Question

pls help me guys hi help

Answer 1

in question number fourth you can use the distance distance formula
AB^2=AC^2
please put the value in this and the answer is one

Answer 2

use formula

${b}^{2} - 4ac$
$0 = ( { - 2 \sqrt{5}) }^{2} - 4(p)(15)$
$0 = 20 - 60p$
$60p = 20$
$p = \frac{1}{3}$
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