Answer:
C) The solution for the given system of equations are A(0,-5) and B(-4,3)
Step-by-step explanation:
The given system of equation are : 
from equation 2, we get y = -5 - 2x .
Put the above value of y in the equation (1).
We get: 
By ALGEBRAIC IDENTITY:

or, 
or, 
⇒ x = 0 or, x = -20/5 = -4
So, the possible values for x are: x = 0 or x = -4
If x = 0, y = -5-2x = -5-2(0) = -5
and if x = -4, y = -5 -2(-4) = -5 + 8 = 3
Hence, the solution for the given system of equations are A(0,-5) and B(-4,3)
Answer:
<h3>
"The answer is the yellow lines in the attached figure.
</h3>
Step-by-step explanation: As shown in the attached figure, regular hexagon FGHIJK and square ABCD shares common centre on the co-ordinate plane and AB || FG.
We are to find the line across which the combined figure will reflect onto itself.
In the attached figure, we see two lines which are yellow in colour. We can easily detect that the figure will be reflected onto itself if these two lines acts as a mirror separately.
Hence these yellow lines are the required lines."
<h3><u>
Hope this helps!</u></h3>
Answer:
1
Step-by-step explanation:
There's a theorem that says "through 2 points it only cross one and only one line". Having this as our start point, we can guarantee that there's only one line that can be drawn through those points