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yoyo
 5 years ago
1) 4k43k>137k1+8
2) r7>9r
3) 30+5x>4(6+8x)
4) 38+5x>7(x+4)
5)77(x7)>4+5x
6) 3(2v5)<13+v
7)5n+7(6n)>4(n+3)
8)x8+3x+2<6(8x7)+4(8x2)
SHOW ALL WORK PLEASE THANK YOU.
yoyo
 5 years ago
1) 4k43k>137k1+8 2) r7>9r 3) 30+5x>4(6+8x) 4) 38+5x>7(x+4) 5)77(x7)>4+5x 6) 3(2v5)<13+v 7)5n+7(6n)>4(n+3) 8)x8+3x+2<6(8x7)+4(8x2) SHOW ALL WORK PLEASE THANK YOU.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yoyo, there a lot of questions there. I'll give you a method by which you can do it yourself. First, ignore the inequality sign and turn it into an equality. Solve the equation for the variable. The values you get here will mark off boundaries on the number line. Now pick one point in each interval you've made and test it with the original inequality. If the answer you get is untrue, that region is not included. If the answer is true, the region is included. For example, in the first one,\[4k43k=137k1+8 \rightarrow k=3 \]On a number line, 3 marks off two regions  all numbers above, and all numbers below. Pick a number from the region above, say, 4. Substituting this into the lefthand side of the inequality gives\[4(4)43(4)=0\]and in the righthand side\[137(4)1+8=8\]We'd have the situation where\[4>8\]...which is TRUE. So those values for k above 3 will satisfy the inequality. Next you pick a number below 3, say 0. LHS = 4 and RHS = 20. We have, in this interval, \[4>20\]which is FALSE.
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