Given:
The inequalities are:
![-2](https://tex.z-dn.net/?f=-2%3Cx%3C3)
![-2\leq x](https://tex.z-dn.net/?f=-2%5Cleq%20x%3C2)
To find:
The integer values that satisfy both inequalities.
Solution:
We have,
![-2](https://tex.z-dn.net/?f=-2%3Cx%3C3)
![-2\leq x](https://tex.z-dn.net/?f=-2%5Cleq%20x%3C2)
For
, the possible integer values are
...(i)
For
, the possible integer values are
...(ii)
The common values of x in (i) and (ii) are
![x=-1,0,1](https://tex.z-dn.net/?f=x%3D-1%2C0%2C1)
Therefore, the integer values -1, 0 and 1 satisfy both inequalities.
Answer:
Step-by-step explanation:
a^2 + b^2 = c^2
where
a and b are the legs
and
c is the hypotenuse
.
the problem gives the "legs" because the LONGEST leg is always the hypotenuse.
a^2 + b^2 = c^2
1071^2 + 1840^2 = c^2
1147041 + 3385600 = c^2
4532641 = c^2
100/6 = 16.66 rounds to 16.7
If A and B are equal:
Matrix A must be a diagonal matrix: FALSE.
We only know that A and B are equal, so they can both be non-diagonal matrices. Here's a counterexample:
![A=B=\left[\begin{array}{cc}1&2\\4&5\\7&8\end{array}\right]](https://tex.z-dn.net/?f=A%3DB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C4%265%5C%5C7%268%5Cend%7Barray%7D%5Cright%5D)
Both matrices must be square: FALSE.
We only know that A and B are equal, so they can both be non-square matrices. The previous counterexample still works
Both matrices must be the same size: TRUE
If A and B are equal, they are literally the same matrix. So, in particular, they also share the size.
For any value of i, j; aij = bij: TRUE
Assuming that there was a small typo in the question, this is also true: two matrices are equal if the correspondent entries are the same.