Answer:
The correct option is A.
Step-by-step explanation:
It is given that DEFG is a parallelogram.
Draw the diagonals DF and EG. Place point H where DF and EG intersect.
In triangle HGD and HEF
,
∠HGD ≅ ∠HEF (Alternate Interior angle)
∠HDG ≅ ∠HFE (Alternate Interior angle)
By the definition of a parallelogram, the opposite sides of a parallelogram are congruent.
DG ≅ EF (Opposite sides of parallelogram)
According to ASA postulate, two triangles are congruent if any two angles and their included side are equal in both triangles.
So, by using ASA criterion for congruence we get,
ΔDGH ≅ ΔFEH
Since corresponding sides of congruent triangles are congruent, therefore
GH ≅ EH (CPCTC)
DH ≅ FH (CPCTC)
Option A is correct.
TO write the function into vertex form, we need to add one zero and the solution is shown below:
f(x)= x² -6x +1
Add one zero in the form of -9 and 9, such as:
f(x) = x² -6x +1 -9 +9
f(x) = (x² -6x +9) +(1-9)
f(x) = (x-3)² + (-8)
f(x) = (x-3)² - 8
This is the vertex form of the given function and only one zero is being added.
Answer:
Graph has been shown in the attached file.
Step-by-step explanation:
We have been given the system of equations
Both the equation represents a straight lines. We can find the x and y intercepts of these lines to graph.
The intercept form of a line is given by
Here a is the x - intercept and b is the y-intercept.
Divide both sides of the equation (1) by 16
Hence, x-intercept = 4 and the point is (4,0)
y-intercept = 4 and the point is (0,4)
Similarly, for the second line
Divide both sides of the equation (2) by -6
x-intercept = -6 and the point is (-6,0)
y-intercept = -1 and the point is (0,-1)
We'll plot these points in the xy- plane and then join to get the graph of these lines.
