Answer:
a) v ∈ ker(<em>L</em>) if only if ![[V]_{E}](https://tex.z-dn.net/?f=%5BV%5D_%7BE%7D)
∈ <em>N</em>(<em>A</em>)
b) w ∈ <em>L</em>(<em>v</em>) if and only if ![[W]_{F}](https://tex.z-dn.net/?f=%5BW%5D_%7BF%7D)
is in the column space of <em>A</em>
<em />
<em>See attached</em>
Step-by-step explanation:
See attached the proof Considering the vector spaces <em>V</em> and <em>W</em> with other bases <em>E</em> and <em>F</em> respectively.
Let <em>L</em> be the Linear transformation form <em>V</em> and <em>W</em> and A is the matrix representing <em>L</em> relative to<em> E</em> and <em>F</em>
1.5 x 10^4 = 15,000
9 x 10^3 = 9,000
15,000 - 9,000 = 6,000 <=== or 6 x 10^3
We have:
(30x²+23x+16)/(cx+3) - 13/(cx+3) = 6x+1
(30x²+23x+16 - 13)/(cx+3) = 6x+1
(30x²+23x+3)/(cx+3) = 6x+1
30x²+23x+3 = (cx+3)(6x+1)
30x²+23x+3 = 6cx²+cx+18x+3
30x² + 23x + 3 - 6cx² - cx - 18x - 3 = 0
(30 - 6c)x² +(5 - c)x = 0
6(5 - c)x² +(5 - c)x = 0
(5 - c)(6x² +x) = 0, and x∈ R\ {3/c} ⇒ 5 - c = 0 ⇒ c = 5.
You probably haven't learnt this but the general equations for circle with centre at the origin is x^2+y^2=R^2 where R is the radius of the circle so R^2=(1737+185)^2.I am lazy to press my calculator so type it in yourself.
