Answer:
Step-by-step explanation:
<ACB = <ECD
These 2 angles are vertically opposite and are equal.
<B = <D
They are both right angles are therefore equal.
The answer is the AA postulate.
A
Note
ASA is a congruence postualate. If S is between two angles that can be shown to be corresponding and equal, then you will have 2 congruent triangles.
SSS if three sides of 1 triangle = 3 sides of a second triangle, then the 2 triangles are congruent. If the the three sides of one triangle are in a ratio with 3 sides of the other triangle, then the triangles could be similar, but that is not the case here.
SAS this is the terminology for congruence as well. We don't know enough to use it for similarity. Some sort of ratio would have to be mentioned to do that.
You are intended to use AA as your answer.
Answer:
hes paid 13,640
Step-by-step explanation:
Answer:
Um.. well I don't know what is your question, but I think you meant square root and the answer is 7 because it is a perfect square.
Step-by-step explanation:
Answer:
it equals 6k but an equivalent expression could be 2k + 4k
Step-by-step explanation:
2k + 4k = 4k + 2k
Answer:
<h3>
f(x) = 5x² + 2x</h3><h3>
g(x) = 6x - 6</h3>
Step-by-step explanation:
![\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}\\\\6(x^2-3x+2)\ne0\ \iff\ x=\frac{3\pm\sqrt{9-8}}{2}\ne0\ \iff\ x\ne2\ \wedge\ x\ne1\\\\\\\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}=\dfrac{x(5x^2-8x-4)}{6(x^2-3x+2)}=\dfrac{x(5x^2-10x+2x-4)}{6(x^2-2x-x+2)}=\\\\\\=\dfrac{x[5x(x-2)+2(x-2)]}{6[x(x-2)-(x-2)]} =\dfrac{x(x-2)(5x+2)}{6(x-2)(x-1)}=\dfrac{x(5x+2)}{6(x-1)}=\dfrac{5x^2+2x}{6x-6}\\\\\\f(x)=5x^2+2x\\\\g(x)=6x-6](https://tex.z-dn.net/?f=%5Cdfrac%7B5x%5E3-8x%5E2-4x%7D%7B6x%5E2-18x%2B12%7D%5C%5C%5C%5C6%28x%5E2-3x%2B2%29%5Cne0%5C%20%5Ciff%5C%20x%3D%5Cfrac%7B3%5Cpm%5Csqrt%7B9-8%7D%7D%7B2%7D%5Cne0%5C%20%5Ciff%5C%20x%5Cne2%5C%20%5Cwedge%5C%20x%5Cne1%5C%5C%5C%5C%5C%5C%5Cdfrac%7B5x%5E3-8x%5E2-4x%7D%7B6x%5E2-18x%2B12%7D%3D%5Cdfrac%7Bx%285x%5E2-8x-4%29%7D%7B6%28x%5E2-3x%2B2%29%7D%3D%5Cdfrac%7Bx%285x%5E2-10x%2B2x-4%29%7D%7B6%28x%5E2-2x-x%2B2%29%7D%3D%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7Bx%5B5x%28x-2%29%2B2%28x-2%29%5D%7D%7B6%5Bx%28x-2%29-%28x-2%29%5D%7D%20%3D%5Cdfrac%7Bx%28x-2%29%285x%2B2%29%7D%7B6%28x-2%29%28x-1%29%7D%3D%5Cdfrac%7Bx%285x%2B2%29%7D%7B6%28x-1%29%7D%3D%5Cdfrac%7B5x%5E2%2B2x%7D%7B6x-6%7D%5C%5C%5C%5C%5C%5Cf%28x%29%3D5x%5E2%2B2x%5C%5C%5C%5Cg%28x%29%3D6x-6)
