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Yuliya22 [10]
3 years ago
7

GEOMETRY NEED HELP A.15 B.12 C.5 D.9

Mathematics
1 answer:
podryga [215]3 years ago
3 0
For the answer it would be A
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5х2-7x-6=0 <br><br> factored form
Vinvika [58]

Answer:

(x - 2)(5x + 3) = 0

Step-by-step explanation:

5x² - 7x - 6 = 0

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 5 × - 6 = - 30 and sum = - 7

The factors are - 10 and + 3

Use these factors to split the x- term

5x² - 10x + 3x - 6 = 0 ( factor the first/second and third/fourth terms )

5x(x - 2) + 3(x - 2) = 0 ← factor out (x - 2) from each term

(x - 2)(5x - 3) = 0 ← in factored form

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3 years ago
The number doing the dividing in a division problem
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Answer:

divisor

Step-by-step explanation:

I just looked it up bro lol

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3 years ago
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I have 3 quarter , 1 nickel , 1 dime , and 3 penny . But I want to divided 2 people .
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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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