To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -5 ± √((5)^2 - 4(-11)(-3)) ] / ( 2(-11) )
x = [-5 ± √(25 - (132) ) ] / ( -22 )
x = [-5 ± √(-107) ] / ( -22)
Since we conclude that √-107 is nonreal, the answer to this question is that there are no real solutions.
12.9 i think, not sure. try it though
X = 50 (Is parallel to the y-axis so there is no intersection point)
Y= 40 (Is parallel to the x-axis so there is no intersection point)
Both no intercept with the other
The answer is 1/(x+4)
Explanation:
You would factor out the denominator
So,
(X-4)(x+4)=x^2-16
So, x-4/(x+4)(x-4)
Then x-4 cancels each other out from the numerator and denominator
Leaving 1/x+4
Answer:
9 over 5
Step-by-step explanation:
