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amorfideBest ResponseYou've already chosen the best response.0
dw:1350320124515:dw
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
any method just explain
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
dw:1350320261447:dw my graph
 one year ago

zugzwangBest ResponseYou've already chosen the best response.1
If x = y, then x = y OR x = y
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
yes i get that part, so this gives dw:1350321371149:dw
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
then figure out x in both cases but i will get 4 answers, it only crosses 3 times, how do i know which one to eliminate?
 one year ago

2leBest ResponseYou've already chosen the best response.0
dw:1350321424876:dw what you drew looks more exponential. The graph for y=4/x is linear with a jump at x=0.
 one year ago

zugzwangBest ResponseYou've already chosen the best response.1
There must be one for which x/2 + 3 is negative. That's an impossibility.
 one year ago

zugzwangBest ResponseYou've already chosen the best response.1
Because x/2 + 3 is the absolute value of something...?
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
dw:1350321988874:dw dw:1350322071606:dw
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
so what would my final answers be
 one year ago

zugzwangBest ResponseYou've already chosen the best response.1
Are you sure about the first part?
 one year ago

zugzwangBest ResponseYou've already chosen the best response.1
I think it should be 3 (+ or ) sqrt (21)
 one year ago

zugzwangBest ResponseYou've already chosen the best response.1
oh wait, sorry... my bad
 one year ago

zugzwangBest ResponseYou've already chosen the best response.1
in any case, 3sqrt(17), try substituting that for x in x/2 + 3
 one year ago

zugzwangBest ResponseYou've already chosen the best response.1
And remember 3sqrt(17) < 3sqrt(16) = 7
 one year ago

zugzwangBest ResponseYou've already chosen the best response.1
No problem... Sorry about my pathetic arithmetic :P
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
still made sense in a way :P
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
\[\rm \left{4 \over x}\right = {x \over 2} + 3 \qquad \Rightarrow \qquad {4 \over x} = \pm\left({x \over 2 }+3 \right)\]Let's solve both.\[\rm {4 \over x}={x \over 2}+3\]so\[8 = \rm x^2 + 6x\]therefore\[\rm x^2 + 6x8=0 \]
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
For the other solution,\[{4 \over \rm x } = \rm {x \over 2}3\]so\[8=\rm x^2  6x\]therefore\[\rm x^2 +6x +8=0\]
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
And as @zugzwang was kind enough to point out, make sure all values of x lead to \[\frac{x}{2}+3\] being nonnegative.
 one year ago
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