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IrinaK [193]
3 years ago
13

Write some general rules used while chatting​

Computers and Technology
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

Well,

Explanation:

Well, this topic is a very broad one and I'm not exactly sure what kind of information you' like provided, but I got a few ideas in mind.

While chatting online, a big letdown for most people is if you curse and especially if you do it excsessivley. Doing so will make you seem toxic and most people would just try to avoid you.

Talking about toxicity, being a toxic person online is petty and quite pathetic. Gloating and raising your ego won't make you any more of a liekable person online, or even in the real world.

Last but not least, another general rule of online chatting is to make sure you know who you're talking too. Talking to strangers online can be very dangerous, in some cases leading to things like pedophilia etc.

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First you go to the 3 dots neer the x you pres it thin you go to downloads and it shod be ther i hop i help if so make me the branliy

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3 years ago
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What’s unique about windows 8 compared to earlier windows
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3 years ago
WAP to find area of circle​
Tatiana [17]

Answer:

program by LET statement

Explanation:

CLS

REM to find the area of circle

LET r=10

LET a=22/7*2*r^2

PRINT "area=;"a

END  

                                                                                         press f5

output:

a=628.57

7 0
3 years ago
(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-
I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit  memory  to add in 2 location:

\to \frac{0}{1}  =2^1  \ (\frac{m}{m}  \ location)

The address of 2-bit  memory to add in 4 location:

\to \frac{\frac{00}{01}}{\frac{10}{11}}  =2^2  \ (\frac{m}{m}  \ location)

similarly,

Complete 'n'-bit memory address' location number is = 2^n.Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

= 2^n \\\\ = 2^{12} \\\\ = 4096

In point b:

\to Let \  Mega= 10^6

              =10^3\times 10^3\\\\= 2^{10} \times 2^{10}

So,

\to 2 \ Mega =2 \times 2^{20}

                 = 2^1 \times 2^{20}\\\\= 2^{21}

The memory position for '2^n' could be 'n' m bits'  

It can use 2^{21} bits to address the memory location of 21.  

That is to say, the 2-mega-location memory needs '21' bits.  

Memory Length = 21 bit Address

In point c:

i^{th} element array addresses are given by:

\to address [i] = B+w \times (i-LB)

_{where}, \\\\B = \text {Base  address}\\w= \text{size of the element}\\L B = \text{lower array bound}

\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address  = 10

\to address  [10] = \$ 52 + 4 \times (10-0)\\

                       =   \$ 52   + 40 \ bytes\\

1 term is 4 bytes in 'MIPS,' that is:

= \$ 52  + 10 \ words\\\\ = \$ 512

In point d:

\to  base \ address = \$ t 5

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are  < 4 byte

Calculating address:

\to address  [5] = \$ t5 + 4 \times (5-0)\\

                     = \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20  \ bytes  \\\\= \$ t5 + 5 \ words  \\\\= \$ t 10  \ words  \\\\

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4 years ago
Help me please, I'm attempting a test<br>​
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