keana
  • keana
HELP! (picture below)
Mathematics
schrodinger
  • schrodinger
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keana
  • keana
anonymous
  • anonymous
Solve for \(C\): \[133 = \dfrac{9}{5} C + 32\] What I did, I used the given Fahrenheit temperature of 133, and substituted its value for the variable \(F\).
keana
  • keana
ok makes sense

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anonymous
  • anonymous
If you want I can check your answer if you would like, once you solve for C.
keana
  • keana
how do i solve for c though...
anonymous
  • anonymous
Isolate the like terms to one side of the equation.
keana
  • keana
alright
anonymous
  • anonymous
\[133 = \dfrac{9}{5} C + 32\] \[133 - 32 = \dfrac{9}{5} C\] \[\cdots\] See where I am going, by isolating the like terms?
keana
  • keana
how do i subtract 9/5 from 133?
anonymous
  • anonymous
They are not like terms, so you cannot. Here is an example: \[44 = 3C + 4\] \[\Rightarrow 44 - 4 = 3C\] \[\Rightarrow 40 = 3C\] \[\Rightarrow \dfrac{40}{3} = \dfrac{3C}{3}\] \[\Rightarrow 13.3333... = C\]
keana
  • keana
so do i subtract 32?
anonymous
  • anonymous
Yes, so \(133 - 32 = 101\). \[\Rightarrow 101 = \dfrac{9}{5} C\] \[\Rightarrow \dfrac{101}{1.8} = \dfrac{1.8C}{1.8}\] To understand where I got \(1.8\), I converted the fraction to a decimal. \(\dfrac{9}{5} = 1.8\) All you have to do is simplify now.
keana
  • keana
i got 56.1111....
anonymous
  • anonymous
Correct. \[C \approx 56.111\]
anonymous
  • anonymous
Remember your question also asks to round to the nearest tenth.

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