Factoring by grouping usually pairs up the first 2 sets of expressions with the second 2 sets. Ours looks like this, then: 
. 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: 
. Now the common expression that can be factored out is the (x-9). When we do that, here's what it looks like: 
. 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, 
. Hopefully, in what I have given you, you can find what you're looking for.
Is it possible for a segment to have more than one bisector? Explain your reasoning.
1 answer:
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ANSWER
EXPLANATION
From the given information, Elena chooses a number from 1 to 10.
The sample space is
S={1,2,3,4,5,6,7,8,9,10}
n(S)=10
The numbers greater than 5 are:
E={6,7,8,9,10}
n(E)=5
The probability that, she chooses a number greater than 5 is:
Substitute the values,