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➷ The correct inequality would be B.
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➶ Hope This Helps You!
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Step-by-step explanation:
For any quadrilateral to be a parallelogram
i) either both the pairs of opposite sides must be equal
ii) Both the pairs of opposite sides must be parallel
iii) Opposite pairs of angles must be equal
iv) Diagonals must bisect each other.
v) A pair of opposite sides must be parallel and equal
Here we are already given that AB || CD
So either we should be given that AD || BC
or we must be given that AB = CD
Here it is given AB = CD as an option.
So Option A) or the first option is the right answer that AB ≅CD is needed to prove ABCD is a parallelogram
Factoring by grouping usually pairs up the first 2 sets of expressions with the second 2 sets. Ours looks like this, then:
![(x^3-9x^2)+(5x-45)=0](https://tex.z-dn.net/?f=%28x%5E3-9x%5E2%29%2B%285x-45%29%3D0)
. If we factor out the common x-squared in the first set of parenthesis, along with factoring out the common 5 in the second set, we get this:
![x^2(x-9)+5(x-9)](https://tex.z-dn.net/?f=x%5E2%28x-9%29%2B5%28x-9%29)
. Now the common expression that can be factored out is the (x-9). When we do that, here's what it looks like:
![(x-9)(x^2+5)](https://tex.z-dn.net/?f=%28x-9%29%28x%5E2%2B5%29)
. I'm not sure how far you are going with this. You could set each of those equal to 0 and solve for x in each case. The first one is easy. If x - 9 = 0, then x = 9. The second one involves the imaginary i since x^2 = -5. In that case,
![x=i \sqrt{5},-i \sqrt{5}](https://tex.z-dn.net/?f=x%3Di%20%5Csqrt%7B5%7D%2C-i%20%5Csqrt%7B5%7D%20%20)
. Hopefully, in what I have given you, you can find what you're looking for.