If it has no real solutions, that means the graph does not intersect the x axis
since we have ax^2+bx+c=0, the parabola opens either up or down
since the vertex is in the second quadrant (x is negative and y is positive in this reigon) and the graph does not cross the x axis, the parabola must open up
if the value of 'a' is positive, then the parabola opens up
so 'a' must be positive
if it is translated to the 4th quadrant, then the vertex is now below the x axis
it will now have 2 x intercepts because the vertex is in the 4th quadrant and look at a graph of a parabola opening up with vertex in 4th quadrant and seehow many time it crosses the x axis
Let's say the numbers are "a" and "b"
thus
set the derivative to 0, and check the critical points, there's only one anyway
and do a first-derivative test, to see if it's a maximum
I don’t think there is a number that does both of those, the factors of 49 are 1, 7, and 49
Answer:
3)5x-4y=-3
6x+4y=14
Step-by-step explanation:
The system of equations is given
5x-4y=-3
3x+2y=7
Multiplying both sides of the lower equation by 2, we get
6x + 4y = 14
That means your answer 3)
