Answer:
Flowchart of an algorithm (Euclid's algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers a and b in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≥ A yields "yes" or "true" (more accurately, the number b in location B is greater than or equal to the number a in location A) THEN, the algorithm specifies B ← B − A (meaning the number b − a replaces the old b). Similarly, IF A > B, THEN A ← A − B. The process terminates when (the contents of) B is 0, yielding the g.c.d. in A. (Algorithm derived from Scott 2009:13; symbols and drawing style from Tausworthe 1977).
Explanation:
Flowchart of an algorithm (Euclid's algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers a and b in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≥ A yields "yes" or "true" (more accurately, the number b in location B is greater than or equal to the number a in location A) THEN, the algorithm specifies B ← B − A (meaning the number b − a replaces the old b). Similarly, IF A > B, THEN A ← A − B. The process terminates when (the contents of) B is 0, yielding the g.c.d. in A. (Algorithm derived from Scott 2009:13; symbols and drawing style from Tausworthe 1977).
Name servers.
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Programmers/Coders write the instructions for a Computer.
When drilling, students in a study session use flashcards to quiz one another. thus, Option B is the correct statement.
<h3>
What do you mean by drill and practice?</h3>
The term drill and practice can be described as a way of practice characterized by systematic repetition of concepts, examples, and exercise problems.
Drill and exercise is a disciplined and repetitious exercise, used as an average of coaching and perfecting an ability or procedure.
Thus, When drilling, students in a study session use flashcards to quiz one another. Option B is the correct statement.
Learn more about drill and practice:
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In a parallel connection, the voltage is same in every branch.
Now, three <span>three resistors connected in parallel.
R1 = 2k ohm
</span>R2 = 6k ohm
R3 = 10k ohm
in parallel, net resisitance =
Now, putting the values, we get, R net = 1.30 k ohm.
Ans, voltage = 100 VDC
Thus, power =
= 7.69 Watt
