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3 years ago

is there an app i can use to make a picture appear in the corner of a video then disappear after being up for a second?

Computers and Technology

2 answers:

Nuetrik [128]3 years ago

7 0

Umm i really do not know i'm sorry

WITCHER [35]3 years ago

6 0

Answer:

vivacut or funimate are good for me

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An ip address in the address range 169.254.x.y, used by a computer when it cannot successfully lease an ip address from a dhcp s

Georgia [21]

Yes that is the correct answer

5 0

3 years ago

Direct Mapped Cache. Memory is byte addressable. Fill in the missing fields based upon the properties of a direct-mapped cache.

xz_007 [3.2K]

The question given is incomplete and by finding it on internet i found the complete question as follows:

Direct Mapped Cache.

Memory is byte addressable

Fill in the missing fields based upon the properties of a direct-mapped cache. Click on "Select" to access the list of possible answers Main Memory Size Cache Size Block Size Number of Tag Bits 3 1) 16 KiB 128 KiB 256 B 20 2) 32 GiB 32 KiB 1 KiB 3) 64 MiB 512 KiB 1 KiB Select] 4 KiB 4) 16 GiB 10 Select ] Select ] 5) 10 64 MiB [ Select ] 6) Select] 512 KiB 7

For convenience, the table form of the question is attached in the image below.

Answers of blanks:

1. 3 bits

2. 20 bits

3. 64 MB

4. 16 MB

5. 64 KB

6. 64 MB

Explanation:

Following is the solution for question step-by-step:

Part 1:

No. of Tag bits = No. of bits to represent

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 128 KB = $2^7 * 2^{10} = 2^{17}$

Cache Memory = 16 KB = $2^4 * 2^{10}= 2^{14}$

Putting values in A:

Tag bits = 17 - 14 = 3 bits

Part 2:

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 32 GB = $2^5 * 2^{30} = 2^{35}$

Cache Memory = 16 KB = $2^5 * 2^{10}= 2^{15}$

Putting values in A:

Tag bits = 35 - 15 = 20 bits

Part 3:

Given:

Tag bits = 7

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

So from equation A

7 = Main Memory size - 19

Main Memory = 7 + 19

Main memory = 26

Main Memory = $2^6 * 2^{20} = 64 MB$

Part 4:

Given that:

Main Memory Size = $2^4 * 2^{30} = 2^{34}$

Tag bits = 10

Cache Memory Bits = 34 - 10 = 24

Cache Memory Size = $2^4 * 2^{20} = 16 MB$

Part 5:

Given that:

Main Memory Size = 64 MB = $2^6 * 2^{20}$

Tag bits = 10

Cache Memory Bits = 26 - 10 = 16

Cache Memory Size = $2^{16} = 2^6 * 2^{10} = 64 KB$

Part 6:

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

Tag Bits = 7

Main Memory Bits = 19 + 7 = 26

Main Memory size = $2^{26} = 2^6 * 2^20 = 64 MB$

i hope it will help you!

6 0

2 years ago

Complete a graphic organizer to explain the responsibilities of a borrower

Blizzard [7]

A borrower is a computer expert, create interest and pay a certain percentage

5 0

2 years ago

Which tab is the "page break" command found on? (there are two!) *

vitfil [10]

Answer:

home and layout

Explanation:

8 0

2 years ago

Please answer immediately!!!

diamong [38]

Answer:

Please check the attachment.

Remember:

All N1, N2, N3N, N4, N5, N6, N7 and each of them are certainly like the server. Also note that Router, repeaters, switch and transmitter are not mentioned in the above diagram. Please assume it in between each location.

Explanation:

Please check the attachment. Also, Remember that Server is at the head office, and rest 7 are the 7 branches. The VOIP, video surveillance, Door security system and the computer networks are being shown in the diagram. And each location has a network of computers joined through VLAN, and each of them are given the IP addresses of Class A, Class B and class C of the iPV4(or IPV6), which can be found by the future network administer, operators through the IP address, that is assigned.

4 0

3 years ago