How many megapixels is in a macbook air 2017 camera

Which of the following is NOT an example of a font style?

erastovalidia [21]

C. Underline because it is not a kind of font. it is an effect.

3 0

3 years ago

Ano kaya ang nararamdaman mo at mangyari sayo kung tatlong araw ka nang hindi kumakain

Feliz [49]

Ang mararamdaman mo ay pananamlay at tila walang lakas. Ang mangyayari sayo ay magkakasakit ka at pag ipinagpatuloy mo parin ang hindi pag kain, baka hindi lang sakit aabotin mo kundi kamatayan.

3 0

3 years ago

The ________ tier of the three-tier architecture consists of computers, phones, and other mobile devices that have browsers that

andrey2020 [161]

Answer:

It is the user tier of the 3-tier architecture that consists of computers, phones, and other mobile devices that have browsers that request and process web pages.

Explanation:

Almost if not all applications that use e-commerce for operations use the 3-tier architecture. 3-tier architecture is simply the arrangement of server tier which consists of all computers that run a web server and a database tier that consists of computers that run a DBMS to store and retrieve data. We also have the user data that consists of computers, phones, and other mobile devices that have browsers that request and process web pages.

When you make a request, for instance to Google in your local browser, the browser sends a request to the internet. This initial interaction between the user and the internet is what is known as the user tier of the 3 tier architecture. Basically, it is the phase where end users access content online through graphical interface browsers and make requests to the web servers and DBMS.

Learn more about 3-tier architecture

brainly.com/question/12627823

#LearnWithBrainly

6 0

3 years ago

P16 Consider the following network. ISP B provides national backbone service to regional ISP A. ISP C provides national backbone

Art [367]

Solution :

Considering the following data :

$\text{ISP B}$ provides the national backbone to the regional $\text{ISP A}$ .

$\text{ISP C}$ provides the national backbone to the regional $\text{ISP D}$ .

Each of the ISP consists of one autonomous systems (AS).

$\text{ISP B}$ would like to hand over the traffic of C through the $\text{West}$ Coast.

The $\text{ISP C}$ would like to receive traffic from B's peering point via East Coast.

One way for $C$ to force $B$ to hand over all of $B$ ’s traffic to $D$ on the east coast is for $C$ to only advertise its route to $D$ via its east coast peering point with $C$ .

7 0

3 years ago

Write a function input_poly(ptr1) that reads in two polynomials at a time with coefficient and exponent from a file cp7_in.txt.

Tanya [424]

Answer:

try this

Explanation:

#include<stdio.h>

#include<malloc.h>

typedef struct poly{ double coff;

int pow;

struct poly *link; }SL;

void create(SL **head)

{

*head=NULL;

}

//Insertion

void ins_beg(SL ** head, double c, int pw)

{

SL *ptr;

ptr=(SL *)malloc(sizeof(SL));

ptr->coff=c;

ptr->pow=pw;

ptr->link=*head;

*head=ptr;

}

void trav(SL **head)

{

SL *ptr;

ptr=*head;

printf("\n\t Cofficient Exponent\n");

while(ptr!=NULL)

{ printf("\t\t%3.2f \t%d\n",ptr->coff, ptr->pow);

ptr=ptr->link;

}

//addition of polynomial............

void mult(SL *pl1, SL *pl2, SL **res)

{

SL trav1, trav2, ptr, new1,*loc;

int x,y;

trav1=*pl1;

while(trav1!=NULL)

{

trav2=*pl2;

while(trav2!=NULL)

{

x=(trav1->coff)*(trav2->coff);

y=(trav1->pow)+(trav2->pow);

//printf("\t%d\t%d\n",x,y);

if(*res==NULL)

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=x;

new1->pow=y;

new1->link=NULL;

*res=new1;

}

else

{

ptr=*res;

while((y<ptr->pow)&&(ptr->link!=NULL))

{

loc=ptr;

ptr=ptr->link;

}

if(y==ptr->pow)

ptr->coff=ptr->coff+x;

else

{

if(ptr->link!=NULL)

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=x;

new1->pow=y;

loc->link=new1;

new1->link=ptr;

}

else

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=x;

new1->pow=y;

ptr->link=new1;

new1->link=NULL;

}

trav2=trav2->link;

}

trav1=trav1->link;

}

//...addition..............

void add(SL *pl1, SL *pl2)

{

SL trav1, ptr, *new1,*loc; trav1=*pl1;

while(trav1!=NULL)

{

loc=ptr=*pl2;

while((trav1->pow<ptr->pow)&&(ptr->link!=NULL))

{

loc=ptr; ptr=ptr->link;

}

if(loc==ptr)

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=trav1->coff;

new1->pow=trav1->pow;

new1->link=*pl2;

*pl2=new1;

}

else

if(trav1->pow==ptr->pow)

{

ptr->coff=ptr->coff+trav1->coff;

//ptr->pow=ptr->pow+trav1->pow;

}

else

if(ptr->link!=NULL)

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=trav1->coff;

new1->pow=trav1->pow;

new1->link=ptr;

loc->link=new1;

}

else

{

new1=(SL *)malloc(sizeof(SL));

new1->coff=trav1->coff;

new1->pow=trav1->pow;

new1->link=NULL;

ptr->link=new1;

}

trav1=trav1->link;

}

int main()

{

SL first, sec, *res;

create(&first);

ins_beg(&first,10.25,0);

ins_beg(&first,4,1);

ins_beg(&first,5,2);

ins_beg(&first,4,4);

printf("\nFirst Polinomial:\n");

trav(&first);

create(&sec);

ins_beg(&sec,11,0);

ins_beg(&sec,6,1);

ins_beg(&sec,4,2);

ins_beg(&sec,3,3);

printf("\nSecond Polinomial:\n");

trav(&sec);

create(&res);

printf("\nMultiplication of two Polinomial:\n");

mult(&first,&sec,&res);

trav(&res);

printf("\nAddition of two Polinomial:\n");

add(&first,&sec);

trav(&sec);

//trav(&first);

return 0;

}

6 0

3 years ago