1. Turn improper
![12 \frac{2}{5} = \frac{62}{5}](https://tex.z-dn.net/?f=12%20%5Cfrac%7B2%7D%7B5%7D%20%20%3D%20%20%5Cfrac%7B62%7D%7B5%7D%20)
![3 \frac{1}{6} = \frac{19}{6}](https://tex.z-dn.net/?f=3%20%5Cfrac%7B1%7D%7B6%7D%20%20%3D%20%20%5Cfrac%7B19%7D%7B6%7D%20)
2. set up the problem
![\frac{62}{5} \times \frac{19}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B62%7D%7B5%7D%20%20%5Ctimes%20%20%5Cfrac%7B19%7D%7B6%7D%20)
now multiple across and simplefy
Answer:
![x = 4](https://tex.z-dn.net/?f=x%20%3D%204)
Step-by-step explanation:
![- x + 4 = \frac{3}{2} x - 6 \\ - 2x + 8 = 3x - 12 \\ 2x - 8 = - 3x + 12 \\ 5x = 20 \\ x = 4](https://tex.z-dn.net/?f=%20-%20x%20%2B%204%20%3D%20%20%5Cfrac%7B3%7D%7B2%7D%20x%20-%206%20%5C%5C%20%20-%202x%20%2B%208%20%3D%203x%20-%2012%20%5C%5C%202x%20-%208%20%3D%20%20-%203x%20%2B%2012%20%5C%5C%205x%20%3D%2020%20%5C%5C%20x%20%3D%204)
Part A
The given expression is:
![(-5(1+2i)+3i(3-4i)](https://tex.z-dn.net/?f=%28-5%281%2B2i%29%2B3i%283-4i%29)
We expand to get:
![-5-10i+9i-12i^2](https://tex.z-dn.net/?f=-5-10i%2B9i-12i%5E2)
Note that ![i^2=-1](https://tex.z-dn.net/?f=i%5E2%3D-1)
![12-5-10i+9i](https://tex.z-dn.net/?f=12-5-10i%2B9i)
![-5-10i+9i+12](https://tex.z-dn.net/?f=-5-10i%2B9i%2B12)
![7-i](https://tex.z-dn.net/?f=7-i)
Part Bi)
The given expression is;
![\sqrt{-50}](https://tex.z-dn.net/?f=%5Csqrt%7B-50%7D)
We simplify to get:
![\sqrt{25\times 2\times -1}](https://tex.z-dn.net/?f=%5Csqrt%7B25%5Ctimes%202%5Ctimes%20-1%7D)
![\sqrt{25\times} \sqr{2}\times \sqrt{-1}](https://tex.z-dn.net/?f=%5Csqrt%7B25%5Ctimes%7D%20%5Csqr%7B2%7D%5Ctimes%20%5Csqrt%7B-1%7D)
Note that:
![\sqrt{-1}=i](https://tex.z-dn.net/?f=%5Csqrt%7B-1%7D%3Di)
![5\sqr{2}i](https://tex.z-dn.net/?f=5%5Csqr%7B2%7Di)
This is now of the form
, where
.
This explains why it is a complex number;
Part bii)
The given expression is :
![\frac{\sqrt{-50} }{\sqrt{5}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B-50%7D%20%7D%7B%5Csqrt%7B5%7D%7D)
We simplify to get
![\sqrt{\frac{-50 }{5}}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B-50%20%7D%7B5%7D%7D)
![\sqrt{-25}](https://tex.z-dn.net/?f=%5Csqrt%7B-25%7D)
![\sqrt{-25}=\sqrt{25}\times \sqrt{-1}](https://tex.z-dn.net/?f=%5Csqrt%7B-25%7D%3D%5Csqrt%7B25%7D%5Ctimes%20%5Csqrt%7B-1%7D)
![\sqrt{-25}=5\times i](https://tex.z-dn.net/?f=%5Csqrt%7B-25%7D%3D5%5Ctimes%20i)
![\sqrt{-25}=5i](https://tex.z-dn.net/?f=%5Csqrt%7B-25%7D%3D5i)
Part C
We want to simplify:
![i^{22}](https://tex.z-dn.net/?f=i%5E%7B22%7D)
We rewrite in terms of ![i^2](https://tex.z-dn.net/?f=i%5E2)
![i^{22}=(i^2)^{11}](https://tex.z-dn.net/?f=i%5E%7B22%7D%3D%28i%5E2%29%5E%7B11%7D)
![i^{22}=(-1)^{11}](https://tex.z-dn.net/?f=i%5E%7B22%7D%3D%28-1%29%5E%7B11%7D)
![i^{22}=-11](https://tex.z-dn.net/?f=i%5E%7B22%7D%3D-11)
Looks like the matrix equation is supposed to be
![C^{-1}(A+X)B^{-1}=I_n](https://tex.z-dn.net/?f=C%5E%7B-1%7D%28A%2BX%29B%5E%7B-1%7D%3DI_n)
where
presumably denotes the
identity matrix.
Since
are all invertible, we have by multiplying on the left by
,
![C(C^{-1}(A+X)B^{-1})=CI_n](https://tex.z-dn.net/?f=C%28C%5E%7B-1%7D%28A%2BX%29B%5E%7B-1%7D%29%3DCI_n)
![(CC^{-1})((A+X)B^{-1})=C](https://tex.z-dn.net/?f=%28CC%5E%7B-1%7D%29%28%28A%2BX%29B%5E%7B-1%7D%29%3DC)
![(A+X)B^{-1}=C](https://tex.z-dn.net/?f=%28A%2BX%29B%5E%7B-1%7D%3DC)
then multiplying on the right by
,
![((A+X)B^{-1})B=CB](https://tex.z-dn.net/?f=%28%28A%2BX%29B%5E%7B-1%7D%29B%3DCB)
![(A+X)(B^{-1}B)=CB](https://tex.z-dn.net/?f=%28A%2BX%29%28B%5E%7B-1%7DB%29%3DCB)
![A+X=CB](https://tex.z-dn.net/?f=A%2BX%3DCB)
and finally subtracting
from both sides to end up with
![X=CB-A](https://tex.z-dn.net/?f=X%3DCB-A)
Answer:
3 over 8
Step-by-step explanation:
It's three ticks to the right of zero, so its positive