Given that <span>Line m is parallel to line n.
We prove that 1 is supplementary to 3 as follows:
![\begin{tabular} {|c|c|} Statement&Reason\\[1ex] Line m is parallel to line n&Given\\ \angle1\cong\angle2&Corresponding angles\\ m\angle1=m\angle2&Deifinition of Congruent angles\\ \angle2\ and\ \angle3\ form\ a\ linear\ pair&Adjacent angles on a straight line\\ \angle2\ is\ supplementary\ to\ \angle3&Deifinition of linear pair\\ m\angle2+m\angle3=180^o&Deifinition of supplementary \angle s\\ m\angle1+m\angle3=180^o&Substitution Property \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7C%7D%0AStatement%26Reason%5C%5C%5B1ex%5D%0ALine%20m%20is%20parallel%20to%20line%20n%26Given%5C%5C%0A%5Cangle1%5Ccong%5Cangle2%26Corresponding%20angles%5C%5C%0Am%5Cangle1%3Dm%5Cangle2%26Deifinition%20of%20Congruent%20angles%5C%5C%0A%5Cangle2%5C%20and%5C%20%5Cangle3%5C%20form%5C%20a%5C%20linear%5C%20pair%26Adjacent%20angles%20on%20a%20straight%20line%5C%5C%0A%5Cangle2%5C%20is%5C%20supplementary%5C%20to%5C%20%5Cangle3%26Deifinition%20of%20linear%20pair%5C%5C%0Am%5Cangle2%2Bm%5Cangle3%3D180%5Eo%26Deifinition%20of%20supplementary%20%5Cangle%20s%5C%5C%0Am%5Cangle1%2Bm%5Cangle3%3D180%5Eo%26Substitution%20Property%0A%5Cend%7Btabular%7D)

</span>
For this case we have the following vectors:

The dot product of two vectors is a scalar.
The point product consists of multiplying component by component and then adding the result of the multiplication of each component.
For the product point of the vectors a and b we have:
Answer:
The product point of the vectors a and b is:
Answer:
Step-by-step explanation:
We can use the distance formula derived from the Pythagorean theorem
D = 
the two points given are
(0, 3) and (-2, -3)

Answer:
1
Step-by-step explanation:
:)..................
27,057,324,083 i am pretty sure