Answer:
a) v ∈ ker(<em>L</em>) if only if
∈ <em>N</em>(<em>A</em>)
b) w ∈ <em>L</em>(<em>v</em>) if and only if
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>
Answer:
1008
Step-by-step explanation:
like the 2nd picture we split it in 3 rectangular prism
=6*7*8=336
=6*7*8=336
=6*7*8=336
V=336+336+336=1008
A. Create a segment EF between two points.
Answer:
5 terms
to the fourth degree
leading coeff of 1
3 turning points
end behavior (when x -> inf, y -> inf. When x -> - inf, y -> -inf)
x intercepts are (0,-4) (0,-2) (0,1) (0,3)
Relative min: (-3.193, -25) (2.193, 25)
Relative max: (-0.5, 27.563)
Step-by-step explanation:
The terms can be counted, seperated by the + and - in the equation given.
The highest exponent is your degree.
The number before the highest term is your leading coeff, if there is no number it is 1.
The turning points are where the graph goes from falling to increasing or vice versa.
End behaviour you have to look at what why does when x goes to -inf and inf.
X int are the points at which the graph crosses the x-axis.
The relative min and max are findable if you plug in the graph on desmos or a graphing calculator.
Four times d to the third minus ten