Round 8.1 down to 8
Round 4.2 down to 4
8(4)= 32
The estimated product is 32.
Hope this helps!
Answer:
![h = \sqrt[3]{\frac{49V}{4}}](https://tex.z-dn.net/?f=h%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B49V%7D%7B4%7D%7D)
Step-by-step explanation:
Represent the volume of the box with V and the dimensions with l, b and h.
The volume (V) is:

Make h the subject of the formula

The surface area (S) of the aquarium is:

Where lb represents the area of the base (i.e. slate):
The cost (C) of the surface area is:



Substitute
for h in the above equation



Differentiate with respect to l and with respect to b


To solve for b and l, we equate both equations and set l to b (to minimize the cost)


By comparison:

becomes

Cross Multiply

Solve for l

![l = \sqrt[3]{\frac{2V}{7}}](https://tex.z-dn.net/?f=l%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B2V%7D%7B7%7D%7D)
Recall that: 
![b = \sqrt[3]{\frac{2V}{7}}](https://tex.z-dn.net/?f=b%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B2V%7D%7B7%7D%7D)
Also recall that:

![h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7BV%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7B2V%7D%7B7%7D%7D%2A%5Csqrt%5B3%5D%7B%5Cfrac%7B2V%7D%7B7%7D%7D%7D)
![h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7BV%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7B4V%5E2%7D%7B49%7D%7D%7D)
Apply law of indices
![h = \sqrt[3]{\frac{49V^3}{4V^2}}](https://tex.z-dn.net/?f=h%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B49V%5E3%7D%7B4V%5E2%7D%7D)
![h = \sqrt[3]{\frac{49V}{4}}](https://tex.z-dn.net/?f=h%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B49V%7D%7B4%7D%7D)
The dimension that minimizes the cost of material of the aquarium is:
![h = \sqrt[3]{\frac{49V}{4}}](https://tex.z-dn.net/?f=h%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B49V%7D%7B4%7D%7D)
Answer:
18
Step-by-step explanation:
24/4 = 6 so....
4*6 is 24 and 3*6 is 18
The half-angle formula for tangent is:
tan(a/2) = (sin a / (1 + cos a)) = ((1 - cos a) / sin a)
Now we can plug in values:
tan(5π/8) = (sin(5π/4) / (1 + cos(5π/4)) = ((1 - cos(5π/4)) / sin(5π/4)
tan(5π/8) = (-√2/2) / (1 + (-√2/2)) = (1 - (-√2/2)) / (-√2/2)
tan(5π/8) = ((-√2/2)) / ((2 - √2)/2) = ((2 + √2)/2) / (-√2/2)
Now we can solve the first half:
(-√2/2)(2 / (2 - √2))
(-√2/2)((4 + 2√2) / 2)
(-√2/2)(2 + √2)
(-2√2 - 2)/2
-√2 - 1
tan(5pi/8) = -√2 - 1