To find q + 3, you need to answer this first:
What is 5-3?
5-3=2.
q would be equal to 2.
so now we have something like this:
1
_ ( 2 + 3) =5
3
9514 1404 393
Answer:
- (c1, c2, c3) = (-2t, 4t, t) . . . . for any value of t
- NOT linearly independent
Step-by-step explanation:
We want ...
c1·f1(x) +c2·f2(x) +c3·f3(x) = g(x) ≡ 0
Substituting for the fn function values, we have ...
c1·x +c2·x² +c3·(2x -4x²) ≡ 0
This resolves to two equations:
x(c1 +2c3) = 0
x²(c2 -4c3) = 0
These have an infinite set of solutions:
c1 = -2c3
c2 = 4c3
Then for any parameter t, including the "trivial" t=0, ...
(c1, c2, c3) = (-2t, 4t, t)
__
f1, f2, f3 are NOT linearly independent. (If they were, there would be only one solution making g(x) ≡ 0.)
1) Always be polite and professional.
2) Specifically address the points that you disagree with.
3) Provide Evidence.
4) Close with a brief summary of your rebuttal.
-4, -3, -2, -1
Consecutive integers are integers that are exactly one unit away from each other. -4 is one unit away from -3, and likewise with -3 and -2 and -2 and -1.
X3 + 4x2 -9x -36
[x3 + 4x2] -9 [ x+4]
take x2 common factor from first praket
x2[x+4] -9 [x+4]
= [x+4] *[x2 -9]
so the zeros of this equation are -4 , 3, -3
