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Len [333]
1 year ago
12

which of the following statements are true about client-side DNS? (Choose all that apply). a. If an APIPA address is assigned, t

hen DNS is the problem b. Client-side DNS should be configured to point towards the DNS server that is authoritative for the domain that client wants to join C.Check out DNS settings using the NSLookup command d.Check out DNS settings using the DIG command e.The cache.dns file has the IP addresses of the 13 root DNS servers f.If a web site can be reached by IP address and not by host name, then DNS or the Hosts file would be the problem
Engineering
1 answer:
Brums [2.3K]1 year ago
5 0

The statements which are true about client-side DNS include all of the following;

B. Client-side DNS should be configured to point towards the DNS server that is authoritative for the domain that client wants to join.

C. Check out DNS settings using the NSLookup command.

D. Check out DNS settings using the DIG command.

E. The cache.dns file has the IP addresses of the 13 root DNS servers.

B. If a web site can be reached by IP address and not by host name, then DNS or the Hosts file would be the problem.

<h3>What is a DNS server?</h3>

In Computer technology, a DNS server can be defined as a type of server that is designed and developed to translate domain names into IP addresses, so as to allow end users access websites and other internet resources through a web browser.

This ultimately implies that, a DNS server simply refers to a type of server that is designed and developed to translate requests for domain names into IP addresses for end users.

Read more on a domain names here: brainly.com/question/19268299

#SPJ1

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End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf
sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα +  ∑Mad    ...............1

put here value and we get

so here I = ( \frac{1}{12} ) × m × L²    ................2

I = ( \frac{1}{12} ) × 5 × 0.8²

I = 0.267 kg-m²  

and

a  is =  r × α    

a  = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x)    ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 -  5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y)    .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81  = 5 × 0.4 × 2²

a(y) = 57.1 N

so

total force at A will be

Fa = \sqrt{a(x)^2+a(y)^2}    ...............5

Fa = \sqrt{(57.1)^2+(5)^2}  

Fa = 57.32 N

3 0
4 years ago
Laws that protect businesses involve
mr_godi [17]
Laws that protect businesses involve contracts, employment law, intellectual property, real estate, bankruptcy, and many other areas of the law.
4 0
3 years ago
Create a class Suzuki with following characteristics Initialize member variable of Car class using the super constructor which c
Ganezh [65]

Answer:

See attached file.

Explanation:

Download txt
5 0
3 years ago
Given two input integers for an arrowhead and arrow body, print a right-facing arrow. Ex: If the input is 0 1, the output is
Lyrx [107]

For printing a right facing with input 01, define the class for right facing arrow and give the value with appropriate space using '%6d'  and then display the right facing arrow.

Further Explanation:

The code to print the arrowhead and arrow body is as follows:

Import java.util.Scanner;

//Define the method.

public class RightfacingArrow

{

//Define the main method.

public static void main(String[] args)

{

//Create an object of Scanner class.

Scanner object = new Scanner(System.in);

//Declare the variables.

int baseChar;

int headChar;

//Prompt the user to enter the input values.

baseChar = object.nextInt();

headChar = object.nextInt();

 

//Display the headchar in position 6 from the  starting.

System.out.printf("%6d ", headChar);

//Display the arrows.

System.out.println(baseChar+""+baseChar+""

+baseChar+""+baseChar+""+baseChar+""

+headChar+""+headChar);

System.out.println(baseChar+""+baseChar

+""+baseChar+""+baseChar+""+baseChar

+""+headChar+""+headChar+""+headChar);

System.out.println(baseChar+""+baseChar

+""+baseChar+""+baseChar+""

+baseChar+""+headChar+""+headChar);

//Display the headchar in position 6 from the  starting.

System.out.printf("%6d ", headChar);

}

}

Output:

If the input is 01, the output will be as below:

    1

0000011

00000111

0000011

    1

Learn more:

1. A company that allows you to license software monthly to use online is an example of ? brainly.com/question/10410011  

2. Prediction accuracy of a neural network depends on _______________ and ______________. brainly.com/question/10599832  

Answer details:

Grade: College Engineering

Subject: Computer Science and Enginnering

Chapter: Java Programming

Keyword:

Java, input, output, programming, statements, if-else, loops, print, scan. right facing arrows, body, main body, char, int, variables, scanner class, abstract class, constructor, desctructor

8 0
3 years ago
Water "bubbles up" h2 = 9 in. above the exit of the vertical pipe attached to three horizontal pipe segments. The total length o
8_murik_8 [283]

The pressure needed at point (1) to produce the flow where water "bubbles up above the exit of the vertical pipe attached to three horizontal pipe segments is 0.750 psi.

<h3>What is Bernoulli's equation?</h3>

The Bernoulli's equation for incompressible fluid can be given as,

P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2

Here, ρ is density of fluid, g is acceleration due to gravity, (P) is the pressure v is velocity, h is height of elevation and subscript (1 and 2) is used for point 1 and 2.

Water "bubbles up" h2 = 9 in.  Above the exit of the vertical pipe attached to three horizontal pipe segments.

The total length of the 1.75-in.- diameter galvanized iron pipe between point (1) and the exit is 21 inches. h3 = 18 in.

Reynolds number is,

R_e=\dfrac{V_3D}{v}\\R_e=\dfrac{4.01\dfrac{0.75}{12}}{1.21\times10^{-15}}\\R_e=2.07\times10^4

The formula for the ratio of pressure to radius, when pressure and velocity at point 2 is zero can be given as,

\dfrac{p_1}{r}=z_2\left(f\dfrac{l}{d}+\sum k_L\right)\dfrac{v^2}{2g}-\dfrac{v_1^2}{2g}

Here, f=0.039 and ∑K(L)=4.5. Put the values,

\dfrac{p_1}{r}=\dfrac{7}{12}\left(0.03\dfrac{21}{0.75}+4.5-1\right)\dfrac{4.01^2}{2\times32.2}\\\dfrac{p_1}{r}=1.73\rm\; ft

Put the value of r we get,

p_1=62.4\times1.73\\p_1=108\rm\; ib/ft^2\\p_1=0.750\rm\; psi

The pressure needed at point (1) to produce the flow where water "bubbles up above the exit of the vertical pipe attached to three horizontal pipe segments is 0.750 psi.

Learn more about the Bernoulli's equation here;

brainly.com/question/7463690

#SPJ1

5 0
2 years ago
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