Question

Pythagorean Theorem:class="latex-formula"> +$b^{2}$ =$c^{2}$ Question 8 options:
$\sqrt{26} \\\\\sqrt{89}$

8
5

Answer 1

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Answer 2

distance formula:

$\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}$

points:

(-4, -1) and (4,3)

$\sqrt{(-4-4)^{2}+(-1-3)^{2}}$

$\sqrt{(-8)^{2}+(-4)^{2}}$

$\sqrt{64+16}$

= $\sqrt{32}$