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VMariaS [17]
3 years ago
14

Many instruction sets contain the instruction NOOP, meaning no operation, which has no effect on the processor state other than

incrementing the program counter.
Suggest some uses of this instruction.

Computers and Technology
1 answer:
lions [1.4K]3 years ago
8 0

The answer & explanation for this question is given in the attachment below.

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Select the three careers in the creative side of digital media.
MrRa [10]

Answer:

I would pick A B and D to be my answer

6 0
3 years ago
Which programming languages are the best choices for desktop applications? Select 3 options.
slega [8]
The three are Python c and html
5 0
3 years ago
Read 2 more answers
Does a triangle with sides of lengths 1.5 ft, 2.5 ft, and 2 ft form a right triangle? How do you know?
timama [110]

Answer:

Yes, the side lengths of 1.5, 2.5, and 2 ft form a right triangle.

Explanation:

Before we solve this problem, we need to know a couple of things.

In a right triangle, you have two legs and a diagonal line that goes from one leg to the other or from one side to the other. This line is called the hypotenuse. It is the longest side of a right triangle. This length can be calculated by using a special formula that is over a thousand years old.

The formula is called the Pythagorean Theorem.

The formula states that if you square one leg and the other leg and add it up. You get the side of hypotenuse which is squared as well.

Here it is in simple terms: a^{2} +b^{2} =c^{2}.

a and b are your two legs.

C is your hypotenuse.

In this case 1.5 and 2 are your legs and 2.5 is your hypotenuse since it is the longest side.

Before we calculate, there is a converse to the Pythagorean Theorem.

If a^{2} +b^{2} \neq c^{2}, then it is not a right triangle.

Lets make 1.5 for a

and 2 for b.

We know 2.5 is for c.

Here it is.

1.5^{2} +2^{2} =2.5^{2}. The / means it is a fraction.

1.5 squared is 9/4 and 2 squared is 4. Add 4 and 9/4 and you get 25/4 squared. Take the square root of 25/4 and you get 5/2.

5/2 is equal to 2.5, so the Pythagorean Theorem applies to this triangle. Therefore, these side lengths do form a right triangle.

Hope this helps!

3 0
3 years ago
Write a procedure named Str_find that searches for the first matching occurrence of a source string inside a target string and r
kirill115 [55]

Answer: Provided in the explanation section

Explanation:

Str_find PROTO, pTarget:PTR BYTE, pSource:PTR BYTE

.data

target BYTE "01ABAAAAAABABCC45ABC9012",0

source BYTE "AAABA",0

str1 BYTE "Source string found at position ",0

str2 BYTE " in Target string (counting from zero).",0Ah,0Ah,0Dh,0

str3 BYTE "Unable to find Source string in Target string.",0Ah,0Ah,0Dh,0

stop DWORD ?

lenTarget DWORD ?

lenSource DWORD ?

position DWORD ?

.code

main PROC

  INVOKE Str_find,ADDR target, ADDR source

  mov position,eax

  jz wasfound           ; ZF=1 indicates string found

  mov edx,OFFSET str3   ; string not found

  call WriteString

  jmp   quit

wasfound:                   ; display message

  mov edx,OFFSET str1

  call WriteString

  mov eax,position       ; write position value

  call WriteDec

  mov edx,OFFSET str2

  call WriteString

quit:

  exit

main ENDP

;--------------------------------------------------------

Str_find PROC, pTarget:PTR BYTE, ;PTR to Target string

pSource:PTR BYTE ;PTR to Source string

;

; Searches for the first matching occurrence of a source

; string inside a target string.

; Receives: pointer to the source string and a pointer

;    to the target string.

; Returns: If a match is found, ZF=1 and EAX points to

; the offset of the match in the target string.

; IF ZF=0, no match was found.

;--------------------------------------------------------

  INVOKE Str_length,pTarget   ; get length of target

  mov lenTarget,eax

  INVOKE Str_length,pSource   ; get length of source

  mov lenSource,eax

  mov edi,OFFSET target       ; point to target

  mov esi,OFFSET source       ; point to source

; Compute place in target to stop search

  mov eax,edi    ; stop = (offset target)

  add eax,lenTarget    ; + (length of target)

  sub eax,lenSource    ; - (length of source)

  inc eax    ; + 1

  mov stop,eax           ; save the stopping position

; Compare source string to current target

  cld

  mov ecx,lenSource    ; length of source string

L1:

  pushad

  repe cmpsb           ; compare all bytes

  popad

  je found           ; if found, exit now

  inc edi               ; move to next target position

  cmp edi,stop           ; has EDI reached stop position?

  jae notfound           ; yes: exit

  jmp L1               ; not: continue loop

notfound:                   ; string not found

  or eax,1           ; ZF=0 indicates failure

  jmp done

found:                   ; string found

  mov eax,edi           ; compute position in target of find

  sub eax,pTarget

  cmp eax,eax    ; ZF=1 indicates success

done:

  ret

Str_find ENDP

END main

cheers i hoped this helped !!

6 0
3 years ago
One or more access points positioned on a ceiling, wall, or other strategic spot in a public place to provide maximum wireless c
Lera25 [3.4K]

Answer:

hotspots.

Explanation:

Hotspots is a small geographical location with at least one access point in an outdoor/indoor setting, that provide people with maximum wireless coverage within that area and its mostly uses a wireless local-area network (WLAN) using a router to connect to an Internet service provider.

7 0
3 years ago
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