Answer:
9
Step-by-step explanation:
Please mark me brainliest
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)
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f1, f2, f3 are NOT linearly independent. (If they were, there would be only one solution making g(x) ≡ 0.)
Just find the average and whatever is more than like say 5 away is an out lier
You can't, it has been done before but there is a fault in the working because the 2 numbers are different. If you want to see a video of someone attempt it, I believe there is a video on the numberphile channel.
Hope this helps :)
F(x) -2 +6 = 4
The answer is 4
