Guest guest Posted November 4, 2004 Report Share Posted November 4, 2004 I was taught a new formula for temp last week. It is 9C=5F-160. It works everytime. > > Dear All, > > The BEST formula is one that you do not have to memorize a bunch for > different conditions. > > C = 5/9 (F-32) > > can be used also as: > > C = 0.556 (F-32) > > or > > 1.8 C = F - 32 > > or > > 1.8 C + 32 = F > > or > > F = 1.8 C + 32 > > > No matter which formula you use you should not have to memorize two > different formulas. > > > You will find the basic ones listed in the Facts and Comparisons. > > In my classroom I have a teacher who teaches two different formulas > for solving for C or F. This is confusing enough for the student. So > when they take finals they screw up the temp questions because they > wonder is it 1.8 C or 1.8 F ...... > > When I teach them ONE formula fits both conversions they NEVER get > them wrong on the finals. Why???? because they know one formula or > no formula. They do not have to worry which one does what. > > My suggestions is solve using only ONE the formula you like best. > NEVER switch. > > > for example: > Solve for C = 40, what is F? > F = 1.8 C + 32 > F = 1.8 x 40 + 32 > F = 72 + 32 > F = 104 > > C = 0.556 (F-32) > 40 = 0.556 (F - 32) > 40/.0556 = 0.556/0.556 (F-32) > 71.94 = F - 32 > 71.94 + 32 = F - 32 + 32 > 103.94 = F > round to: > 104 = F > > C = 5/9 (F-32) > 40 = 5/9 (F -32) > 40 x 9/5 = 5/9 x 9/5 (F- 32) > 72 = F - 32 > 72 + 32 = F -32 + 32 > 104 = F > > > Conversely Solve for F = 40, what is C? Use the SAME formula: > F = 1.8 C + 32 > 40 = 1.8 C + 32 > 40 - 32 = 1.8 C + 32 - 32 > 8 = 1.8 C > 8/1.8 = 1.8 C/1.8 > 4.44 = C > > > C = 0.556 (F-32) > C = 0.556 (40 - 32) > C = 0.556 x 8 > C = 4.448, rounded to 4.45 > > > C = 5/9 (F-32) > C = 5/9 (40-32) > C = 5/9 x 8 = 40/9 > C = 4.44 > > While some of you may prefer the decimals I prefer the fractions. > The point here is you do not need to memorize TWO different formulas > for solving for C or F. USE ONLY ONE formula for solving for either. > Less chance of error, less confusion, less memorization. > > Respectfully, > Jeanetta Mastron CPhT BS Chem > Pharm Tech Educator Quote Link to comment Share on other sites More sharing options...
Guest guest Posted November 7, 2004 Report Share Posted November 7, 2004 Dear Kimness2000, Yes that is a variation of the SAME formulas that have been posted. They ALL work every time. However it is my duty to teach/introduce an algebraic formula to my students in the class room. On the job or for one's own purpose using this formula is great. I think the MAIN point I make is this: it is not necessary to change formula just because you are looking for F and in another problem for T. Use the ONE formula and 'plug' in what you KNOW. Example: Change 7 degrees C to F 9C = 5F - 160 9 x 7 = 5F - 160 63 = 5F - 160 63 + 160 = 5F 223 = 5F 223/5 = 5F/5 44.5 = F AND Change 50 degrees F to C 9C = 5F - 160 9C = (5 x 50) - 160 9C = 250 - 160 9C = 90 9C/9 = 90/9 C = 10 I hope this helps others as well. Again it does NOT matter which formula that you use. However you will make LESS errors if you stick to ONE formula for solving for either degrees F or C. Good Luck, Jeanetta Mastron CPhT BSChem Founder/Owner of this site. kimness2000 <gkdtk@...> wrote: I was taught a new formula for temp last week. It is 9C=5F-160. It works everytime. Quote Link to comment Share on other sites More sharing options...
Guest guest Posted November 7, 2004 Report Share Posted November 7, 2004 Dear All, I have atypo in the following excerpt frm my previous email/post: " I think the MAIN point I make is this: it is not necessary to change formula just because you are looking for F and in another problem for T. Use the ONE formula and 'plug' in what you KNOW. " It SHOULD READ: " I think the MAIN point I make is this: it is not necessary to change formula just because you are looking for F and in another problem for C. Use the ONE formula and 'plug' in what you KNOW. " Respectfully, Jeanetta Mastron CPhT BS Chem Jeanetta Mastron <rxjm2002@...> wrote: Dear Kimness2000, Yes that is a variation of the SAME formulas that have been posted. They ALL work every time. However it is my duty to teach/introduce an algebraic formula to my students in the class room. On the job or for one's own purpose using this formula is great. I think the MAIN point I make is this: it is not necessary to change formula just because you are looking for F and in another problem for T. Use the ONE formula and 'plug' in what you KNOW. Example: Change 7 degrees C to F 9C = 5F - 160 9 x 7 = 5F - 160 63 = 5F - 160 63 + 160 = 5F 223 = 5F 223/5 = 5F/5 44.5 = F AND Change 50 degrees F to C 9C = 5F - 160 9C = (5 x 50) - 160 9C = 250 - 160 9C = 90 9C/9 = 90/9 C = 10 I hope this helps others as well. Again it does NOT matter which formula that you use. However you will make LESS errors if you stick to ONE formula for solving for either degrees F or C. Good Luck, Jeanetta Mastron CPhT BSChem Founder/Owner of this site. kimness2000 <gkdtk@...> wrote: I was taught a new formula for temp last week. It is 9C=5F-160. It works everytime. Quote Link to comment Share on other sites More sharing options...
Guest guest Posted November 7, 2004 Report Share Posted November 7, 2004 typos! typos! typos! damn those fingernails! Jeanetta Jeanetta Mastron <rxjm2002@...> wrote: Dear All, I have atypo in the following excerpt frm my previous email/post: " I think the MAIN point I make is this: it is not necessary to change formula just because you are looking for F and in another problem for T. Use the ONE formula and 'plug' in what you KNOW. " It SHOULD READ: " I think the MAIN point I make is this: it is not necessary to change formula just because you are looking for F and in another problem for C. Use the ONE formula and 'plug' in what you KNOW. " Respectfully, Jeanetta Mastron CPhT BS Chem Jeanetta Mastron <rxjm2002@...> wrote: Quote Link to comment Share on other sites More sharing options...
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