Check the Wronskian determinant:

The determinant is not zero, so the solutions are indeed linearly independent.
Answer:
Range is 32
Step-by-step explanation:
11,12,13,17,17,17,17,19,20,32,43
43-11 = 32
<span>a.The graph of f(x)= x^2 is widened</span>
Distance between two points is
Root of X2 - X1 squared + Y2 - Y1 squared
So here X2 is 0 X1 is 0 as the origin is 0,0
Root of Y2-Y1 squared is what we are left with
Root of -2- -3 squared
Root of 1 squared
Root of 1
Which is 1
Distance is 1
Answer:
The answer is 2.78
Step-by-step explanation: