Step-by-step explanation:
area of equilateral ∆ =

here a = 3
area of ∆ = 3.9 in^2
area of circle

= 19.63 in^2
area of shared region = 19.63-3.9= 15.73 in^2
<span>x in (-oo:+oo)
((5-9*x)^1)/2 = ((4*x+3)^1)/2 // - ((4*x+3)^1)/2
((5-9*x)^1)/2-(((4*x+3)^1)/2) = 0
(5-9*x)/2-((4*x+3)/2) = 0
(5-9*x)/2+(-1*(4*x+3))/2 = 0
5-1*(4*x+3)-9*x = 0
2-13*x = 0
(2-13*x)/2 = 0
(2-13*x)/2 = 0 // * 2
2-13*x = 0
2-13*x = 0 // - 2
-13*x = -2 // : -13
x = -2/(-13)
x = 2/13
x = 2/13</span>
<h3>
Answer: True</h3>
Explanation:
This theorem doesn't have a name unfortunately. So searching it out is a bit tricky if you don't know the right way to word things. Luckily it wasn't too hard of a find, and I managed to track it down in a linear algebra textbook.
Check out the screenshot below for the snippet of the theorem and the corresponding proof. The book simply refers to it as "theorem 1.9", which again, is an unfortunate choice of naming.
Answer:
False.
Step-by-step explanation:
The graph that is shown is correct for the equation x= -3. The graph for y= -3 would run vertically, not horizontally.