The minimum surface area that such a box can have is 380 square
<h3>How to determine the minimum surface area such a box can have?</h3>
Represent the base length with x and the bwith h.
So, the volume is
V = x^2h
This gives
x^2h = 500
Make h the subject
h = 500/x^2
The surface area is
S = 2(x^2 + 2xh)
Expand
S = 2x^2 + 4xh
Substitute h = 500/x^2
S = 2x^2 + 4x * 500/x^2
Evaluate
S = 2x^2 + 2000/x
Differentiate
S' = 4x - 2000/x^2
Set the equation to 0
4x - 2000/x^2 = 0
Multiply through by x^2
4x^3 - 2000 = 0
This gives
4x^3= 2000
Divide by 4
x^3 = 500
Take the cube root
x = 7.94
Substitute x = 7.94 in S = 2x^2 + 2000/x
S = 2 * 7.94^2 + 2000/7.94
Evaluate
S = 380
Hence, the minimum surface area that such a box can have is 380 square
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Answer:
C. √2 - 1
Step-by-step explanation:
If we draw a square from the center of the large circle to the center of one of the small circles, we can see that the sides of the square are equal to the radius of the small circle (see attached diagram)
Let r = the radius of the small circle
Using Pythagoras' Theorem 
(where a and b are the legs, and c is the hypotenuse, of a right triangle)
to find the diagonal of the square:
So the diagonal of the square = 
We are told that the radius of the large circle is 1:
⇒ Diagonal of square + r = 1
Using the quadratic formula to calculate r:
As distance is positive, 
only
Answer:
diameter = IJ
radius = KH or IH or JH
chord = LM
length HK = 2
Step-by-step explanation:
if IJ is the diameter and the radius is KH then r must be 2 as the radius is always half the diameter
4000 milliliters are in 4 liters
I believe your answer will be reflection
hope this helps
