...1994
t = year to model (ie: When t = 1, the year I am modeling is 1995. When t = 2, the year I am modeling is 1996)
2,200,000[e^(0.595476617 * t)] = A
===============================================================================================================...
Actually I am not a student at all - I am self-taught and lack many skills when it comes to efficient programming.
However I must mention that the options in my form are grouped into seperate frames (three seperate frames, one for each textbox).
If you feel so inclined I would greatly...
Hello PC888, thanks for asking!
I was having the same question myself! The higher the coffee temperature, the shorter it would take the coffee to cool . . . that seemed way wrong to me also! I beleive this results because the constants in this logarithmic and exponential function calculation...
Actually I am not a student at all - I am self-taught and lack many skills when it comes to efficient programming.
However I must mention that the options in my form are grouped into seperate frames (three seperate frames, one for each textbox).
If you feel so inclined I would greatly...
...inputted into textbox "Temperature of Coffee" the same in the formula calculation.
End If
If optT_Coffee_F = True Then
sngT_Coffee = (5 / 9) * (txtT_Coffee.Text - 32) 'Converts fahrenheit temperature inputted into textbox "Temperature of Coffee" to celsius in the formula calculation.
End...
...inputted into textbox "Temperature of Coffee" the same in the formula calculation.
End If
If optT_Coffee_F = True Then
sngT_Coffee = (5 / 9) * (txtT_Coffee.Text - 32) 'Converts fahrenheit temperature inputted into textbox "Temperature of Coffee" to celsius in the formula calculation.
End...
...temperature inputted into textbox "Temperature of Coffee" the same in the formula calculation.
If optT_Coffee_F = True Then
sngT_Coffee = (5 / 9) * (txtT_Coffee.Text - 32) 'Converts fahrenheit temperature inputted into textbox "Temperature of Coffee" to celsius in the formula...
...temperature inputted into textbox "Temperature of Coffee" the same in the formula calculation.
If optT_Coffee_F = True Then
sngT_Coffee = (5 / 9) * (txtT_Coffee.Text - 32) 'Converts fahrenheit temperature inputted into textbox "Temperature of Coffee" to celsius in the formula...
Sorry all!
I just needed to divide my solution with log(e)
Wohoo! Applied Learning!
Although I beleive Visual Basic should include a Natural Logarithm function!
Sorry all!
I just needed to divide my solution with log(e)
Wohoo! Applied Learning!
Although I beleive Visual Basic should include a Natural Logarithm function!
I understand how to calculate my problem and cancel out everything perfect while using the natural logarithm, but I am unfamiliar with substituting the LN with log because of the cancelation differences.
Code example:
txtTime.Text = LN((sngT_Coffee - sngT_Ambient) / (75)) / (-0.1)
Note for my...
I am sorry johnwm. I understand how to calculate my problem and cancel out everything perfect while using the natural logarithm, but I am unfamiliar with substituting the LN with log because of the cancelation differences.
Code example:
txtTime.Text = LN((sngT_Coffee - sngT_Ambient) / (75)) /...
I am having a difficult time finding information on how to use the Natural Logarithm in my Visual Basic v6.0 code. If anyone is familiar using the Natural Logarithm with Visual Basic, please give me a hand.
Here is my code example:
txtTime.Text = LN((sngT_Coffee - sngT_Ambient) / (75)) /...
How can I use the Natural Logarithm in my code? I am having a difficult time finding information on how to use the Natural Logarithm in my Visual Basic code.
If anyone is familiar with using the Natural Logarithm with Visual Basic, please give me a hand.
Here is my code example:
txtTime.Text...
...so much.
Here is the code to my fully working application! Thanks to your help my current and future applications will be much more flexible!
'**************************************************************************************
'
' Project: Encoding Video Bitrate...
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