What is the distance (5/3 ,-16/9),(-1/3 , 19/9) ?

The distance (5/3 ,-16/9),(-1/3 , 19/9) is (sqrt(1549))/9

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distance (5/3 ,-16/9),(-1/3 , 19/9)

distance

(\frac{5}{3}, - \frac{16}{9}), (- \frac{1}{3}, \frac{19}{9})

\frac{1549}{9}

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{5}{3}, - \frac{16}{9})

and

(- \frac{1}{3}, \frac{19}{9})

= (- \frac{1}{3} - \frac{5}{3})^{2} + (\frac{19}{9} - (- \frac{16}{9}))^{2}

Simplify

= \frac{1549}{9}

f (x) = - 3 cos (3 x)

f (x) = sin (e^{- t})

f (x) = 5^{x - 3}

f (x) = 3 x^{2} - 6 x - 9

f (x) = \frac{4 x + 6}{x + 3}

What is the distance (5/3 ,-16/9),(-1/3 , 19/9) ?
The distance (5/3 ,-16/9),(-1/3 , 19/9) is (sqrt(1549))/9