This is the same as showing the following system of equations doesn't have a solution:
or in matrix form,
The quickest way to check if there is a solution is to check whether the coefficient matrix is invertible. If its determinant is 0, then it is not invertible.
And the quickest way to show that the determinant is 0 is by observing that the third row is a linear combination of the first two rows:
(-2, 9, 6) - (-3, 2, 1) = (-2 + 3, 9 - 2, 6 - 1) = (1, 7, 5)
So there are indeed no such scalars <em>c₁</em>, <em>c₂</em>, and <em>c₃</em>.
Answer:
Step-by-step explanation:
We are given f(x) and g(x)
1. (f+g)(x)
(f+g)(x) = f(x) + g(x)
=
=
Domain : All real numbers as it there exists a value of (f+g)(x) f every x .
2. (f-g)(x)
(f-g)(x) = f(x)-g(x)
=
=
Domain : All real numbers as it there exists a value of (f-g)(x) f every x .
Part 3 .
Domain : In this case we see that the function is not defined for values of x for which the denominator becomes 0 or less than zero . Hence only those values of x are defined for which
or
Hence taking square roots on both sides and solving inequality we get.
1/3<span> is </span>bigger <span>because the three in </span>1/3<span> is less than 1/4 &1/5. </span>
And a quotient. D.
In a division you have:
Quotient = Dividend / Divisor
The result is called the Quotient, the number at the top of the division is called the dividend and the number below is the divisor.