Answer:
Kayla is correct The center is a fixed point in the middle of the sphere
Step-by-step explanation:
In mathematics we have certain habit of rules for notation of points, coordinates, segments, angles and so on.
Usually we denote points, by letters even more we denote with the first letter of the object we are denoting
Occasionally, we also denote segments as radius in a circle and in a sphere, with letters, that is r stands for radius, h stands for height, in most cases we denote point for capital letters ( in a segment)
When we denote radius, with small letter it should be placed at the center or over the segment we are traying to denote.
For points we only need to place the letter close to the to the point we want to denote.
Therefore Kayla is correct when says that c stand for " the center of the sphere"
Answer:
3
Step-by-step explanation:
If 4 | 1 = 41, that means that when you're looking at the number, you basically see it without the "|". If you see it that way, the smallest number would be 03.
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
The number of ways of choosing 4 digits from 10 digits without replacement ( order is important) is 
. This is the number of all possible permutations of choosing 4 from 10. The first 4 numbers of your phone number is a unique 4 digit number(4 different digits). The required probability is
I think it’s B but I’m not sure
