I will give brainliest to the first person who can answer correctly.

One can find mountain ranges, elevation levels, and deserts on a physical map.

prohojiy [21]

Answer:

True

Explanation:

There is a quizlet about this that I took

4 0

3 years ago

Read 2 more answers

Create a function names minElement() that takes an array of numbers as a paaramter and returns the min value of the array. The a

Vika [28.1K]

Answer:

#include<iostream>

using namespace std;

//create the function

int minElement(int arr[],int n){

int minimum = arr[0]; //store the first value of array

//for loop for iterate the each element

for(int i=0;i<n;i++){

if(arr[i]<minimum){ //compare

minimum=arr[i];

}

return minimum;

}

int main(){

int arr[]={4,1,7,9,2,6};

int array_size = 6;

int result = minElement(arr,array_size); //calling

cout<<"The min value is: "<<result<<endl;

}

Explanation:

First include the library iostream in the c++ programming for input/output.

then, create the function minElement() for find the minimum value in the array.

it required the one loop for iterate the each element in the array and then compare with first element in the array which is store in the variable minimum.

if array element is less than the minimum then, update the value in the minimum variable.

after complete the loop, return the result.

for capture the result create the main function and call the function with parameter array.

and finally display the result.

8 0

3 years ago

What is wrong with my code...

stiv31 [10]

I'm not sure of the problem that you had in the first place, but I can point out that in your "bigger" method, if a number is greater than one then it is bigger, however the else statement says that the number *can* also be equal[==] to the second argument, so for example bigger(1,1) it would check if 1 > 1 and return false, so it will return that b[the second 1] is bigger! Hope this helps :D

3 0

3 years ago

A coworker is concerned about the veracity of a claim because the sender of an email denies sending it. The coworker wants a way

Aleks04 [339]

<u>Some recommend way to prove the authenticity of an email:</u>

The coworker as truthful to the company or organization he or she has to check where the same mail is sent long back to the same sender.

If search on emails sent item if he sends it will be well and good. Some time backup the mail will be help to proven authenticity of an email.

Sometimes once mail is downloaded into ms-outlook or lotus domino some time mail persists in the mail server.

He or she can coordinate with the mail server administrator to possible get a log file of sending mail and he can proceed as a record to management.

5 0

4 years ago

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

Evgen [1.6K]

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 }.

7 0

3 years ago