2/3 = 6/9 So,
6/9 + 2/9 = 8/9
Hello,
(t*s)(x)=t(x)*s(x)=(4x²-x+3)*(x-7)
Answer D
y might be 1 and I got that cuz 0y+3ey=3e and replace y with 1 and you get 0•1+3e•1=3e and to be honest it kinda looks like it makes sense
- Discriminant Formula: b² - 4ac, with a = x^2 coefficient, b = x coefficient, and c = constant
So firstly, using our equation plug in the values into the discriminant formula and solve as such:
(-7)² - 4 × 3 × 4
49 - 48
1
So our discriminant is 1. <u>Since 1 is positive and a perfect square, this means that there are 2 real, rational solutions.</u>
Consider that,
x^2+4x+4 = (x+2)(x+2)
x^2+7x+10 = (x+2)(x+5)
Dividing those expressions leads to
(x^2+4x+4)/(x^2+7x+10) = (x+2)/(x+5)
The intermediate step that happened is that we have (x+2)(x+2) all over (x+2)(x+5), then we have a pair of (x+2) terms cancel as the diagram indicates (see below). This is where the removable discontinuity happens. Specifically when x = -2. Plugging x = -2 into (x+2)/(x+5) produces an output, but it doesn't do the same for the original ratio of quadratics. So we must remove x = -2 from the domain.