Answer:
The maximum height of the prism is 
Step-by-step explanation:
Let
x------> the height of the prism
we know that
the area of the rectangular base of the prism is equal to
so
-------> inequality A
------> equation B
-----> equation C
Substitute equation B in equation C
------> equation D
Substitute equation B and equation D in the inequality A
-------> using a graphing tool to solve the inequality
The solution for x is the interval---------->![[0,12]](https://tex.z-dn.net/?f=%5B0%2C12%5D)
see the attached figure
but remember that
The width of the base must be 
meters less than the height of the prism
so
the solution for x is the interval ------> ![(9,12]](https://tex.z-dn.net/?f=%289%2C12%5D)
The maximum height of the prism is
Answer:
f = -5, s = -6
Step-by-step explanation:
F= first number
S= second number
4f - s = -14
f + 3s = -23
3(4f - s = -14)
12f - 3s = -42
+ f + 3s = -23
13f = -65
————
13
f = -5
4f - s = -14
4(-5) - s = -14
-20 - s = -14
+20 +20
-s = 6
———
-1
s = -6
5∛x + 4 = 44
5∛x = 40
∛x = 8
x = 8^3
x = 512
answer
C. 512
Answer:
Actually it's not polygon. it's a nonagon. With r=8.65mm″, the law of cosines gives us side a:
a=√{b²+c²−2bc×cos40°}
a=√{149.645−149.645cos40°}
Area Nonagon = (9/4)a²cos40°
=9/4[149.645−149.645cos40°]cot20°
=336.70125[1−cos(40°)]cot(20°)
Applying an identity for the cos(40°) does not get us very far…
= 336.70125[1−(cos2(20°)−1)]cot(20°)
= 336.70125[2−cos2(20°)]cot(20°)
= 336.70125[2−(1−sin2(20°))]cot(20°)
= 336.70125[1+sin2(20°)]cos(20°)sin(20°)
= 336.70125[cot(20°)+sin(20°)cos(20°)]mm²
