Allen is given three lengths of rope 4 feet , 8 feet , and 15 feet . Can Allen form a triangle with side lengths of 4 feet 8 feet and 15 feet using these three pieces of rope why or why not ?

Allen is given three lengths of rope 4 feet , 8 feet , and 15 feet . Can Allen form a triangle with

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Janznznz4012

12.07.2022

Mathematics

$\displaystyle 8$

Step-by-step explanation:

we would like to compute the following limit

$\displaystyle \lim_{x \to 16} \left( \frac{x - 16}{ \sqrt{x} - 4} \right)$

if we substitute 16 directly we'd end up

$\displaystyle = \frac{16 - 16}{ \sqrt{16} - 4}$

$\displaystyle = \frac{0}{ 0}$

which isn't a good answer now notice that we have a square root on the denominator so we can rationalise the denominator to do so multiply the expression by √x+4/√x+4 which yields:

$\displaystyle \lim_{x \to 16} \left( \frac{x - 16}{ \sqrt{x} - 4} \times \frac{ \sqrt{x} + 4 }{ \sqrt{x} + 4 } \right)$

simplify which yields:

$\displaystyle \lim_{x \to 16} \left( \frac{(x - 16)( \sqrt{x} + 4)}{ x - 16} \right)$

we can reduce fraction so that yields:

$\displaystyle \lim_{x \to 16} \left( \frac{ \cancel{(x - 16)}( \sqrt{x} + 4)}{ \cancel{x - 16} } \right)$

$\displaystyle \lim _{x \to 16} \left( \sqrt{x } + 4\right)$

now it's safe enough to substitute 16 thus

substitute:

$\displaystyle = \sqrt{16} + 4$

simplify square root:

$\displaystyle = 4 + 4$

simplify addition:

$\displaystyle = 8$

hence,

$\displaystyle \lim_{x \to 16} \left( \frac{x - 16}{ \sqrt{x} - 4} \right) = 8$

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