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muminat
2 years ago
5

Help what is the equation

Mathematics
1 answer:
Hitman42 [59]2 years ago
3 0

Answer:

y=-5x

Step-by-step explanation:

take two points, I took 0,4 and 1-,1 and do

y2-y1 over x2-x1

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How to do this question plz answer me step by ​
Jet001 [13]

Answer:

481.92

Step-by-step explanation:

First find the increase

466.98 * 3.2%

466.98 * .032

14.94336

Add this to the original amount

466.98+14.94336

481.92336

Round to 2 decimal places

481.92

7 0
3 years ago
NOTE: Angles not necessarily drawn to scale.
Anastasy [175]

The answer is 13!! :)

7 0
3 years ago
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A quadrilateral has vertices (2, 0), (0, –2), (–2, 4), and (–4, 2). Which special quadrilateral is formed by connecting the midp
HACTEHA [7]
In order to know what the polygon is, you have to plot the coordinates. After plotting, it is obviously shown a form of a rectangle. After connecting the midpoints of the sides, it formed a (D) rhombus, not a kite.
8 0
3 years ago
Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

  • f(a) = a c {1} if a has an odd number of 1's
  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

  • If x has an even number of 1's, then the last digit of y has to be 0, and f(x) = x c {0} = y
  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

3 0
3 years ago
I will make you the brainleist!!!! Please explain and no guessing. Thank you!! Ps due in 10
aniked [119]
Is this even a question? its 66/100 you always use the first examining over 100.... makes sense
5 0
2 years ago
Read 2 more answers
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