What is the distance (16/3 , 1/5),(1/3 , 6/5) ?

The distance (16/3 , 1/5),(1/3 , 6/5) is square root of 26

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distance (16/3 , 1/5),(1/3 , 6/5)

distance

(\frac{16}{3}, \frac{1}{5}), (\frac{1}{3}, \frac{6}{5})

26

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{16}{3}, \frac{1}{5})

and

(\frac{1}{3}, \frac{6}{5})

= (\frac{1}{3} - \frac{16}{3})^{2} + (\frac{6}{5} - \frac{1}{5})^{2}

Simplify

= 26

- 3 sin (x - \frac{π}{4})

lo g_{7} (- x)

f (x) = \frac{1}{x + 2}

6 sin (t + 4)

f (x) = 4 x^{3} - 6 x^{2} + 5 x - 5

What is the distance (16/3 , 1/5),(1/3 , 6/5) ?
The distance (16/3 , 1/5),(1/3 , 6/5) is square root of 26