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Alexxx [7]
2 years ago
6

For the following exercises, determine whether the relation is a function.

Mathematics
1 answer:
elixir [45]2 years ago
7 0

Answer:

The given relation $\{(a, b),(c, d),(e, d)\}$ is a function.

Step-by-step explanation:

A relation $\{(a, b),(c, d),(e, d)\}$ is given.

It is required to determine whether the given relation is a function.

To determine whether the given function is a relation, identify the domain and range and then check whether the given relation is a function.

Step 1 of 1

The given relation is {(a, b),(c, d),(e, d)}.

The set of the first components of each ordered pair is called the domain.

From the relation, the domain is {a, c, e}.

The set of the second components of each ordered the range.

From the relation, the range is {b, d, d}.

The given relation is a function. But it is not a one-to-one function.

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Answer:

I'll setup the problem and leave the computation to you

Step-by-step explanation:

The equation to calculate fixed payments

P =  \frac{r(PV)}{1- {(1 + r)}^{ - n} }

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PV = present value (or the amount borrowed)

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r = .25/4 (4 months = quarter of a year)

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if you have questions, put them in the comment

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3 years ago
1.64 as a mixed number in simplest form
garri49 [273]
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3 years ago
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Find the value of this expression if x = 3.<br> x2 + 3/X
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Step-by-step explanation:

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I'm doing Pre-algebra and I don't know how would I figure out the 13th term of the pattern; 100, 91, 82, 73.... without writing
Over [174]
The way we would answer the question would be to express it as

100-9(n-1) with n being the nth term of the pattern.

Why is this? We know we subtract 9 each time and START from 100.
Remember though that 100 is the first term in the sequence. So, we would have (n-1)*9 to subtract 0 from our 1st term and 9 from our second term.

Also, we have 9*(n-1) because we are just subtracting multiples of 9. It is always going to subtract by 9.


If you needed the answer here it is...

The answer to the problem would be 
100-9(13-1)
= 100-9(12)
= 100-108
= -8

5 0
3 years ago
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