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safia21

  • 5 years ago

algebra can you help me with # 2 and # 3 thanks

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  1. safia21
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    the cooking one?

  3. safia21
    • 5 years ago
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    yes please

  4. anonymous
    • 5 years ago
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    cook has 4 quarts that is 50% chicken stock. so at the moment it is 50% of 4 = 2 quarts chicken stock. so for example if she adds 3 quarts of chicken stock she will have 4+3= 7 quarts of liquid of which 2 + 3 = 5 quarts is chicken stock, and the percent will be 5/7 * 100. this is not what you want obviously, i am just trying to explain where the equation will come from. if she adds x quarts of chicken stock she will have 2+x quarts of chicken stock and 4+x quarts of liquid. you want \[\frac{2+x}{4+x}=.75\] \[2+x=.75(4+x)\] last equation says your two quarts of chicken stock plus your x quarts must be 75% of the total liquid. now we solve multiply out \[x+2=.75x+.75\times 4= .75x+3\] \[.25x=1\] subtract .75x from both sides and subtract 2 from both sides \[x=\frac{1}{.25}=\frac{100}{25}=4\]

  5. safia21
    • 5 years ago
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    and #3 thanks

  6. anonymous
    • 5 years ago
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    ok #3 looks like the previous one we did so let me be careful and not put the 1 on the wrong side like i did last time.

  7. anonymous
    • 5 years ago
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    keep seeming to put the 1 on the wrong side. you have two rates, r and r +1 \[\frac{12}{r}=\frac{12}{r+1}+1\] slower person's time is one less more than faster persons. \[\frac{12}{r}=\frac{12+r+1}{r+1}=\frac{r+13}{r+1}\] cross multiply\[12(r+1)=r(r+13)\] \[12r+12=r^2+13r\] \[r^2+r-12=0\] \[(r+4)(r-3)=0\] \[r=3\] \[r=-4\] so r = 3

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