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.)
They are all not functions
7(x - 3) = 4(x + 5)
7x - 21 = 4x + 20
7x - 4x = 20 + 21
3x = 41
x = 41/3
(x - 3) - 2(x + 6) = -5
x - 3 - 2x - 12 = -5
-3x - 15 = -5
-3x = -5 + 15
-3x = 10
x = -10/3
Answer: 6.50x + 3.50 < 23
Step-by-step explanation:
I believe this is correct because the total of the one-time fee and the money per game she played, should be less than $23
Answer:
3-5(-3)
3 minus -15
-12
Step-by-step explanation:
