In DTP, a good rule of thumb is to use as many special effects as you need to make your documents look interesting.

1 answer:

5 0

It all depends back on who youre present it t

younger ages (Primary-Middle),
Its Ok to use plenty of special effects, but dont ovedue your DTP with effects, so that they still can get the point.

Older ages (High & above)
Older people usually more serious than older ages to pursue their careers, so less effect, more points

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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

Binary search requires that the data to search be in order. True or false

iren [92.7K]

Hello Pennstatealum 

Question: Binary search requires that the data to search be in order.
True or false

Answer: True

Hope This Helps
-Chris

6 0

3 years ago

Convert the following hexadecimal numbers to decimal: (a) FF (b) F0A2 (c) 0F100 (d) 100

Hatshy [7]

Answer:

a) 255

b) 61602

Explanation:

You need to understand the decimal equivalent of hexadecimal numbers, from 0 to 9 numbers are represented the same way, from 10 to 15 we use the alphabet, meaning 10 equals A in hexadecimal base, 11-B, 12-C, 13-D, 14-E, and 15-F.

For your first exercise you'll enumerate the number's positions fromright to lef begining with 0:

a. F F

position 1 0

Now you'll multiply your hexadecimal number (using the decimal equivalent for your letters) for the base (16) elevated to the number of the position:

$F*16^{0}=15*1=15\\F*16^{1}=15*16=240$