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3 years ago

What is the difference of expository and reflexive documentary

1 answer:

3 0

The reflexive documentary mode focuses on the relationship between the filmmaker and the audience. While expository documentaries set up a specific point of view or argument about a subject and a narrator often speaks directly to the viewer.

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If a*b = 2a - 56, calculate the value of<br>3 * 4

tino4ka555 [31]

Answer:

A=34

Explanation:

Rewrite the equation as 2a−56=3⋅4

2a−56=3⋅4

Multiply 3 by 4

2a−56=12

Move all terms not containing a to the right side of the equation.

2a=68

Divide each term by 2

and simplify.

a=34

3 0

3 years ago

______ devices are high-performance storage servers that are individually connected to a network to provide storage for the comp

natali 33 [55]

NAS - network attached storage

3 0

4 years ago

The next generation ip version and successor to ipv4 is called what? ipv5 ipv6 iana ssl

Olenka [21]

I study IT,it was Ipv 6 in our textbook

4 0

4 years ago

The net force on a vehicle that is accelerating at a rate of 1.2 m/s2 is 1500 newtons. What is the mass of the vehicle to the ne

Rudiy27

Answer:

The mass of the vehicle is 1250kg

Explanation:

Given

$Net\ Force = 1500N$

$Acceleration = 1.2m/s^2$

Required

Determine the vehicle's mass

This question will be answered using Newton's second law which implies that:

$Net\ Force (F) = Mass (m) * Acceleration (a)$

In other words:

$F = ma$

Substitute values for F and a

$1500 = m * 1.2$

Make m the subject

$m = \frac{1500}{1.2}$

$m = 1250kg$

<em>Hence, the mass of the vehicle is 1250kg</em>

6 0

3 years ago

The instruction format of a computer is given by one indirect bit, opcode bits, register address bits, immediate operand bits an

oksian1 [2.3K]

Answer:

Following are the answer to this question:

Explanation:

The memory size is 1 Giga Bytes which is equal to $2^{30}$

$\texttt{Number of address bits \ \ \ \ \ \ \ \ \ \ \ \ \ \ Number of addresses}\\\\ \ \ \ \ 30 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2^{30}= 1073741824$

B) $\texttt{I \ \ \ Opcode \ \ \ Register address \ \ \ Immediate operand \ \ \ Memory address}\\\\\\\textt{1 \ bit \ \ \ \ \ \ 128 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \ bits \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 24 \ bits \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 30 \ bits}\\\\\\$ $= 2^7 \\\\ = 7 \ bits$

calculating the register Bits:

$= 64-(1+7+24+30)\\\\=64 -62\ \ bits\\\\= 2\ \ bits\\$

Immediate value size while merging the additional benefit with the address field:

$= 2^{24} + 2^{30}\\\\= 2^{54}\\\\$ $\texttt{Range before combining(-,+) 24 bits \ \ \ \ \ \ Range after combining( -,+)54bits}\\\\$ $\textt{-2^{12} from + (2^{12}-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2^{27} from + (2^{27}- 1)}$

$= \frac{24}{2} = 12\\\\= \frac{54}{2} = 27$

The range is accomplished by dividing the bits by 2 into the two sides of the o and the number is one short to 0.

8 0

3 years ago