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anonymous
 5 years ago
Solve the inequality (x/2) > (8)/(x + 4) + 3 for domain
anonymous
 5 years ago
Solve the inequality (x/2) > (8)/(x + 4) + 3 for domain

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0is the "3" in the denominator??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 3 is added on after the fraction so I suppose its + 3/1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x 8  >  + 3 2 x+4 get common denoms...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(x^2 +4x) > 16 + 6x +24 (x^2 +4x) > 6x + 8 x^2 +4x 6x > 8 x^2 2x >8 x^2 2x 8 > 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(x4)(x+2) >0 when x = 2 or 4; the equation = 0; so lets draw a number line: <.......2...........4..........>  + +   +  +  +

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this tells us that when: x = 2, everything to the left is () and to the right is (+) x = 4, everything to the left is () and to the right is (+) We multiply the signs together to get the bottom stuff; our answer is everything that is (+); becasue that is greater than zero. (inf,2) OR (4,inf)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm that doesnt appear to be one of the answers but ill double check the work

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0try x= 2 (2/2) > (8)/(2 + 4) + 3 1 > 8/2 + 3 1 > 4 + 3 1> 1... 2 is one of the "options" there :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0try x= 4 (4/2) > (8)/(4 + 4) + 3 2 > 8/8 + 3 2 > 1 + 3 2> 2... 4 is the other "option" there :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we know zero is between them, so lets see if we include that or exclude that area... try x= 0 (0/2) > (8)/(0 + 4) + 3 0 > 8/4 + 3 0 > 2 + 3 0> 1 is a false statement so the area that has "0" in it is thrown out; that only leaves: (inf,2) U (4,inf)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0These are the answers available: A. ( inf, 4) U (2, 4) B. ( inf, 4) U [2, 4] C. (4, 2] U [4, inf) D. ( inf, 4] U [2, 4) E. (4, 2) U (4, inf)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Check the problem, did you typo a sign?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x 8  >  + 3 2 x+4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i see an error on my part :) for starters 0 IS > 1, so zero is a solution

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0never mind that comment, its my senility taking riit :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0root... its like my fingers are having a stroke ....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0there is no "equal to" line in the equaiton, so it aint gonna have any [...] in it;

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I think it narrows down to "E" 6/2 = 3 3 > 8 + 3 3 > 5 is true, so 6 is a solution so "E" is wrong. You got something messed up in either the answers or the questions; because the 2 dont match up.... unless you can see an error in my calculations

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and 4 is not an option; becasue that zeros out the denom....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I plugged it into my calculator and im trying to eye it from the graph

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0no 4 and no 0 A.  B.  C. (4, 2] U [4, inf) D.  E. (4, 2) U (4, inf)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0there is no "greater than or EQUAL TO" sign in it so no "[4" 4/2 > 8/8 +3 2> 1 +3 2>2 ...... bad statement A.  B.  C.  D.  E. (4, 2) U (4, inf) test for "E"?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x = 6 6/2 = 8(6+4) + 3 3> 8/2 + 3 3 > 4+3 = 7...... 6 is a bad number and anything on that end... the answer is "E" then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I plugged it into my calculator and im trying to eye it from the graph

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I keep retrying it but it keeps turning out wrong! grrr

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0use a pencil and paper ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did, ive solved it 4 times now :/

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we know that 2 and 4 are boundaries; we know that zero is not an option; we know that {4 is not an option becasue there is no ">=" in the equation; that leaves "E" as the only possible shot. the only thing we need to determine is if it ends at 4 or not; so plug in a 6 and see if it hold :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.06 is bad value, so "E" is correct.
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