To get which design would have maximum area we need to evaluate the area for Tyler's design. Given that the design is square, let the length= xft, width=(120-x)
thus:
area will be:
P(x)=x(120-x)
P(x)=120x-x²
For maximum area P'(x)=0
P'(x)=120-2x=0
thus
x=60 ft
thus for maximum area x=60 ft
thus the area will be:
Area=60×60=3600 ft²
Thus we conclude that Tyler's design is the largest. Thus:
the answer is:
<span>Tyler’s design would give the larger garden because the area would be 3,600 ft2. </span>
Answer:
Step-by-step explanation:
Use the change-of-base rule.
log₂(62) = ln(62)/ln(2)
Answer:
y = 4 sin(2π/11 x) + 2
Step-by-step explanation:
y = A sin(2π/T x + B) + C
where A is the amplitude,
T is the period,
B is the phase shift,
and C is the midline.
A = 4, T = 11, and C = 2. We'll assume B = 0.
y = 4 sin(2π/11 x) + 2
Answer:
Step-by-step explanation:
The function is 
Changing functional notation of a(m) to y:
Now, interchanging m and y:
Now, solving for y:
![e^{0.423y}=\frac{m}{0.6735}\\ln[e^{0.423y}]=ln[\frac{m}{0.6735}]\\0.423y=ln(\frac{m}{0.6735})\\y=\frac{ln(\frac{m}{0.6735})}{0.423}](https://tex.z-dn.net/?f=e%5E%7B0.423y%7D%3D%5Cfrac%7Bm%7D%7B0.6735%7D%5C%5Cln%5Be%5E%7B0.423y%7D%5D%3Dln%5B%5Cfrac%7Bm%7D%7B0.6735%7D%5D%5C%5C0.423y%3Dln%28%5Cfrac%7Bm%7D%7B0.6735%7D%29%5C%5Cy%3D%5Cfrac%7Bln%28%5Cfrac%7Bm%7D%7B0.6735%7D%29%7D%7B0.423%7D)
Thus, the inverse function is:
