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:
-3
Step-by-step explanation:
Allison has 5 negative tiles
Next she needs to remove 2 negative tiles
What is left is 3 negative tiles
(-5) -(-2) = (-5+2)=-3
Answer:
We have 7n < 6 - 48;
7n < - 42;
n < -6;
Then, 3n - 4 ≤ 8;
3n ≤ 12;
n ≤ 4;
Step-by-step explanation:
Answer:
The point 
is not a solution of the system of inequalities
Step-by-step explanation:
we have
-----> inequality A
-----> inequality B
we know that
If a ordered pair is a solution of the system of inequalities
then
the ordered pair must be satisfy the inequalities of the system
Verify
For 
substitute the value of x and the value of y in the inequalkity A and in the inequality B
Inequality A
-------> is not true
therefore
The point is not a solution of the system of inequalities
