The figure is NOT unique.
Imagine the following quadrilaterals:
Rectangle
Square
We know that:
Both quadrilaterals have at least two right angles.
However, they are not unique because they depend on the lengths of their sides.
Answer:
The figure described is not unique.
Answer: 60
Step-by-step explanation:
9/15 = 0.6 = 1% 1% x 100 = 100% so 0.6 x 100 = 60
since we know those two triangles are similar then we can use proportions.
![\cfrac{AE}{AB}=\cfrac{AD}{AC}\implies \cfrac{14-8}{2x}=\cfrac{14}{2x+4}\implies \cfrac{6}{2x}=\cfrac{14}{2x+4}\implies \cfrac{3}{x}=\cfrac{14}{2x+4} \\\\\\ 6x+12=14x\implies 12=8x\implies \cfrac{12}{8}=x\implies \cfrac{3}{2}=x \\\\[-0.35em] ~\dotfill\\\\ AB=2x+4\implies AB=2\left( \frac{3}{2} \right)+4\implies AB=3+4\implies AB=7](https://tex.z-dn.net/?f=%5Ccfrac%7BAE%7D%7BAB%7D%3D%5Ccfrac%7BAD%7D%7BAC%7D%5Cimplies%20%5Ccfrac%7B14-8%7D%7B2x%7D%3D%5Ccfrac%7B14%7D%7B2x%2B4%7D%5Cimplies%20%5Ccfrac%7B6%7D%7B2x%7D%3D%5Ccfrac%7B14%7D%7B2x%2B4%7D%5Cimplies%20%5Ccfrac%7B3%7D%7Bx%7D%3D%5Ccfrac%7B14%7D%7B2x%2B4%7D%20%5C%5C%5C%5C%5C%5C%206x%2B12%3D14x%5Cimplies%2012%3D8x%5Cimplies%20%5Ccfrac%7B12%7D%7B8%7D%3Dx%5Cimplies%20%5Ccfrac%7B3%7D%7B2%7D%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20AB%3D2x%2B4%5Cimplies%20AB%3D2%5Cleft%28%20%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%2B4%5Cimplies%20AB%3D3%2B4%5Cimplies%20AB%3D7)
To find a positive and a negative angle coterminal<span> with a given </span>angle<span>, you can add and subtract if the </span>angle<span> is measured in degrees or if the </span>angle<span> is measured in radians</span>
The binomial (2 · x + y)⁷ in expanded form by 128 · x⁷ + 448 · x⁶ · y + 672 · x⁵ · y² + 560 · x⁴ · y³ + 280 · x³ · y⁴ + 84 · x² · y⁵ + 14 · x · y⁶ + y⁷.
<h3>How to expand the power of a binomial</h3>
Herein we have the seventh power of a binomial, whose expanded form can be found by using the binomial theorem and Pascal's triangle. Hence, we find the following expression for the expanded form:
(2 · x + y)⁷
(2 · x)⁷ + 7 · (2 · x)⁶ · y + 21 · (2 · x)⁵ · y² + 35 · (2 · x)⁴ · y³ + 35 · (2 · x)³ · y⁴ + 21 · (2 · x)² · y⁵ + 7 · (2 · x) · y⁶ + y⁷
128 · x⁷ + 448 · x⁶ · y + 672 · x⁵ · y² + 560 · x⁴ · y³ + 280 · x³ · y⁴ + 84 · x² · y⁵ + 14 · x · y⁶ + y⁷
Then, the binomial (2 · x + y)⁷ in expanded form by 128 · x⁷ + 448 · x⁶ · y + 672 · x⁵ · y² + 560 · x⁴ · y³ + 280 · x³ · y⁴ + 84 · x² · y⁵ + 14 · x · y⁶ + y⁷.
To learn more on binomials: brainly.com/question/12249986
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