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Softa [21]
3 years ago
15

What are the maximum data rates of 802.11 legacy, 802.11a, 802.11b, and 802.11g?

Mathematics
1 answer:
lina2011 [118]3 years ago
7 0
802.11 legacy = 2 Mbps<span>
802.11a = </span>54 Mbps<span>
802.11b = </span>11 Mbps<span>
802.11g = </span><span>54 Mbps</span>
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What is the value of x and how do you know
Nookie1986 [14]

Answer:

x=35°

Step-by-step explanation:

So first, recall that the interior angles of a triangle must total 180°.

The sum of the angles for the given triangle can be described by:

110+x+x\\=2x+110

Since the total must equal 180°, set the expression equal to 180°.

2x+110=180

To find the value of x, we just need to solve for x.

To start, subtract 110 from both sides. The 110s on the left cancels:

(2x+110)-110=(180)-110\\2x=70

Now, divide both sides by 2. The 2s on the left cancel.

\frac{(2x)}{2}=\frac{(70)}{2}\\ x=35 \textdegree

Therefore, the value of x is 35°.

5 0
3 years ago
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AysviL [449]
The answer is C
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4 0
2 years ago
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Write an equation in slope for a line that passes through (3,4) (and has a y intercept of -8)
Vlad [161]

Answer:

Part 1) The equation in slope intercept form is y=4x-8

Step-by-step explanation:

Part 1) Write an equation in slope-intercept form for a line that passes through (3,4) (and has a y intercept of -8)

we know that

The equation of a line in slope intercept form is equal to

y=mx+b

where

m is the slope

b is the y-intercept

we have

b=-8

point\ (3,4)

substitute

4=m(3)-8

solve for m

4+8=3m

3m=12

m=4

therefore

y=4x-8

5 0
3 years ago
Complete the equation and tell which property you used (18x2)x5=18x(2x____)
tangare [24]
<span>Associative Property.
(18x2)x5=18x(2x5)
All you have to do is move the parentheses but keep the numbers the same.</span>
4 0
3 years ago
What is the vertex of a parabola defined by the equation <br> x = 5y2?
QveST [7]

Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0

3 0
3 years ago
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