Answer:
12 possibilities
Step-by-step explanation:
In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.
The same thing occurs in the second urn, as all balls have different labels.
The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).
For the first urn, we have a combination of 4 choose 2:
C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities
For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.
In total we have 6 + 6 = 12 possibilities.
Area will change by a factor of 3.5² = 12.25
I believe it is GCF because 9v^3 can be factored out of the expression, becoming:
9v^3 (4 - 3v^2)
9v^3 (-3v^2 + 4)
The quadratic in parentheses can be solved, but with the quadratic equation or by graphing.
<h3>#End behaviour:-</h3>
#1
#2
<h3>#Degree:-</h3>
Find nodes
#1
#2
It's a parabola so it's the graph of a quadratic equation.
<h3>Real zeros</h3>
#1
#2
Answer:
-1 179/250
Step-by-step explanation:
