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Nesterboy [21]
3 years ago
9

4) Two processes, A and B, each need three records, 1, 2, and 3 in a database. If A asks for them in the order 1,2,3, and B asks

for them in the same order, deadlock is not possible. With three resources, there are 3! (factoral 3*2*1) or six possible combinations in which each process can request them. What fraction of all the combinations is guaranteed to be deadlock free?
Computers and Technology
1 answer:
vfiekz [6]3 years ago
4 0

Answer:

Explanation:

If A request for 1 and B also request for 1, then A succeed and B is going to block. This will avoid the deadlock.

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simplify the expression below and state the value of m for which the simplified expression is not defined 2m² + m - 15/ m² - 9​
motikmotik

Answer:

  • The simplified expression is: \frac{2m-5}{m-3}
  • The simplified expression is undefined for m=3

Explanation:

The given expression is:

\frac{2m^2+m-15}{m^2-9}

The numerator can be siplified by using factorization and denominator will be simplified using the formula

a^2-b^2 = (a+b)(a-b)

So,

= \frac{2m^2+6m-5m-15}{(m)^2-(3)^2}\\=\frac{2m(m+3)-5(m+3)}{(m-3)(m+3)}\\=\frac{(2m-5)(m+3)}{(m-3)(m+3)}\\=\frac{2m-5}{m-3}

A fraction is undefined when the denominator is zero. In order to find the value of m on which the simplified fraction will be undefined we will put denominator equal to zero.

So,

m-3 = 0 => m = 3

Hence,

  • The simplified expression is: \frac{2m-5}{m-3}
  • The simplified expression is undefined for m=3
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