Which postulate proves the two
1 answer:
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Answer: The correct answer is:
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The given "graph" in the bottom right, lowest corner
Step-by-step explanation:
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Note: When there is only one (1) equation give for a graph;
and/or: only one (1) "inequality given";
we look for the symbol.
If the symbol is "not" an "equals" symbol (i.e. not an: = symbol) ;
we check for the type of "inequality" symbol.
If there is a: "less than" (<) ; or a "greater than" (>) symbol; the graph of the "inequality" will have "dashed lines" (since there will be a "boundary").
If there is an "inequality" that is a: "less than or equal to" (≤) ;
or a: "greater than or equal to" (≥) ;
→ then there will be not be a dashed line when graphed;
but rather—a "solid line" ; since "less than or equal to" ;
or "greater than or equal to" —is similar to:
"up to AND including"; or: "lesser/fewer than AND including".).
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Note: We are given the "inequality" :
→ " y < x² + 4x " .
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Note that we have a "less than" symbol (< ) ; so the graph will have a:
"solid line" [and not a "dotted line".].
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Note that all of the graphs among our 4 answer choices have "dotted lines".
Not that all values (all x and y coordinated) within the "shaded portion" of the corresponding graph are considered part of the graph.
As such, given any point within the shaded part, the x and y coordinates must match the inequality (i.e. the given inequality must be true when one puts in the "x-coordinate" and "y-coordinate" into the "given inequality" :
→ " y < x² + 4x " .
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Likewise, we can take any point within the "white, unshaded" portion of any of the graph, and take the "x-coordinate" and "y-coordinate" of that point, and the inequality: → " y < x² + 4x " ; will not hold true when the "x-coordinate" and "y-coordinate" values of that point— are substituted into the "inequality".
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{Note: Answer is continued on images attached.}.
Wishing you the best!
