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
The answer is 3) 5+r/14
Let's go segment by segment:
<span>"five plus" means addition, so we have 5 +
</span><span>"the quotient of r and 14" is the division result, so sign / must be present: 14/r
</span>
Therefore "five plus the quotient of r and 14" is 5 + 14/r
Answer:
solution:-We know that for any two finite sets A and B, n(A∪B)=n(A)+n(B)−n(A∩B).
Here, it is given that n(A)=20,n(B)=30 and n(A∪B)=40, therefore,
n(A∪B)=n(A)+n(B)−n(A∩B)
⇒40=20+30−n(A∩B)
⇒40=50−n(A∩B)
⇒n(A∩B)=50−40
⇒n(A∩B)=10
Hence, n(A∩B)=10
Step-by-step explanation:
hope it helps you friend ☺️
8xy-7xy-3xy-3y²-2y²+8y²+5x²+12x²
then simplify by combining like terms -2xy+3y²+17x²
