Majken is turning 18 years old soon and wants to calculate how many seconds she will have been alive.

To exclude members of a group from the basic permissions for a folder, which type of permission would you assign? Deny Allow Mod

oksano4ka [1.4K]

Answer:

Deny

Explanation:

Under the New Technology System, the Deny permission is applied when the administrator wants to overrule the permission given to members of a group. The Allow and Deny options help in regulating access to the components of the system. Only authorized groups are granted access to the files.

Most times, the Deny permission is considered before the Allow permission. If the user is denied access to some files and granted access to some, the deny permission is most times considered first.

4 0

3 years ago

Which principle of design does a designer apply when they create rhythmic movement in a garment by using recurring pleats or det

WITCHER [35]

Answer:

Repetition, and regular rhythm

Explanation:

Its certainly repetition method, and repetition method is of three types which are repetition, pattern and rhythm.

Remember repetition can be boring at times as we need to read same thing again and again. However, repetition sometimes can be good like the one we are referring to, and the other one can be like placing the logo or menu at same place in all web page. And its since it creates a habit in us for reading the web pages, And hence we can understand faster, and it looks please to our eyes hence as well.

4 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

1. an image can be stored either in a vector graphic file or in a bitmap file true or false

lions [1.4K]

Answer:

1 true 2 true 3 false 4 true

Explanation:

7 0

3 years ago

Look at the logic program below. Set a breakpoint after each commented instruction. You can also single step through the program

Goshia [24]

Using programming knowledge, it is possible to use programming logic, we can differentiate and find the terms

<h3>Writing the code and understanding the programming logic:</h3>

.data

sourceword:

.word 0xAB

wordmask:

.word 0xCF

operand:

.long 0xAA

.text

.globl _start

_start:

movw sourceword, %ax # ax=0x00AB ( move the word value of sourceword =0x00AB to ax register.)

movw %ax, %bx # bx=0x00AB ( copy the value of ax to bx register.)

xorw %cx, %cx # cx=0000 ( cx=cx XOR cx, that is clear cx register.)

andw wordmask, %ax # ax=0x008B ( ax=ax AND wordmask= 0x00AB AND 0x00CF= 0x008B)

orw wordmask, %bx # bx=0x00EF (bx=bx OR wordmask= 0x00AB OR 0x00CF= 0x00EF.)

xorw wordmask, %cx # cx=0x00CF (cx=cx XOR wordmask= 0x0000 XOR 0x00CF=0x00CF.)

movl operand, %eax # eax=0x000000AA. ( copy the 32-bit value of operand=0x000000AA to eax.)

movl %eax, %ebx # ebx=0x000000AA.( copy the value of eax to ebx.)

movl %eax, %ecx # ecx=0x000000AA. ( copy the value of eax to ecx.)

shll $3,%eax # eax=0x00000550. ( logical shift left of eax=0x000000AA by 3 bits.)

# 0x000000AA=00000000000000000000000010101010

# =>00000000000000000000010101010000=0x00000550.

rorl $4,%ebx # ebx=0xA000000A. (Rotate right ebx=0x000000AA by 4 bits.)

# 0x000000AA=00000000000000000000000010101010

3 =>10100000000000000000000000001010= 0xA000000A.

sarl %ecx # here the count value to shift the bits not mentioned.

# It is the arithmetic shift right instruction. the shifted left bits filled with MSB bit.

See more about logic program at brainly.com/question/14970713

#SPJ1

5 0

2 years ago