Sodium bicarbonate (NaHCO3) in pancreatie juice cmulsifies fats- activ.pdf
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Sodium bicarbonate ( NaHCO 3 ) in pancreatie juice cmulsifies fats. activates pepsin. neutralizes acid chyme. stimulates the release of insulin. Question 8 (1 point) E. coli are beneficial to humans because they: convert pepsinogen to pepsin. produce vitamins and amino acids. absotb water from the large intestine. synthesize urea from the breakdown of amino acids..
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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.
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}
public SLLNode(T el, SLLNode<T> ptr) {
info = el; next = ptr;
}
}
protected SLLNode<T> head, tail;
public SLL() {
head = tail = null;
}
public boolean isEmpty() {
return head == null;
}
public void addToHead(T el) {
head = new SLLNode<T>(el,head);
if (tail == null)
tail = head;
}
public void addToTail(T el) {
if (!isEmpty()) {
tail.next = new SLLNode<T>(el);
tail = tail.next;
}
else head = tail = new SLLNode<T>(el);
}
public T deleteFromHead() { // delete the head and return its info;
if (isEmpty())
return null;
T el = head.info;
if (head == tail) // if only one node on the list;
head = tail = null;
else head = head.next;
return el;
}
public T deleteFromTail() { // delete the tail and return its info;
if (isEmpty())
return null;
T el = tail.info;
if (head == tail) // if only one node in the list;
head = tail = null;
else { // if more than one node in the list,
SLLNode<T> tmp; // find the predecessor of tail;
for (tmp = head; tmp.next != tail; tmp = tmp.next);
tail = tmp; // the predecessor of tail becomes tail;
tail.next = null;
}
return el;
}
public void delete(T el) { // delete the node with an element el;
if (!isEmpty())
if (head == tail && el.equals(head.info)) // if only one
head = tail = null; // node on the list;
else if (el.equals(head.info)) // if more than one node on the list;
head = head.next; // and el is in the head node;
else { // if more than one node in the list
SLLNode<T> pred, tmp;// and el is in a nonhead node;
for (pred = head, tmp = head.next;
tmp != null && !tmp.info.equals(el);
pred = pred.next, tmp = tmp.next);
if (tmp != null) { // if el was found;
pred.next = tmp.next;
if (tmp == tail) // if el is in the last node;
tail = pred;
}
}
}
@Override
public String toString() {
if(head == null)
return "[ ]";
String str = "[ ";
SLLNode<T> tmp = head;
while(tmp != null){
str += tmp.info + " ";
tmp = tmp.next;
}
return str+"]";
}
public boolean contains(T el) {
if(head == null)
return false;
SLLNode<T> tmp = head;
while(tmp != null){
if(tmp.info.equals(el))
return true;
tmp = tmp.next;
}
return false;
}
public int size(){
if(head == null)
return 0;
int count = 0;
SLLNode<T> p = head;
while(p != null) {
count++;
p = p.next;
}
return count;
}
}
Given the code above, solve the following tasks (please show the solution for every task
clearly):
For part (c) this is the code for the test program:
public class SLL_Driver {
public static void main(String[] args) {
SLL<String> myList = new SLL<String>();
String[] cityNames = {"Jubail", "Riyadh", "Abha", "Dammam", "Taif"};
for(int i = 0; i < cityNames.length; i++)
myList.addToHead(cityNames[i]);
System.out.println("The list is: " + myList);
System.out.println("It is " + myLis.
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