Answer:
the gradient is the same= 4
the y intecept is different, one is positive 1 and the other is negative 3
Answer:
The answer is below
Step-by-step explanation:
Two polygons are said to be congruent if they have the same size and shape that is their corresponding angles and sides are equal.
Hence since Quadrilaterals ABCD is congruent to EFGH, then their corresponding angles and sides are equal.
In quadrilateral ABCD:
∠A + ∠B + ∠C + ∠D = 360° (sum of angles in a quadrilateral)
Substituting:
47 + 39 + 112 + ∠D = 360
∠D + 198 = 360
∠D = 360 - 198
∠D = 162°
The image of Quadrilaterals ABCD and EFGH is not given but let us assume that they have the same orientation, hence:
∠A = ∠E = 47°
∠B = ∠F = 39°
∠C = ∠G = 112°
∠D = ∠H = 162°
something noteworthy is that the independent and squared variable in this case will be the "x", namely the graph of that quadratic is a vertical parabola.
![\bf f(x) = (x+2)(x-4)\implies 0=(x+2)(x-4)\implies x = \begin{cases} -2\\ 4 \end{cases} \\\\\\ \boxed{-2}\rule[0.35em]{7em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}\stackrel{\downarrow }{1}\rule[0.35em]{10em}{0.25pt}\boxed{4}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%20%3D%20%28x%2B2%29%28x-4%29%5Cimplies%200%3D%28x%2B2%29%28x-4%29%5Cimplies%20x%20%3D%20%5Cbegin%7Bcases%7D%20-2%5C%5C%204%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cboxed%7B-2%7D%5Crule%5B0.35em%5D%7B7em%7D%7B0.25pt%7D0%5Crule%5B0.35em%5D%7B3em%7D%7B0.25pt%7D%5Cstackrel%7B%5Cdownarrow%20%7D%7B1%7D%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D%5Cboxed%7B4%7D)
so the parabola has solutions at x = -2 and x = 4, and its vertex will be half-way between those two guys, namely at x = 1.
since this is a vertical parabola, its axis of symmetry, the line that splits its into twin sides, will be a vertical line, and it'll be the x-coordinate of the vertex, since the vertex hasa a coordinate of x = 1, then the axis of symmetry is the vertical line of x = 1.
I assume you mean:
[x^(5/6)]/[x^(1/6)]
The rule for dividing similar bases with different exponents is:
(a^b)/(a^c)=a^(b-c)
In this case we have:
x^(5/6-1/6)
x^(4/6)
x^(2/3)
