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Naya [18.7K]
2 years ago
14

How to find only f on a graph that defines a function

Mathematics
1 answer:
puteri [66]2 years ago
3 0

Answer:

Read Below

Step-by-step explanation:

we can represent a function using a graph. Graphs display many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. We typically construct graphs with the input values along the horizontal axis and the output values along the vertical axis.

The vertical line test can be used to determine whether a graph represents a function. A vertical line includes all points with a particular

x

value. The

y

value of a point where a vertical line intersects a graph represents an output for that input

x

value. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that

x

value has more than one output. A function has only one output value for each input value.

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7 1/6 = 5 2/5 + p<br><br> p = ???
Crank

\huge\textbf{Hey there!}

\mathbf{7\dfrac{1}{6} = 5\dfrac{2}{5} + p}

\rightarrow\mathbf{7 \dfrac{1}{6} = 5\dfrac{2}{5} + p}

\mathbf{\rightarrow p + 5\dfrac{2}{5}= 7\dfrac{1}{6}}

\mathbf{\rightarrow {p + \dfrac{27}{5}=\dfrac{43}{6}}}

\text{SUBTRACT }\rm{\dfrac{27}{5}}\text{ to BOTH SIDES}

\mathbf{p + \dfrac{27}{5} - \dfrac{27}{5} = \dfrac{43}{6}- \dfrac{27}{5}}

\text{CANCEL out: }\rm{\dfrac{27}{5} - \dfrac{27}{5}}\text{ because it gives you 0}

\text{KEEP: }\rm{\dfrac{43}{6} - \dfrac{27}{5}}\text{ because it help solve for the p-value}

\text{NEW EQUATION: }\rm{p = \dfrac{43}{6} - \dfrac{27}{5}}

\large\textsf{SIMPLIFY IT!}

\mathbf{p = \dfrac{53}{30} \approx p = 1\dfrac{23}{30}}

\huge\boxed{\textsf{Therefore, your answer is: \boxed{\mathsf{p = 1\dfrac{23}{30}}}}}\huge\checkmark

\huge\textbf{Good luck on your assignment \&}\\\huge\textbf{enjoy your day!}

~\frak{Amphitrite1040:)}

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