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lilavasa [31]
3 years ago
6

The following algorithm should output the t times table in the format:

Computers and Technology
1 answer:
kow [346]3 years ago
5 0

t = INT(INPUT(“Enter the times table you want to output:”))

FOR x = 1 TO 12

p = t * x

PRINT ( x + " × " + t + " = " + p)

NEXT x

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Write a C# solution for the following problem. Submit your .cs file to this link. Sample output(s) attached. == Create an Employ
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Answer:

See attached file for detailed solution.

Explanation:

See attached file for explanation.

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3 years ago
In modern web design, color, font, font size, and style should be declared using:
inn [45]
They should be declared using style sheets (CSS).
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- Consider the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = { {B, C} -> {D}, {B} -&
Olenka [21]

Answer:

The key of R is {A, B}

Explanation:

A key can be seen as a minimal set of attributes whose closure includes all the attributes in R.

Given that the closure of {A, B}, {A, B}+ = R, one key of R is {A, B} But in this case, it is the only key.

In order for us to to normalize R intuitively into 2NF then 3NF, we have to make use of these approaches;

First thing we do is to identify partial dependencies that violate 2NF. These are attributes that are

functionally dependent on either parts of the key, {A} or {B}, alone.

We can calculate

the closures {A}+ and {B}+ to determine partially dependent attributes:

{A}+ = {A, D, E, I, J}. Hence {A} -> {D, E, I, J} ({A} -> {A} is a trivial dependency)

{B}+ = {B, F, G, H}, hence {A} -> {F, G, H} ({B} -> {B} is a trivial dependency)

To normalize into 2NF, we remove the attributes that are functionally dependent on

part of the key (A or B) from R and place them in separate relations R1 and R2,

along with the part of the key they depend on (A or B), which are copied into each of

these relations but also remains in the original relation, which we call R3 below:

R1 = {A, D, E, I, J}, R2 = {B, F, G, H}, R3 = {A, B, C}

The new keys for R1, R2, R3 are underlined. Next, we look for transitive

dependencies in R1, R2, R3. The relation R1 has the transitive dependency {A} ->

{D} -> {I, J}, so we remove the transitively dependent attributes {I, J} from R1 into a

relation R11 and copy the attribute D they are dependent on into R11. The remaining

attributes are kept in a relation R12. Hence, R1 is decomposed into R11 and R12 as

follows: R11 = {D, I, J}, R12 = {A, D, E} The relation R2 is similarly decomposed into R21 and R22 based on the transitive

dependency {B} -> {F} -> {G, H}:

R2 = {F, G, H}, R2 = {B, F}

The final set of relations in 3NF are {R11, R12, R21, R22, R3}

4 0
3 years ago
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lidiya [134]

Answer:

<u>the first friend</u>

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Consider also, IT experts often acknowledge that in terms of speed of execution, it is proven that, "dual-core systems" are <em>faster</em> (even twice faster) than a "single-core system". The other friends were wrong because they disagreed with a widely accepted fact that dual-core is faster than single-core; and of course, <em>we know that without them being faster they can't run twice the applications and twice the data. </em>

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nirvana33 [79]

Answer:

it's slide master view.

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