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goblinko [34]
3 years ago
8

Let C be the nullspace of the 1 x 9 matrix H1 (6.2.6) In other words, let C be the parity check code of length 9 (8 data bits, 1

parity check bit) (a) Use our standard methods to find a basis B for C. (b) What is the dimension of C? In general, what will the dimension of the parity check code of length n be? (c) Let G be the matrix whose columns are the vectors in the basis B from part (a). For a given message bitstring m, explain why the encoding procedure of Example 6.2.8 is equivalent to transmitting x-Gm.
Mathematics
1 answer:
Roman55 [17]3 years ago
5 0

Answer:hahahha

Step-by-step explanation:shsbshbss

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2. Write the equation of a circle that has a diameter of 12 units if its center is at (4,7).
ahrayia [7]

Answer:

The Answer is B

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Here diameter = 12, thus radius = 12 ÷ 2 = 6 and (h, k) = (2.5, - 3.5), thus

(x - 2.5)² + (y - (- 3.5))² = 6², that is

(x - 2.5)² + (y + 3.5)² = 36 ← equation of circle

5 0
3 years ago
Given GH with endpoints (-7,3) andH (7. - 11), what are the coordinates of the<br> mielpoint of GH?
Y_Kistochka [10]

Answer:

Midpoint is (0, -4)

Step-by-step explanation:

x-coordinate of midpoint = (-7 + 7)/2 = 0/2 = 0

y-coordinate of midpoint = (3 - 11)/2 = -8/2 = -4

Midpoint is (0, -4)

8 0
3 years ago
Read 2 more answers
Rationalize 1 plus root 2 divided by 1 minus root 2
yulyashka [42]
The rationalized form of 1+ √2/1-√2 is -2√2-3

6 0
3 years ago
Give a real-life situation for the sum-15 + 10​
iren [92.7K]

Answer:

I owed a friend 15 bucks till someone payed for me 2/3 of what I owed so I only had to pay him back 5 bucks.

8 0
3 years ago
Read 2 more answers
Help me to do 10(b) answer​
antiseptic1488 [7]

Answer:

Step-by-step explanation:

Big circle:

R = radius = diameter ÷2 = 42 ÷ 2 = 21 cm

Area of big circle= πR²

                            =\dfrac{22}{7}*21*21=22*3*21\\\\\\= 1386 \ cm^{2}

Small circle:

Diameter of small circle  = radius of big circle = 21 cm

r = 21/2 = 10.5 cm

Area of small circle = πr²

                                =\dfrac{22}{7}*10.5*10.5 = 22 * 1.5*10.5\\\\\\= 346 .5 \ cm^{2}

Area of shaded region = area of big circle - area of small circle

= 1386 - 346.5

= 1039.5 cm²

                                 

3 0
3 years ago
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