What is the distance (4/5 , 1/10),(1/5 , 15/10) ?

The distance (4/5 , 1/10),(1/5 , 15/10) is (sqrt(58))/5

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distance (4/5 , 1/10),(1/5 , 15/10)

distance

(\frac{4}{5}, \frac{1}{10}), (\frac{1}{5}, \frac{15}{10})

\frac{58}{5}

Solution steps

The distance between

(x_{1}, y_{1}), (x_{2}, y_{2})

is:

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

The distance between

(\frac{4}{5}, \frac{1}{10})

and

(\frac{1}{5}, \frac{15}{10})

= (\frac{1}{5} - \frac{4}{5})^{2} + (\frac{15}{10} - \frac{1}{10})^{2}

Simplify

= \frac{58}{5}

- x + 4 - 1

f (x) = x^{3} - 5 x^{2} - 8 x + 4

f (x) = \frac{x ( x + 3 )}{x ^{2} + x - 6}

\frac{- x - 3}{x + 2}

f (x) = 5 - 3 x

What is the distance (4/5 , 1/10),(1/5 , 15/10) ?
The distance (4/5 , 1/10),(1/5 , 15/10) is (sqrt(58))/5