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
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Answer:
two hundredths
Step-by-step explanation:
6 is in the ones place. 3 is in the tenths place. 2 is in the hundredths place
5 is in the thousandths place.
No, because there are TWO values of y for a certain value of x
say x=5, then y=2 and y=-2
That does not satisfy the concept of the function.
If 68 is a prime number, then the only factors it has are 1 and 68.
If it has any other factors besides 1 and 68, then it's NOT prime.
Right away, without any higher math, you can look at just the last digit
in 68 . The last digit is '8'. That tells you that '68' is an even number,
and THAT tells you that '2' must be one of its factors. So '68' is not a
prime number.
The factors of 68 are 1, 2, 4, 17, 34, and 68 .
68 has four more factors besides 1 and 68, so it's not a prime number.
Evaluating the given sequence, it is evident that the next number is twice the number prior to it. Thus, the given is a geometric sequence with first term (a1) equal to 1 and common ratio of 2. The geometric series may be calculated by the equation,
Sn = a1 x (1 - r^n) / (1 - r)
where Sn is the sum of n terms in this case, n = 11.
Substituting the known values,
<span> Sn = 1 x (1 - 2^11) / (1 - 2) = 2047
</span>
Thus, S11 is 2047.
