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3 years ago

Which hypervisor works on older pcs without hardware virtualization support?

2 answers:

7 0

It is called Virtual Box. VirtualBox has a to a great degree secluded plan with very much characterized inside programming interfaces and a customer/server outline. This makes it simple to control it from a few interfaces without a moment's delay: for instance, you can begin a virtual machine in a regular virtual machine GUI and after that control that machine from the charge line, or potentially remotely.

patriot [66]3 years ago

5 0

The hypervisor which works on older PC's without hardware virtualization support is called VirtualBox. It is a software allows you to run operating systems in special environment which is called virtual machine. It means that you can run another OS without re-installing existing one.

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Consider the following declaration: double[] sales = new double[50]; int j; Which of the following correctly initializes all the

vlada-n [284]

Answer:

The answer is "None of these".

Explanation:

In the given question an array "sales[]" is declared, which contains 50 double type elements, and in the next line, an integer variable j is defined, which uses a for loop. In this question two options is given, in which both are not correct, that can be described as follows:

In option (i), A loop is defined that, uses variable j which starts with 0 and ends with 48, So total elements are 48 that's why it is not correct.
In option (ii), A loop will use variable j that, starts with 1 and ends with 49, That's why it is not correct.

3 0

3 years ago

To provide for unobtrusive validation, you can install the ____________________ package for unobtrusive validation.

NARA [144]

Answer:

AspNet.ScriptManager.jQuery?

Explanation:

Unobtrusive validation means we can perform a simple client-side validation without writing a lot of validation code by adding suitable attributes and also by including the suitable script files.

One of the benefits of using a unobtrusive validation is that it help to reduce the amount of the Java script that is generated. We can install the AspNet.ScriptManager.jQuery? for the unobtrusive validation.

When the unobtrusive validation is used, validation of the client is being performed by using a JavaScript library.

4 0

3 years ago

Microsoft acknowledged that if you type a res:// url (a microsoft-devised type of url) which is longer than ____ characters in i

cluponka [151]

Which is longer than 20 characters

5 0

3 years ago

Refer to the exhibit. An administrator is examining the message in a syslog server. What can be determined from the message? Thi

ad-work [718]

Answer:

This is a notification message for a normal but significant condition

Explanation:

Syslog represents the standard for logging message, it sends messages through UDP port 514. Familiar syslog facilities includes IP, OSPF protocols, etc. The messages from syslog are both about facility and level.

A syslog server is a means through which network devices sends messages about events into a logging server which is called syslog server. Since a syslog protocol supports so many devices, it can also be used to log a good number of events.

8 0

3 years ago

Read 2 more answers

Determine and prove whether an argument in English is valid or invalid. About Prove whether each argument is valid or invalid. F

yawa3891 [41]

Answer:

Each understudy on the respect roll got an A.

No understudy who got a confinement got an A.

No understudy who got a confinement is on the respect roll.

No understudy who got an A missed class.

No understudy who got a confinement got an A.

No understudy who got a confinement missed class

Explanation:

M(x): x missed class

An (x): x got an A.

D(x): x got a confinement.

¬∃x (A(x) ∧ M(x))

¬∃x (D(x) ∧ A(x))

∴ ¬∃x (D(x) ∧ M(x))

The conflict isn't considerable. Consider a class that includes a lone understudy named Frank. If M(Frank) = D(Frank) = T and A(Frank) = F, by then the hypotheses are overall evident and the end is counterfeit. Toward the day's end, Frank got a control, missed class, and didn't get an A.

Each understudy who missed class got a confinement.

Penelope is an understudy in the class.

Penelope got a confinement.

Penelope missed class.

M(x): x missed class

S(x): x is an understudy in the class.

D(x): x got a confinement.

Each understudy who missed class got a confinement.

Penelope is an understudy in the class.

Penelope didn't miss class.

Penelope didn't get imprisonment.

M(x): x missed class

S(x): x is an understudy in the class.

D(x): x got imprisonment.

Each understudy who missed class or got imprisonment didn't get an A.

Penelope is an understudy in the class.

Penelope got an A.

Penelope didn't get repression.

M(x): x missed class

S(x): x is an understudy in the class.

D(x): x got a repression.

An (ax): x got an A.

H(x): x is on the regard roll

An (x): x got an A.

D(x): x got a repression.

∀x (H(x) → A(x)) a

¬∃x (D(x) ∧ A(x))

∴ ¬∃x (D(x) ∧ H(x))

Real.

1. ∀x (H(x) → A(x)) Hypothesis

2. c is a self-self-assured element Element definition

3. H(c) → A(c) Universal dispatch, 1, 2

4. ¬∃x (D(x) ∧ A(x)) Hypothesis

5. ∀x ¬(D(x) ∧ A(x)) De Morgan's law, 4

6. ¬(D(c) ∧ A(c)) Universal dispatch, 2, 5

7. ¬D(c) ∨ ¬A(c) De Morgan's law, 6

8. ¬A(c) ∨ ¬D(c) Commutative law, 7

9. ¬H(c) ∨ A(c) Conditional character, 3

10. A(c) ∨ ¬H(c) Commutative law, 9

11. ¬D(c) ∨ ¬H(c) Resolution, 8, 10

12. ¬(D(c) ∧ H(c)) De Morgan's law, 11

13. ∀x ¬(D(x) ∧ H(x)) Universal hypothesis, 2, 12

14. ¬∃x (D(x) ∧ H(x)) De Morgan's law, 13

4 0

3 years ago