Step-by-step explanation: need to find a basis for the solutions to the equation Ax = 0. To do this ... 0 0 0 1 −3. ⎤. ⎦. From this we can read the general solution, x = ⎡. ⎢. ⎢. ⎢. ⎢. ⎣ ... two vectors are clearly not multiples of one another, they also give a basis. So a basis ... 4.4.14 The set B = {1 − t2,t − t2,2 − 2t + t2} is a basis for P2.
Write the set of points from -6 to 0 but excluding -4 and 0 as a union of intervals
First we take the interval -6 to 0. In that -4 and 0 are excluded.
So we split the interval -6 to 0.
Start with -6 and go up to -4. -4 is excluded so we break at -4. Also we use parenthesis for -4.
Interval becomes [-6,-4) . It says -6 included but -4 excluded.
Next interval starts at -4 and ends at 0. -4 and 0 are excluded so we use parenthesis not square brackets
(-4,0)
Now we take union of both intervals
[-6,-4) U (-4,0) --- Interval from -6 to 0 but excluding -4 and 0
Answer:
(x-2)(y+1)
Step-by-step explanation:
Y = 1/2x+5 I think that’s it
Step-by-step explanation:
Consider x⅓ =a
then the following expression can be written as
a²+a-2
a²+2a-a-2
a(a+2)-1(a+2)
(a+2)(a-1)
a= -2 or, a= 1
Putting the value of a
x⅓ = -2 and, x⅓ =1
