Iam an isosceles triangle. the midpoint of my non-congruent side is the point (7, −2). this side is a horizontal line with a length of 4 and my vertex opposite this side is at the point (7, −7). oh, i almost forgot to tell you. i am upside down. where am i? (find the coordinates of each base vertex.) 2 answers a small and large x-value

So, we can picture out the problem that this is an inverted triangle. So, the base of the triangle is along y = -2. All we have to find are the x coordinates. The main formula we will use here is the distance formula for computing distance between any 2 points.

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

The solution is as follows:

D between (7,-2) and (7,-7).
d = √[(7 - 1)² + (-7 - -2)²] = 5 units
Let's solve the hypotenuse.
h = √(5² + (4/2)²) = √29
So, the distance of the hypotenuse is the distance between the top vertex (7,-7) and one bottom side vertex (x,-2) is √29.
√29 = √[(x - 7)² + (-2 - -7)²]
Solving for x,
x = 5
Since the base is 4 units in length, the other base vertex is at x = 5 - 4 = 1
Therefore, the vertex of the bases of the isosceles triangle are at (5,-2) and (1,-2).

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

0 = ( 2 - 4 ) ( p + q )0 = ( 2 - 4 ) ( 2 + 5 i ),   by multiplying these0 = ( 4 + 10 i - 8 - 20 i )0 = 4 - 8 + 10 i - 20 i0 = - 4 -10 i   0 = 2 ( - 2 - 5 i )0 / 2 = - 2 - 5 i- 2 - 5 i = 0- 5 i = 2i = 2 / - 5

The answer would be c