Helloo, i don’t see the image seen above, is that the only picture?
First we need to find k ( rate of growth)
The formula is
A=p e^kt
A future bacteria 4800
P current bacteria 4000
E constant
K rate of growth?
T time 5 hours
Plug in the formula
4800=4000 e^5k
Solve for k
4800/4000=e^5k
Take the log for both sides
Log (4800/4000)=5k×log (e)
5k=log (4800/4000)÷log (e)
K=(log(4,800÷4,000)÷log(e))÷5
k=0.03646
Now use the formula again to find how bacteria will be present after 15 Hours
A=p e^kt
A ?
P 4000
K 0.03646
E constant
T 15 hours
Plug in the formula
A=4,000×e^(0.03646×15)
A=6,911.55 round your answer to get 6912 bacteria will be present after 15 Hours
Hope it helps!
Answer:
d=54
Step-by-step explanation:
22 + 15 = d - 17
37=d-17
54=d
d=54
Step-by-step explanation:
(cos 10° − sin 10°) / (cos 10° + sin 10°)
Rewrite 10° as 45° − 35°.
(cos(45° − 35°) − sin(45° − 35°)) / (cos(45° − 35°) + sin(45° − 35°))
Use angle difference formulas.
(cos 45° cos 35° + sin 45° sin 35° − sin 45° cos 35° + cos 45° sin 35°) / (cos 45° cos 35° + sin 45° sin 35° + sin 45° cos 35° − cos 45° sin 35°)
sin 45° = cos 45°, so dividing:
(cos 35° + sin 35° − cos 35° + sin 35°) / (cos 35° + sin 35° + cos 35° − sin 35°)
Combining like terms:
(2 sin 35°) / (2 cos 35°)
Dividing:
tan 35°
No. It is not. Only what's inside counts.
