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3 years ago

What are the first and the last physical memory addressesaccessible using

Computers and Technology

1 answer:

Misha Larkins [42]3 years ago

6 0

Answer

For First physical memory address,we add 00000 in segment values.

For Last physical memory address,we add 0FFFF in segment values.

NOTE-For addition of hexadecimal numbers ,you first have to convert it into binary then add them,after this convert back it in hexadecimal.

a)1000

For First physical memory address, we add 00000 in segment value

We add 0 at the least significant bit while calculating.

10000 +00000 = 10000 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

we add 0FFFF in segment value

10000 +0FFFF =1FFFF (from note)

b)0FFF

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

0FFF0 +00000=0FFF0 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

we add 0FFFF in segment value

0FFF0 +0FFFF =1FFEF (from note)

c)0001

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

00010 +00000=00010 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

we add 0FFFF in segment value

00010 +0FFFF =1000F (from note)

d) E000

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

E0000 +00000=E0000 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

we add 0FFFF in segment value

E0000 +0FFFF =EFFFF (from note)

e) 1002

For First physical memory address,we add 00000 in segment value

We add 0 at the least significant bit while calculating.

10020 +00000=10020 (from note)

For Last physical memory address,We add 0 at the least significant bit while calculating.

we add 0FFFF in segment value

10020 +0FFFF =2001F (from note)

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Keeping in mind the role the order of precedence plays in equations, what would Excel display as the result of the following equation?

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3 years ago

Which is a connectionless protocol in the transport layer? What are the small chunks of data called?

xeze [42]

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User Datagram Protocol (UDP) , transport-layer segment.

Explanation:

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A continuous and differentiable function f(x) with the following properties: f(x) is decreasing at x=−5 f(x) has a local minimum

butalik [34]

The continuous and differentiable function where f(x) is decreasing at x = −5 f(x) has a local minimum at x = −2 f(x) has a local maximum at x = 2 is given as: y = 9x - (1/3)x³ + 3.

<h3>What is a continuous and differentiable function?</h3>

The continuous function differs from the differentiable function in that the curve obtained is a single unbroken curve in the continuous function.

In contrast, if a function has a derivative, it is said to be differentiable.

<h3>What is the solution to the problem above?</h3>

It is important to note that a function is differentiable when x is set to a if the function is continuous when x = a.

Given the parameters, we state that

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x = -5

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the local maximum is given as

x = 3

Thus, x = -3 ; alternatively,

x = 3. With this scenario, we can equate both to zero.

Hence,

x + 3 = 0;

3-x = 0.

To get y' we must multiply both equations to get:

y' = (3-x)(x + 3)

y' = 3x + 9 - x² - 3x

Collect like terms to derive:

y' = 3x - 3x + 9 - x²; thus

y' = 9-x²

When y' is integrated, the result is

y = 9x - (x³/3) + c

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F (-5) < 0

This means that:

9 x -5 - (-5³/3) + c < 0
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Collecting like terms we have:
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c < 3.33

Substituting C into

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2 years ago

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Answer:

See explaination

Explanation:

program code.

/* PRE PROCESSOR DIRECTIVES */

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int saturating_add(int firstNumber, int secondNumber)

{

FOR BETTER UNDERSTANDING, LETS TAKE TEST CASE,

WHERE firstNumber = 5 AND secondNumber = 10

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sizeof(firstNumber) VALUE IS 4, SO USING BINARY LEFT SHIFT OPERATOR TO THREE PLACES,

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int addition = firstNumber + secondNumber;

MASK INTEGER VARIABLE IS TAKEN

mask BIT IS LEFT SHIFTED TO 31 PLACES => 2^31 IS THE NEW VALUE

int mask = 1 << (w - 1);

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int msbFirstNumber = firstNumber & mask;

/* SECOND NUMBER MOST SIGNIFICANT BIT IS CALCULATED BY USING AND OPERATOR */

int msbSecondNumber = secondNumber & mask;

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int msbAddition = addition & mask;

/* POSITIVE OVERFLOW IS DETERMINED */

int positiveOverflow = ~msbFirstNumber & ~msbSecondNumber & msbAddition;

/* NEGATIVE OVERFLOW IS DETERMINED */

int negativeOverflow = msbFirstNumber & msbSecondNumber & !msbAddition;

/* THE CORRESPONDING VALUE IS RETURNED AS PER THE SATURATING ADDITION RULES */

(positiveOverflow) && (addition = TMax);

(negativeOverflow) && (addition = TMin);

return addition;

}

/* MAIN FUNCTION STARTS HERE */

int main(){

/* TEST CASE */

int sum = saturating_add(5, 10);

/* DISPLAY THE RESULT OF TEST CASE */

printf("The Sum Is : %d\n\n",sum);

}

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