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mixas84 [53]
3 years ago
11

Write a MIPS assembly language program that

Computers and Technology
1 answer:
8_murik_8 [283]3 years ago
5 0

Answer:

123456789098765432

Explanation:

asdfghjklokjhgfdsasertyujn nbgfdfvbnjujhgvfcdec vgfredfghjmk

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Help with scripting/coding questions!
user100 [1]

Answer:

Explanation:

For the first questions the variable's name is <u>Today,</u> and is declared like a date variable, this will be returned a date value.

For the second question, where the browser will show today's date because with the variable Today, we get the date and with the document.write(Today) is shown the variable information.

With the third question, the result will be "Hello!", because the August is the 8 month, and July the 7, in this case the IF is False and for that will show "Hello!".

5 0
3 years ago
PLEASE HELP I HAVE BEEN STUCK ON THIS QUESTION FOR 5m NOW
Nataly_w [17]

Answer:

They are personal, formal and informal.

Explanation:

Hope this helps!

3 0
3 years ago
Read 2 more answers
The basic information in a database program is stored in a table format similar to a:
Ierofanga [76]

Can you add the options?

7 0
3 years ago
Is the method a network interface uses to access the medium and send data frames and the structure of these frames?
Alja [10]
Network technology <span>Is the method a network interface uses to access the medium and send data frames and the structure of these frames.</span><span />
3 0
3 years ago
Suppose we perform a sequence of stack operations on a stack whose size never exceeds kk. After every kk operations, we make a c
Ainat [17]

Answer:

The actual cost of  (n) stack operations will be two ( charge for push and pop )

Explanation:

The cost of (n) stack operations involves two charges which are actual cost of operation of the stack which includes charge for pop ( poping an item into the stack) which is one and the charge for push ( pushing an item into the stack ) which is also one. and the charge for copy which is zero. therefore the maximum size of the stack operation can never exceed K because for every k operation there are k units left.

The amortized cost of the stack operation is constant 0 ( 1 ) and the cost of performing an operation on a stack made up of n elements will be assigned as 0 ( n )

7 0
3 years ago
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