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cluponka [151]
2 years ago
5

True / False An instruction’s opcode generally indicates the number and type of its operands.

Computers and Technology
1 answer:
liq [111]2 years ago
7 0

Answer:

True

Explanation:

An opcode is the part of instruction which specifies the operation to be performed by the instruction.

In general, the opcode also provides information about the number and type of operands.

For example, let us consider the MIPS instructions for addition.

  • ADD reg_dest, reg_src1, reg_src2

This instruction adds the contents of registers reg_src1 and reg_src2 and stores the result in reg_dest.

  • Whereas, ADDI reg_src, reg_dest, value

This instruction adds the value to the content of reg_src and stores the result in reg_dest.

As we can see the opcode type indicates the operand type, number and semantics.

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The term eof represents
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The text discusses three approaches to combining overall cost leadership and differentiation competitive advantages. Which of th
dedylja [7]

Answer:

C. deriving benefits from highly focused and high technology markets

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circular_prime A number is called a circular prime if all rotations of its digits form a prime. For example, the number 197 is a
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Answer:

function out=circular_primes(no)

prim=primes(no);% find the all prime number till the given number

pr=0;

nos=[];

po=[];

for p=1:length(prim)

   n=prim(p); % picking up each prime no one by one

   n=num2str(n);% change into string for rotation of no

   d=length(n); % length of string

   if d>1          % take nos greater than 10 beacuase below 10 no need for rotation

       for h=1:d

           a=n(1);

           for r=1:d % for rotation right to left

               if r==d  5 % for the last element of string

                   n(d)=a;

               else

               n(r)=n(r+1); %shifting

               end

           end

           

       s=str2num(n); % string to number

       nos=[nos,s]; % store rotated elements in array

       end

       if nos(end)==no   %if given no is also a circular prime we need smaller

           break;

       end

       for gr=1:length(nos) % checking rotated nos are prime or not

           p1=isprime(nos(gr));

           po=[po,p1];    %storing logical result in array

       end

           if sum(po(:))==length(nos) %if all are prime the length and sum are must be equal

               

               pr=pr+1;

               out=pr;

           else

               out=pr;

           end

       po=[];

       nos=[];

   else  

       s=str2num(n); %numbers less than 10

       f=isprime(s);

       if f==1

           pr=pr+1;

           out=pr;

       else

           out=pr;

       end

   end

       

      end

end

Explanation:

3 0
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