The given statement is false because it isn't an empty set!
<u>Step-by-step explanation:</u>
We have following sets of inequalities:
From 
we get ,
Therefore solution set is x=2.
Now, for 
we get ,
Therefore solution set is x>2.
For 
we get ,
Therefore solution set is x<2.
Now, the union of x=2, x<2 & x>2 is -∞<x<∞. i.e. all possible values of x. And so above statement is false because it isn't an empty set!
Answer:
Step-by-step explanation:
The wording on this is not the best. It sounds like the 1 zero has even multiplicity (that's because of where the modifier is). On top of that it has an odd power. You could try this. y =x*(x^2+1)^2
The problem is not with the power. It gives x^5. The problem is with the multiplicity of the one place where it crosses. (X^2 + 1) does factor, but it gives a complex root. I'm not sure that's allowed. However, it is the best I can do.
One hundred and eighty erasers
Answer:
5 < x <21
Step-by-step explanation:
The side lengths of a triangle :
The sum of two side lengths selected at random must be bigger than the third side length and their difference must be smaller than the third side length
If we call the third side x than
x must be smaller than 13 + 8 and x must be bigger than 13 - 8 which means
5 < x <21
Answer:
Step-by-step explanation:
9x + 2 + 5 = 14x - 8
9x +7 = 14x -8
7 +8 = 14x - 9x
5x = 15
x = 15/ 5
x = 3
