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)
11 = w - 12
11 + 12 = w -12 + 12
w = 11 + 12
w = 23
Answer:
:^
Step-by-step explanation:
- sorry I couldn't type the equation so I took a screenshot
a) Since both limits are <em>distinct</em> and do not exist, we conclude that x = - 1 is not part of the domain of the <em>rational</em> function.
b) The function
is equivalent to the function
.
<h3>How to determine whether a limit exists or not</h3>
According to theory of limits, a function f(x) exists for x = a if and only if
. This criterion is commonly used to prove continuity of functions.
<em>Rational</em> functions are not continuous for all value of x, as there are x-values that make denominator equal to 0. Based on the figure given below, we have the following <em>lateral</em> limits:


Since both limits are <em>distinct</em> and do not exist, we conclude that x = - 1 is not part of the domain of the <em>rational</em> function.
In addition, we can simplify the function by <em>algebra</em> properties:


The function
is equivalent to the function
.
To learn more on lateral limits: brainly.com/question/21783151
#SPJ1
Answer:
90 ounces
Step-by-step explanation:
72 x 15/12 = 90