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Inessa [10]
3 years ago
12

Special programs that facilitate communication between a device and the os are called

Computers and Technology
1 answer:
beks73 [17]3 years ago
8 0
Drivers

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Rr lyrae stars pulsate, but with shorter periods and lower luminosities than cepheids. true false
KIM [24]

True, RR Lyrae stars pulsate, but with shorter periods and lower luminosities than Cepheids.

The pulsation periods of the fundamental-mode RR Lyrae stars range from about 0.3 to 1.0 days (Population II pulsators with longer periods are categorized as type II Cepheids), but the vast majority of them have generations longer than 0.45 days.

<h3>Why do RR Lyrae stars pulsate?</h3>

RR Lyrae stars pulse like Cepheid variables, but the nature and histories of these stars is thought to be rather different. Like all variables on the Cepheid fluctuation strip, pulsations are caused by the κ-mechanism, when the ambiguity of ionised helium varies with its temperature.

To learn more about RR Lyrae, refer

brainly.com/question/13946889

#SPJ4

8 0
2 years ago
What is the difference between an electronic notebook and electronic flash cards?
Ivenika [448]

Answer:

A) Electronic notebooks store images, videos, notes, and voice recordings in one area, while electronic flash cards are study tools with information written on both sides.

Explanation:

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7 0
3 years ago
What kind of software consists of programs designed to make users more productive and/or assist them with personal tasks?
Sunny_sXe [5.5K]
It is application software
6 0
2 years ago
What is the exact number of bytes in system of memory that contains (a) 64K bytes (In Binary)(b) 32M bytes
VashaNatasha [74]

Answer:

64 K bytes = 65536 bytes

32 M bytes = 33554432 bytes

Explanation:

The question expect the number of bytes in binary instead of decimal. So this is important to understand that:

  • 1K bytes = 1024 bytes (in binary)

Therefore,

  • 64 Kb = 64 x 1024 =  65536 bytes

Using the similar calculation logic, we know

  • 1M bytes = 1024 x 1024 = 1048576 bytes (in binary)

Therefore,

  • 32 M bytes = 32 x 1048576 =  33554432 bytes
5 0
3 years ago
A(n) _____ is a common output device for hard copy.
Sidana [21]
The answer is d it is d it is d I think I think, I’m not sure though
8 0
3 years ago
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