you want log5(12)
x = log5 (12)
or
5^x = 12
log10 (5^x) = log10 (12) = 1.0792
or
x log10 (5) = log10 (12) = 1.0792
x (.699) = 1.0792
x = 1.0792/.699
1) given that x= log(small5)3 + log(small5)4, find algebraically the value of x
2)a mug of tea cools according to the law T(small t)= T(small 0)e to the power^-kt where T(small 0) is the initial temp and T(small c)is the temp after t minutes. all temps are in celciius.
a) a particular mug of tea is cooled from 100*C to 75*C in a quater of an hour. therefore clculate value of k.
b) by how many degrees will the temperature of this tea fall in the next quater of an hour.
3 answers
2)a mug of tea cools according to the law T(small t)= T(small 0)e to the power^-kt where T(small 0) is the initial temp and T(small c)is the temp after t minutes. all temps are in celciius.
Tt = To e^(-kt)
a) a particular mug of tea is cooled from 100*C to 75*C in a quater of an hour. therefore clculate value of k.
75 = 100 e^-(.25k) if t in hours and not minutes
.75 = e^-.25k
ln .75 = -.25k
k = -.288/-.25
b) by how many degrees will the temperature of this tea fall in the next quater of an hour.
T = 75 e^-1.15 (.25)
I will leave that to you
Tt = To e^(-kt)
a) a particular mug of tea is cooled from 100*C to 75*C in a quater of an hour. therefore clculate value of k.
75 = 100 e^-(.25k) if t in hours and not minutes
.75 = e^-.25k
ln .75 = -.25k
k = -.288/-.25
b) by how many degrees will the temperature of this tea fall in the next quater of an hour.
T = 75 e^-1.15 (.25)
I will leave that to you
thanks alot damon ...i luv u!!!!
4 answers