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PtichkaEL [24]
3 years ago
15

Normal wear and road conditions can take their toll on a car’s steering and suspension system, altering __________. A. wheel ali

gnment B. gear ratios C. tire air pressure D. brake operation
Computers and Technology
2 answers:
OLga [1]3 years ago
7 0

The answer is A: wheel alignment  

Bad road conditions and normal wear can take their toll on a car’s steering and suspension system altering the wheel alignment settings and throwing them out of specifications. When your car’s tires start wandering or pulling to one side, make sure to align your tires. A car’s wheels are aligned to achieve maximum tire life, enhance the comfort of your car, and improve fuel economy and handling. Wheel alignment involves adjusting the wheels so that they are parallel to each other. and perpendicular to the ground.



irakobra [83]3 years ago
5 0
Come on man you really don't know this its gear ratios.
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The _____ feature automatically corrects typing, spelling, capitalization, or grammar errors as you type them.
sergejj [24]
Spell-check feature is the one that auto corrects any misspelled words
5 0
3 years ago
11)When, if ever, will the geometric average return exceed the arithmetic average return for a given set of returns?A) When the
grigory [225]

Answer: E. Never

geometric average return can NEVER exceed the arithmetic average return for a given set of returns

Explanation:

The arithmetic average return is always higher than the other average return measure called the geometric average return. The arithmetic return ignores the compounding effect and order of returns and it is misleading when the investment returns are volatile.

Arithmetic returns are the everyday calculation of the average. You take the series of returns (in this case, annual figures), add them up, and then divide the total by the number of returns in the series. Geometric returns (also called compound returns) involve slightly more complicated maths.

6 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
Please help guys I'm so lost ​
PilotLPTM [1.2K]

Answer:

The correct answer is C ( W * 5 )

4 0
3 years ago
Read 2 more answers
Each of the following programs has errors. Find as many as you can. 65. // Find the error in this program. #include using namesp
Serggg [28]

Answer:

There are two error in this program--

  1. In header file inclusion, file is not defined.
  2. In the statement "result = ++(num1 + num2);" , bracket is fixed after the increment operator.

Explanation:

  • For the first error, the user needs to add the file because "#include" is used to add the library for the program which states about the function and symbols used in the program.
  • The second error is because there must be a variable with the increment operator ( increment operator is being used to increase the value of a variable by 1), but there is a small brace fix in between the operator and operands.
7 0
3 years ago
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