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slamgirl [31]
3 years ago
5

Suddenly, though, just as the procedure was about to be relayed to the astronauts, the pressure . . . the temperature . . . drop

ped! And . . . stayed down. The frozen slug, apparently, had melted! (Probably due to the extreme heat in the fuel line.) Problem solved.
—Team Moon: How 400,000 People

Landed Apollo 11 on the Moon,

Catherine Thimmesh


Why does the author include the NASA scientists’ attempts to fix the fuel line?


to show that not all problems need to be solved

to show that the scientists’ hard work had not paid off

to show that many solutions can be failures

to show that the situation was tense for everyone


i wiil give you brainliest answer!
Computers and Technology
2 answers:
nydimaria [60]3 years ago
6 0

Answer:The answer is D.

Explanation:I did the assignment on edge

konstantin123 [22]3 years ago
5 0

Answer:

D)to show that the situation was tense for everyone

Explanation:

got it right on edge

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The term ____ refers to a wide variety of different database technologies that were developed in order to overcome some of the l
ella [17]
Database management systems are used to store and to access information.
In <span>the Relational Database Management System </span>(RDBMS), all the data is in the form of simple columns and rows in a table. Structured Query Language (SQL) is is the standard query language for RDBMS. Some of the disadvantages of RDBMS are: not enough storage area to handle data such as images, digital and audio/video, do not provide good support for nested structures,not much efficient and effective integrated support.
<span>The term object oriented database management systems (OODBMS)  refers to a wide variety of different database technologies that were developed in order to overcome some of the limitations of relational databases.</span>





5 0
3 years ago
Which of the following are agricultural industry clusters? Human population systems, computer technoloy biotechnology or food pr
kondor19780726 [428]

Biotechnology

Food Products and Processing Systems.

Further explanation

Career clusters help prepare learners with knowledge that they need to use towards achieving their career goals. One such career cluster is the Agriculture, Food, and Natural Resources career cluster which is divided into seven pathways. They include:

  • Food Products and Processing Systems.
  • Biotechnology Systems
  • Agribusiness Systems
  • Plant Systems
  • Animal Science
  • Environmental Service Systems
  • Natural Resource Systems

This brings us to our answers above. Those working in the Food Products and Processing Systems pathway are responsible in discovering new food sources and coming up with ways to process and store food set out by industry regulation. On the other hand, biotechnology systems deal with techniques that use science to solve problems concerning living organisms and anyone thinking about pursuing this pathway ought to demonstrate competence in the application of biotech in the context of Agriculture, Food, and Natural Resources.

Learn about Agriculture, Food, and Natural Resources career cluster

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#LearnWithBrainly

7 0
3 years ago
How can social media be used to find out about activities?
ICE Princess25 [194]
Hello RandomHacker


Using social media can inform you when events are happening near by. 

For example: Your favorite singer of band is going on tour and they are doing it near you, social media advertises this and you can find out in time to go.

Hope this helps
-Chris

3 0
3 years ago
Read 2 more answers
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
Define the following terms: data model, database schema, database state, internal schema, conceptual schema, external schema, da
PolarNik [594]

Answer:

Following are the solution to the given terms:

Explanation:

Data Model: A set of concepts that are ideal for describing the layout, computational tools, relations, semantics, constraints on accuracy.

Database Scheme: It is also known as the overall design of the database.

Database State: It is the data in the database at a particular moment in time.

Internal Schema: The physical structure of the database is defined.

Conceptual Schema: It hides the details of physical storage.

External Schema: It describes the part of the database the user group sees. In other words, we can say that it is a view of the database.

Data Independence: The ability to modify the schema at a single level while modifying the schema at a higher level is often recognized.

DDL: It stands for the Data Definition Language, which is used by DBA and dev use for creating With particular scheme.  

DML: It refers to the Data Manipulation Language, which is used to manipulate, add, recover, insert, remove, and alter data.  

SDL: It stands for the storage definition language, in which the internal schema of the database is applied for storage description.  

VDL: The view definition language specifies the user to view and mapping their conceptual schema.

Query Language: It is just an independent, dynamic, high-level dialect used during manipulation.  

Host Language: It is also the machine language, that would be incorporated into the general programming language by low-level data handling commands.  

Data sublanguage: It uses for embedding in the last language.  

Database Utility: It is a program, which helps to handle the database with DBA's modules.  

Catalog: The entire structure and constraints of the database is defined.

client/server architecture: The design of the database server system that device usability: model user, model server (landless storage) software Government computer design in which many users (remote processors) demand a central server to receive service (host computer).  

Three-tier architecture: It is the application, data, user interface.

n-tier architecture: multiple levels. The nth level must only be able to handle the n+1 application and transmit it to the n-1 point.

3 0
3 years ago
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