Respuesta :

dalendrk
[tex]\dfrac{9}{5}C+32=F\ \ \ \ |subtract\ 32\ from\ both\ sides\\\\\dfrac{9}{5}C=F-32\ \ \ \ |multiply\ both\ sides\ by\ \dfrac{5}{9}\\\\\boxed{C=\frac{5}{9}(F-32)}[/tex]
Ckaranja
F=9/5C+32
F-32=9/5C
9/5C=F-32
9C=5(F-32)
C=[tex] \frac{5(F-32)}{9} [/tex]
, end="") y += 5(a) $$$$$(b) $$(c) $$$$$$$$$$$$$$$(d) Nothing wil
when determining the nature of a drug inquiry, how much detail required is also considered the ___ of the inquiry
pleeeeease help meeeeeee
a solid substance a is soluble in water to the extent of 10 mg/ml of water at 25oc and 100 mg/ml of water at 100oc. you have a sample that contains 100 mg of a