Is 5.78778777 a rational or irrational number?..
1 answer:
You might be interested in
Answer:
E is not a subspace of
Step-by-step explanation:
E is not a subspace of
In order to see this, we must find two points (a,b), (c,d) in E such that (a,b) + (c,d) is not in E.
Consider
(a,b) = (1,1)
(c,d) = (-1,-1)
It is easy to see that both (a,b) and (c,d) are in E since 1*1>0 and (1-)*(-1)>0.
But (a,b) + (c,d) = (1-1, 1-1) = (0,0)
and (0,0) is not in E.
By the way, it can be proved that in any vector space all sub spaces must have the vector zero.
Te x intercepts are where the graph crosses the x axis or wher y=0
the y intercept is where the graph crosses the y axis or where x=0
to find vertex, here is a hack
the x coordinate of the vertex for an equation in form ax^b+bx+c=y is -b/2a
so
y=3x^2+12x+7
-b/2a=-12/2(3)=-12/6=-2
sub that back
y=3(-2)^2+12(-2)+7
y=3(3)-24+7
y=9-17
y=-8
vertex is (-2,-8)
intercepts
x intercept is where y=0
0=3x^2+12x+7
using quadratic formula
x=(-6-√15)/3 and (-6+√15)/3
xints at ((-6-√15)/3,0) and ((-6+√15)/3,0)
yint is where x=0
set x=0
y=7
yint is at y=7 or (0,7)
vertex at (-2,-8)
xints at x=
and 
or at the points 
and 
yint at y=7 or at (7,0)
the y intercept is where the graph crosses the y axis or where x=0
to find vertex, here is a hack
the x coordinate of the vertex for an equation in form ax^b+bx+c=y is -b/2a
so
y=3x^2+12x+7
-b/2a=-12/2(3)=-12/6=-2
sub that back
y=3(-2)^2+12(-2)+7
y=3(3)-24+7
y=9-17
y=-8
vertex is (-2,-8)
intercepts
x intercept is where y=0
0=3x^2+12x+7
using quadratic formula
x=(-6-√15)/3 and (-6+√15)/3
xints at ((-6-√15)/3,0) and ((-6+√15)/3,0)
yint is where x=0
set x=0
y=7
yint is at y=7 or (0,7)
vertex at (-2,-8)
xints at x=
yint at y=7 or at (7,0)