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Eva8 [605]
3 years ago
12

Write a setInterval() function that increases the count by 1 and displays the new count in counterElement every 300 milliseconds

. Call clearInterval() to cancel the interval when the count displays 5.
var count = 0;


var counterElement = document.getElementById("counter");


counterElement.innerHTML = count;



/* Your solution goes here*/
Computers and Technology
1 answer:
Alina [70]3 years ago
8 0

Answer:

  1. var count = 0;
  2. var counterElement = document.getElementById("counter");
  3. counterElement.innerHTML = count;
  4. var a = setInterval(
  5.    function(){  
  6.        count++;
  7.        counterElement.innerHTML = count;
  8.        if(count == 5){
  9.            clearInterval(a);
  10.        }
  11.    }
  12.        , 300);

Explanation:

The solution code is given from Line 6 - 15.  

setInterval function is a function that will repeatedly call its inner function for an interval of time. This function will take two input, an inner function and the interval time in milliseconds.  

In this case, we define an inner function that will increment count by one (Line 8) and then display it to html page (Line 9). This inner function will repeatedly be called for 300 milliseconds. When the count reaches 5, use clearInterval to stop the innerFunction from running (Line 11 - 13).  

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Determine whether or not the following pairs of predicates are unifiable. If they are, give the most-general unifier and show th
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Answer:

a) P(B,A,B), P(x,y,z)

=> P(B,A,B) , P(B,A,B}  

Hence, most general unifier = {x/B , y/A , z/B }.

b. P(x,x), Q(A,A)  

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q.

c. Older(Father(y),y), Older(Father(x),John)

Thus , mgu ={ y/x , x/John }.

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))  

This is not unifiable as x cannot be bound for both A and B.

e) P(f(x), x, g(x)), P(f(y), A, z)    

=> P(f(A), A, g(A)), P(f(A), A, g(A))  

Thus , mgu = {x/y, z/y , y/A }.

Explanation:  

Unification: Any substitution that makes two expressions equal is called a unifier.  

a) P(B,A,B), P(x,y,z)  

Use { x/B}  

=> P(B,A,B) , P(B,y,z)  

Now use {y/A}  

=> P(B,A,B) , P(B,A,z)  

Now, use {z/B}  

=> P(B,A,B) , P(B,A,B}  

Hence, most general unifier = {x/B , y/A , z/B }  

b. P(x,x), Q(A,A)  

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q  

c. Older(Father(y),y), Older(Father(x),John)  

Use {y/x}  

=> Older(Father(x),x), Older(Father(x),John)  

Now use { x/John }  

=> Older(Father(John), John), Older(Father(John), John)  

Thus , mgu ={ y/x , x/John }  

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))  

Use { y/x }  

=> Q(G(x,z),G(z,x)), Q(G(x,x),G(A,B))

Use {z/x}  

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))  

This is not unifiable as x cannot be bound for both A and B  

e) P(f(x), x, g(x)), P(f(y), A, z)  

Use {x/y}  

=> P(f(y), y, g(y)), P(f(y), A, z)  

Now use {z/g(y)}  

P(f(y), y, g(y)), P(f(y), A, g(y))  

Now use {y/A}  

=> P(f(A), A, g(A)), P(f(A), A, g(A))  

Thus , mgu = {x/y, z/y , y/A }.

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Answer:

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It is one of which every component is arranged in a series connection. Hence series circuit will have same current at all points of the circuit. The voltage drop across each component in the circuit adds up to sum of voltage source across each component and of an equivalent component value. Breaking of the series circuit will make entire circuit to stop working. Suppose consider the three bulbs are connected in series connection and if even one bulb burns out or broken then all the three bulbs will stop working as well. In series circuit components like current (I) is sum of all the element and Voltage is sum of all the voltage drops and resistance is the sum of individual resistances.

Explanation:

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Answer: b) GNATT chart

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