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anonymous
 4 years ago
3x / x1 < 2
anonymous
 4 years ago
3x / x1 < 2

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Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1b is easy. 3x2<1 and 3x2>1 x<1 and x>1/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dn't get it..can you please attach?? :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its the first question on attachment..please have a look..thanks :)

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.13x2<1 means 3x2 is less than 1 unit from 0 which means 3x2 <1 and 3x2>1 Just solve each part.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.03 x / ( x1) < 2 this is the question... :'(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0humm, I have problem with a and c :(

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1I think just by inspection that x <1 is the solution to the first one.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this question is really confusing :@..

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1The third one, let's just take the log of both sides and solve.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0humm I am trying with log..

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1\[(2x1)\log_{10}2<(3x2)\log_{10}3 \]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1\[(2x1)(.301)<(3x2)(.477)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are we allowed to use the log value...(if i'm not allowed to bring my calculator then how will i solve).. :(

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1The logs you need would have to be provided.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1Are calculators forbidden?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not sure..bt, my friend told me.. :'(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks man for your efforts... :)

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1for the first one, I found these steps: 1. Move all terms to one side of the inequality sign so that one side is 0. 2. Replace the inequality sign with an equal sign and solve the equation. These solutions are critical values 3. find all values that result in division by 0. These are critical values 4. plot the critical values on a number line. 5. Test each interval defined by the critical values. If an interval satisfies the inequality then it is part of the solution.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{3x}{x1}2=0\]

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1I found three critical numbers: 2, 2/5 and 1

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1And the inequality is true in all intervals except x>1 so just as I thought, the solution is x<1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hummm...but, its too complicated....are we allowed to use equal sign for inequality? I'm not sure about this process...

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1It is only as an aid to find the critical numbers. It is not part of the solution.
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