2006-12-06 20:46:04 · 12 answers · asked by drew t 1 in Science & Mathematics ➔ Mathematics
60*(9/5)+32
=140*F
10(9/5)+32
=50*F
2006-12-06 20:48:42 · answer #1 · answered by raj 7 · 0⤊ 0⤋
The formula is 9/5 of C + 32º The reverse is 5/9 of F -32º
60 C= 9/5 C = 9x12 = 108 +32 = 140 F
10C = 9x2 = 18 +32 = 50F
This from memory- but I am pretty sure this is correct.
2006-12-07 05:02:16 · answer #2 · answered by jamartx2000 3 · 0⤊ 0⤋
F = C(9/5)+32
So 60 C = 140 F
10 C = 50 F
2006-12-07 04:53:03 · answer #3 · answered by Srinivas c 2 · 0⤊ 0⤋
To convert centegrade to Fahrenheit multiply by 9 and divide by 5 and then add 32
To convert from F to C subtract 32 multiply by 5 and divide by 9
So 60 deg C = 140 deg F 10 deg C = 50 deg F
Brendan E
2006-12-07 04:56:35 · answer #4 · answered by Brendan E 2 · 0⤊ 0⤋
60deg C=60x9/5+32=140 deg F.
10 =10x9/5+32= 50.
VR
2006-12-07 04:49:06 · answer #5 · answered by sarayu 7 · 0⤊ 0⤋
60 degrees Celsius = 140 degrees Fahrenheit
10 degrees Celsius = 50 degrees Fahrenheit
2006-12-07 04:49:52 · answer #6 · answered by Cinna B 2 · 0⤊ 0⤋
C/5 = (F-32)/9
2006-12-07 04:54:33 · answer #7 · answered by Anonymous · 0⤊ 0⤋
to calibrate a thermometer we use:
lowest temp i.e.freezing point of water = 0C =32 F
highest temp, i.e. boiling point of water =100C =212F
diff: 100C = (212-32=180F)
hence any temp t' in F above freezing pt. is, t' = (180/100)tC +32
=((9C/5)+32)F
where tC is the temp given in C.
in ur case t=60C hence t' = 140F
t=10C => t' = 50F
2006-12-07 05:02:15 · answer #8 · answered by anami 3 · 0⤊ 0⤋
60degC = 140degF
the formula is F = (C*9/5) + 32
Similarly, 10degC = (10 * 1.8) + 32 = 50degF
2006-12-07 05:17:45 · answer #9 · answered by cooldude 3 · 0⤊ 0⤋
F = (9/5)*C + 32
C = (5/9)*(F-32)
-40°C = -40°F
Plug 'n play âº
Doug
2006-12-07 04:48:55 · answer #10 · answered by doug_donaghue 7 · 0⤊ 0⤋