The containers must be spheres of radius = 6.2cm
<h3>
How to minimize the surface area for the containers?</h3>
We know that the shape that minimizes the area for a fixed volume is the sphere.
Here, we want to get spheres of a volume of 1 liter. Where:
1 L = 1000 cm³
And remember that the volume of a sphere of radius R is:
Then we must solve:
![V = \frac{4}{3}*3.14*R^3 = 1000cm^3\\\\R =\sqrt[3]{ (1000cm^3*\frac{3}{4*3.14} )} = 6.2cm](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%7D%7B3%7D%2A3.14%2AR%5E3%20%3D%201000cm%5E3%5C%5C%5C%5CR%20%3D%5Csqrt%5B3%5D%7B%20%20%281000cm%5E3%2A%5Cfrac%7B3%7D%7B4%2A3.14%7D%20%29%7D%20%3D%206.2cm)
The containers must be spheres of radius = 6.2cm
If you want to learn more about volume:
brainly.com/question/1972490
#SPJ1
Answer:
Inequality : 4x+8 < 120
x<28
Step-by-step explanation:
Hi, to answer this question we have to apply the next formula:
Perimeter of a rectangle (P)= 2 length +2 width
Where:
Width = x
Length = x+4
Replacing with the values given
P = 2 (x+4) + 2x
P = 2x+8+2x
P=4x+8
Since the perimeter of the rectangle must be less than 120:
4x+8 < 120
Solving for x
4x<120-8
4x<112
x<112/4
x<28
Feel free to ask for more if needed or if you did not understand something.
Hello,
Answer C if x≠0
(x^5-x^4+x²)/(-x²)=-x²(x^3-x²+1)/x²=-(x^3-x²+1)=-x^3+x²-1
