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posledela
3 years ago
7

Edge Algebra 2 Q4 Answers:

Mathematics
1 answer:
Gnoma [55]3 years ago
6 0

Answer:

Step-by-step explanation:

1. C

2. A

3. B

4. A

5. D

6. C

7. A

8. C

9. B

10. D

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2 complementary angles are in the ratio 7:8.find the angles
slavikrds [6]
Complementary angles add up to 90 degrees
7:8....added = 15

7/15 * 90 = 630/15 = 42 <== one angle is 42 degrees
8/15 * 90 = 720/15 = 48 <== other angle is 48 degrees
3 0
4 years ago
a polygon has an area of 225 square meters. A similar polygon is created with an area that is triple the first. By what factor i
GuDViN [60]
The area of the original polygon is:
 A = 225 m ^ 2
 The similar polygon area is:
 A '= (k ^ 2) * (A)
 Substituting values:
 3 * 225 = (k ^ 2) * (225)
 Clearing k we have:
 k ^ 2 = 3
 k = (3) ^ (1/3)
 Answer:
 
The length of each side increased by:
 
k = (3) ^ (1/3)
6 0
3 years ago
Find FJ plsssssssssssss
Whitepunk [10]

Answer:

the value of FJ would end up being 10

7 0
3 years ago
Which graph represents the equation (x-3)^2 + y^2 =16
Goryan [66]

Answer:

C

Step-by-step

Correct on Edge

6 0
3 years ago
Read 2 more answers
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
4 years ago
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