c. A destructor to free any allocated objects.d. A copy constructo.pdf

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c. A destructor to free any allocated objects. d. A copy constructor that allocates its own new member storage and copies the contents of member variables. e. An assignment operator that deallocates old storage before allocating new storage and copying all the member variables. Solution c. A destructor to free any allocated objects. d. A copy constructor that allocates its own new member storage and copies the contents of member variables. e. An assignment operator that deallocates old storage before allocating new storage and copying all the member variables..

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Use of the plates within a specific time is necessary as time of incubation is related to growth of the bacterial cells which grows with time initially and after a particular period starts declining. Moisture can increase the process of diffusion and also can spread bacterial cells into a zone which is known as zone of inhibition. Solution Use of the plates within a specific time is necessary as time of incubation is related to growth of the bacterial cells which grows with time initially and after a particular period starts declining. Moisture can increase the process of diffusion and also can spread bacterial cells into a zone which is known as zone of inhibition..

The five terrestrial adaptations of the seed plants include, reducti.pdf

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The five terrestrial adaptations of the seed plants include, reduction of the gametophyte generation, heterospory, ovules, and pollen. Pollen grains were an important adaptation because the evolution of pollen allowed for pollination and contributed to the diversity of seed plants. Pollen grains are male gametophytes and are carried by wind, water, or a pollinator. The whole structure is protected from desiccation and can reach the female organs without dependence on water. Male gametes reach female gametophyte and the egg cell gamete though a pollen tube: an extension of a cell within the pollen grain. Seeds can survive harsh conditions through dormancy, are distributed far from their parent sporophyte, and are multicellular. Seeds contain a diploid embryo that will germinate into a sporophyte. Storage tissue to sustain growth and a protective coat give seeds their superior evolutionary advantage. Several layers of hardened tissue prevent desiccation, and free reproduction from the need for a constant supply of water. Furthermore, seeds remain in a state of dormancy—induced by desiccation and the hormone abscisic acid—until conditions for growth become favorable. Whether blown by the wind, floating on water, or carried away by animals, seeds are scattered in an expanding geographic range, thus avoiding competition with the parent plant. All these various reasons make seeded plants to be domonant in plant communities today. Solution The five terrestrial adaptations of the seed plants include, reduction of the gametophyte generation, heterospory, ovules, and pollen. Pollen grains were an important adaptation because the evolution of pollen allowed for pollination and contributed to the diversity of seed plants. Pollen grains are male gametophytes and are carried by wind, water, or a pollinator. The whole structure is protected from desiccation and can reach the female organs without dependence on water. Male gametes reach female gametophyte and the egg cell gamete though a pollen tube: an extension of a cell within the pollen grain. Seeds can survive harsh conditions through dormancy, are distributed far from their parent sporophyte, and are multicellular. Seeds contain a diploid embryo that will germinate into a sporophyte. Storage tissue to sustain growth and a protective coat give seeds their superior evolutionary advantage. Several layers of hardened tissue prevent desiccation, and free reproduction from the need for a constant supply of water. Furthermore, seeds remain in a state of dormancy—induced by desiccation and the hormone abscisic acid—until conditions for growth become favorable. Whether blown by the wind, floating on water, or carried away by animals, seeds are scattered in an expanding geographic range, thus avoiding competition with the parent plant. All these various reasons make seeded plants to be domonant in plant communities today..

SELF THEORY OF LATE ADULT HOOD it describes the changes of a adult.pdf

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SELF THEORY OF LATE ADULT HOOD : it describes the changes of a adult human being when they are getting aged.according to this theory 1.as the age of the particular person grows they will be more dependent on the others for their needs. 2.some feelings like guilt,shame and depression develop as a result of their dependency. 3.many of the people will be feeling lonely when their loving partner passes away which will have a negative impact on their health. 4.according to a theory stated by erikson people with a positive attitude will feel integrated where as people with negative attitude or thoughts will be desperate. STRATIFICATION THEORIES : These are related to the persons social categeory his social status and the individual choices.this stratification is going to limit in many ways.many theories have been stated they are as follows Disengagement theory : the person disengages himself or herself away from the society and become passive. Activity theory : states that elderly people have to remain active in various social activities like spending time with relatives,friends and various community groups and will be withdrawn only as a result of age. DYNAMIC THEORY : these show the change and readjustment rather than ongoing selfor the legacy of stratification.these dynamic theories states that each persons life will be chaning and is purely self propelled within specific special contexts which will be ever changing. Solution SELF THEORY OF LATE ADULT HOOD : it describes the changes of a adult human being when they are getting aged.according to this theory 1.as the age of the particular person grows they will be more dependent on the others for their needs. 2.some feelings like guilt,shame and depression develop as a result of their dependency. 3.many of the people will be feeling lonely when their loving partner passes away which will have a negative impact on their health. 4.according to a theory stated by erikson people with a positive attitude will feel integrated where as people with negative attitude or thoughts will be desperate. STRATIFICATION THEORIES : These are related to the persons social categeory his social status and the individual choices.this stratification is going to limit in many ways.many theories have been stated they are as follows Disengagement theory : the person disengages himself or herself away from the society and become passive. Activity theory : states that elderly people have to remain active in various social activities like spending time with relatives,friends and various community groups and will be withdrawn only as a result of age. DYNAMIC THEORY : these show the change and readjustment rather than ongoing selfor the legacy of stratification.these dynamic theories states that each persons life will be chaning and is purely self propelled within specific special contexts which will be ever changing..

PV = PMTiThe goal is to solve the formula for i, doing so by isol.pdf

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Normal gene regulation in MAT locus YeastThere are 3 regulatory p.pdf

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Normal gene regulation in MAT locus Yeast: There are 3 regulatory protein in yeast mating control namely a1, alpha 1 & alpha 2. These 2 and alpha cells will have a opposite mating control genes.Thay both expose an suitable opposite attracting pheromones. a cells activate genes - Produce a factor and cell surface receptors - which binds with the alpha factor and provokes the signalling process and there by repress the same..This happens the same viceversa from alpha to a cells. alpha 1 - protein activator. alpha 2 - protein repressor. a1 - has no effect on its own, but forms complex with alpha 2 proteins and turns off different sets of genes. a2 expressed in haploid \"a\" cells - it does affect the regulation part bacause, this a2 protein shows no growth, and also mating/sporulation defects. It lacks to provoke the signalling of a factors and cell surface receptors, thereby fails to activate or repress the MAT locus gene regulation. So it\'s presence does not initiate any alteration in the gene regulation of MAT locus Solution Normal gene regulation in MAT locus Yeast: There are 3 regulatory protein in yeast mating control namely a1, alpha 1 & alpha 2. These 2 and alpha cells will have a opposite mating control genes.Thay both expose an suitable opposite attracting pheromones. a cells activate genes - Produce a factor and cell surface receptors - which binds with the alpha factor and provokes the signalling process and there by repress the same..This happens the same viceversa from alpha to a cells. alpha 1 - protein activator. alpha 2 - protein repressor. a1 - has no effect on its own, but forms complex with alpha 2 proteins and turns off different sets of genes. a2 expressed in haploid \"a\" cells - it does affect the regulation part bacause, this a2 protein shows no growth, and also mating/sporulation defects. It lacks to provoke the signalling of a factors and cell surface receptors, thereby fails to activate or repress the MAT locus gene regulation. So it\'s presence does not initiate any alteration in the gene regulation of MAT locus.

My answer 23SolutionMy answer 23.pdf

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Ionic compounds have electropositive and electronegative elements pr.pdf

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Ionic compounds have electropositive and electronegative elements present and high melting points due to strong ionic bonding. Polar molecules have weak intermolecular dipole-dipole interaction and dispersion forces, and low melting points. Nonpolar molecules have weak intermolecular forces, and low melting points. LiF - ionic compound ionic bonding BeF2 - ionic compound ionic bonding BF3 - nonpolar molecule dispersion force CF4 - nonpolar molecule dispersion force NF3 - polar molecule dispersion force dipole-dipole interaction OF2 - polar molecule dispersion force dipole-dipole interaction F2 - nonpolar molecule dispersion force Solution Ionic compounds have electropositive and electronegative elements present and high melting points due to strong ionic bonding. Polar molecules have weak intermolecular dipole-dipole interaction and dispersion forces, and low melting points. Nonpolar molecules have weak intermolecular forces, and low melting points. LiF - ionic compound ionic bonding BeF2 - ionic compound ionic bonding BF3 - nonpolar molecule dispersion force CF4 - nonpolar molecule dispersion force NF3 - polar molecule dispersion force dipole-dipole interaction OF2 - polar molecule dispersion force dipole-dipole interaction F2 - nonpolar molecule dispersion force.

Grey matter in the spinal cord is known as the grey column which tra.pdf

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Grey matter in the spinal cord is known as the grey column which travels down the spinal cord and is distributed in three grey columns that are presented in an \"H\" shape. 1. Anterior grey column. It contains motor neurons. These synapse with interneurons and the axons of cells that have travelled down the pyramidal tract. These cells are responsible for the movement of muscles. 2. Posterior grey column . The posterior grey column contains the points where sensory neurons synapse. These receives sensory information from the body, including fine touch, proprioception, and vibration. This information is sent from receptors of the skin, bones, and joints through sensory neurons whose cell bodies lie in the dorsal root ganglion. This information is then transmitted in axons up the spinal cord in spinal tracts, including the dorsal column-medial lemniscus tract and the spinothalamic tract. 3. Lateral grey column. The lateral grey column is composed of sympathetic preganglionic visceral motor neurons which are part of the autonomic nervous system. It primarily involved with activity in the sympathetic division of the autonomic motor system. Solution Grey matter in the spinal cord is known as the grey column which travels down the spinal cord and is distributed in three grey columns that are presented in an \"H\" shape. 1. Anterior grey column. It contains motor neurons. These synapse with interneurons and the axons of cells that have travelled down the pyramidal tract. These cells are responsible for the movement of muscles. 2. Posterior grey column . The posterior grey column contains the points where sensory neurons synapse. These receives sensory information from the body, including fine touch, proprioception, and vibration. This information is sent from receptors of the skin, bones, and joints through sensory neurons whose cell bodies lie in the dorsal root ganglion. This information is then transmitted in axons up the spinal cord in spinal tracts, including the dorsal column-medial lemniscus tract and the spinothalamic tract. 3. Lateral grey column. The lateral grey column is composed of sympathetic preganglionic visceral motor neurons which are part of the autonomic nervous system. It primarily involved with activity in the sympathetic division of the autonomic motor system..

Date Description Debit Credit 2014 Jan. 4 Delivery Truck 28000 Cash 28000 Nov 2 Truck Repair Expense 635 Cash 635 Dec 31 Depreciation Expense-Delivery Truck 11680 Delivery truck 11680 Depreciation as per double declining method = 2*(cost – residual value)/ life of asset = 2*(28000 – 4640) / 4 = 11680 2015 Jan. 6 Delivery Truck 49400 Cash 49400 April 1 Depreciation Expense-Delivery Truck 1947 Delivery truck 1947 Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset * ( period used/12 month) = 2*[(28000 – 11680 )- 4640 / 3 * 3/ 12 = 1947 April 1 Cash 15180 Delivery truck 14373 Profit on sale of truck 807 Nov 2 Truck Repair Expense 470 Cash 470 Dec 31 Depreciation Expense-Delivery Truck 15972 Delivery truck 15972 Depreciation as per double declining method = 2*(cost – residual value)/ life of asset = 2*(49400-9470) / 5 = 15972 2016 July. 1 Delivery Truck 55400 Cash 55400 Oct 2 Depreciation Expense-Delivery Truck 8984 Delivery truck 8984 Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset * ( period used/12 month) = 2*[(49400- 15972 )-9470 / 4 * 9/ 12 = 8984 Oct 2 Cash 17498 Loss on sale of truck 6946 Delivery truck 24444 Dec 31 Depreciation Expense-Delivery Truck 10611 Delivery truck 10611 Depreciation as per double declining method = 2*(cost – residual value)/ life of asset = 2*(55400-12955) / 8 = 10611 Date Description Debit Credit 2014 Jan. 4 Delivery Truck 28000 Cash 28000 Nov 2 Truck Repair Expense 635 Cash 635 Dec 31 Depreciation Expense-Delivery Truck 11680 Delivery truck 11680 Depreciation as per double declining method = 2*(cost – residual value)/ life of asset = 2*(28000 – 4640) / 4 = 11680 2015 Jan. 6 Delivery Truck 49400 Cash 49400 April 1 Depreciation Expense-Delivery Truck 1947 Delivery truck 1947 Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset * ( period used/12 month) = 2*[(28000 – 11680 )- 4640 / 3 * 3/ 12 = 1947 April 1 Cash 15180 Delivery truck 14373 Profit on sale of truck 807 Nov 2 Truck Repair Expense 470 Cash 470 Dec 31 Depreciation Expense-Delivery Truck 15972 Delivery truck 15972 Depreciation as per double declining method = 2*(cost – residual value)/ life of asset = 2*(49400-9470) / 5 = 15972 2016 July. 1 Delivery Truck 55400 Cash 55400 Oct 2 Depreciation Expense-Delivery Truck 8984 Delivery truck 8984 Depreciation as per double declining method next year = 2*(WDV – residual value)/ life of asset * ( period used/12 month) = 2*[(49400- 15972 )-9470 / 4 * 9/ 12 = 8984 Oct 2 Cash 17498 Loss on sale of truck 6946 Delivery truck 24444 Dec 31 Depreciation Expense-Delivery Truck 10611 Delivery truck 10611 Depreciation as per double declining method = 2*(cost – residual value)/ life of asset = 2*(55400-12955) / 8 = 10611 Solution Date Description Debit Credit 2014 Jan. 4 Delivery Truck 28000 Cash 28000 Nov 2 Truck Repair Expense 635 Cash 635 Dec 31 Depreciation Expense-Delivery Truck 11680 Delive.

Barrier to entryThe government intervention to the barriers to en.pdf

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Barrier to entry: The government intervention to the barriers to entry to exist as a result of, while others occur naturally within the business world. Often, existing firms within an industry lobby for the government to erect new barriers to entry. In this is done to done by the protect the integrity of the industry and prevent fly-by-night operations from setting up shop and hawking inferior products and services. In reality, firms favor barriers to entry when already comfortably ensconced in an industry to limit competition and claim a larger share of the industry\'s revenue. Other barriers to entry occur naturally, often evolving over time as certain industry players establish dominance. A company can use this technology, for example, to build a barrier to entry, to build in switching costs, and even, sometimes, to completely change the basis of competition. some companies have seized the advantage, while others, more complacent, have ended up playing the difficult and expensive game of catch-up ball. He also points out that it is important for executives to make this competitive analysis in assessing where IS fits in their companies, since in some cases it appropriately plays a support role and can add only modestly to the value of a company’s products, while in other settings it is at the core of their competitive survival. The computer’s main purpose is to cut order-entry costs and to provide more flexibility to customers in the time and process of order submission. The system yields a larger competitive advantage, adding value for customers and a substantial rise in their sales. The resulting sharp increase in the company’s market share forces a primary competitor into a corporate reorganization and a massive systems development effort to contain the damage. Natural Barriers to Entry: In industries where customers incur high costs switching from one brand to another, this becomes a de facto barrier to entry for new firms, as they face difficulty enticing prospective customers to pay the money required to make a chahange. Brand identity and customer loyalty serve as barriers to entry for outsiders. Barriers to entry can also form naturally as the dynamics of an industry take shape of the natural barrier. Barrier option: A barrier option is a type of option whose payoff depends on whether or not the underlying asset has reached or exceeded a predetermined price. A barrier option can be a knock-out, meaning it can expire worthless if the underlying exceeds a certain price, limiting profits for the holder but limiting losses for the writer. It can also be a knock-in, meaning it has no value until the underlying reaches a certain price. BREAKING DOWN \'Barrier Option\': Barrier options are considered a type of exotic option because they are more complex than basic American or European options. Barrier options are also considered a type of path-dependent option because their value fluctuates as the underlying\'s value changes during the opt.

Answer D all of the aboveLipids like triglycerides are semi-solid.pdf

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An example of when a one-way ANOVA could be used is if you want to d.pdf

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An example of when a one-way ANOVA could be used is if you want to determine if there is a difference in the mean height of stalks of three different types of seeds. Since there is more than one mean, you can use a one-way ANOVA since there is only one factor that could be making the heights different. Now, if take these three different types of seeds, and then add the possibility that four different types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of the stalks could be different for a combination of several reasons: The types of seed could cause the change, the types of fertilizer could cause the change, and/or there is an interaction between the type of seed and the type of fertilizer. There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then you can use a two-way ANOVA. Solution An example of when a one-way ANOVA could be used is if you want to determine if there is a difference in the mean height of stalks of three different types of seeds. Since there is more than one mean, you can use a one-way ANOVA since there is only one factor that could be making the heights different. Now, if take these three different types of seeds, and then add the possibility that four different types of fertilizer is used, then you would want to use a two-way ANOVA. The mean height of the stalks could be different for a combination of several reasons: The types of seed could cause the change, the types of fertilizer could cause the change, and/or there is an interaction between the type of seed and the type of fertilizer. There are two factors here (type of seed and type of fertilizer), so, if the assumptions hold, then you can use a two-way ANOVA..

10.dry and hot conditions11.C4 plants12.bundle sheet cells14.p.pdf

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10.dry and hot conditions 11.C4 plants 12.bundle sheet cells 14.plasmodesmata 15)PEP carboxylase 16) bundle sheet cells 17) pineapple 18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if takes place carboxylase activity will involve 19)corn Solution 10.dry and hot conditions 11.C4 plants 12.bundle sheet cells 14.plasmodesmata 15)PEP carboxylase 16) bundle sheet cells 17) pineapple 18) oxygenase activity in C3 plants in general photo respiration does not occur in C4 plants if takes place carboxylase activity will involve 19)corn.

1. Off Page refference arrow2. Connector Symbol3. Process4.Del.pdf

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1) the values of potassium, chlorine and HCO3 values are abnormal an.pdf

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1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic alkalosis which is causing high HCO3 values. 2) the pH is alkalotic and it is abnormal due to metabolic alkalosis Solution 1) the values of potassium, chlorine and HCO3 values are abnormal and this is due to metabolic alkalosis which is causing high HCO3 values. 2) the pH is alkalotic and it is abnormal due to metabolic alkalosis.

The intermediate is the compound A. There are a c.pdf

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The intermediate is the compound A. There are a couple of reasons for that: 1. The oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic. Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom (the one that bears OH and then gets into the cycle) does not change in the reaction (retention of the configuration). That supports the formation of the intermediate A. If the intermediate B participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion of the configuration). Solution The intermediate is the compound A. There are a couple of reasons for that: 1. The oxygen (on the phosphorus in the 5\'-adenilyc acid) with negative charge is more nucleophilic. Therefore, it would attack DCC to form the intermediate. 2. Configuration of the carbon atom (the one that bears OH and then gets into the cycle) does not change in the reaction (retention of the configuration). That supports the formation of the intermediate A. If the intermediate B participated in the reaction the oxygen atom in the cyclic AMP would have pointed up (inversion of the configuration)..

When ionic solids dissolve in water theions that are adjacent to each.pdf

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When ionic solids dissolve in water theions that are adjacent to each other in the solid become surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to large ions. If this formation of the hydratedions were the only factor than we would expect all ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we need to condsider the other factors in thesolution process. the solution process is governed bythe solute-solute, solvent-solvent and solute-solventintermolecular attractive forces. So far we have only consideredhydration of the ions by the water molecules, that is thesoluteÐsolvent interactions. When we consider the soluteinteractions we begin to see some of the problems that can arise.The ions in a crystal are strongly attracted to each other and todissolve it is necessary to overcome the electrostatic attractionbetween the oppositely charged ions. The lattice energy of a solidis a measure of the strength of those electrostatic attractions.The lattice energy works to keep the ions in the solid state andionic compounds with large lattice energies are insoluble in waterwhile compounds with small lattice energies are soluble. Whenpotassium iodide dissolves in water the ions is the potassiumiodide solid must be separated; KI(s) -ÐH2O -->K+(g) + I-(g) ÆH1 = +632kJ/mol Now the gaseous ions are distributed inwater according to the equation; H2O(l) + K+(g) +I-(g) -Ð-> K+(aq) + IÐ(aq)ÆH2 + ÆH3 = -617 kJ The sum of these two equations yields theoverall solution process for KI dissolving in water. The value ofÆH2 + ÆH3 is call the hydrationenergy. In this example the hydration energy is not as large, inabsolute terms as the energy required to separate the ions in thesolute and the heat of solution isendothermic. Solution When ionic solids dissolve in water theions that are adjacent to each other in the solid become surroundedby the water molecules (hydrated). The attraction force that occursbetween the ion and water is called an ion Ðdipole forces. Thepolar water molecules orient themselves so that the partiallycharged ends of the molecule are opposite the charge of the ions.So water molecules are oriented with their hydrogen atoms pointedat the anion and the oxygen atoms pointed at the cation. Thisprocess is called hydration. Hydration is more favored for smallions as compared to large ions. If this formation of the hydratedions were the only factor than we would expect all ionic compoundsto dissolve in water. However, that is not the case, and theproblem is that we need to condsider the other factors in.

#include iostream #include fstream #include map #include.pdf

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#include #include #include #include #include using namespace std; //Used from int index_of_largest(const int array[], int startIndex, const int size); void selectSort(int a[], const int size); void swap(int& i1, int& it2); void fileOption(); void inputOption(); void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size); void printFileScreen(const int uniqueArray[], const int countArray[], const int size); int main() { // int size; char userInput[5]; // cout << \"Read file from Input file? (Y/N) \"; cin >> userInput; if(toupper(userInput[0]) == \'Y\') { fileOption(); } else { inputOption(); } return 0; } void selectSort(int array[], const int size) { int smallestIndex; for(int i = 0;i < size - 1;i++) { smallestIndex = index_of_largest(array , i , size); swap(array[i] , array[smallestIndex]); } } void swap(int& i1, int& i2) { int temp; temp = i1; i1 = i2; i2 = temp; } int index_of_largest(const int array[], int startIndex, const int size) { int min = array[startIndex] , index_of_min = startIndex; for(int i = startIndex + 1;i < size;i++) if(array[i] > min) { min = array[i]; index_of_min = i; } return index_of_min; } void unique(const int inputArray[], int uniqueArray[], int countArray[], const int size) { int number; int j = 0; int index; for(int i = 0;i < size;i++) { int flag = 0; number = inputArray[i]; //for each element in inputArray, check if it is there in uniqueArray for (int k = 0; k < j; ++k) { if(number == uniqueArray[k]) { flag = 1; index = k; } } if(flag == 0) { //if so, increment the countArray by 1 countArray[j]++; // copy the element to uniqueArray uniqueArray[j] = number; j++; } else countArray[index]++; } } void fileOption() { int inputArray[50]; int countArray[50] = {0}; int uniqueArray[50]= {0}; ifstream in_stream; char fileName[16]; cout << \"What file would you like to obtain input from? \"; cin >> fileName; in_stream.open(fileName); if(in_stream.fail()) { cout << \"Error: Could not open file!\" << endl; exit(1); } int size = 0; while(in_stream >> inputArray[size]) { size++; } selectSort(inputArray , size); unique(inputArray , uniqueArray , countArray , size); printFileScreen(uniqueArray , countArray , size); } void inputOption() { int inputArray[50]; int countArray[50] = {0}; int uniqueArray[50] = {0}; int size; cout << \"How many numbers would you like to enter? \"; cin >> size; cout << \"Enter numbers:\" << endl; for(int i = 0;i < size;i++) { cin >> inputArray[i]; } selectSort(inputArray , size); unique(inputArray , uniqueArray , countArray , size); printFileScreen(uniqueArray , countArray , size); } void printFileScreen(const int uniqueArray[], const int countArray[], const int size) { //cout << \"jaja\"; ofstream output_file; output_file.open(\"lab3Exercise2.txt\"); if(output_file.fail()) { cout << \"Error: Could not open file!\" << endl; exit(1); } output_file << \"N\\t\" << \"count\ \"; for(int i = 0;i < size;i++) { if(uniqueArray[i] != 0) { output_file << uniqueArray[i] << \"\\t\"; outpu.

r=(13)d(I-)dt .pdf

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on the phosphorus. Solution .pdf

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You have the priority for 1 right (the OH), howev.pdf

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You have the priority for 1 right (the OH), however the positions 2 and 3 as you have labeled are really reversed because, both 2 and 3 are directly connected to a CH2OH group, however the top groups (the one you labeled CH3 top) is much larger and therefore gets priority 2 while the CH2OH at the bottom gets priority 3. Also, keep in mind the way this is set up, the H is projected outwards so you get the opposite configuration of what you see (if it looks like an S you have an R). Therefore 1 = OH on the right 2 = large group at the top 3 = CH2OH at bottom Follow this and you go counterclockwise, which is R. But keep in mind the H is in the front of the molecule, therefore the R is really an S. Hope this helped! Solution You have the priority for 1 right (the OH), however the positions 2 and 3 as you have labeled are really reversed because, both 2 and 3 are directly connected to a CH2OH group, however the top groups (the one you labeled CH3 top) is much larger and therefore gets priority 2 while the CH2OH at the bottom gets priority 3. Also, keep in mind the way this is set up, the H is projected outwards so you get the opposite configuration of what you see (if it looks like an S you have an R). Therefore 1 = OH on the right 2 = large group at the top 3 = CH2OH at bottom Follow this and you go counterclockwise, which is R. But keep in mind the H is in the front of the molecule, therefore the R is really an S. Hope this helped!.

a electrolytic solution because it is an ionic co.pdf

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conjugated dienes are those in which two double b.pdf

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Fluorine is the most reactive and most electroneg.pdf

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Fluorine is the most reactive and most electronegative of the elements. Fluorine also has a very high affinity for gaining an electron to form the fluoride ion in an exothermic manner. As such, if not the strongest, fluorine is one of the most powerful oxidants known. Iodine on the other hand is considerably less electronegative than fluorine and it\'s electron affinity is considerably less exothermic than that of fluorine. Solution Fluorine is the most reactive and most electronegative of the elements. Fluorine also has a very high affinity for gaining an electron to form the fluoride ion in an exothermic manner. As such, if not the strongest, fluorine is one of the most powerful oxidants known. Iodine on the other hand is considerably less electronegative than fluorine and it\'s electron affinity is considerably less exothermic than that of fluorine..

hey dear neutron and protons do not occupy orbita.pdf

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at 100 degree KOH needs less Ea which cannot be p.pdf

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2024.06.01 Introducing a competency framework for languag learning materials ...

Sandy Millin

http://sandymillin.wordpress.com/iateflwebinar2024 Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error. Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials. This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.

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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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