What is the volume of a hemisphere with a radius of 3.8in, rounded to the nearest tenth of a cubic inch
1 answer:
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Answer:
height = 14.5 ft
width = length = 3.2 ft
Step-by-step explanation:
Let the side of the square base is a and the height of the box is h.
The material required to make the box is equal to the total surface area of the box.
Area of base = a²
Cost of square base is $ 2 per ft²
So, the cost of making base = 2a²
Area of four walls + area of top = 2 (a² + ah + ah) + a² = 3a² + 2ah
Cost of making walls and the top is $ 1 per ft²
So, the cost of making walls and the top = 1 x ( 3a² + 2ah)
Total cost of making the box = 2a² + 3a² + 2ah = 5a² + 2ah
According to the question
144 = 5a² + 2ah
.... (1)
The volume of the box is V = length x width x height
V = a x a x h
V = 72 a - 2.5 a³ from equation (1)
Differentiate both sides with respect to a.
dV/da = 72 - 7.5 a²
Put, it equal to zero for maxima and minima
7.5 a² = 72
a = 3.2 ft
Put in equation (1)
h = 14.5 ft
So, the dimensions of the box are
height = 14.5 ft
width = length = 3.2 ft
Answer: second option.
Step-by-step explanation:
In order to solve this exercise, it is necessary to remember the following properties of logarithms:
In this case you have the following inequality:
So you need to solve for the variable "x".
The steps to do it are below:
1. You need to apply to both sides of the inequality:
2. Now you must apply the properties shown before:
3. Then, rounding to the nearest ten-thousandth, you get: