Some basic formulas involving triangles
\ a^2 = b^2 + c^2 - 2bc \textrm{ cos } \alphaa 2 =b 2+2 + c 2
−2bc cos α
\ b^2 = a^2 + c^2 - 2ac \textrm{ cos } \betab 2=
m_b^2 = \frac{1}{4}( 2a^2 + 2c^2 - b^2 )m b2 = 41(2a 2 + 2c 2-b 2)
b
Bisector formulas
\ \frac{a}{b} = \frac{m}{n} ba =nm
\ l^2 = ab - mnl 2=ab-mm
A = \frac{1}{2}a\cdot b = \frac{1}{2}c\cdot hA=
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
\iits whatever A = prA=pr with r we denote the radius of the triangle inscribed circle
\ A = \frac{abc}{4R}A=
4R
abc
- R is the radius of the prescribed circle
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
Answer:
1/2 = 3/6
Step-by-step explanation:
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When you say "which," it sounds like it should be multiple choice. Anyways, here's the simplified form of
![\sqrt{\frac{1}8}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B1%7D8%7D%7D)
![\sqrt{\frac{1}8}} = \frac{\sqrt{1}}{\sqrt{8}} = \frac{1}{\sqrt{8}} * \frac{\sqrt{8}}{\sqrt{8}}=\boxed{\frac{\sqrt{8}}8}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B1%7D8%7D%7D%20%3D%20%5Cfrac%7B%5Csqrt%7B1%7D%7D%7B%5Csqrt%7B8%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B8%7D%7D%20%2A%20%5Cfrac%7B%5Csqrt%7B8%7D%7D%7B%5Csqrt%7B8%7D%7D%3D%5Cboxed%7B%5Cfrac%7B%5Csqrt%7B8%7D%7D8%7D)
(When simplifying fractions, you should <em>never</em> have a square root on the bottom. Multiply by the square root to cancel it out)
Answer:
131
Step-by-step explanation:
khan academy
Answer:
im stuck on that question too funny bc im in your class and caught u cheating
Step-by-step explanation: