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Nezavi [6.7K]
3 years ago
12

Yo wus up my people, my names Pablo Tablo, I'm your avg joe looking for a very specific shawty, like 5'2 and blonde or brunette,

a spicy one who smelling sweet..
Hook me up fast, I'm thirsty as a bass: (3.2.3) - 824 - 1834

Peace ma dudes,
Don't forget, I need her fast
Mathematics
1 answer:
DochEvi [55]3 years ago
6 0

Answer:

I- ayo bud are you that desperate-? :3

Step-by-step explanation:

:3

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What is the inverse of the function f(x) = x + 3​
xenn [34]

Answer:

the answer would be f(x)=x-3

Step-by-step explanation:

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3 years ago
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What is the formula for the following geometric sequence? -5, -10, -20, -40, ... an = -5 · 2n - 1 an = -5 · (-2)n - 1 an = -2 ·
vazorg [7]
-5 * 2^{n-1}
8 0
3 years ago
HURRY QUICK!! Apply the distributive property to create an equivalent expression. 1/2 ( 6e - 3f - 3/4 )
tensa zangetsu [6.8K]

Answer:

1 2/6      3 3/4

Step-by-step explanation:

7 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:

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  • 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
Simplify. Assume that all variables are positive.
mihalych1998 [28]

Answer: -3x^{2} y\sqrt[3]{2y}

Step-by-step explanation:

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