The equation below shows the relationship between the temperature in degrees Celsius, C, and degrees Fahrenheit, F:
To isolate the term containing C, subtract 32 from both sides. Can you do that?
How can i subtract 32 from f? @ospreytriple
It becomes F-32.
f-32=9/5 c+32 -32 f-32=(9/5)c Is that correct?
Now you should have\[F-32=\frac{ 9 }{ 5 }C\]How should you get rid of the 9/5 and isolate C?
ahhh ok :P u move 9/5 to the other side making it -9/5 correct?
Not quite. The C is being MULTIPLIED by 9/5. What is the inverse operation of multiplication?
division :D
Right. So you need to divide both sides by 9/5. Can you do that?
f=160/9+9/5c correct?
Let's back up and take it in smaller steps. Instead of thinking a C as being multiplied by 9/5, we can think of C as being MULTIPLIED by 9 and DIVIDED by 5. OK?
I'm confused its a variable how do i multiply that
I'm just trying to say that\[\frac{ 9 }{ 5 }C\]and\[\frac{ 9\times C }{ 5 }\]are the same thing. Does that make sense?
Remember, your job is to isolate C.
ahhhhh ok so i just divide 9/5 on that one side theoretically right?
waitttttt that is just 1 and 4/5 :D
Right. But you have to divide BOTH sides by 9/5. What do you get?
\[\frac{ F-32 }{ \frac{ 9 }{ 5 } }=\frac{ \frac{ 9 }{ 5 }C }{ \frac{ 9 }{ 5 }}\]You're doing this. Now simplify.
so 32 divided by 9/5 is 160/9 or 17.7? so its f- 160/9 and 81c/25 ?
What do you get when you divide any number by itself? Like 2/2=?
0 :D
Nope. Try again.
What is 1/1, or 2/2, or 3/3, or 1.8/1.8?
\[F-32=\frac{9C}{5}\] Instead of diving by 9/5 all at once you can instead multiply by 5 first \[5(F-32)=\frac{5 \times9C}{5}\]
It is much simpler to understand like this
Thanks @Nishant_Garg . If you read above, we've looked at that.
Yeah the next step you need to figure out @Sparklestaraa
What do you think would happen if we multiply 9C with 5 and divide by 5 aswell, sound like knitting a sweater and tearing it back apart!
Right now we're struggling with 5/5. Any additional insight you can provide are appreciated.
Sorry folks, I have to leave. Good luck @Sparklestaraa.
are you there @sparklesstaraa
|dw:1424117394206:dw|
Join our real-time social learning platform and learn together with your friends!