Hello,
Please, see the attached file.
Thanks.
Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
If (x+4)(x+9) then either x+4 = 0 or x+9 = 0
This is by the zero product property
Here is your answer
REASON:
Theorem used: The diagonals of a parallelogram bisect each other.
Let diagonals AC and BD bisect each other at O
So, OA=OC
Now,
3x=4x-6 [OA=3x and OC=4x-6]
4x-3x= 6
x= 6
HOPE IT IS USEFUL
