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lbvjy [14]
3 years ago
9

Which statements about viruses are true? select the four statements that are true. select the four statements that are true. all

viral genomes contain both dna and rna. hiv contains two identical strands of dna. hiv contains reverse transcriptase. a retrovirus contains rna. the capsid enters the host cell if the virus is enveloped. all rna-containing viruses are retroviruses. enveloped viruses bud from the host cell?
Computers and Technology
1 answer:
navik [9.2K]3 years ago
4 0
<span>A) All viral genomes contain both DNA and RNA. FALSE. Viruses contain the smallest necessary amount of genetic information packaged in a capsule composed of proteins. Containing both DNA and RNA would be a redundancy that would unnecessarily increase the size of the virus and make it more difficult for the virus to penetrate a cell. B) A retrovirus contains RNA. TRUE. By definition a retrovirus contains single-stranded positive-sense RNA instead of DNA. C) HIV contains two identical strands of DNA. FALSE. HIV is a retrovirus and, therefore, contains RNA instead of DNA. D) HIV contains reverse transcriptase. TRUE. HIV is a retrovirus that contains an RNA genome. All retroviruses require reverse transcriptase to convert their genome from RNA to DNA once the virus has been entered the cell. E) The capsid enters the host cell if the virus is enveloped. TRUE. Viruses are to large to enter a cell by passive diffusion or active transport. The only remaining major form a transport into the cell is endocytosis, or being enveloped. F) All RNA-containing viruses are retroviruses. FALSE. By definition a retrovirus contains single-stranded positive-sense RNA. Viruses can contain RNA in other forms (ie double-stranded) and would, therefore, not be considered a retrovirus. G) Enveloped viruses bud from the host cell. TRUE. Viruses are to large to enter a cell by passive diffusion or active transport. The only remaining major form a transport into the cell is exocytosis. For a virus to exit the host cell and infect other cells, they must exit by exocytosis, or budding.</span>
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Answer:

8

Explanation:

Gear reduction = driven gear teeth / driving gear teeth

First stage reduction :

Driven gear teeth = 36

Driving gear teeth = 12

36 /12 = 3

Second stage reduction :

Driven gear teeth = 60

Driving gear teeth = 12

60 /12 = 5

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3 + 5 = 8

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Databars show the relative magnitude of values in a dataset. Sparkliness are tiny charts that reside in a cell in excel. These charts are used to show the trend over the time or variation in the dataset.

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lesya [120]

Answer:

If there’s one topic that trips people up (both new and experienced) in the networking industry, it is that of Subnetting.

One of the reasons this happens is that one has to perform (mental) calculations in decimal and also binary. Another reason is that many people have not had enough practice with subnetting.

In this article, we will discuss what Subnetting is, why it came about, its usefulness, and how to do subnetting the proper way. To make this article as practical as possible, we will go through many examples.

Note: While subnetting applies to both IPv4 and IPv6, this article will only focus on IPv4. The same concepts explained here can be applied to IPv6. Moreover, subnetting in IPv6 is more of a want rather than a necessity because of the large address space.

IP address network

For example, any traffic with a destination IP address of 192.168.1.101 will be delivered to PC1, while traffic addressed to 192.168.1.250 will be delivered to SERVER.

Note: This is an oversimplification of things just for understanding sake and refers to Unicast (one-to-one) IPv4 addresses. Traffic sent to Multicast (one-to-many) and Broadcast (one-to-all) IP addresses can be delivered to multiple devices. Also, features like Network Address Translation (NAT) allow one IP address to be shared by multiple devices.

To help your understanding of IP addresses and subnetting, you need to resolve the following fact in your head: Computers think in binary, that is, 0s and 1s. Therefore, even though we see an IP address represented like 192.168.1.250, it is actually just a string of bits – 32 bits in total for IPv4 addresses.

To make them more readable for humans, IPv4 addresses are represented in dotted decimal notation where the 32 bits are divided into 4 blocks of 8 bits (also known as an octet), and each block is converted to a decimal number.

For example, 01110100 in binary is 116 in decimal:

A unicast IPv4 address such as 192.168.1.250 can be divided into two parts: Network portion and Host ID. So what does this mean? Well, IPv4 addresses were originally designed based on classes: Class A to Class E. Multicast addresses are assigned from the Class D range while Class E is reserved for experimental use, leaving us with Class A to C:

Class A: Uses the first 8 bits for the Network portion leaving 24 bits for host IDs. The leftmost bit is set to “0”.

Class B: Uses the first 16 bits for the Network portion leaving 16 bits for host IDs. The two leftmost bits are set to “10”.

Class C: Uses the first 24 bits for the Network portion leaving 8 bits for host IDs. The three leftmost bits are set to “110”.

Note: The range of Class A is actually 1-126 because 0.x.x.x and 127.x.x.x are reserved.

With these classes, a computer/device can look at the first three bits of any IP address and determine what class it belongs to. For example, the 192.168.1.250 IP address clearly falls into the Class C range.

Looking at the Host ID portion of the classes, we can determine how many hosts (or number of individual IP addresses) a network in each class will support. For example, a Class C network will ideally support up to 256 host IDs i.e. from 00000000 (decimal 0) to 11111111 (decimal 255). However, two of these addresses cannot be assigned to hosts because the first (all 0s) represents the network address while the last (all 1s) represents the broadcast address. This leaves us with 254 host IDs. A simple formula to calculate the number of hosts supported

Explanation: Final answer is Start address: 192.168.58.0 + 1 = 192.168.58.1

End address: 192.168.58.16 – 2 = 192.168.58.14

Broadcast address: 192.168.58.16 – 1 = 192.168.58.15

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That's standard for document size by default.

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