Factor Factor -2bk 2 + 6bk - 2b. ill give ya 14 points !
2 answers:
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Answer:
See attached. (Both inequalities graph to the same region.)
Step-by-step explanation:
To graph an inequality, you first replace the inequality symbol by an equal sign. Then you graph the line (or curve) described by the equation. If the inequality symbol includes the "or equal to" case, then the line (or curve) is solid and is part of the solution set. If not, then the line (or curve) is dashed and is <em>not part of the solution set</em>.
Here, the line x+y=5, will bound the solution set, but is not part of it.
Note that the line -x -y = -5 (from the second inequality) is the same ilne.
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After you have graphed the boundary of the solution region, you need to determine which side of the boundary has the solutions. I find it easy to do this by locating an x- or y-term with a positive coefficient and seeing which side of the inequality symbol that lies on. Here, both the x-term and the y-term lie on the "is greater than" side of the symbol, so the half-plane is shaded above and to the right of the line, where x- and y-values are greater than they are on the line.
In the second inequality, you can multiply the whole thing by -1 to get
... x + y > 5 . . . . same as the first inequality
or you can add (x+y) to both sides to get
... 0 < -5 +x +y
Once again, the x- and y-values that satisfy this are greater than those that are on the boundary line.
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The graph seems to show only one equation. Both are there, but the dashed blue line sits on top of the dashed red line, so you can't see it. The purple shading indicates both red and blue shading in the same area.
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Of course you graph the boundary line the way you would any linear equation. In this form, it is not too difficult to see that both the x- and y-intercepts are 5. (When x=0, y=5; and when y=0, x=5 for the boundary line's equation x+y=5.) Draw the dotted line through these intercept points, add shading above, and you have your graph.
the denominator cannot be zero, because the division by zero is not defined, therefore:
Therefore the domain of (f o g)(x) is: