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IgorC [24]
2 years ago
11

For what values of b is the relation R:{(b^2, 5), (5b, 6)} NOT a function?

Mathematics
1 answer:
Roman55 [17]2 years ago
3 0

Answer:

b=0,b=5

Step-by-step explanation:

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True or False: Exponential growth and exponential decay graphs are always<br> functions
adell [148]

Answer:

true

Step-by-step explanation:

8 0
3 years ago
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Could the inverse of a non-function be a function? Explain or give an example.
Kitty [74]

Answer:

The inverse of a non-function mapping is not necessarily a function.

For example, the inverse of the non-function mapping \lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\! is the same as itself (and thus isn't a function, either.)

Step-by-step explanation:

A mapping is a set of pairs of the form (a,\, b). The first entry of each pair is the value of the input. The second entry of the pair would be the value of the output.  

A mapping is a function if and only if for each possible input value x, at most one of the distinct pairs includes x\! as the value of first entry.

For example, the mapping \lbrace (0,\, 0),\, (1,\, 0) \rbrace is a function. However, the mapping \lbrace (0,\, 0),\, (1,\, 0),\, (1,\, 1) \rbrace isn't a function since more than one of the distinct pairs in this mapping include 1 as the value of the first entry.

The inverse of a mapping is obtained by interchanging the two entries of each of the pairs. For example, the inverse of the mapping \lbrace (a_{1},\, b_{1}),\, (a_{2},\, b_{2})\rbrace is the mapping \lbrace (b_{1},\, a_{1}),\, (b_{2},\, a_{2})\rbrace.

Consider mapping \lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\!. This mapping isn't a function since the input value 0 is the first entry of more than one of the pairs.

Invert \lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\! as follows:

  • (0,\, 0) becomes (0,\, 0).
  • (0,\, 1) becomes (1,\, 0).
  • (1,\, 0) becomes (0,\, 1).
  • (1,\, 1) becomes (1,\, 1).

In other words, the inverse of the mapping \lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\! would be \lbrace (0,\, 0),\, (1,\, 0),\, (0,\, 1),\, (1,\, 1) \rbrace\!, which is the same as the original mapping. (Mappings are sets. There is no order between entries within a mapping.)

Thus, \lbrace (0,\, 0),\, (0,\, 1),\, (1,\, 0),\, (1,\, 1) \rbrace\! is an example of a non-function mapping that is still not a function.

More generally, the inverse of non-trivial ellipses (a class of continuous non-function \mathbb{R} \to \mathbb{R} mappings, including circles) are also non-function mappings.

3 0
2 years ago
-5y - 6bStep 1 of 2: Identify the first term of the algebraic expression. Indicate whether theterm is a variable term or a const
ira [324]

Answer:

a. Variable term

b. Variable term

Explanation:

a) We were given the algebraic expression:

-5y-6b

The first term of the algebraic expression is:

-5y

The first term is a variable term

The variable is "y" and its coefficient is "-5"

b) We were given the algebraic expression:

-5y-6b

The second term of the algebraic expression is:

-6b

The second term is a variable term

The variable is "b" and the coefficient is "-6"

6 0
1 year ago
What expressions represent 5/6 × 5
pickupchik [31]

Answer:

5 \div 6 \times 5 = 25\div 6

<h2>25/6 <em>is </em><em>the </em><em>Answer</em></h2>
7 0
3 years ago
16.
KatRina [158]

Answer:

D

Step-by-step explanation:

is the correct answer i believe.

3 0
3 years ago
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