Expanding the limit, we get (x^2+2x∆x+∆x^2-2x-2∆x+1-x^2+2x-1)/<span>∆x
Crossing the 1s , the 2xs, and the x^2s out, we get
(2x</span>∆x+∆x^2-2∆x)/<span>∆x
Dividing the </span><span>∆x, we get
2x+</span><span>∆x-2.
Making the limit of </span><span>∆x=0, we get 2x-2.</span>
Answer:
Hi, there the answer is 3
Step-by-step explanation:
2x^2 + x + 3=0 has only complex roots.
The determinant is 1-4*2*3 = -23
-23 (or any determinant) is the part under the square root sign, If that determinant is negative, knowing you cannot take the square root of a negative number, we know the answers must be complex.
Answer:
Brainelist?
Step-by-step explanation:
7-4i
just use a online calculator
