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sweet [91]
3 years ago
14

G = {(5,3), (2, 3), (6,4)} Is G^-1 a function and why?

Mathematics
1 answer:
sleet_krkn [62]3 years ago
5 0

Answer:

The inverse relation G^(-1) is not a function. Why not? Because the y value y = 3 is paired up with more than one x value (x = 5, x = 2). The inverse relation G^(-1) is the set shown below

{(3,5), (3,2), (4,6)}

All I've done is swap the (x,y) values for each ordered pair to form the inverse relation. As you can see, x = 3 leads to multiple y value outputs which is why this relation is not a function. So in short, the answer is choice C. To have the inverse relation be a function, each value in the original domain must map to exactly one value in the range only. However that doesn't happen as the domain values map to an overlapping y value (y = 3).

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Hi,

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7 0
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Determine whether
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Answer:

(a) and (b) are not equivalent

(c) is equivalent

Step-by-step explanation:

Given

\frac{25^m}{5}

Required

Determine an equivalent or nonequivalent expression

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We have:

25^{m-1

Apply law of indices

25^{m-1} = \frac{25^m}{25}

<em>This is not equivalent to </em>\frac{25^m}{5}<em></em>

(b)\ 25^{2m - 1}

We have:

25^{2m - 1}

Apply law of indices

25^{2m - 1} = \frac{25^{2m}}{25}

<em>This is not equivalent to </em>\frac{25^m}{5}<em></em>

<em></em>

<em />(c)\ 5^{2m-1}<em />

We have:

<em />5^{2m-1}<em />

Apply law of indices

<em />5^{2m-1} = \frac{5^{2m}}{5^1}<em />

<em />5^{2m-1} = \frac{5^{2m}}{5}<em />

<em>Evaluate the numerator</em>

<em />5^{2m-1} = \frac{25m}{5}<em />

<em />

<em>This is an equivalent expression</em>

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