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Now it is time to look at solving some more hard inequalities. In this section we will be solving (single) inequalities which involve polynomials of degree at least two. Or, to put it into other terms, the polynomials won't be linear any more. Just as we saw while solving equations the procedure that we have for solving linear inequalities won't work here.
As it's easier to see the procedure as we work an example let's do that. As along with the linear inequalities, we are looking for all of the values of the variable which will make the inequality true. It means that our solution will approximately certainly involve inequalities as well. The procedure that we're going to go through will give the answers in that form.
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i cant figure this out
simplify the following expressions for the given values x3 -x2/x2 -x x=3 how do i do this to get answer need step by step instruction
y=-x-2 y=-5x+2
How to solve the complex RAE?
Now, let's get back to parabolas. There is a basic procedure we can always use to get a pretty good sketch of a parabola. Following it is. 1. Determine the vertex. We'll discus
for what value of x is \2x-3\-4
(3x+2y+2z) + (3x+5y+3z)
There are two given points ( x 1 , y 1 ) and ( x 2 , y 2 ), the distance between these points is prearranged by the formula: Don't allow the subscripts fright you. Th
We have to give one last note on interval notation before moving on to solving inequalities. Always recall that while we are writing down an interval notation for inequality that t
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