No nothing is even it is supposed to have 2 pairs of congruent and parallel sides
First, you add like terms. In this case, the like terms are 6 and 3.
4x + 6 + 3 = 17
4x + 9 = 17
Then, subtract 9 on each side:
4x + 9 - 9 = 17 - 9
4x = 8
Then divide by 4 on each side to get the x by itself:
4x/4 = 8/4
x = 2
The first quadrant is the upper right hand corner of the graph where both y and y are positive
the second quadrant in the upper left hand corner includes negative values of x and positive values of y
the third quadrant the lower he as corner includes negative values of both x and y
the fourth quadrant the lower right handed corner the x is positive and the y is negative
hope this helps, can i have brainliest please
Step-by-step explanation:
vector AB(3-(-6); 5-7)
vector AB(9;-2)
AB=
=
M is the midpoint of AB
we have B(-5;10) and M(1;7)
let A(x;y)
(x-5)/2 = 1 ⇒ x-5 = 2⇒ x = 7
(10=y)/2 = 7⇒ 10+y = 14 ⇒y= 4
so : A(7;4)
the center of the circle is the midponit of the line joining both ends of the diameter
let A(x;y) be the other end
(-2+x)/2 = 2 ⇒ -2+x = 4⇒ x= 6
(5=y) = -1 ⇒ 5+y = -2 ⇒ y= -7
so the coordinates of the other end are (6; -7)
A,B and C are collinear such as AB=BC so b is the midpoint of AC
(-5+1)/2 = y ⇒ y = -4/2 ⇒ y = -2
((-3=x)/2 = 7 ⇒ -3+x = 14 ⇒ x = 17
so x= 17 and y = -2