F=ar^t if the half-life is 96 years...
.5=r^96
ln(.5)=96lnr
ln(.5)/96=lnr
r=e^((ln.5)/96)
f=800e^(336(ln.5)/96)
f=70.71
So about 71mg will remain after 336 years.
Answer:
x=-5
Step-by-step explanation:
-3x+25=40. You want to isolate x so you have to subtract 25 from both sides. -3x=15. We then want x to be isolated so we have to divide both sides by -3 to make x by itself. x=-5. Therefore, x=-5.
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Check the picture below, that's just an example of a parabola opening upwards.
so the cost equation C(b), which is a quadratic with a positive leading term's coefficient, has the graph of a parabola like the one in the picture, so the cost goes down and down and down, reaches the vertex or namely the minimum, and then goes back up.
bearing in mind that the quantity will be on the x-axis and the cost amount is over the y-axis, what are the coordinates of the vertex of this parabola? namely, at what cost for how many bats?

![\bf \left( -\cfrac{-7.2}{2(0.06)}~~,~~390-\cfrac{(-7.2)^2}{4(0.06)} \right)\implies (60~~,~~390-216) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{\textit{number of bats}}{60}~~,~~\stackrel{\textit{total cost}}{174})~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20-%5Ccfrac%7B-7.2%7D%7B2%280.06%29%7D~~%2C~~390-%5Ccfrac%7B%28-7.2%29%5E2%7D%7B4%280.06%29%7D%20%5Cright%29%5Cimplies%20%2860~~%2C~~390-216%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%28%5Cstackrel%7B%5Ctextit%7Bnumber%20of%20bats%7D%7D%7B60%7D~~%2C~~%5Cstackrel%7B%5Ctextit%7Btotal%20cost%7D%7D%7B174%7D%29~%5Chfill)
Hello, I'm from Indonesia. I hope that answer your question.
F + V - E = 2....solving for F
V = 16
E = 37
just a matter of subbing and solving
F + 16 - 37 = 2
F - 21 = 2
F = 2 + 21
F = 23