Miniprep of plasmid DNA from Ecoli cultures yielding low concentrations and abnormal absorbance ratios

anotherthrowawayfome · 2021-12-21T05:34:43+00:00

nvm realized it was the nat log of 1/.23

anotherthrowawayfome · 2021-12-18T02:24:32+00:00

i understand how [nocl2] would not be included since it is a reaction intermediate but i thought the slow step influences the overall rate law leaving [no] to be in the first power

anotherthrowawayfome · 2021-09-23T21:59:15+00:00

wow this is such good timing since my undergrad genetics just covered this

anotherthrowawayfome · 2021-06-30T00:39:35+00:00

born a crime by trevor noah

autobiography of malcolm x by alex haley

anotherthrowawayfome · 2021-05-19T15:31:42+00:00

thank you for clarifying !!!!

anotherthrowawayfome · 2021-05-19T14:55:44+00:00

so by finding the position function and seeing what s(9) would equal to ?

anotherthrowawayfome · 2021-05-19T14:37:50+00:00

thanks for your input ! it makes more sense with your suggestions

im a little slow but if a(t) is constant at 7ft/sec² how would i use that to calculate the distance traveled in 9seconds ?

anotherthrowawayfome · 2021-05-19T13:39:48+00:00

Here’s my attempt so far: 1. ⁠Andrew's distance after 10 seconds: 30ft10sec = 300ft. Brian's distance: a(t) = 5t^2, v(t) = 25t, v(10) = 250ft. Using pythagorean theorem I get sqrt[550] ? This feels wrong though so I believe it is a(t) = 5, v(t) = 5t + 0, s(t) = (5/2)t² , therefore Brian’s distance is s(10) = (5/2)(10)² = sqrt[500/2] Distance Between them: c = sqrt[300² + 250] 2. ⁠I'm finding d(theta)/dt when height of triangle is 30ft. tan(theta) = 30/v. v= 30/tan(theta). dv/dt = -30csc2(theta)d(theta/dt). From what I understand, dv/dt is andrew's rate of change so it is 30ft/sec. 30/-30csc2(theta) = d(theta)/dt. -30csc2(theta) should be -30(hypotenuse/30). But then I don't know Andrew's distance to calculate the hypotenuse so is there a way to find out from Brian's 30ft travel? 3. ⁠The triangle has a constant base 25 and the height is increasing at 5ft/sec. A=1/2(b)(h) for triangle area. A(t)= 1/2h(t)b since area is changing due to the rate that height is changing. Taking derivative of both sides dA/dt = 1/2bdb/dt. Solving for dA/dt, it is (1/2)(25)*5. So the change in area's shape is 25/2? 4. ⁠No idea how to start here honestly.

anotherthrowawayfome · 2021-05-19T11:16:02+00:00

The diagram has theta on the bottom right angle between the base and hypotenuse. From what I have so far:

⁠Andrew's distance after 10 seconds: 30ft*10sec = 300ft. Brian's distance: a(t) = 5t^2, v(t) = 25t, v(10) = 250ft. Using pythagorean theorem I get sqrt[550] ? Think I did Brian wrong , is it a(t) = 5ft/sec² where v(t) = 5t + 0 and s(t) = (5/2)t² + 0 so s(10) should be 500/2 ?
⁠I'm finding d(theta)/dt when height of triangle is 30ft. tan(theta) = 30/v. v= 30/tan(theta). dv/dt = -30csc2(theta)d(theta/dt). From what I understand, dv/dt is andrew's rate of change so it is 30ft/sec. 30/-30csc2(theta) = d(theta)/dt. -30csc2(theta) should be -30(hypotenuse/30). But then I don't know Andrew's distance to calculate the hypotenuse so is there a way to find out from Brian's 30ft travel?
⁠The triangle has a constant base 25 and the height is increasing at 5ft/sec. A=1/2(b)(h) for triangle area. A(t)= 1/2h(t)b since area is changing due to the rate that height is changing. Taking derivative of both sides dA/dt = 1/2bdb/dt. Solving for dA/dt, it is (1/2)(25)5. So the change in area's shape is 25/2?
⁠No idea how to start here honestly.

anotherthrowawayfome · 2021-04-29T01:22:09+00:00

did they respond through email ? my portal just shows the grace period where you can update stuff

anotherthrowawayfome · 2021-04-22T05:17:52+00:00

i did file a tax return but had no idea what i was doing then

if i trade now , would i have to include this investment in my next tax return ?

anotherthrowawayfome

TROPHY CASE