Answer:
No
Step-by-step explanation:
Complex answer: For a point to be in quadrant II, it must have a negative x value and a positive y value.
Simple answer: The 1 would have to be negative to be in quadrant II, and it isn't
(0,-3) (1, -1) (2, 1) ....
Answer:
why are you asking me ask jose
Answer:
<em>2 solutions</em>
Step-by-step explanation:
Given the expression
2m/2m+3 - 2m/2m-3 = 1
Find the LCM of the expression at the left hand side:
2m(2m-3)-2m(2m+3)/(2m+3)(2m-3) = 1
open the bracket
4m²-6m-4m²-6m/(4m²-9) = 1
Cross multiply
4m²-6m-4m²-6m = 4m² - 9
-12m = 4m² - 9
4m² - 9+12m = 0
4m² +12m-9 = 0
<em>Since the resulting equation is a quadratic equation, it will have 2 solutions since the degree of the equation is 2</em>
