All categories

3 years ago

How do i play Fortnite on controller?

Engineering

2 answers:

Andru [333]3 years ago

7 0

Answer:

Use bluetooth it works easily

Explanation:

neonofarm [45]3 years ago

4 0

Answer:

use Bluetooth if you have an unwired controller, but If its wired just hook it up.

You might be interested in

For some transformation having kinetics that obey the Avrami equation (Equation 11.17), the parameter n is known to have a value

Nastasia [14]

Explanation:

Below is an attachment containing the solution.

5 0

3 years ago

Create a program named IntegerFacts whose Main() method declares an array of 10 integers.Call a method named FillArray to intera

musickatia [10]

Answer:

C# codee

using System;

class IntegerFacts

{

static void Main()

{

int[] x = new int[10];

int sum = 0;

int siz = 0, high = 0, low = 0, avg = 0;

siz = FillArray(x);

Statistics(x, ref high, ref low, ref sum, ref avg, siz);

Console.Write("The highest value is " + high);

Console.Write("\nThe lowest value is " + low);

Console.Write("\nThe sum of the values is " + sum);

Console.Write("\nThe average is " + avg);

}

static void Statistics (int [] b, ref int h, ref int l, ref int s, ref int a, int size)

{

if (size == 0)

h = l = s = a = 0;

else

{

int i =0;

l = 999;

h = 0;

s = 0;

for (; i < size;++i)

{

if(b[i] > h)

h = b[i];

if(b[i] < l)

l = b[i];

s += b[i];

}

a = s / size;

}

static int FillArray (int[] a)

{

int i = 0, count = 0;

int intTemp =0 ;

string temp;

for(i = 0 ; i < 10 ; i++)

{

Console.Write("Enter element number"+(i+1) +" : ");

temp = Console.ReadLine();

if (int.TryParse(temp, out intTemp)) {

if (intTemp != 999)

{

a[i] = intTemp;

count++;

}

else

break;

}

else

{

Console.Write("\n\nOops.. You entered a wrong number. Try again \n\n");

i--;

}

return count;

}

Explanation:

The output of the above program is given in the attached file.

8 0

4 years ago

Write a function which multiplies the values in odd position values by 10. Odd positions in this case refers to the first value

xxMikexx [17]

Answer:

Using linkedlist on C++, we have the program below.

Explanation:

#include<iostream>

#include<cstdlib>

using namespace std;

//structure of linked list

struct linkedList

{

int data;

struct linkedList *next;

};

//print linked list

void printList(struct linkedList *head)

{

linkedList *t=head;

while(t!=NULL)

{

cout<<t->data;

if(t->next!=NULL)

cout<<" -> ";

t=t->next;

}

//insert newnode at head of linked List

struct linkedList* insert(struct linkedList *head,int data)

{

linkedList *newnode=new linkedList;

newnode->data=data;

newnode->next=NULL;

if(head==NULL)

head=newnode;

else

{

struct linkedList *temp=head;

while(temp->next!=NULL)

temp=temp->next;

temp->next=newnode;

}

return head;

}

void multiplyOddPosition(struct linkedList *head)

{

struct linkedList *temp=head;

while(temp!=NULL)

{

temp->data = temp->data*10; //multiply values at odd position by 10

temp = temp->next;

//skip odd position values

if(temp!= NULL)

temp = temp->next;

}

int main()

{

int n,data;

linkedList *head=NULL;

// create linked list

head=insert(head,20);

head=insert(head,5);

head=insert(head,11);

head=insert(head,17);

head=insert(head,23);

head=insert(head,12);

head=insert(head,4);

head=insert(head,21);

cout<<"\nLinked List : ";

printList(head); //print list

multiplyOddPosition(head);

cout<<"\nLinked List After Multiply by 10 at odd position : ";

printList(head); //print list

return 0;

}

5 0

3 years ago

A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °

SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)= $\frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3} (Btu/s)/(ft.F)$

Thermal Conductivity is SI units:

$k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K$

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

$A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J$

In Btu:

$A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu$

PArt B:

At midpoint Length=L/2=0.1 m

$Q=\frac{k*A*(T_1-T_2)}{L}$

On rearranging:

$T_2=T_1-\frac{Q*L}{KA}$

$T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C$

4 0

3 years ago

How many people made machines

olganol [36]

Answer:

The total number of people whom have made machines is not a recorded figure? Need to be more specific :/

Explanation:

Sorry not very helpful, your question is REALLY broad

3 0

3 years ago

Read 2 more answers