anonymous
  • anonymous
need help solving this inequality please. x >-7
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
...isn't that already solved? You know the values of x. That's about as simple as it gets
anonymous
  • anonymous
it's a question on an exam prep....
anonymous
  • anonymous
hm... anything more to it than just "solve..."?

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anonymous
  • anonymous
So what are the values of x that satisfy the relation x > -7
anonymous
  • anonymous
like choices? Otherwise i think im at a loss
myininaya
  • myininaya
(-7,infinity)
anonymous
  • anonymous
how would I graph it....thats the other part of the question...sorry guys
anonymous
  • anonymous
\[x \in (-7, \infty)\] Graph the relation? or the solutions?
anonymous
  • anonymous
on a number line, draw it with an open circle on -7, with an arrow to the right
anonymous
  • anonymous
Thanks Bwah!
anonymous
  • anonymous
on a cartesian plane, a vertical line at x=-7, dotted (non inclusive). shade to the right
myininaya
  • myininaya
x=-7 is a vertical line so since we have x>-7 and not x>=-7 then you draw a broken vertical line at x=7 and since we have x>-7 shade everything to the right 0f -7
anonymous
  • anonymous
no prob, glad to help
anonymous
  • anonymous
how about 3-10x>33
anonymous
  • anonymous
well, the process is the same as any equation u just have an inequality sign at the end
anonymous
  • anonymous
Depends on which dimensions he has to include in the graph. In \(R^1\) it's just a number line. In \(R^2\) it is a field comprised of all (x,y) pairs such that \(y\in (-\infty, \infty) \text{ and } x \in (-7, \infty)\) . In \(R^3\) it would be a boxy thing, and so on.
anonymous
  • anonymous
I think I am lost on the process :)
anonymous
  • anonymous
How would you solve 3-10x = 33?
anonymous
  • anonymous
-10x>30 x<-3 (if i recall correctly, u switch the sign if u divide or multiply by a negative number)
anonymous
  • anonymous
subtract 3 from both sides?
anonymous
  • anonymous
Indeed. You can do the same thing to inequalities.
anonymous
  • anonymous
Then what would you do?
anonymous
  • anonymous
divide by 10......I think I am seeing the light :)
anonymous
  • anonymous
Yep. The only rule is that if you divide or multiply by a negative you have to change the direction of the inequality. \(-x < 5 \implies x > 5\)
anonymous
  • anonymous
Ack
anonymous
  • anonymous
\[-x < 5 \implies x > -5\] rather
anonymous
  • anonymous
Thanks so very much!
myininaya
  • myininaya
I don't know if you have ever though of like if we have 5>2 then and we multiply both sides by negative one or divide by negative one then it becomes -5<-2 since -2 is bigger than -5
myininaya
  • myininaya
any any negative number

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