Answer:
about 8.2 cm
Step-by-step explanation:
The key is to realize that the volume never changes.
The formula for the volume of a sphere is (4/3)*pi*radius^3
22/7 is being substituted in place of pi in this problem
The volume of one of the small spheres is
(4/3)*(22/7)*2^3 is about 33.524 cm^3
64 of those spheres would have a volume of 33.524*64, or about 2145 cm^3
Now, the problem is to find a sphere with a volume of 2145 cm^3
Volume = (4/3)*(22/7)*radius^3
plug in and solve
2145 = 4.1905*r^3
r^3=511.87
r is about equal to 8.2 centimeters
Answer:
Answer is explained below in the explanation section.
Step-by-step explanation:
Solution:
For this question, we need to have a number line, we can not make the number line here. As it will be very long. Instead, I have attached a number line from -11 to +11 for your understanding and explained it below. Please refer to the attachment below for the number line.
Number Line:
A number line is a line which contains all the numbers from negative infinity to positive infinity divided by a single number zero. It has two sides, left hand side and right hand side. A side left to zero contains all the negative numbers and a side right to 0 contains, all the positive numbers.
So,
We are asked to add 34 in (-212), we will get 178 but with the negative sign, and it will lie on the left hand side of the number line.
SO,
34 + (-212) = 34 - 212 = -178
Hence,
Our sum which is 178 will lie on the left hand side of the number line.
Answer:
x meters= 35
5x meters= 175
6x - 8 = 202
Step-by-step explanation:
x + 5x + (6x - 8)= 412
12x - 8= 412
12x = 412 + 8
12x = 420
x= 35
5x = 35× 5
= 175
6x - 8= 6(35) - 8
= 210 - 8
= 202
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).
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