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Scilla [17]
4 years ago
11

Which relation represents a function

Mathematics
1 answer:
BARSIC [14]4 years ago
3 0

Answer:

A relation is a function if each element of the domain is paired with exactly one element of the range. If given a graph, this means that it must pass the vertical line test.

Step-by-step explanation:

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zhenek [66]

Answer:    \bold{y=\dfrac{1\pm \sqrt{4x+13}}{2}}

<u>Step-by-step explanation:</u>

Inverse is when you swap the x's and y's and solve for y:

y+\dfrac{13}{4}=\bigg(x-\dfrac{1}{2}\bigg)^2\\\\\\x+\dfrac{13}{4}=\bigg(y-\dfrac{1}{2}\bigg)^2\rightarrow \text{(swapped the x's and y's)}\\\\\\\sqrt{x+\dfrac{13}{4}}=\sqrt{\bigg(y-\dfrac{1}{2}\bigg)^2}\rightarrow \text{(took square root of both sides)}\\\\\\\sqrt{\dfrac{4x+13}{4}}=\pm \bigg(y-\dfrac{1}{2}\bigg)}\rightarrow \text{(created common denominator in radical)}\\\\\\ \dfrac{\sqrt{4x+13}}{2}=\pm \bigg(y-\dfrac{1}{2}\bigg)}\rightarrow \text{(simplified radical)}

\pm \dfrac{\sqrt{4x+13}}{2}=y-\dfrac{1}{2}}\rightarrow \text{(divided both sides by}\ \pm )\\\\\\\dfrac{1}{2}\pm \dfrac{\sqrt{4x+13}}{2}=y\quad \rightarrow \text{(added}\ \dfrac{1}{2}\ \text{to both sides)}\\\\\\\dfrac{1\pm \sqrt{4x+13}}{2}=y\quad \rightarrow \text{(combined numerators)}

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3 years ago
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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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olya-2409 [2.1K]

Answer:

the last one

(-15)x12

hope this helps!

3 0
3 years ago
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