No. x cannot equal 2 and 7
        
             
        
        
        
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:
with what know question was there
 
        
             
        
        
        
Answer:
The correct answer is 15 cm.
Step-by-step explanation:
Let the width of the required poster be a cm.
We need to have a 6 cm margin at the top and a 4 cm margin at the bottom. Thus total margin combining top and bottom is 10 cm.
Similarly total margin combining both the sides is (4+4=) 8 cm.
So the required printing area of the poster is given by {( a-10 ) × ( a - 8) } 
This area is equal to 125  as per as the given problem.
 as per as the given problem.
∴ (a - 10) × (a - 8) = 125
⇒  - 18 a +80 -125 =0
 - 18 a +80 -125 =0
⇒  - 18 a -45 = 0
 - 18 a -45 = 0
⇒ (a-15) (a-3) = 0
By law of trichotomy the possible values of a are 15 and 3.
But a=3 is absurd as a  4.
 4.
Thus the required answer is 15 cm.
 
        
                    
             
        
        
        
Step-by-step explanation:
Assuming figure to be trapezoid, 
Perimeter=6+4+7+4yd
 =21 yd
Area of trapezoid=((6+7)/2) x ((4)^2-(0.5)^2)
 =(13/2)x(16-0.25)
 =6.5 x 15.75
 =102.375 Sq yards