The formula for fractional exponents is ![\sqrt[n]{m} = x^{ \frac{m}{n} }](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bm%7D%20%3D%20%20x%5E%7B%20%5Cfrac%7Bm%7D%7Bn%7D%20%7D%20)
For this particular problem the expression would be written as![\sqrt[2]{ x^{3} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B2%5D%7B%20x%5E%7B3%7D%20%7D%20)
For this particular problem the expression would be written as
Which of the following points is a solution of the inequality y < |x - 2|?
1 answer:
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Answer:
The set of polynomial is Linearly Independent.
Step-by-step explanation:
Given - {f(x) =7 + x, g(x) = 7 +x^2, h(x)=7 - x + x^2} in P^2
To find - Test the set of polynomials for linear independence.
Definition used -
A set of n vectors of length n is linearly independent if the matrix with these vectors as columns has a non-zero determinant.
The set is dependent if the determinant is zero.
Solution -
Given that,
f(x) =7 + x,
g(x) = 7 +x^2,
h(x)=7 - x + x^2
Now,
We can also write them as
f(x) = 7 + 1.x + 0.x²
g(x) = 7 + 0.x + 1.x²
h(x) = 7 - 1.x + 1.x²
Now,
The coefficient matrix becomes
A =
Now,
Det(A) = 7(0 + 1) - 1(7 - 7) + 0
= 7(1) - 1(0)
= 7 - 0 = 7
⇒Det(A) = 7 ≠ 0
As the determinant is non- zero ,
So, The set of polynomial is Linearly Independent.