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Nimfa-mama [501]
3 years ago
6

What kind of service is typically not offered by hosting service​

Computers and Technology
1 answer:
Finger [1]3 years ago
7 0

Answer:

I think the answer is Dynamic DNS

Explanation:

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What are the four major software packages of microsoft​
Reika [66]

Answer:

» Microsoft word ( word processing )

» Microsoft powerpoint ( presentation )

» Microsoft access ( database mamagement )

» Microsoft excel ( spread sheets )

Explanation:

.

3 0
3 years ago
Have fire have ....<br><br>babi​ from babi098
Genrish500 [490]

Answer:

yessss I tooo have a fire at my home

lolololololololololo

ok have a great day what is your name ?

4 0
3 years ago
Edhesive 6.8 lesson practice answers
FromTheMoon [43]

Answer:

1.) 25 ; 15 ; 15

2.) 50 ; 15 ; 50

Explanation:

In the first function written :

The variable val was initially decaled or assigned a value of 25 and that was what was printed first.

However, after the example function was written, the val variable was finally assiagned a value of 15 within the function. However, it was also declared that the global variable takes uonthe val value. Hence, the val variable initially assigned a value, of 25 changes to 15 globally.

For the second code :

From the top:

Val was assigned a value of 50 ;

Hence,

print(val) gives an output of 50

Within the function definition which prints the value of val that is assigned a value of 25 within the function.

Since tbe global variable isnt reset.

Printing Val again outputs 50;since ito is outside the function.

6 0
2 years ago
URLs are the global ______ of resources on the Internet.
adell [148]
I believe the answer might be addresses 
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4 0
3 years ago
Read 2 more answers
Suppose you have a 16-bit machine with a page size of 16B. Assume that any unsigned 16-bit integer can be a memory address. Also
irinina [24]

Answer:

2^11

Explanation:

Physical Memory Size = 32 KB = 32 x 2^10 B

Virtual Address space = 216 B

Page size is always equal to frame size.

Page size = 16 B. Therefore, Frame size = 16 B

If there is a restriction, the number of bits is calculated like this:  

number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]

where

physical memory size = 32KB  which is the restriction

n bit machine = frame size = 16

Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11

7 0
3 years ago
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