Begin by subtracting 1/8 from both sides to get
![- \frac{2}{x} =5- \frac{1}{8}](https://tex.z-dn.net/?f=-%20%5Cfrac%7B2%7D%7Bx%7D%20%3D5-%20%5Cfrac%7B1%7D%7B8%7D%20)
and get a common denominator of 8 on the right:
![- \frac{2}{x} = \frac{40}{8} - \frac{1}{8}](https://tex.z-dn.net/?f=-%20%5Cfrac%7B2%7D%7Bx%7D%20%3D%20%5Cfrac%7B40%7D%7B8%7D%20-%20%5Cfrac%7B1%7D%7B8%7D%20)
. Now simplify and cross-multiply: 39x = -16 andx = -16/39
If m ACD = 30 => m DCB = 60.
In triangle ACD:
![sin 30^{o}=sin \frac{AD}{AC} =\ \textgreater \ \frac{1}{2} = \frac{8}{AC} =\ \textgreater \ AC=16 ](https://tex.z-dn.net/?f=sin%2030%5E%7Bo%7D%3Dsin%20%20%5Cfrac%7BAD%7D%7BAC%7D%20%3D%5C%20%5Ctextgreater%20%5C%20%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%3D%20%20%5Cfrac%7B8%7D%7BAC%7D%20%3D%5C%20%5Ctextgreater%20%5C%20%20AC%3D16%0A)
![CD= 8 \sqrt{3}](https://tex.z-dn.net/?f=CD%3D%208%20%5Csqrt%7B3%7D%20)
![BC= 16\sqrt{3}](https://tex.z-dn.net/?f=BC%3D%2016%5Csqrt%7B3%7D%20)
AC^{2}+CB^{2}=AB^{2} => AB=
![\sqrt{1024}](https://tex.z-dn.net/?f=%20%5Csqrt%7B1024%7D%20)
=32.
=> P=16+32+
![16 \sqrt{3}](https://tex.z-dn.net/?f=16%20%5Csqrt%7B3%7D%20)
=48 +
![16 \sqrt{3}](https://tex.z-dn.net/?f=16%20%5Csqrt%7B3%7D%20)
.
Answer:
(-2, 5) < x
Step-by-step explanation:
K= 23/7 simplified to 3 2/7