sorry to be so off topic but does someone know of a tool to log back into an instagram account? I was stupid forgot my password. I love any help you can offer me!
@Case Kaiser I really appreciate your reply. I found the site through google and I'm waiting for the hacking stuff atm. Looks like it's gonna take quite some time so I will reply here later with my results.
This is sort of a trivial problem, because I said to determine the rate at half of the Vmax, which was 40 umol/s. The rate is obviously 20 umol/s. However, something more subtle and important here is that when the [S] is equal to the Km of the enzyme, you are always at half of the Vmax. This example proved it mathematically.
For an enzyme that follows simple MichaelisMenten kinetics, what is the value of Vmax if
V0
is equal to 1 µmol/minute at 1/10 KM?
what is the answer
sorry to be so off topic but does someone know of a tool to log back into an instagram account?
I was stupid forgot my password. I love any help you can offer me!
@Tate Albert instablaster :)
@Case Kaiser I really appreciate your reply. I found the site through google and I'm waiting for the hacking stuff atm.
Looks like it's gonna take quite some time so I will reply here later with my results.
@Case Kaiser It worked and I actually got access to my account again. I am so happy:D
Thanks so much, you really help me out !
@Tate Albert You are welcome :)
The question literally says half of vmax is 40. so isnt the answer 40?
This is sort of a trivial problem, because I said to determine the rate at half of the Vmax, which was 40 umol/s. The rate is obviously 20 umol/s. However, something more subtle and important here is that when the [S] is equal to the Km of the enzyme, you are always at half of the Vmax. This example proved it mathematically.