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2 years ago

Help with Similarity Triangles Geometry. I dont understand how to solve for x on this one.

Mathematics

1 answer:

valkas [14]2 years ago

6 0

since we know those two triangles are similar then we can use proportions.

$\cfrac{AE}{AB}=\cfrac{AD}{AC}\implies \cfrac{14-8}{2x}=\cfrac{14}{2x+4}\implies \cfrac{6}{2x}=\cfrac{14}{2x+4}\implies \cfrac{3}{x}=\cfrac{14}{2x+4} \\\\\\ 6x+12=14x\implies 12=8x\implies \cfrac{12}{8}=x\implies \cfrac{3}{2}=x \\\\[-0.35em] ~\dotfill\\\\ AB=2x+4\implies AB=2\left( \frac{3}{2} \right)+4\implies AB=3+4\implies AB=7$

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a) Since both limits are distinct and do not exist, we conclude that x = - 1 is not part of the domain of the rational function.

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<h3>How to determine whether a limit exists or not</h3>

According to theory of limits, a function f(x) exists for x = a if and only if $\lim_{x\to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$ . This criterion is commonly used to prove continuity of functions.

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$\frac{x}{x^{2}+ x} = \frac{x}{x\cdot (x + 1)} = \frac{1}{x + 1}$

$g(x) = \frac{1}{x + 1}$

The function $f(x) = \frac{x}{x^{2}+ x}$ is equivalent to the function $g(x) = \frac{1}{x + 1}$ .

To learn more on lateral limits: brainly.com/question/21783151

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3 years ago

Help with Similarity Triangles Geometry. I dont understand how to solve for x on this one.​

Help with Similarity Triangles Geometry. I dont understand how to solve for x on this one.