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4 years ago

11

High capacity circuits are limited by the least capable network link, which is typically the user connection. As such, it is imp

ortant to provide capable broadband access to end users.
A. True
B. False

Computers and Technology

1 answer:

Mariana [72]4 years ago

4 0

The answer is “true”

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What is the best programing language to use for building video games?

Alexxandr [17]

hey

I'm going to college for game design.

one of the best languages and the one I'm studying in is c# it is used in unity and many other game engines but there are many more. Just to list a few c++, Java, and many more it is up to you. if you would like more info about this just let me know By the way what game are you planning to make that is one of the most important factors

Hope this helps

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3 years ago

wariber [46]

Answer:

It would definitely be true studied a lot of this stuff.

Explanation:

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3 years ago

Create an interactive program to use class a LightsOut class to allow a user to play a game. Each step in the game will require

expeople1 [14]

Answer:

9+10=21

Explanation:

wyd boy

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3 years ago

[30 points, will mark Brainliest] Which of the following is the lowest hexadecimal value? Explain why. Options to chose; F2, 81,

larisa86 [58]

Answer:

The answer to this question is given below in the explanation section.

Explanation:

First, we need to convert these hexadecimal numbers into decimal numbers, then we can easily identify which one is the lowest hexadecimal.

The hexadecimal numbers are F2, 81, 3C, and 39.

F2 = (F2)₁₆ = (15 × 16¹) + (2 × 16⁰) = (242)₁₀

81 = (81)₁₆ = (8 × 16¹) + (1 × 16⁰) = (129)₁₀

3C = (3C)₁₆ = (3 × 16¹) + (12 × 16⁰) = (60)₁₀

39 = (39)₁₆ = (3 × 16¹) + (9 × 16⁰) = (57)₁₀

The 39 is the lowest hexadecimal number among the given numbers.

Because 39 hex is equal to 57 decimal.

39 = (39)₁₆ = (3 × 16¹) + (9 × 16⁰) = (57)₁₀

5 0

3 years ago

Find the propagation delay for a signal traversing the in a metropolitan area through 200 km, at the speed of light in cable (2.

Ede4ka [16]

Answer:

t= 8.7*10⁻⁴ sec.

Explanation:

If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.

As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.

Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:

$v = \frac{(xf-xo)}{(t-to)}$

If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:

$t =\frac{xf}{v} =\frac{2e5 m}{2.3e8 m/s} =8.7e-4 sec.$

⇒ t = 8.7*10⁻⁴ sec.

4 0

3 years ago