Question

C=5/p(F-32) ...Solve for "F"

Answer 1

P=9

Answer 2

given :
\[\large{C = \frac{5}{p(F-32)}}\]

given P = 9

so :
\[\large{\frac{5}{9(F-32)}}\]

Answer 3

m i right? @gabbbdee

Answer 4

Yes the bottom one is the correct equation.

Answer 5

wait... no

Answer 6

its C=5/9 (F-32) and i have to solve for F

Answer 7

ok so it is :
\[\large{C=\frac{5}{9}(F-32)}\]

Answer 8

right?

Answer 9

Do I multiply "(F-32)" by 5/9 first?

Answer 10

yes.

Answer 11

No ... first step will be :
\[\large{C\times \frac{9}{5}=\frac{5}{9}(F-32) \times \frac{9}{5}}\]

Answer 12

I just multiplied both sides by 9/5

Answer 13

why do you flip the fraction?

Answer 14

just to cancel 5/9 there

Answer 15

just for example:
\[\large{3=\frac{a}{3}}\]
\[\large{3\times 3 = \frac{a}{3} \times 3}\]
\[\large{9 = \frac{a}{\cancel{3}^1}\times \cancel{3}^1}\]
\[\large{9=a}\]

Answer 16

that above was an example

Answer 17

ohhh okay, and after i muliply both sides whats next? can you go through the problem and explain it?

Answer 18

ok
\[\large{C\times \frac{9}{5} = \frac{5}{9} \times \frac{9}{5} (F-32)}\]
\[\large{C\times \frac{9}{5} = \frac{\cancel{5}^1}{\cancel{9}^1}\times \frac{\cancel{9}^1}{\cancel{5}^1} (F-32)}\]
\[\large{\frac{9C}{5}=F-32}\]

Answer 19

\[\large{9C=(F-32)5}\]
can you expand 5(F-32) ?

Answer 20

@gabbbdee please reply

Answer 21

sorry if i expand it be 5F-160 of i do the rainbow thing and the equation will equal -32 if im right? O.o

Answer 22

@mathslover

Answer 23

\[\large{9C=5F-160}\]
\[\large{9C+160=5F}\]
\[\large{\frac{9C+160}{5}=F}\]

Answer 24

THANK YOU!

Answer 25

@nincompoop lol