Given scalene triangle DEF, explain how to find locus of points equidistant from DE EF DF using incenter

1 answer:

5 0

Points equidistant from DE EF are in the bisector of angle DEF
points equidistant from EF DF are in the bisector of angle EFD
the sought after point is the intersection of bisectricess of triangle

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Match each given domain of the function f(x)=2/2.5 with its corresponding range.

mamaluj [8]

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$\text{Domain:}\{40,\ 45,\ 50,\ 55,\ 60\}\\\\f(40)=\dfrac{40}{2.5}=16\\\\\text{Range:}\{16,\ 18,\ 20,\ 22,\ 24\}\\----------------------------\\\text{Domain:}\{52,\ 56,\ 63,\ 67,\ 71\}\\\\f(16)=\dfrac{52}{2.5}=20.8\\\\\text{Range:}\{20.8,\ 22.4,\ 25.2,\ 26.8,\ 28.4\}\\----------------------------$

$\text{Domain:}\{37,\ 46,\ 49,\ 55,\ 57\}\\\\f(37)=\dfrac{37}{2.5}=14.8\\\\\text{Range:}\{14.8,\ 18.4,\ 19.6,\ 22,\ 22.8\}\\----------------------------\\\text{Domain:}\{44,\ 51,\ 59,\ 66,\ 73\}\\\\f(44)=\dfrac{44}{2.5}=17.6\\\\\text{Range:}\{17.6,\ 20.4,\ 23.6,\ 26.4,\ 29.2\}\\----------------------------$