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Stels [109]
3 years ago
11

Geometry, please answer question ASAP

Mathematics
2 answers:
zaharov [31]3 years ago
6 0
Answer:

m∠A = 101*

Step-by-step explanation:

m∠A = (25x + 1)*

m∠B = 82*

m∠C = (25x - 2)*

m∠D = (20x - 1)*

Find m∠A. Key: solve for x

(25x + 1) + 82 + (25x - 2) + (20x - 1) = 360*

Subtract 82 from both sides

(25x + 1) + (25x - 2) + (20x - 1) = 278*
…

x = 4

Now substitute x for 4 in A measure

(25x + 1)

(25(4) + 1)

(100 + 1)

101

m∠A = 101*
Amanda [17]3 years ago
5 0

Answer:

D. 101

Step-by-step explanation:

(25x+1)+(25x-2)+(20x-1)+82=360

70x=360-1+2+1-82

70x=280

x=4

A=25x+1=101

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Find the additive inverse or opposite of -4/3
Vikentia [17]
It would be positive 4/3 or the inverse would be -3/4
5 0
3 years ago
The ratio of the number of Sam's CDs to the number of Alice's CD's is 7 to 8. Sam has 84 CDS.
Alex

Answer:

180 CDs

Step-by-step explanation:

So if Sam has 84 CDs, and the ratio from Sam to Alice is 7:8, we can do 84/7=12 so each number is equal to 12. Then we add up 7 and 8 and get 15, so we do 15*12 and we get 180.

7 0
3 years ago
Multiply using the distributive property 12(5y+4).
iragen [17]

Answer:

12(5y+4)

12(9y)

108y

hope it helps and I hope I did it right

Step-by-step explanation:

5 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
Someone plzzzzzz!!!!!
Alex777 [14]

Answer:

yes

Step-by-step explanation:

5 0
3 years ago
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