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ANEK [815]
3 years ago
9

Which one if these sets of ordered pairs is not a function ? help please

Mathematics
1 answer:
abruzzese [7]3 years ago
7 0
A function has only one y-value for each x-value. If you find a set that has an x-value used more than once, it is not a function, because that x-value is now paired with more than one y-value.

The fourth answer is the only one that has the same x-value used more than once.

Side note: Don’t worry if a y-value is used more than once. That’s totally fine. It’s all about the x-values.
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What is 79.35 divided by 5
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79.35 divided by 5 equals: <u><em>15.87</em></u>
<em>~~~~~~~~~~~~~~~~~~~~~~~~~~~</em>
To check:
    243
    15.87
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3 years ago
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2 years ago
Use the Distributive Property to rewrite
Stella [2.4K]

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c

Step-by-step explanation:

6 0
2 years ago
PLEASE HELP ASAP!!!!
Setler [38]

Piecewise Function is like multiple functions with a speific/given domain in one set, or three in one for easier understanding, perhaps.

To evaluate the function, we have to check which value to evalue and which domain is fit or perfect for the three functions.

Since we want to evaluate x = -8 and x = 4. That means x^2 cannot be used because the given domain is less than -8 and 4. For the cube root of x, the domain is given from -8 to 1. That meand we can substitute x = -8 in the cube root function because the cube root contains -8 in domain but can't substitute x = 4 in since it doesn't contain 4 in domain.

Last is the constant function where x ≥ 1. We can substitute x = 4 because it is contained in domain.

Therefore:

\large{  \begin{cases} f( - 8 ) =   \sqrt[3]{ - 8}  \\ f(4) = 3 \end{cases}}

The nth root of a can contain negative number only if n is an odd number.

\large{  \begin{cases} f( - 8 ) =   \sqrt[3]{ - 2 \times -  2 \times   - 2}  \\ f(4) = 3 \end{cases}} \\  \large{  \begin{cases} f( - 8 ) =  - 2\\ f(4) = 3 \end{cases}}

Answer

  • f(-8) = -2
  • f(4) = 3
6 0
2 years ago
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