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In-s [12.5K]
3 years ago
6

Which equation is equivalent to the first equation of the

Mathematics
1 answer:
Greeley [361]3 years ago
5 0
The number of solutions of the system of equations is 904 086
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Due Soon Need Help Geometry!
AveGali [126]

 

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)

​  

5 0
2 years ago
Find the midpoint of each line segment
devlian [24]

Answer:

(\frac{7}{2},\frac{1}{2})

Step-by-step explanation:

For this you use the midpoint formula

(\frac{x_{2} +x_{1} }{2} ,\frac{y_{2} +y_{1} }{2} )

the points would be (2,1) and (5,2)

The midpoint would be (\frac{7}{2},\frac{1}{2})

5 0
2 years ago
The ratio of the area to the circumference is<br> A. 2r<br> B.r/2<br> C. 2/r
weqwewe [10]
It's nothing because e=mc^2
4 0
3 years ago
Read 2 more answers
A bit hard for me !!!!!
Annette [7]
Where they intersect is (-1,2) hope this helps
4 0
2 years ago
Read 2 more answers
Find x. (More info in the pic)
jolli1 [7]
There is no picture
6 0
3 years ago
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