Not necessarily.
and
may be linearly dependent, so that their span forms a subspace of
that does not contain every vector in
.
For example, we could have
and
. Any vector
of the form
, where
, is impossible to obtain as a linear combination of these
and
, since
unless
and
.
You can reword the two equations as:
-5x-y=15 (Divide original value by 3)
-2x+6y=6
Then use elimination to find x:
-30x-6y=90 (Multiply by 6 to get y values to be same to cancel out)
-2x+6y=6
You're left with:
-32x=96. Which can then be solved to find x which is -3.
Then plug back in
-2x+6y=6
Now to: -2(-3)+6y=6.
Which reduces to 6+6y=6. So y=0.
To graph them, just reword the equations (yes once again) so that y is in front.
y=-5x-15 and y=(1/3)x+1
Answer:
i think n(a) =10
Step-by-step explanation:
hope this is helpful
Answer:
B. Not a function
Step-by-step explanation:
A) distributive is where it distributes a number through a set of parenthesis.
