wastes from the body fluids.
Lymphatic system function is to produce T- helper cells and B- helpe.pdf
Lymphatic system function is to produce T- helper cells and B- helpe.pdf
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b) 0.2 M HI HI is a strong acid, which can completely ionized. LiOH is a base (ionized as Li+ and OH-) CH3COH is a very weak acid, which is only barely ionized. Solution b) 0.2 M HI HI is a strong acid, which can completely ionized. LiOH is a base (ionized as Li+ and OH-) CH3COH is a very weak acid, which is only barely ionized..
b) 0.2 M HI HI is a strong acid, which can comple.pdf
b) 0.2 M HI HI is a strong acid, which can comple.pdf
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Integer version of the member functions are as below int operate(int &a,int &b); int myOps is also integer member function that overrides the operate function. a and b are called as member variables Solution Integer version of the member functions are as below int operate(int &a,int &b); int myOps is also integer member function that overrides the operate function. a and b are called as member variables.
Integer version of the member functions are as belowint operate(in.pdf
Integer version of the member functions are as belowint operate(in.pdf
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I have done it under neatbeans IDE and just added ToggleGroup for grouping radion buttons inorder to select one at time. package tempconverter; import javafx.application.Application; import javafx.beans.value.ChangeListener; import javafx.beans.value.ObservableValue; import javafx.event.ActionEvent; import javafx.event.EventHandler; import javafx.geometry.Pos; import javafx.scene.Scene; import javafx.scene.control.Button; import javafx.scene.control.Label; import javafx.scene.control.RadioButton; import javafx.scene.control.ScrollPane; import javafx.scene.control.TextArea; import javafx.scene.control.TextField; import javafx.scene.control.Toggle; import javafx.scene.control.ToggleGroup; import javafx.scene.layout.BorderPane; import javafx.scene.layout.StackPane; import javafx.scene.layout.VBox; import javafx.stage.Stage; public class Tempconverter extends Application { /** * @param args the command line arguments executes project */ final ToggleGroup group1 = new ToggleGroup(); final ToggleGroup group2 = new ToggleGroup(); public static void main(String[] args) { launch(args); } @Override public void start(Stage primaryStage) throws Exception { BorderPane pane = new BorderPane(); //the big picture BorderPane p1 = new BorderPane();//top BorderPane p2 = new BorderPane(); //bottom Label temp = new Label(\"Enter a temperature\"); //enter temp label Label convertTemp = new Label(\"Converted temperature\"); //converted temp label Label input = new Label(\"Input Scale\");//input scale label Label output = new Label(\"Output Scale\");//Output Scale label TextField tempText = new TextField(); //text field to enter temp TextField convertText = new TextField(); //output textfield for the converted temp VBox scaleIn = new VBox(15); VBox scaleOut = new VBox(15); RadioButton c = new RadioButton(\"celcius\"); //button for celcius on input side c.setToggleGroup(group1); RadioButton f = new RadioButton(\"Fahrenheit\");//button for Fahrenheit on input side f.setToggleGroup(group1); RadioButton k = new RadioButton(\"kelvin\");//button for kelvin on inputside k.setToggleGroup(group1); RadioButton c1 = new RadioButton(\"celcius\");//button for celcius on outputside c1.setToggleGroup(group2); RadioButton f1 = new RadioButton(\"Fahrenheit\"); //button for f on output side f1.setToggleGroup(group2); RadioButton k1 = new RadioButton(\"kelvin\"); //button for kelvin on output side k1.setToggleGroup(group2); p1.setStyle(\"-fx-border-color:red\"); //set top border pane to red p1.setRight(tempText); p1.setLeft(temp); pane.setTop(p1); pane.setBottom(p2); p2.setLeft(convertTemp); p2.setRight(convertText); p2.setStyle(\"-fx-border-color:red\"); input.setPrefWidth(100); output.setPrefWidth(100); c.setPrefWidth(100); f.setPrefWidth(100); k.setPrefWidth(100); c1.setPrefWidth(100); f1.setPrefWidth(100); k1.setPrefWidth(100); scaleIn.setStyle(\"-fx-border-color:red\"); scaleIn.getChildren().addAll(input, c, f, k); scaleIn.setAlignment(Pos.CENTER); pane.setLeft(scaleIn); TextArea.
I have done it under neatbeans IDE and just added ToggleGroup for gr.pdf
I have done it under neatbeans IDE and just added ToggleGroup for gr.pdf
annaimobiles
Fossil record is complete for the following reasons: Solution Fossil record is complete for the following reasons:.
Fossil record is complete for the following reasonsSolutionFo.pdf
Fossil record is complete for the following reasonsSolutionFo.pdf
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operating systems , ch-05, (CPU Scheduling), 3rd level, College of Computers, Seiyun University. انظمة التشغيل لطلاب المستوى الثالث بكلية الحاسبات بجامعة سيئون المحاضرة 05
OS-operating systems- ch05 (CPU Scheduling) ...
OS-operating systems- ch05 (CPU Scheduling) ...
Dr. Mazin Mohamed alkathiri
Program code examples (known also as worked examples) play a crucial role in learning how to program. Instructors use examples extensively to demonstrate the semantics of the programming language being taught and to highlight the fundamental coding patterns. Programming textbooks allocate considerable space to present and explain code examples. To make the process of studying code examples more interactive, CS education researchers developed a range of tools to engage students in the study of code examples. These tools include codecasts (codemotion,codecast,elicasts), interactive example explorers (WebEx, PCEX), and tutoring systems (DeepTutor). An important component in all types of worked examples is code explanations associated with specific code lines or code chunks of an example. The explanations connect examples with general programming knowledge explaining the role and function of code fragments or their behavior. In textbooks, these explanations are usually presented as comments in the code or as explanations on the margins. The example explorer tools allow students to examine these explanations interactively. Tutoring systems, which engage students in explaining the code, use these model explanations to check student responses and provide scaffolding. In all these cases, to make a worked example re-usable beyond its presentation in a lecture, the explanations have to be authored by instructors or domain experts i.e., produced and integrated into a specific system. As the experience of the last 10 years demonstrated, these explanations are hard to obtain. Those already collected are usually “locked” in a specific example-focused system and can’t be reused. The purpose of this working group is to support broader re-used of worked examples augmented with explanations. Our current plan is to develop а standard approach to represent explained examples. This approach will enable an example created for any of the existing systems to be explored in a standard format and imported into any other example-focused system. We plan to follow a successful experience of the PEML working group focused on re-using programming exercises.
SPLICE Working Group:Reusable Code Examples
SPLICE Working Group:Reusable Code Examples
Peter Brusilovsky
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HCl is a polar-covalent compound and also Acetonitrile they are soluble dueto hydrogen - bonding between them. So they have London-dispersion forces between them. Solution HCl is a polar-covalent compound and also Acetonitrile they are soluble dueto hydrogen - bonding between them. So they have London-dispersion forces between them..
HCl is a polar-covalent compound and also Acetoni.pdf
HCl is a polar-covalent compound and also Acetoni.pdf
annaimobiles
for a compound to be soluble in water 1. it should have a large dipolemoment or 2. it should be ionic or 3. it should form H bond with water in first pair CH3CH2CH3 is non polar and is not soluble but CH2OHCHOHCH2OH is soluble in water by making H bonds. in second pair Though CH3Cl has little dipolemoment HCl has very higher dipolemomet because of charge separation and is readily soluble in water Solution for a compound to be soluble in water 1. it should have a large dipolemoment or 2. it should be ionic or 3. it should form H bond with water in first pair CH3CH2CH3 is non polar and is not soluble but CH2OHCHOHCH2OH is soluble in water by making H bonds. in second pair Though CH3Cl has little dipolemoment HCl has very higher dipolemomet because of charge separation and is readily soluble in water.
for a compound to be soluble in water 1. it shou.pdf
for a compound to be soluble in water 1. it shou.pdf
annaimobiles
e) II and III have the loest, equal stability Solution e) II and III have the loest, equal stability.
e) II and III have the loest, equal stability .pdf
e) II and III have the loest, equal stability .pdf
annaimobiles
wait doing Solution wait doing.
wait doingSolutionwait doing.pdf
wait doingSolutionwait doing.pdf
annaimobiles
TRUE The Reynolds number of falling 2.5 micrometer flyash particles in air exceed that of same size of soot particle by more than 50% because the soot particle is a outcome produced due to in complete combustion of carbon particles and very lighter in density compared to the flyash. Solution TRUE The Reynolds number of falling 2.5 micrometer flyash particles in air exceed that of same size of soot particle by more than 50% because the soot particle is a outcome produced due to in complete combustion of carbon particles and very lighter in density compared to the flyash..
TRUE The Reynolds number of falling 2.5 micrometer flyash particles .pdf
TRUE The Reynolds number of falling 2.5 micrometer flyash particles .pdf
annaimobiles
the electron rich molecule is a central Xe with 4 F singlebonded to it, 1 lone pair, and 1 oxygen molecule double bonded toit (if this is confusing the molecule is called xenonoxytetrafluoride so you can google it or whatever) XeF6+H2O XeOF4 +2HF Solution the electron rich molecule is a central Xe with 4 F singlebonded to it, 1 lone pair, and 1 oxygen molecule double bonded toit (if this is confusing the molecule is called xenonoxytetrafluoride so you can google it or whatever) XeF6+H2O XeOF4 +2HF.
the electron rich molecule is a central Xe with 4 F singlebonded.pdf
the electron rich molecule is a central Xe with 4 F singlebonded.pdf
annaimobiles
The given formulae themselves are explicit. Explicit means direct i.e., the ones having a clear expression of one variable in terms of several others which is the case here, as an is expressed as a direct function of n and several lower order terms. By direct or clear we mean that expression which represents, the variable (on the LHS of the equation) that is expressed, as a function of other variables does not involve the expressed variable (in the RHS of the equation). Solution The given formulae themselves are explicit. Explicit means direct i.e., the ones having a clear expression of one variable in terms of several others which is the case here, as an is expressed as a direct function of n and several lower order terms. By direct or clear we mean that expression which represents, the variable (on the LHS of the equation) that is expressed, as a function of other variables does not involve the expressed variable (in the RHS of the equation)..
The given formulae themselves are explicit. Explicit means direct i..pdf
The given formulae themselves are explicit. Explicit means direct i..pdf
annaimobiles
The answer is Meiosis, It occurs in human body at reporductive organs only. For healing and growth Mitosis plays a crucial role. Solution The answer is Meiosis, It occurs in human body at reporductive organs only. For healing and growth Mitosis plays a crucial role..
The answer is Meiosis, It occurs in human body at reporductive organ.pdf
The answer is Meiosis, It occurs in human body at reporductive organ.pdf
annaimobiles
Structure of Hemoglobin Hemoglobin is a chromo protein and is found in red blood cells, it’s a conjugated protein (heme as prosthetic group + globin as the protein part apoprotein). Adults have 14.0 to 16.0 gm% of Hb and 90 mg/kg of Hb will be produced and destroyed in the body on daily basis. Molecular weight of Hb is 67,000 and 3.4 mg of iron present in each gram of Hb. The combination of iron with a porphyrin ring produces the heme. Structure of Heme Heme is derived from porphyrin and porphyrins are cyclic compounds and are formed when 4 pyrrole rings fuse and are linked by methenyl bridges and the four rings are named as I,II,III, IV and Alpha, beta, gamma and delta are the bridges. To the side chain of Porphyrins, four pyrrole rings are attached. We can find one ferrous atom (Fe++) co-ordinated at the centre of the of protoporphyrin IX tetra pyrrole ring. Structure of Globin We can find the tetramer of globin polypeptide chains and each Hb molecule will have 4 Heme units and the subunits of hemoglobin are found to be arranged in a tetrahedral array. And this arrangement will give tight spherical overall appearance (which allows the polar residues being on the exposed surface and keeps the non-polar interactions internal). A molecule of hemoglobin is known to transport up to four oxygen molecules and here iron ion interacts with oxygen molecule to form oxyhemoglobin. Oxyhemoglobin blood is bright red and interactions between the iron–oxygen are very weak and thus can easily be separated without disturbing the heme unit/ the oxygen molecule (completely reversible binding). Deoxyhemoglobin is the hemoglobin molecule without oxygen and dark red.. Primary structure of hemoglobin 141 AA residues will be present in linear sequence of alpha chain contains and non- (, and ) chains will be of 146 amino acids in length (here the beta chain will have valine and histidine as their first residues and Tyr b145 and His b146 found at C-terminal residues). Only 10 residue difference between the delta chain and the beta chain. Secondary structure of hemoglobin We can find nearly 75 percent of the amino acids in or chains in a helical arrangement and 8 helical areas will be found in the chains. Tertiary structure of or chains Sphere type o structure will result during the tertiary folding of each globin chain and this folding brings the Polar or charged side chains directed towards the outer surface of the subunit and non-polar structures directed inwards making Hb water soluble, Heme pocket will be created and is open-toped cleft in the surface and this folding will bring Hb in correct orientation to allow these bonds to form. Quaternary structure of hemoglobin Finally the Hb tetramer will be formed composed of two identical dimers ()1and ()2. These two polypeptide chains are held together tightly (though hydrophobic and ionic interactions andy hydrogen bonding). The two dimers can move with respect to each other. T and R forms of Hb T form (taut structure.
Structure of HemoglobinHemoglobin is a chromo protein and is found.pdf
Structure of HemoglobinHemoglobin is a chromo protein and is found.pdf
annaimobiles
Solution : Selection Sort after two iterations: 1 14 8 9 5 16 2 1 2 14 9 8 16 5 program: class Sorting { public static void main(String args[]) { int a[]=new int[]{8,14,2,9,5,16,1}; int i,j,k; for(i=0;i<2;i++) { for(j=i+1;j<7;j++) { if(a[i]>a[j]) { int temp=a[i]; a[i]=a[j]; a[j]=temp; } } for(k=0;k<7;k++) { System.out.print(a[k]+\" \"); } System.out.println(); } } } INsertion Sort: after 2 iterations 8 14 2 9 5 16 1 2 8 14 9 5 16 1 class Sorting { public static void main(String args[]) { int a[]=new int[]{14,8,2,9,5,16,1}; int i,j,k; for (j = 1; j < 7; j++) { int temp = a[j]; i = j-1; while ( (i > -1) && ( a [i] > temp ) ) { a [i+1] = a [i]; i--; } a[i+1] = temp; for(k=0;k<7;k++) { System.out.print(a[k]+\" \"); } System.out.println(); } } } Bubble Sort: after 1 Iteration: 8 2 9 5 14 1 16 class Sorting { public static void main(String args[]) { int a[]=new int[]{14,8,2,9,5,16,1}; int i,j,k,temp; for(i=0; i < 7; i++){ for(j=1; j < (7-i); j++){ if(a[j-1] > a[j]){ //swap the elements! temp = a[j-1]; a[j-1] = a[j]; a[j] = temp; } } for(k=0;k<7;k++) { System.out.print(a[k]+\" \"); } System.out.println(); } } }.
SolutionSelection Sort after two iterations1 14 8 9 5 16 2 1.pdf
SolutionSelection Sort after two iterations1 14 8 9 5 16 2 1.pdf
annaimobiles
Starch agar consists of starch. The secretion of amylase enzyme by some microbes cause hydrolysis of starch around the colony. The plate flooded with iodine solution after incubation period. The amylase prodding colonies appear unstained and form a color zone because they’ve hydrolyzed starch. The colonies that don’t produce amylase, still have starch in their surrounding and appear dark-blue or black (color of starch-iodine complex) by reaction of starch with iodine. There is no clearance zone around the amylase-negative colonies. Thus, the media differentiates amylase producers from non-amylase produces and can be treated as a differential media. Solution Starch agar consists of starch. The secretion of amylase enzyme by some microbes cause hydrolysis of starch around the colony. The plate flooded with iodine solution after incubation period. The amylase prodding colonies appear unstained and form a color zone because they’ve hydrolyzed starch. The colonies that don’t produce amylase, still have starch in their surrounding and appear dark-blue or black (color of starch-iodine complex) by reaction of starch with iodine. There is no clearance zone around the amylase-negative colonies. Thus, the media differentiates amylase producers from non-amylase produces and can be treated as a differential media..
Starch agar consists of starch. The secretion of amylase enzyme by s.pdf
Starch agar consists of starch. The secretion of amylase enzyme by s.pdf
annaimobiles
Solution : find the solution below and please give the feedback and please let me know if you have any query regarding this question. Time (hrs) Intensity(in/hr) Intensity (cm/hr) F value Infiltration value(f) 0.25 1.2 3.048 1.137 2.503671944 0.5 2.4 6.096 1.6759 1.907607852 0.75 4.8 12.192 2.1167 1.645712666 1 3.6 9.144 2.5073 1.490595461 1.25 1.4 3.556 2.866 1.385389044 1.5 0.6 1.524 3.2022 1.308180314 Total rainfall 3.2 8.128 2.56028932 (Total Infiltration) Type of soil silty loam Features: Porosity 0.501 residual porosity 0.015 effective porosity 0.486 suction head(cm) 16.68 Conductivity(k) (cm/hr) 0.65 Degree os saturation 0.6 delta =(effective porosity(1-degree of saturation)) 0.1944 si 3.242592 F (F-si*LN((1+(F/si))-k*t)=0 Time (hrs) Intensity(in/hr) Intensity (cm/hr) F value Infiltration value(f) 0.25 1.2 3.048 1.137 2.503671944 0.5 2.4 6.096 1.6759 1.907607852 0.75 4.8 12.192 2.1167 1.645712666 1 3.6 9.144 2.5073 1.490595461 1.25 1.4 3.556 2.866 1.385389044 1.5 0.6 1.524 3.2022 1.308180314 Total rainfall 3.2 8.128 2.56028932 (Total Infiltration) Type of soil silty loam Features: Porosity 0.501 residual porosity 0.015 effective porosity 0.486 suction head(cm) 16.68 Conductivity(k) (cm/hr) 0.65 Degree os saturation 0.6 delta =(effective porosity(1-degree of saturation)) 0.1944 si 3.242592 F (F-si*LN((1+(F/si))-k*t)=0.
Solution find the solution below and please give the feedback and p.pdf
Solution find the solution below and please give the feedback and p.pdf
annaimobiles
C BOTH of them Solution C BOTH of them.
C BOTH of them .pdf
C BOTH of them .pdf
annaimobiles
public class DoubleArraySeq implements Cloneable { // Private Instance Variables private double[] data; private int manyItems; private int currentIndex; //Constructor Methods /** * Initialize an empty sequence with an initial capacity of 10. **/ public DoubleArraySeq() { try{ //Set a default capacity for new DoubleArraySeq\'s. int INITAL_CAPACITY = 10; //Set each instance variable to its initial value. data = new double[INITAL_CAPACITY]; manyItems = 0; currentIndex = 0; }//end try catch (OutOfMemoryError e){ throw new OutOfMemoryError (\"There is not enough memory to create a new sequence!\"); }//end catch }//end DoubleArraySeq() method /** * Initialize an empty sequence with a specified initial capacity. Note that the addAfter and addBefore methods work * efficiently (without needing more memory) until this capacity is reached. **/ public DoubleArraySeq(int initialCapacity) { try{ //Set each instance variable to its initial value. data = new double[initialCapacity]; currentIndex = 0; manyItems = 0; }//end try catch (OutOfMemoryError e){ throw new OutOfMemoryError (\"There is not enough memory to create a new sequence of capacity \" + initialCapacity + \"!\"); }//end catch }//end DoubleArraySeq(int initialCapacity) method // Accessor Methods /** **/ public boolean isCurrent() { return (currentIndex < manyItems); }//end isCurrent() method /** * Accessor method to get the current element of this sequence. **/ public double getCurrent() { //Confirm that there is a current element first. if (isCurrent()) return data[currentIndex]; else throw new IllegalStateException(\"There is no current element! Please specify a current element first.\"); }//end getCurrent() method /** * Accessor method to get the current capacity of this sequence. **/ public int getCapacity() { //Returns the number of indexes in the array. return data.length; }//end getCapacity() method /** * Accessor method to get the available capacity (number of empty indexes) of this sequence. * The available capacity (number of empty indexes) of this sequence. **/ public int getAvailCapacity() { //Returns the number of empty indexes in the array. return data.length - manyItems; }//end getAvailCapacity() method /** **/ public int size() { //Returns the number of elements in the sequence. return manyItems; }//end size() method // Setter Methods /** * A method to move forward, so the current element is now the next element in this sequence. **/ public void advance() { if (isCurrent()) currentIndex++; else throw new IllegalStateException (\"There is no current element! Advance may not be called.\"); }//end advance() method /** * A method to set the current element at the front of this sequence. **/ public void start() { if (manyItems > 0) currentIndex = 0; else throw new IllegalStateException(\"This sequence is empty!\"); }//end start() method /** * A method that makes the last element of the sequence the current element. **/ public void setCurrentLast() { if (manyItems > 0) currentIndex = ma.
public class DoubleArraySeq implements Cloneable { Priva.pdf
public class DoubleArraySeq implements Cloneable { Priva.pdf
annaimobiles
plese post the question clearly not able to understand Solution plese post the question clearly not able to understand.
plese post the question clearly not able to understandSolution.pdf
plese post the question clearly not able to understandSolution.pdf
annaimobiles
Lymphatic system function is to produce T- helper cells and B- helper cells,it is majorly concerned about productions of Antibodies that fights aganish infections.T he B& T cells produce memory cells which can recognise the germ if it affects second time .It plays a important role in immune response that protects body from infections Both cardiovascular and lymphatic systems are considered about transport of fluids entire body The difference is cardiovascular system considered about transporting materials like oxygen,hormones,salts etc....whereas lymphatic system considered about removal of wastes from the body fluids Solution Lymphatic system function is to produce T- helper cells and B- helper cells,it is majorly concerned about productions of Antibodies that fights aganish infections.T he B& T cells produce memory cells which can recognise the germ if it affects second time .It plays a important role in immune response that protects body from infections Both cardiovascular and lymphatic systems are considered about transport of fluids entire body The difference is cardiovascular system considered about transporting materials like oxygen,hormones,salts etc....whereas lymphatic system considered about removal of wastes from the body fluids.
Lymphatic system function is to produce T- helper cells and B- helpe.pdf
Lymphatic system function is to produce T- helper cells and B- helpe.pdf
annaimobiles
b) 0.2 M HI HI is a strong acid, which can completely ionized. LiOH is a base (ionized as Li+ and OH-) CH3COH is a very weak acid, which is only barely ionized. Solution b) 0.2 M HI HI is a strong acid, which can completely ionized. LiOH is a base (ionized as Li+ and OH-) CH3COH is a very weak acid, which is only barely ionized..
b) 0.2 M HI HI is a strong acid, which can comple.pdf
b) 0.2 M HI HI is a strong acid, which can comple.pdf
annaimobiles
Integer version of the member functions are as below int operate(int &a,int &b); int myOps is also integer member function that overrides the operate function. a and b are called as member variables Solution Integer version of the member functions are as below int operate(int &a,int &b); int myOps is also integer member function that overrides the operate function. a and b are called as member variables.
Integer version of the member functions are as belowint operate(in.pdf
Integer version of the member functions are as belowint operate(in.pdf
annaimobiles
I have done it under neatbeans IDE and just added ToggleGroup for grouping radion buttons inorder to select one at time. package tempconverter; import javafx.application.Application; import javafx.beans.value.ChangeListener; import javafx.beans.value.ObservableValue; import javafx.event.ActionEvent; import javafx.event.EventHandler; import javafx.geometry.Pos; import javafx.scene.Scene; import javafx.scene.control.Button; import javafx.scene.control.Label; import javafx.scene.control.RadioButton; import javafx.scene.control.ScrollPane; import javafx.scene.control.TextArea; import javafx.scene.control.TextField; import javafx.scene.control.Toggle; import javafx.scene.control.ToggleGroup; import javafx.scene.layout.BorderPane; import javafx.scene.layout.StackPane; import javafx.scene.layout.VBox; import javafx.stage.Stage; public class Tempconverter extends Application { /** * @param args the command line arguments executes project */ final ToggleGroup group1 = new ToggleGroup(); final ToggleGroup group2 = new ToggleGroup(); public static void main(String[] args) { launch(args); } @Override public void start(Stage primaryStage) throws Exception { BorderPane pane = new BorderPane(); //the big picture BorderPane p1 = new BorderPane();//top BorderPane p2 = new BorderPane(); //bottom Label temp = new Label(\"Enter a temperature\"); //enter temp label Label convertTemp = new Label(\"Converted temperature\"); //converted temp label Label input = new Label(\"Input Scale\");//input scale label Label output = new Label(\"Output Scale\");//Output Scale label TextField tempText = new TextField(); //text field to enter temp TextField convertText = new TextField(); //output textfield for the converted temp VBox scaleIn = new VBox(15); VBox scaleOut = new VBox(15); RadioButton c = new RadioButton(\"celcius\"); //button for celcius on input side c.setToggleGroup(group1); RadioButton f = new RadioButton(\"Fahrenheit\");//button for Fahrenheit on input side f.setToggleGroup(group1); RadioButton k = new RadioButton(\"kelvin\");//button for kelvin on inputside k.setToggleGroup(group1); RadioButton c1 = new RadioButton(\"celcius\");//button for celcius on outputside c1.setToggleGroup(group2); RadioButton f1 = new RadioButton(\"Fahrenheit\"); //button for f on output side f1.setToggleGroup(group2); RadioButton k1 = new RadioButton(\"kelvin\"); //button for kelvin on output side k1.setToggleGroup(group2); p1.setStyle(\"-fx-border-color:red\"); //set top border pane to red p1.setRight(tempText); p1.setLeft(temp); pane.setTop(p1); pane.setBottom(p2); p2.setLeft(convertTemp); p2.setRight(convertText); p2.setStyle(\"-fx-border-color:red\"); input.setPrefWidth(100); output.setPrefWidth(100); c.setPrefWidth(100); f.setPrefWidth(100); k.setPrefWidth(100); c1.setPrefWidth(100); f1.setPrefWidth(100); k1.setPrefWidth(100); scaleIn.setStyle(\"-fx-border-color:red\"); scaleIn.getChildren().addAll(input, c, f, k); scaleIn.setAlignment(Pos.CENTER); pane.setLeft(scaleIn); TextArea.
I have done it under neatbeans IDE and just added ToggleGroup for gr.pdf
I have done it under neatbeans IDE and just added ToggleGroup for gr.pdf
annaimobiles
Fossil record is complete for the following reasons: Solution Fossil record is complete for the following reasons:.
Fossil record is complete for the following reasonsSolutionFo.pdf
Fossil record is complete for the following reasonsSolutionFo.pdf
annaimobiles
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plese post the question clearly not able to understandSolution.pdf
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