3 years ago

The following equation is solved using the Pythagorean Theorem: 4² + b² =6²

1 answer:

Sloan [31]3 years ago

3 0

16 + b^2 = 36
b^2 = 20
b = 4.472

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Consider the expression

$\dfrac{x^3-1}{(x+1)^3+1}$

Factorize the numerator and denominator a difference/sum of cubes:

$\dfrac{(x-1)(x^2+x+1)}{((x+1)+1)((x+1)^2-(x+1)+1)}$

Expand the denominator:

$\dfrac{(x-1)(x^2+x+1)}{(x+2)((x^2+2x+1)-(x+1)+1)}=\dfrac{(x-1)(x^2+x+1)}{(x+2)(x^2+x+1)}=\dfrac{x-1}{x+2}$

since x = 2020, and clearly 2020² + 2020 + 1 ≠ 0, so we can cancel the factor of x ² + x + 1. This leaves us with

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Llana [10]

Answer:

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Step-by-step explanation:

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lora16 [44]

Answer:

18 units over 210 units of time

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the function reaches its limit at 116 and starts at 0 at -94

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