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Verdich [7]
3 years ago
9

What Intel socket recommends the use of a liquid cooling system?

Computers and Technology
1 answer:
Komok [63]3 years ago
7 0
The socket which Intel recommends that one should use with a liquid cooling system is LGA 2011. LGA 2011, also known as socket R is a CPU socket manufactured by Intel. It was released into the market in November 2011 and it replaced LGA 1366 and LGA 1567 in the performance and high end desk tops and server platforms. Socket R has 2011 pins that touch contact points on the underside of the processor.
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Write the line of code to calculate the area of a circle with radius 3 and store it in a variable called
zloy xaker [14]

Answer:

area = 2.14 * 3 ** 2

Explanation:

No clue what language you use but this should be pretty universal. To find the area of a circle you use pi(r) ^ 2 or pi times radius squared

\pi r^{2}

In python it looks like this:

>>> area = 3.14 * 3 ** 2

>>> print(area)

28.26<em> <Printed text</em>

3 0
3 years ago
(1) Prompt the user to enter two words and a number, storing each into separate variables. Then, output those three values on a
Sergeu [11.5K]

Answer:

import java.util.Scanner;

public class num3 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter Favorite color");

       String word1 = in.next();

       System.out.println("Enter Pet's name");

       String word2 = in.next();

       System.out.println("Enter a number");

       int num = in.nextInt();

       //Point One

       System.out.println("You entered: "+word1+" "+word2+

               " "+num);

       //Point Two

       String passwordOne = word1+"_"+word2;

       String passwordTwo = num+word1+num;

       System.out.println("First password: "+passwordOne);

       System.out.println("Second password: "+passwordTwo);

       //Point Three

       int len_passwrdOne = passwordOne.length();

       int len_passwrdTwo = passwordTwo.length();

       System.out.println("Number of characters in "+passwordOne+" :" +

               " "+len_passwrdOne);

       System.out.println("Number of characters in "+passwordTwo+" :" +

               " "+len_passwrdTwo);

   }

}

Explanation:

  • This question is solved using java programming language
  • The scanner class is used to receive the three variables (Two string and one integer)
  • Once the values have been received and stored in the respective variables (word1, word2 and num), Use different string concatenation to get the desired output as required by the question.
  • String concatenation in Java is acheived with the plus (+) operator.
3 0
3 years ago
Which of the following terms describes surgery through a small incision in the abdomen?
ra1l [238]
Which of the following terms describes surgery through a small incision in the abdomen? Laparoscopy. Laparoscopy is super done by using a fiber-optic instrument that is inserted through the abdomen. When this type of instrument is used, the incision is small and there are smaller cuts made to the person being operated on. 
4 0
3 years ago
Hey yall! Its spoopy season! I am having a live stream on the Spoon app under the username teendragonqueen️‍ if yall wanna join!
Aleks [24]

Answer:

Ok sounds cool.

Explanation:

3 0
4 years ago
Read 2 more answers
Find the root using bisection method with initials 1 and 2 for function 0.005(e^(2x))cos(x) in matlab and error 1e-10?
frutty [35]

Answer:

The root is:

c=1.5708

Explanation:

Use this script in Matlab:

-------------------------------------------------------------------------------------

function  [c, err, yc] = bisect (f, a, b, delta)

% f the function introduce as n anonymous function

%       - a y b are the initial and the final value respectively

%       - delta is the tolerance or error.

%           - c is the root

%       - yc = f(c)

%        - err is the stimated error for  c

ya = feval(f, a);

yb = feval(f, b);

if  ya*yb > 0, return, end

max1 = 1 + round((log(b-a) - log(delta)) / log(2));

for  k = 1:max1

c = (a + b) / 2;

yc = feval(f, c);

if  yc == 0

 a = c;

 b = c;

elseif  yb*yc > 0

 b = c;

 yb = yc;

else

 a = c;

 ya = yc;

end

if  b-a < delta, break, end

end

c = (a + b) / 2;

err = abs(b - a);

yc = feval(f, c);

-------------------------------------------------------------------------------------

Enter the function in matlab like this:

f= @(x) 0.005*(exp(2*x)*cos(x))

You should get this result:

f =

 function_handle with value:

   @(x)0.005*(exp(2*x)*cos(x))

Now run the code like this:

[c, err, yc] = bisect (f, 1, 2, 1e-10)

You should get this result:

c =

   1.5708

err =

  5.8208e-11

yc =

 -3.0708e-12

In addition, you can use the plot function to verify your results:

fplot(f,[1,2])

grid on

5 0
3 years ago
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