Answer:
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola
y=5−x^2. What are the dimensions of such a rectangle with the greatest possible area?
Width =
Height =
Width =√10 and Height 
Step-by-step explanation:
Let the coordinates of the vertices of the rectangle which lie on the given parabola y = 5 - x² ........ (1)
are (h,k) and (-h,k).
Hence, the area of the rectangle will be (h + h) × k
Therefore, A = h²k ..... (2).
Now, from equation (1) we can write k = 5 - h² ....... (3)
So, from equation (2), we can write
![A =h^{2} [5-h^{2} ]=5h^{2} -h^{4}](https://tex.z-dn.net/?f=A%20%3Dh%5E%7B2%7D%20%5B5-h%5E%7B2%7D%20%5D%3D5h%5E%7B2%7D%20-h%5E%7B4%7D)
For, A to be greatest ,

⇒ ![h[10-4h^{2} ]=0](https://tex.z-dn.net/?f=h%5B10-4h%5E%7B2%7D%20%5D%3D0)
⇒ 
⇒ 
Therefore, from equation (3), k = 5 - h²
⇒ 
Hence,
Width = 2h =√10 and
Height = 
I think n is 9 hope i helped
Answer:
The simplified expression is 16^-1/6
Step-by-step explanation:.
16^5/4•16^1 /4/ (16^1/2)^2
The dot means multiplication. The expression is rewritten as
16^5/4 × 16^1/4 / (16^1/2)^2
Recall the following rules of indices
1) y^a × y^b = y^(a+b)
2) (y^a)^b = y^ab
y^b/ y^a = y^(b-a)
Applying these rules of indices
16^5/4 × 16^1/4= 16^((5/4×1/4) = 16^((5/16)
(16^1/2)^2= 16^ 1/2×2 = 16^1 = 16
Therefore
16^5/4 × 16^1/4 / (16^1/2)^2
= 16^(5/16) /16
=16^(5/6) × 16^-1
= 16^ (5/6)-1
= 16^-1/6)
,The simplified expression is 16^-1/6