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.)
Answer:
11
Step-by-step explanation:

Answer:
<em>C. Line Graph</em>
Step-by-step explanation:
<em>The reason is that because like you would make the line graph put each month on it and then mark the amount you earned that month and then you can see which one is the highest</em>
Answer:
8
Step-by-step explanation: