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anonymous
 5 years ago
need help solving this inequality please. x >7
anonymous
 5 years ago
need help solving this inequality please. x >7

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...isn't that already solved? You know the values of x. That's about as simple as it gets

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's a question on an exam prep....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hm... anything more to it than just "solve..."?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what are the values of x that satisfy the relation x > 7

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like choices? Otherwise i think im at a loss

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how would I graph it....thats the other part of the question...sorry guys

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x \in (7, \infty)\] Graph the relation? or the solutions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on a number line, draw it with an open circle on 7, with an arrow to the right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on a cartesian plane, a vertical line at x=7, dotted (non inclusive). shade to the right

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0x=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
 5 years ago
Best ResponseYou've already chosen the best response.0no prob, glad to help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, the process is the same as any equation u just have an inequality sign at the end

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Depends 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
 5 years ago
Best ResponseYou've already chosen the best response.0I think I am lost on the process :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How would you solve 310x = 33?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.010x>30 x<3 (if i recall correctly, u switch the sign if u divide or multiply by a negative number)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0subtract 3 from both sides?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Indeed. You can do the same thing to inequalities.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then what would you do?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0divide by 10......I think I am seeing the light :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep. 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[x < 5 \implies x > 5\] rather

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0any any negative number
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