180|2
90|2
45|3
15|3
5|5
1
90|2
45|3
15|3
5|5
1
Please help, DUE TODAY. If correct I will mark Brainliest.
1 answer:
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The volume of a box with dimensions x by x by x is 
.
So if we are given the volume V of a cube-shaped box, the side length of thit is![\sqrt[3]{V}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7BV%7D%20)
.
So we calculate the cubic roots of the volumes we have, and we add their heights.
To calculate the cubic root of 729, we can factorize it, and group the perfect cubes together, as follows:
, which we recognize as 
so
similarly 1,331 can be found to be
.
Thus we have 2 boxes with side length equal to 11 m and one with side length equal to 9 m.
Answer: h= 11+11+9 = 31 (meters)
So if we are given the volume V of a cube-shaped box, the side length of thit is
So we calculate the cubic roots of the volumes we have, and we add their heights.
To calculate the cubic root of 729, we can factorize it, and group the perfect cubes together, as follows:
so
similarly 1,331 can be found to be
Thus we have 2 boxes with side length equal to 11 m and one with side length equal to 9 m.
Answer: h= 11+11+9 = 31 (meters)
Given:
The temperature of the first 6 days in march is:
23F, 58F, 62F, 64F, 64F, 65F
23 F is removed from the data set.
To find:
How much greater is the range of the original 6 temperatures than the range of the 5 temperatures.
Solution:
Range of a data set is:
Range = Maximum value - Minimum value
The given data set is:
23 F, 58 F, 62 F, 64 F, 64 F, 65 F
The range of given data set is:
Range = 65 - 23
= 42
If the temperature 23 F is removed from the data set, then the new data set is:
58 F, 62 F, 64 F, 64 F, 65 F
The range of this new data set is:
Range = 65 - 58
= 7
Now, the difference between original range and new range is:
The original range is 35 more than the new range. Therefore, the correct option is a.