Question

$$\left. \begin{array} { l } { \text { a n i n e i n t h e o \right.$$

Answer

p=0

Solution


Remove parentheses.
\[completeretnombercente\times \frac{AM}{AD}=\frac{AN}{AC}-\frac{MN}{BC}\]
Regroup terms.
\[ccoommpltttrrnnbeeeeee\times \frac{AM}{AD}=\frac{AN}{AC}-\frac{MN}{BC}\]
Simplify  \(ccoommpltttrrnnbeeeeee\times \frac{AM}{AD}\)  to  \(\frac{{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}beeeeeeAM}{AD}\).
\[\frac{{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}beeeeeeAM}{AD}=\frac{AN}{AC}-\frac{MN}{BC}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}b{e}^{6}AM}{AD}=\frac{AN}{AC}-\frac{MN}{BC}\]
Regroup terms.
\[\frac{{e}^{6}AM{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}b}{AD}=\frac{AN}{AC}-\frac{MN}{BC}\]
Multiply both sides by \(AD\).
\[{e}^{6}AM{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}b=(\frac{AN}{AC}-\frac{MN}{BC})AD\]
Divide both sides by \({e}^{6}\).
\[AM{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}b=\frac{(\frac{AN}{AC}-\frac{MN}{BC})AD}{{e}^{6}}\]
Rewrite \(\frac{AN}{AC}-\frac{MN}{BC}\) in the form \({a}^{2}-{b}^{2}\), where \(a=0\) and \(b=0\).
\[AM{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}b=\frac{({0}^{2}+{0}^{2})AD}{{e}^{6}}\]
Simplify  \({0}^{2}\)  to  \(0\).
\[AM{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}b=\frac{(0+0)AD}{{e}^{6}}\]
Simplify  \(0+0\)  to  \(0\).
\[AM{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}b=\frac{0AD}{{e}^{6}}\]
Simplify  \(0AD\)  to  \(0\).
\[AM{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}b=\frac{0}{{e}^{6}}\]
Simplify  \(\frac{0}{{e}^{6}}\)  to  \(0\).
\[AM{c}^{2}{o}^{2}{m}^{2}pl{t}^{3}{r}^{2}{n}^{2}b=0\]
Solve for \(p\).
\[p=0\]