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Yuki888 [10]
3 years ago
15

How to create a photo album in PowerPoint 2013

Computers and Technology
1 answer:
DerKrebs [107]3 years ago
8 0

Access the Insert tab of the Ribbon, and click the down-arrow within Photo Album button. This brings up a small menu -- choose the New Photo Album option, This opens the Photo Album dialog box Now click the ‘File/Disk….’ button , which opens the Insert New Pictures dialog box Navigate to where your photos are, and select the desired pictures:  To select all pictures in your folder, press the keyboard shortcut Ctrl + A.  To select adjacent pictures, click on the first picture, then press and hold the Shift key and select the last picture in the sequence. To select multiple non-adjacent pictures, press and hold the Ctrl key while clicking on each picture you want to insert.  When the selection is made, click the Insert button. This will add the selected pictures in the Pictures in album list in the Photo Album dialog box.  The Photo Album dialog box provides several options such as re-ordering the pictures, making picture adjustments, changing picture layouts, and applying Themes. click the Create button, above to insert the pictures you have selected.  A seperate presentation will be created for the Photo Album, Save your presentation.


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4 years ago
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Tcecarenko [31]

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Read 2 more answers
Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 1024-byte 4-way set as
Cerrena [4.2K]

Answer:

The answer is The Cache Sets (S) = 32, Tag bits (t)=24, Set index bits(s) = 5 and Block offset bits (b) = 3

Explanation:

Solution

Given Data:

Physical address = 32 bit (memory address)

Cache size = 1024 bytes

Block size = 8 bytes

Now

It is a 4 way set associative mapping, so the set size becomes 4 blocks.

Thus

Number of blocks = cache size/block size

=1024/8

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= 32

Hence the number of sets = 32

←Block ←number→

Tag → Set number→Block offset

←32 bit→

Now, =

The block offset = Log₂ (block size)

=Log₂⁸ = Log₂^2^3 =3

Then

Set number pc nothing but set index number

Set number = Log₂ (sets) = log₂³² =5

The remaining bits are tag bits.

Thus

Tag bits = Memory -Address Bits- (Block offset bits + set number bits)

= 32 - (3+5)

=32-8

=24

So,

Tag bits = 24

Therefore

The Cache Sets = 32

Tag bits =24

Set index bits = 5

Block offset bits = 3

Note: ←32 bits→

Tag 24 → Set index  5→Block offset 3

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Answer:

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