If the result is neither negative nor positive, the simplest option is saying that 0 is neither negative nor positive.
For a number to be subtracted from it's square and equal 0, the only possible number is 1.
1 squared is 1, and 1-1=0.
Answer:
136 square millimeters
Step-by-step explanation:
The answer is four and five eights!
Answer:
The mean of the remaining 3 boys is 51.
Step-by-step explanation:
The mean mark of the entire group (
), dimensionless, is:
(Eq. 1)
(
)
(Eq. 1b)
Where
is the mark of the i-th boy, dimensionless.
In addition, we know the following mean marks from statement:
(Eq. 2)
(
)
(Eq. 2b)
(Eq. 3)
Where:
- Mean mark of the first 7 boys, dimensionless.
- Mean mark of the remaining 3 boys, dimensionless.
By applying sum properties in (Eq. 1b) and using (Eq. 2b) and (Eq. 3), we obtain the mean of the remaining 3 boys:
![\frac{1}{10}\cdot [\Sigma_{i = 1}^{7}x_{i}+\Sigma_{i=8}^{10}x_{i}] = 58](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B10%7D%5Ccdot%20%5B%5CSigma_%7Bi%20%3D%201%7D%5E%7B7%7Dx_%7Bi%7D%2B%5CSigma_%7Bi%3D8%7D%5E%7B10%7Dx_%7Bi%7D%5D%20%3D%2058)
![\frac{1}{10}\cdot [7\cdot (61)+3\cdot p_{3}] = 58](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B10%7D%5Ccdot%20%5B7%5Ccdot%20%2861%29%2B3%5Ccdot%20p_%7B3%7D%5D%20%3D%2058)




The mean of the remaining 3 boys is 51.
Since you are giving no numbers, in order to solve you would multiply the length, width, and height of the jewelry box