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3 years ago

What is the slope of a line perpendicular to the line passing through E(-1,6) and F(-2,-10)?

Mathematics

1 answer:

STatiana [176]3 years ago

7 0

Answer:

Step-by-step explanation:

E(-1,6) & F(-2,-10)

$Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\\ \ \ \ = \frac{-10-6}{-2-[-1]}\\$

$=\frac{-10-6}{-2+1}\\\\=\frac{-16}{-1}\\\\= 16$

Slope of the line perpendicular to this line = -1/m = -1/16

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Evaluate lim x approaches to 2 : (sqrt(6-x)-2)/(sqrt(3-x)-1)

solong [7]

Answer:

The answer is "0.5".

Step-by-step explanation:

Given:

$\to \lim_{x\to 2} \frac{(\sqrt{(6-x)}-2)}{(\sqrt{(3-x)}-1)}\\\\ \to \lim_{x\to 2} \frac{\frac{d(\sqrt{(6-x)}-2)}{dx}}{\frac{(\sqrt{(3-x)}-1)}{dx}}\\\\$

$\to \lim_{x\to 2} \frac{\frac{-1}{2\sqrt{(6-x)} }}{\frac{-1}{2\sqrt{(3-x)}}}\\\\$

$\to \lim_{x\to 2} \frac{\sqrt{3-x}}{\sqrt{6-x}}\\\\ \to \frac{\sqrt{3-2}}{\sqrt{6-2}}\\\\ \to \frac{\sqrt{1}} {\sqrt{4}}\\\\ \to \frac{1}{2}\\\\ \to 0.5$

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