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3 years ago

Which of the following groups would an animal with an exoskeleton, segmented body, and jointed appendages belong to???

Computers and Technology

2 answers:

DerKrebs [107]3 years ago

8 0

D) All Arthropoda have skeletons

Annette [7]3 years ago

7 0

All arthropods have exoskeletons, like insects, spiders and crustaceans.

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A company has an Aruba solution. The company wants to host a guest login portal with this solution, and the login portal must gi

mariarad [96]

Answer: D. Choose ClearPass or the other external captive portal option for the guest WLAN.

Explanation:

The Aruba Instant On mobile app allows the configuration and monitoring of a network from anywhere. It provides an affordable solution to small businesses and they're reliable and secure.

Based on the criteria given in the question, the network administrator can meet the criteria by choosing a ClearPass or the other external captive portal option for the guest WLAN.

8 0

3 years ago

Maeve has created the website, but she is not able to include more than one blank space at a time. Even if she hits the SPACE bu

GenaCL600 [577]

Answer:

For the browser to display multiple spaces, Maeve must use   in her code to create more spaces.

Explanation:

  is known as hard space or fixed space. NBSP stands for Non-Breaking Space. The statement will continue without breaking into another line.

Maeve can easily use the   many times to get the required space she needs.

5 0

3 years ago

Assume you are given three variables, revenue, expenses, and profit, all of type Money (a structured type with two int fields, d

Andre45 [30]

Answer:

if(revenue.cents - expenses.cents < 0){

profit.dollars = revenue.dollars - expenses.dollars - 1;

profit.cents = 1 - revenue.cents - expenses.cents;

}

else{

profit.dollars = revenue.dollars - expenses.dollars;

profit.cents = revenue.cents - expenses.cents;

}

Explanation:

We know that profit is given as: revenue - expenses from the question.

From the given expression above;

if(revenue.cents - expenses.cents < 0)

then profit.dollar will be revenue.dollars - expenses.dollars - 1; the 1 is to be carry over to the cent part. And the profit.cent will be 1 - revenue.cents - expenses.cents;

else the profit.dollars and the profit.cent is computed directly without needing to carry over:

profit.dollars = revenue.dollars - expenses.dollars;

profit.cents = revenue.cents - expenses.cents;

7 0

2 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

A type of address translation in which a gateway has a pool of public ip addresses that it is free to assign to a local host whe

krok68 [10]

<span>Dynamic Network Address Translation (DNAT)</span>

7 0

2 years ago