Answer:
12y + 120
Step-by-step explanation:
<u>First of, what is the distributive property?</u>
The distributive property is the property that helps distribute numbers under the parenthesis.
For example, lets use {x+1}4. You distribute the 4 towards the x and the one like this:
4(x) + 1(4)
4x + 4
Now, let's go back to your problem
<u>Solution:</u>
{y+10}12
12(y) +12(10)
= 12y + 120
<em>Hope this helps! Make sure to give me the brainliest answer since it would be greatly appreciated! Thank You! </em>
Answer:
yes
Step-by-step explanation:
in 13, both x and y are increasing by 1 correspondingly, and, in 14, both sides are multiplying by 2 together.
a.
![XA^{-1}=A^3](https://tex.z-dn.net/?f=XA%5E%7B-1%7D%3DA%5E3)
![(XA^{-1})A=A^3A](https://tex.z-dn.net/?f=%28XA%5E%7B-1%7D%29A%3DA%5E3A)
![X(A^{-1}A)=A^4](https://tex.z-dn.net/?f=X%28A%5E%7B-1%7DA%29%3DA%5E4)
![X=A^4](https://tex.z-dn.net/?f=X%3DA%5E4)
b.
![AXB=(BA)^2](https://tex.z-dn.net/?f=AXB%3D%28BA%29%5E2)
![A^{-1}(AXB)B^{-1}=A^{-1}(BA)^2B^{-1}](https://tex.z-dn.net/?f=A%5E%7B-1%7D%28AXB%29B%5E%7B-1%7D%3DA%5E%7B-1%7D%28BA%29%5E2B%5E%7B-1%7D)
![(A^{-1}A)X(BB^{-1})=A^{-1}(BA)^2B^{-1}](https://tex.z-dn.net/?f=%28A%5E%7B-1%7DA%29X%28BB%5E%7B-1%7D%29%3DA%5E%7B-1%7D%28BA%29%5E2B%5E%7B-1%7D)
![X=A^{-1}(BA)^2B^{-1}](https://tex.z-dn.net/?f=X%3DA%5E%7B-1%7D%28BA%29%5E2B%5E%7B-1%7D)
c.
![(A^{-1}X)^{-1}=(AB^{-1})^{-1}(AB^2)](https://tex.z-dn.net/?f=%28A%5E%7B-1%7DX%29%5E%7B-1%7D%3D%28AB%5E%7B-1%7D%29%5E%7B-1%7D%28AB%5E2%29)
![X^{-1}A=(BA^{-1})(AB^2)](https://tex.z-dn.net/?f=X%5E%7B-1%7DA%3D%28BA%5E%7B-1%7D%29%28AB%5E2%29)
![X^{-1}A=B(A^{-1}A)B^2](https://tex.z-dn.net/?f=X%5E%7B-1%7DA%3DB%28A%5E%7B-1%7DA%29B%5E2)
![X^{-1}A=B^3](https://tex.z-dn.net/?f=X%5E%7B-1%7DA%3DB%5E3)
![(XX^{-1})A=XB^3](https://tex.z-dn.net/?f=%28XX%5E%7B-1%7D%29A%3DXB%5E3)
![XB^3=A](https://tex.z-dn.net/?f=XB%5E3%3DA)
![X(B^3(B^3)^{-1})=A(B^3)^{-1}](https://tex.z-dn.net/?f=X%28B%5E3%28B%5E3%29%5E%7B-1%7D%29%3DA%28B%5E3%29%5E%7B-1%7D)
![X=A(B^3)^{-1}](https://tex.z-dn.net/?f=X%3DA%28B%5E3%29%5E%7B-1%7D)
d. Not totally sure what the equation is supposed to be, but I guess it's
![ABXA^{-1}B^{-1}=A](https://tex.z-dn.net/?f=ABXA%5E%7B-1%7DB%5E%7B-1%7D%3DA)
![ABX(BA)^{-1}=A](https://tex.z-dn.net/?f=ABX%28BA%29%5E%7B-1%7D%3DA)
![((AB)^{-1}(AB))X((BA)^{-1}(BA))=(AB)^{-1}A(BA)](https://tex.z-dn.net/?f=%28%28AB%29%5E%7B-1%7D%28AB%29%29X%28%28BA%29%5E%7B-1%7D%28BA%29%29%3D%28AB%29%5E%7B-1%7DA%28BA%29)
![X=(AB)^{-1}A(BA)](https://tex.z-dn.net/?f=X%3D%28AB%29%5E%7B-1%7DA%28BA%29)
![X=(B^{-1}A^{-1})A(BA)](https://tex.z-dn.net/?f=X%3D%28B%5E%7B-1%7DA%5E%7B-1%7D%29A%28BA%29)
![X=B^{-1}(A^{-1}A)(BA)](https://tex.z-dn.net/?f=X%3DB%5E%7B-1%7D%28A%5E%7B-1%7DA%29%28BA%29)
![X=(B^{-1}B)A](https://tex.z-dn.net/?f=X%3D%28B%5E%7B-1%7DB%29A)
![X=A](https://tex.z-dn.net/?f=X%3DA)