Suppose we wish to determine whether or not two given polynomials with complex coefficients have a common root. Given two first-degree polynomials a0 + a1x and b0 + b1x, we seek a single value of x such that
Solving each of these equations for x we get x = -a0/a1 and x = -b0/b1 respectively, so in order for both equations to be satisfied simultaneously we must have a0/a1 = b0/b1, which can also be written as a0b1 - a1b0 = 0. Formally we can regard this system as two linear equations in the two quantities x0 and x1, and write them in matrix form as
Hence a non-trivial solution requires the vanishing of the determinant of the coefficient matrix, which again gives a0b1 - a1b0 = 0.
Now consider two polynomials of degree 2. In this case we seek a single value of x such that
Hope this helped, Hope I did not make it to complated
Please give me Brainliest
Answer:
For 
the function f(x) is continuous on 
.
Step-by-step explanation:
We have the following function
For the function f(x) to be continuous on it is sufficient to have continuity at x = 6, we need to ensure that as x approaches 6, the left and right limits match, this means that
,
which holds if and only if
namely if .
12*11*10"9*8 whatever that turns out to be lol
Answer:
yes it is
Step-by-step explanation:
it is since there is a common ratio between each term
The correct answer is choice D. To find this you will find the probability of each event happening and then multiply them together.
Green: 5/30
Yellow: 8/29 (you subtract one from the denominator because you already picked a green the first time)
5/30 x 8/29 = 40/870
