3 years ago

Segment BD segment BD is a perpendicular bisector of segment AC segment AC . AD=5x+8AD=5x+8 and DC=43DC=43 .

Mathematics

1 answer:

Irina-Kira [14]3 years ago

3 0

I hope this helps you

5x+8=43

5x=35

x=7

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Ad is tangent to circle o at d. find ab. round to the nearest tenth if necessary

Naddik [55]

Check the picture below.

$\bf 0=AB^2+6AB-121\implies \stackrel{\textit{using the quadratic formula}}{AB=\cfrac{-6 \pm \sqrt{6^2-4(1)(-121)}}{2(1)}}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{520}}{2}\implies AB=\cfrac{-6 \pm \sqrt{4\cdot 130}}{2}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{2^2\cdot 130}}{2}\implies AB=\cfrac{-6 \pm 2\sqrt{130}}{2}
\\\\\\
AB=-3\pm\sqrt{130}\implies AB=
\begin{cases}
\boxed{-3+\sqrt{130}}\\\\
-3-\sqrt{130}
\end{cases}$

since the distance AB cannot be a negative value, thus is not -3-√(130).