Answer:
Step-by-step explanation: So there are 148 students total and we know the fifth graders filled up 12 rows and since each row has a number of 8 seats we should multiply 8 times 12, which is equal to 96. Now in order to find the number of rows needed for the fourth graders we should first, subtract 178 minus 96; that is equal to 82. Next, we should divide 82 by the number of seats in each row since what we are trying to find id the number of rows needed for the fourth graders. So, we should divide 82 by 8 which is equal to 10. 25.
The number of rows needed is 11 since all students need to have a seat; the answer cannot be 10 because that means a few students would not have a seat.
Answer:
Step-by-step explanation:
33.414−33.36
=33.414−33.36
=33.414+−33.36
=0.054
Here we go ~
Let's calculate volume of Cylinder ~
That's all, ask me if you have any queries ~
Answer:
A) Verified
B) Proved
Step-by-step explanation:
a) Let's verify it for 2 x 2 matrix,
![A=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
and ![B=\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D)
![AB=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\left[\begin{array}{ccc}e&f\\g&h\end{array}\right]=\left[\begin{array}{ccc}a.e+b.g&a.f+b.h\\c.e+d.g&c.f+d.h\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da.e%2Bb.g%26a.f%2Bb.h%5C%5Cc.e%2Bd.g%26c.f%2Bd.h%5Cend%7Barray%7D%5Cright%5D)
![(AB)^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]](https://tex.z-dn.net/?f=%28AB%29%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7B%28a.e%2Bb.g%29%28c.f%2Bd.h%29-%28a.f%2Bb.h%29%28c.e%2Bd.g%29%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc.f%2Bd.h%26-%28a.f%2Bb.h%29%5C%5C-%28c.e%2Bd.g%29%26a.e%2Bb.g%5Cend%7Barray%7D%5Cright%5D)
![A^{-1}=\frac{1}{a.d-b.c} \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7Ba.d-b.c%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%26-b%5C%5C-c%26a%5Cend%7Barray%7D%5Cright%5D)
![B^{-1}=\frac{1}{e.h-f.g} \left[\begin{array}{ccc}h&-f\\-g&e\end{array}\right]](https://tex.z-dn.net/?f=B%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7Be.h-f.g%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dh%26-f%5C%5C-g%26e%5Cend%7Barray%7D%5Cright%5D)
![B^{-1}A^{-1}=\frac{1}{(a.e+b.g)(c.f+d.h)-(a.f+b.h)(c.e+d.g)}\left[\begin{array}{ccc}c.f+d.h&-(a.f+b.h)\\-(c.e+d.g)&a.e+b.g\end{array}\right]](https://tex.z-dn.net/?f=B%5E%7B-1%7DA%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7B%28a.e%2Bb.g%29%28c.f%2Bd.h%29-%28a.f%2Bb.h%29%28c.e%2Bd.g%29%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc.f%2Bd.h%26-%28a.f%2Bb.h%29%5C%5C-%28c.e%2Bd.g%29%26a.e%2Bb.g%5Cend%7Barray%7D%5Cright%5D)
So it is proved that the results are same.
b) Now, let's prove it for any n x n matrix.
Answer:
The correct answers on edg are 2) AE and CE are always equal 3) BE and DE are always equal 4) Segment AC and segment BD always bisect each other
Step-by-step explanation:
