Step-by-step explanation:
<h3><em>To</em><em> </em><em>write</em><em> </em><em>8</em><em> </em><em>more</em><em> </em><em>than</em><em> </em><em>follow</em><em> </em><em>this</em><em>:</em><em>-</em></h3><h3><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>k</em><em>+</em><em>8</em></h3>
We assume you intend
![D(h)=35e^{-0.15h}](https://tex.z-dn.net/?f=D%28h%29%3D35e%5E%7B-0.15h%7D)
You want to evaluate this function for h=0 and h=8.
![D(0)=35e^{0}=35\\D(8)=35e^{-0.15\cdot 8}\approx 10.54](https://tex.z-dn.net/?f=D%280%29%3D35e%5E%7B0%7D%3D35%5C%5CD%288%29%3D35e%5E%7B-0.15%5Ccdot%208%7D%5Capprox%2010.54)
The amount injected was 35 milligrams; 8 hours later 10.54 milligrams remained.
A = 5 + t
t + 3 = 2/3a
t + 3 = 2/3(5 + t)
t + 3 = 10 + 2t / 3
3t + 9 = 10 + 2t
t = 1
a = 5 + 1
a = 6
Answer:
36
Step-by-step explanation: